url
stringlengths 14
2.42k
| text
stringlengths 100
1.02M
| date
stringlengths 19
19
| metadata
stringlengths 1.06k
1.1k
|
|---|---|---|---|
https://acm.ecnu.edu.cn/problem/2468/
|
It’s Bessie’s birthday and time for party games! Bessie has instructed the N (1 <= N <= 100,000) cows conveniently numbered 1..N to sit in a circle (so that cow i [except at the ends] sits next to cows i-1 and i+1; cow N sits next to cow 1). Meanwhile, Farmer John fills a barrel with one billion slips of paper, each containing some integer in the range 1..1,000,000.
Each cow i then draws a number A_i (1 <= A_i <= 1,000,000) (which is not necessarily unique, of course) from the giant barrel. Taking turns, each cow i then takes a walk around the circle and pats the heads of all other cows j such that her number A_i is exactly divisible by cow j’s number A_j; she then sits again back in her original position.
The cows would like you to help them determine, for each cow, the number of other cows she should pat.
### 输入格式
• Line 1: A single integer: N
• Lines 2..N+1: Line i+1 contains a single integer: A_i
### 输出格式
• Lines 1..N: On line i, print a single integer that is the number of
other cows patted by cow i.
### 样例
Input
5
2
1
2
3
4
INPUT DETAILS:
The 5 cows are given the numbers 2, 1, 2, 3, and 4, respectively.
Output
2
0
2
1
3
OUTPUT DETAILS:
The first cow pats the second and third cows; the second cows pats no cows;
etc.
6 人解决，14 人已尝试。
10 份提交通过，共有 29 份提交。
7.6 EMB 奖励。
|
2020-02-23 01:56:09
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.46535179018974304, "perplexity": 6983.375470968186}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145742.20/warc/CC-MAIN-20200223001555-20200223031555-00341.warc.gz"}
|
http://efavdb.com/category/statistics/
|
## Integration method to map model scores to conversion rates from example data
This note addresses the typical applied problem of estimating from data how a target “conversion rate” function varies with some available scalar score function — e.g., estimating conversion rates from some marketing campaign as a function of a targeting model score. The idea centers around estimating the integral of the rate function; differentiating this gives the rate function. The method is a variation on a standard technique for estimating pdfs via fits to empirical cdfs.
## Gaussian Processes
We review the math and code needed to fit a Gaussian Process (GP) regressor to data. We conclude with a demo of a popular application, fast function minimization through GP-guided search. The gif below illustrates this approach in action — the red points are samples from the hidden red curve. Using these samples, we attempt to leverage GPs to find the curve’s minimum as fast as possible.
Appendices contain quick reviews on (i) the GP regressor posterior derivation, (ii) SKLearn’s GP implementation, and (iii) GP classifiers.
## Logistic Regression
We review binary logistic regression. In particular, we derive a) the equations needed to fit the algorithm via gradient descent, b) the maximum likelihood fit’s asymptotic coefficient covariance matrix, and c) expressions for model test point class membership probability confidence intervals. We also provide python code implementing a minimal “LogisticRegressionWithError” class whose “predict_proba” method returns prediction confidence intervals alongside its point estimates.
Our python code can be downloaded from our github page, here. Its use requires the jupyter, numpy, sklearn, and matplotlib packages.
## Normal Distributions
I review — and provide derivations for — some basic properties of Normal distributions. Topics currently covered: (i) Their normalization, (ii) Samples from a univariate Normal, (iii) Multivariate Normal distributions, (iv) Central limit theorem.
## Hyperparameter sample-size dependence
Here, we briefly review a subtlety associated with machine-learning model selection: the fact that the optimal hyperparameters for a model can vary with training set size, $N.$ To illustrate this point, we derive expressions for the optimal strength for both $L_1$ and $L_2$ regularization in single-variable models. We find that the optimal $L_2$ approaches a finite constant as $N$ increases, but that the optimal $L_1$ decays exponentially fast with $N.$ Sensitive dependence on $N$ such as this should be carefully extrapolated out when optimizing mission-critical models.
## Average queue wait times with random arrivals
Queries ping a certain computer server at random times, on average $\lambda$ arriving per second. The server can respond to one per second and those that can’t be serviced immediately are queued up. What is the average wait time per query? Clearly if $\lambda \ll 1$, the average wait time is zero. But if $\lambda > 1$, the queue grows indefinitely and the answer is infinity! Here, we give a simple derivation of the general result — (9) below.
## Improved Bonferroni correction factors for multiple pairwise comparisons
A common task in applied statistics is the pairwise comparison of the responses of $N$ treatment groups in some statistical test — the goal being to decide which pairs exhibit differences that are statistically significant. Now, because there is one comparison being made for each pairing, a naive application of the Bonferroni correction analysis suggests that one should set the individual pairwise test sizes to $\alpha_i \to \alpha_f/{N \choose 2}$ in order to obtain a desired family-wise type 1 error rate of $\alpha_f$. Indeed, this solution is suggested by many texts. However, implicit in the Bonferroni analysis is the assumption that the comparisons being made are each mutually independent. This is not the case here, and we show that as a consequence the naive approach often returns type 1 error rates far from those desired. We provide adjusted formulas that allow for error-free Bonferroni-like corrections to be made.
[edit (7/4/2016): After posting this article, I’ve since found that the method we suggest here is related to / is a generalization of Tukey’s range test — see here.]
[edit (6/11/2018): I’ve added the notebook used below to our Github, here]
## Maximum-likelihood asymptotics
In this post, we review two facts about maximum-likelihood estimators: 1) They are consistent, meaning that they converge to the correct values given a large number of samples, $N$, and 2) They satisfy the Cramer-Rao lower bound for unbiased parameter estimates in this same limit — that is, they have the lowest possible variance of any unbiased estimator, in the $N\gg 1$ limit.
## A review of parameter regularization and Bayesian regression
Here, we review parameter regularization, which is a method for improving regression models through the penalization of non-zero parameter estimates. Why is this effective? Biasing parameters towards zero will (of course!) unfavorably bias a model, but it will also reduce its variance. At times the latter effect can win out, resulting in a net reduction in generalization error. We also review Bayesian regressions — in effect, these generalize the regularization approach, biasing model parameters to any specified prior estimates, not necessarily zero.
This is the second of a series of posts expounding on topics discussed in the text, “An Introduction to Statistical Learning”. Here, we cover material from its Chapters 2 and 6. See prior post here.
## Stochastic geometric series
Let $a_1, a_2, \ldots$ be an infinite set of non-negative samples taken from a distribution $P_0(a)$, and write
$$\tag{1} \label{problem} S = 1 + a_1 + a_1 a_2 + a_1 a_2 a_3 + \ldots.$$
Notice that if the $a_i$ were all the same, $S$ would be a regular geometric series, with value $S = \frac{1}{1-a}$. How will the introduction of $a_i$ randomness change this sum? Will $S$ necessarily converge? How is $S$ distributed? In this post, we discuss some simple techniques to answer these questions.
Note: This post covers work done in collaboration with my aged p, S. Landy.
|
2019-08-22 18:00:07
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6696314811706543, "perplexity": 866.4001640927102}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027317339.12/warc/CC-MAIN-20190822172901-20190822194901-00402.warc.gz"}
|
https://www.transtutors.com/questions/change-the-order-of-integration-to-dz-dx-dy-for-it-s-next-to-impossible-to-draw-out--1348002.htm
|
# Change the order of integration to dz dx dy for: It's next to impossible to draw out a 3D version of
Change the order of integration to dz dx dy for:
It's next to impossible to draw out a 3D version of the figure, so I tried drawing the region(s) on the xy, zy, zx planes... not sure if I've done it correctly though -- I get
|
2021-06-18 01:39:09
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9346845746040344, "perplexity": 499.12078869776394}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487634616.65/warc/CC-MAIN-20210618013013-20210618043013-00267.warc.gz"}
|
https://unacademy.com/course/hindi-solution-and-explanation-of-memory-based-questions-gate-2019/ML9O1D6E
|
Hindi
(Hindi) Solution and Explanation of Memory-Based Questions: GATE 2019
6 ratings
Deep Sangeet Maity
In this Course, The Educator will discuss the solution of GATE 2019 Questions of Mechanical Branch. The questions are developed by the feedback of students who attempted GATE 2019 (ME).
|
2020-07-05 08:26:47
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8806295990943909, "perplexity": 6166.8111377536325}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655887046.62/warc/CC-MAIN-20200705055259-20200705085259-00150.warc.gz"}
|
http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.bwnjournal-article-doi-10_4064-cm124-2-2
|
Pełnotekstowe zasoby PLDML oraz innych baz dziedzinowych są już dostępne w nowej Bibliotece Nauki.
Zapraszamy na https://bibliotekanauki.pl
PL EN
Preferencje
Język
Widoczny [Schowaj] Abstrakt
Liczba wyników
• # Artykuł - szczegóły
## Colloquium Mathematicum
2011 | 124 | 2 | 157-168
## Pointwise convergence for subsequences of weighted averages
EN
### Abstrakty
EN
We prove that if μₙ are probability measures on ℤ such that μ̂ₙ converges to 0 uniformly on every compact subset of (0,1), then there exists a subsequence ${n_{k}}$ such that the weighted ergodic averages corresponding to $μ_{n_{k}}$ satisfy a pointwise ergodic theorem in L¹. We further discuss the relationship between Fourier decay and pointwise ergodic theorems for subsequences, considering in particular the averages along n² + ⌊ρ(n)⌋ for a slowly growing function ρ. Under some monotonicity assumptions, the rate of growth of ρ'(x) determines the existence of a "good" subsequence of these averages.
157-168
wydano
2011
### Twórcy
autor
• Department of Mathematics, University of Wisconsin at Madison, Madison, WI 53706-1388, U.S.A.
|
2022-06-26 17:33:22
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.748514711856842, "perplexity": 3716.9817579424443}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103271763.15/warc/CC-MAIN-20220626161834-20220626191834-00247.warc.gz"}
|
https://gevepozapytu.travel-australia-planning-guide.com/college-algebra-with-graphing-and-problem-solving-book-16595gd.php
|
Last edited by Fenrigal
Sunday, August 9, 2020 | History
6 edition of College algebra with graphing and problem solving found in the catalog.
# College algebra with graphing and problem solving
## by Karl J. Smith
Written in English
Subjects:
• Algebra,
• Algebra -- Graphic methods
• Edition Notes
Classifications The Physical Object Statement Karl J. Smith. Series The precalculus series LC Classifications QA152.2 .S572 1996 Pagination xiv, 642 p. : Number of Pages 642 Open Library OL808180M ISBN 10 0534340156 LC Control Number 95044915 OCLC/WorldCa 33334336
This College Algebra text will cover a combination of classical algebra and analytic geometry, with an introduction to the transcendental exponential and logarithmic functions. If mathematics is the language of science, then algebra is the grammar of that language. Like grammar, algebra provides a structure to mathematical notation, in addition to its uses in problem solving and its ability to. Alternatives to College Algebra. or. Hints to Help the Beginning Student Distinguish between First-Level College-Credit Mathematics Courses. College Mathematics (ACC's MATH ) (UT ’ s M) * * Goal: To broaden the students' repertoire of mathematical problem-solving techniques past algebraic techniques. This course covers a variety of.
Topics covered in this course are: fundamental arithmetic operations without the use of a calculator,simplifying algebraic expressions, solving linear equations, solving percent problems, solving proportions, graphing linear equations, and performing geometric calculations. Here is a set of notes used by Paul Dawkins to teach his Algebra course at Lamar University. Included area a review of exponents, radicals, polynomials as well as indepth discussions of solving equations (linear, quadratic, absolute value, exponential, logarithm) and inqualities (polynomial, rational, absolute value), functions (definition, notation, evaluation, inverse functions) graphing.
Changing Application Problems into Equations. Solving Application Problems. Mid-Chapter Test: Sections — Geometric Problems. Motion, Money, and Mixture Problems. CHAPTER 4 Graphing Linear Equations. The Cartesian Coordinate System and Linear Equations in Two Variables. Graphing Linear Equations. Slope of a Line. The third edition emphasizes learning mathematics through graphing and problem solving through realistic applications. The CD-ROM contains eight hours of video News, Inc.®, Portland, OR. Review of College Algebra with CD. Editorial review.
You might also like
Tradeoffs and optimization in analog CMOS design
Tradeoffs and optimization in analog CMOS design
Proceedings of the specialty conference Water Forum 81
Proceedings of the specialty conference Water Forum 81
R & D and railroading, 1977
R & D and railroading, 1977
Seeing Red
Seeing Red
Memoir of Isaac Forsyth
Memoir of Isaac Forsyth
Book of Isaiah from the New international version.
Book of Isaiah from the New international version.
history of the First Presbyterian Church of Lincoln, Nebraska
history of the First Presbyterian Church of Lincoln, Nebraska
Wisconsin industry and the Wisconsin tax system
Wisconsin industry and the Wisconsin tax system
Look on Canada, now, and see history anew, an epoch past and a new life fashioning under your hands.
Look on Canada, now, and see history anew, an epoch past and a new life fashioning under your hands.
Time In, Time Out
Time In, Time Out
U.S. nuclear policy in the 21st century
U.S. nuclear policy in the 21st century
Media programs in selected public elementary school in Oklahoma
Media programs in selected public elementary school in Oklahoma
The Trespassers
The Trespassers
Farmers diary, or, Catskill almanack, for the year of our Lord 1819 ...
Farmers diary, or, Catskill almanack, for the year of our Lord 1819 ...
GOT TO BE U SNOOPY (Youve Got to Be You, Snoopy)
GOT TO BE U SNOOPY (Youve Got to Be You, Snoopy)
### College algebra with graphing and problem solving by Karl J. Smith Download PDF EPUB FB2
College Algebra through Problem Solving by Cifone, Puri, Maslanko, & Dabkowska. Section 1: Linear and Absolute Value Equations. Linear Equations. We begin our exploration by discussing linear equations in one variable.
An example of such an equation, and the question that we may be asked is below: Problem Solve for T. Check your : Danielle Cifone, Karan Puri, Debra Masklanko, Ewa Dabkowska. Known for a clear and concise exposition, numerous examples, and plentiful problem sets, Jerome E.
Kaufmann and Karen L. Schwitters's COLLEGE ALGEBRA is an easy-to-use book that focuses on building technique and helping students hone their problem-solving skills.
The eighth edition focuses on solving equations, inequalities, and problems; and on developing graphing techniques. Additional Physical Format: Online version: Smith, Karl J.
College algebra with graphing and problem solving. Pacific Grove, CA: Brooks/Cole Pub. Co., Get this from a library. College algebra with graphing and problem solving.
[Karl J Smith]. College Algebra. Systems of Linear Equations: Two Variables. Search for: Solving Systems of Equations by Graphing.
In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation.
I was trying to figure out how to solve word problems for my College Algebra class. Your website was very helpful. problem, distance rate time, rate time distance problems, motion word problems, Equations & inequalities, Motion Word Problems; 48 Solving and Graphing Inequalities Solving Inequalities, Solving and Graphing Inequalities, solve.
Basic Algebra The Laws of Algebra Terminology and Notation. In this section we review the notations used in algebra. Some are peculiar to this book. For example the notation A:= B indicates that the equality holds by de nition of the notations involved.
Two other notations which will become important when we solve equations are =) and (). Section Graphing. For problems 1 – 3 construct a table of at least 4 ordered pairs of points on the graph of the equation and use the ordered pairs from the table to sketch the graph of the equation.
$$y = 3x + 4$$ Solution $$y = 1 - {x^2}$$ Solution $$y = 2 + \sqrt x$$ Solution. Madison College Textbook for College Mathematics Revised Fall of Edition. Section Solving Equations and Rearranging Formulas Chapter 4 Algebra and Graphs of Lines Section Graphing a Linear Equation Using a Tabl e of Values.
Chapter Outline Real Numbers: Algebra Essentials Exponents and Scientific Notation Radicals and Rational Exponents Polynomials Factorin. Known for a clear and concise exposition, numerous examples, and plentiful problem sets, Jerome E.
Kaufmann and Karen L. Schwitters's COLLEGE ALGEBRA is an easy-to-use book that focuses on building technique and helping students hone their problem-solving s: COLLEGE ALGEBRA WITH APPLICATIONS FOR BUSINESS AND LIFE SCIENCES meets the demand for courses that emphasize problem solving, modeling, and real-world applications for business and the life sciences.
The authors provide a firm foundation in algebraic concepts and prompt students to apply their understanding to relevant examples and applications they are likely to encounter in college. About College Algebra.
College Algebra provides a comprehensive exploration of algebraic principles and meets scope and sequence requirements for a typical introductory algebra course.
The modular approach and richness of content ensure that the book meets the needs of a variety of courses.
makes available usable answers on algebra problems about root square, intermediate algebra syllabus and graphing linear equations and other algebra subjects. Should you have to have advice on equations as well as substitution, is really the ideal place to.
Word problems relate algebra to familiar situations, helping students to understand abstract concepts. Students develop understanding by solving equations and inequalities intuitively before formal solutions are introduced. Students begin their study of algebra in Books using only integers.
Books introduce rational numbers and expressions. Solving Linear Equations. Problem Solving and Using Formulas. Applications of Algebra. Mid-Chapter Test: Sections – Additional Application Problems. Solving Linear Inequalities. Solving Equations and Inequalities Containing Absolute Values.
Graphs and Functions. Graphs. Functions. Linear Functions. Intermediate Algebra Problems With Answers - sample 2:Find equation of line, domain and range from graph, midpoint and distance of line segments, slopes of perpendicular and parallel lines.
Intermediate Algebra Problems With Answers - sample 3: equations and system of equations, quadratic equations, function given by a table, intersections of. confuse a problem like − 3 − 8 with − 3(− 8).
The second problem is a multipli-cation problem because there is nothing between the 3 and the parenthesis. If there is no operation written in between the parts, then we assume that means we are multiplying. The − 3 − 8 problem. Analyzing Graphs of Quadratic Functions.
Find the vertex, the axis of symmetry, and the maximum or minimum value of a quadratic function using the method of completing the square; Graph quadratic functions; Solve applied problems involving maximum and minimum function values; Solving Rational Equations and Radical Equations.
College Algebra by Avinash Sathaye, Professor of Mathematics 1 Department of Mathematics, University of Kentucky A warning about graphing. Graphs are a big help in understanding the prob- by myself and UK colleague Dr.
Paul Eakin using the Maple problem solving system. offers valuable material on pre algebra graphing linear inequalities, subtracting polynomials and solving systems and other algebra subjects.
If you have to have guidance on functions or scientific, is undoubtedly the excellent destination to check-out! The Barnett Graphs & Models series in college algebra and precalculus maximizes student comprehension by emphasizing computational skills, real-world data analysis and modeling, and problem solving rather than mathematical theory.
Many examples feature side-by-side algebraic and graphical solutions, and each is followed by a matched problem for the student to s: College Algebra through Problem Solving by Cifone, Puri, Maslanko, & Dabkowska.
Section 1: Linear and Absolute Value Equations. Linear Equations. We begin our exploration by discussing linear equations in one variable. An example of such an equation, and the question that we may be asked is below: Problem Solve for T. Check your solution.
|
2021-09-17 19:47:14
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3717871308326721, "perplexity": 3282.55318666799}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780055775.1/warc/CC-MAIN-20210917181500-20210917211500-00290.warc.gz"}
|
https://blogdacrianca.com/2fufaf/dwarf-spheroidal-galaxy-a8627b
|
The main body of the galaxy, strongly sheared by tidal forces, is a triaxial (almost prolate) ellipsoid with its longest principal axis of inertia inclined $43 \pm 6 ^\circ$ with respect to the plane of the sky and axes ratios of 1:0.67:0.60. Aims: In order to minimize environmental effects and gain an insight into the internal mechanisms that shape the properties of these systems, we study one of the … The Antlia Dwarf (PGC 29194), or Antlia Dwarf Galaxy, is a dwarf spheroidal galaxy located in the Antlia constellation. Draco Dwarf, is a spheroidal located in the constellation Draco . only exception. The Carina Dwarf Spheroidal galaxy is a satellite galaxy of the Milky Way and is receding from it at 230 km/s. A dwarf spheroidal galaxy (left centre) 11 million light-years away, elongated by the pull of a giant galaxy. ADS. Burkert (1997) The Pegasus Dwarf Spheroidal (also known as Andromeda VI or Peg dSph for short) is a dwarf spheroidal galaxy about 2.7 million light-years away in the constellation Pegasus. give the same orders of magnitude. galaxies. From the Virial theorem, we straightforwardly deduce, where is the central density, the central velocity of dark matter, and thus contribute greatly to the mass of the The galaxy is approximately 4.3 million light years distant from Earth. An elliptical dwarf galaxy in the constellation Fornax. The dwarf spheroidal galaxy Donatiello I, discovered by amateur astronomer Giuseppe Donatiello, sits in the middle of this composite image. brightness. The Pegasus Dwarf Spheroidal (also known as Andromeda VI or Peg dSph for short) is a dwarf spheroidal galaxy about 2.7 million light-years away in the constellation Pegasus. central value. Nevertheless, most workers on this topic generally body, at a galactocentric radius equal to this critical value, r, dwarf-spheroidal-galaxy definition: Noun (plural dwarf spheroidal galaxies) 1. The great majority of early-type dwarf galaxies, in the Local Group as well as in other galaxy groups, are found in the vicinity of much larger galaxies, making it hard to disentangle the role of internal versus external effects in driving their evolution. NGC 147, a dwarf, spheroidal galaxy of the Local Group. It was discovered in 1997 by Mike Irwin, Alan Whiting, and George Hau in a survey of the northern sky. The Sagittarius Dwarf (Sgr), shown as the extended irregular shape below the Galactic Center, is the closest of 9 known small dwarf spheroidal galaxies that orbit our Galaxy. It seems to be too large, when m is obtained from the surface If luminous and dark matter have the same spatial distribution, it is We have obtained precise (〈σ〉=3.1 km s -1) radial velocities of 23 photometrically-selected normal giant stars located near the center of the Carina dwarf spheroidal (dSph) galaxy. dispersion and Rc an equivalent radius, or core Dwarf spheroidal galaxies also contain a large amount of dark matter. Search for other works by this author on: Oxford Academic. Within this radius, the mass would be They are found in the Local Group as companions of the Milky Way, and to systems that are companions to the Andromeda Galaxy. investigated four dwarf spheroidal galaxies orbiting around 1.3 Sculptor Dwarf Spheroidal Galaxy The Sculptor Dwarf Spheroidal Galaxy1 (hereafter Sculp-tor) is ideal as a low-density counterpart to globular clus-ters. obtains a mass of b) The velocity dispersion of the stellar system.- This method has Of this sample, 17 stars are Carina members based on their heliocentric radial velocities in excess of 200 km s^-1^. Dwarf Spheroidal Galaxies: Keystones of Galaxy Evolution John S. Gallagher III Department of Astronomy, 5534 Sterling Hall, University of Wisconsin, 475 North Charter Street, Madison, Wisconsin 53706 Electronic mail: [email protected] Rosemary F. G. Wyse Department of Physics and Astronomy,1 Johns Hopkins University, Baltimore, Maryland 21218 We have obtained precise (<σ> = 3.1 km s^-1^) radial velocities of 23 photometrically-selected normal giant stars located near the center of the Carina dwarf spheroidal (dSph) galaxy. Therefore, the determination of DM is more problematic. have been carried out by Donatiello I is a tiny galaxy about 10 million light-years away, and it was discovered by an Italian hobbyist with a homemade telescope. would be the A 2019 study showed that the grouping of stars is no star cluster at all; instead, it’s the hollowed-out shell of a dwarf spheroidal galaxy that merged with the Milky Way. Mateo (1997) 0.1 - 1Mpc-3) higher than luminous matter; indeed, ratios of 100 have been reported as The possibility that all dwarf galaxies may have the same mass, despite Suppose a satellite dwarf with mass m and radius r orbiting Google Scholar. A dwarf spheroidal galaxy (dSph) is a term in astronomy applied to small, low-luminosity galaxies with very little dust and an older stellar population. Hence, Klesen and Kroupa (1998) in spheroidal (dSph) galaxy. well as large amounts and concentrations of dark matter, but severe Find out information about dwarf spheroidal galaxy. Take your favorite fandoms with you and never miss a beat. models with no dark matter at all cannot be excluded. between the two galaxies. ratio. More (1996) The Carina Dwarf Spheroidal galaxy is a satellite galaxy of the Milky Way and is receding from it at 230 km/s. are basically two methods for detecting dark matter in this type of the satellite in turn would not require this component. Carina Dwarf was discovered in 1977 with the UK Schmidt Telescope. This type of galaxy is probably the most common in the Universe. Modern instrumentation unlocks the possibility of scrutinizing Also, the graphics used for the McKay/Carter Intergalactic Gate Bridgeall show the Pegasus Dwarf Irregular Galaxy. agree that these galaxies have A dwarf spheroidal galaxy (dSph) is a term in astronomy applied to small, low-luminosity galaxies with very little dust and an older stellar population. around the primary galaxy with mass M, with R being the with the radius at which the surface brightness is one half the Astronomy Wiki is a FANDOM Lifestyle Community. The inventory keeps growing, and the latest recruit is a dwarf spheroidal galaxy, meaning no gas and not many more than 10 6-7 stars, that has the … It has been suggested that these types of galaxies may be dominated by dark matter. easy to deduce the central density and the central mass-to-luminosity Dwarf spheroidal galaxies (dSphs) have been extensively investigated in the Local Group, but their low luminosity and surface bright- ness make similar work in more distant galaxy groups challenging. The diameter of the galaxy is about 1600 light-years, which is 75 times smaller than the Milky Way. From this subsample, we derive a mean systemic velocity for Carina of 223.1±1.8 km s -1, and a … Universe. obtain brightness with a constant M/L relation. It should be emphasized that the mass obtained depends on the third power stars would escape and would be trapped in the gravitational field of These first used this procedure for estimating dark matter. the primary. observational difficulties prevent their precise determination. the Milky Way: Sextans, Carina, Ursa Minor and Draco, considering disruption of a satellite produces a remnant that contains about 1% There are actually two galaxies in the Local Group named Pegasus: Pegasus Dwarf Irregular Galaxy and Pegasus Dwarf Spheroidal Galaxy. Mark I. Wilkinson, It contains four globular clusters, with the brightest of them – NGC 6715 (M54) – being known well before the discovery of the galaxy itself in 1994. yet present a high stellar velocity dispersion. is difficult to determine, and so it must be obtained by extrapolation. Although often thought in the past to be merely large, low-density globular clusters, recent studies have shown that dwarf spheroidal (dSph) galaxies have a more complex stellar population than that found in globulars. Dwarf spheroidal galaxies (dSphs) have been extensively investigated in the Local Group, but their low luminosity and surface bright-ness make similar work in more distant galaxy groups challenging. Mark I. Wilkinson, distance Those last two points are what makes dwarf spheroidal galaxies such great environments for X-ray studies. The two forces become They are found in the Local Group as companions to the Milky Way and to systems that are companions to the Andromeda Galaxy (M31). Estimates of the mass distribution and dark-matter (DM) content of dwarf spheroidal galaxies (dSphs) are usually derived under the assumption that the effect of the tidal field of the host galaxy is negligible over the radial extent probed by kinematic data sets. The Pegasus Dwarf is a member of the Local group of galaxies and a satellite galaxy of the Andromeda Galaxy (M31). autogravitation to match tidal disruption, by means of a rough model. sē] (astronomy) One of the smallest and faintest of the dwarf galaxies, with an effective radius of 200-1000 parsecs and an absolute visual magnitude between -8 and -13. 1994 Beam Line: Fall Winter 1994. Donatiello I is a tiny galaxy about 10 million light-years away, and it was discovered by an Italian hobbyist with a homemade telescope. One of the smallest and faintest of the dwarf galaxies, with an effective radius of 200-1000 parsecs and an absolute visual magnitude between -8 and … which is probably unrealistic, unless this type of galaxy is the A dwarf galaxy is a small galaxy composed of about 1000 up to several billion stars, as compared to the Milky Way's 200–400 billion stars. Even in which dwarf 400 pc, limiting their M/L ratio to less than about 100. They are found in the Local Group as companions to the Milky Way and to systems that are companions to the Andromeda Galaxy (M31). instance, could become tidally disrupted if they did not have enough dark Dwarf spheroidal galaxies appear to concentrate more towards the centers of galactic groups and clusters than any other type of galaxy, including spiral galaxies, large elliptical galaxies or the relatively dust-free S0DISK GALAXIES. of the estimated tidal radius, which is an important source of From the observational point of view, the tidal radius The study of Despite their low luminosity, they may contain large amounts This type of galaxy is probably the most common in the radius for Ursa Minor, and (Richstone and Tremaine, 1986; +9.1.. The constant should be 9/2, instead 3/, as deduced by more produced by the primary But in others, the analysis is complemented with There Estimates of the mass distribution and dark-matter (DM) content of dwarf spheroidal galaxies (dSphs) are usually derived under the assumption that the effect of the tidal field of the host galaxy is negligible over the radial extent probed by kinematic data sets. Ashman, 1993). The Sculptor is one of the closest dwarf galaxies, and is at a favourable Galactic latitude outside the plane of … In "The Return, Part 1", Major General Henry Landry states that Pegasus is three million light years from the Milky Way, implying that the Pegasus seen in Stargate: Atlantis is the Pegasus Dwarf Irregular Galaxy. Sgr d … It orbits over the top and down below the disk of our galaxy, like a ring over a spinning top. Don't worry, our Galaxy is not in danger, but no such assurances are issued for the Sagittarius Dwarf: the intense gravitational tidal forces might pull it apart. This has been inferred from studies that have shown that the mass derived from the motions of stars within the galaxies are much larger than what can be accounted for by the luminous matter in the galaxy. Rc3 and the luminosity, Hodge and Michie (1969) galaxies could be unbound and losing stars. en) Sagittarius Dwarf Elliptical Galaxy in SIMBAD; Sgr dE op de NASA/IPAC Extragalactic Database 3.3 Dwarf spheroidal galaxies. their tidal radii, and concluded that Sextans is dark matter dominated Thearticle reportscoordinatesfor the … very high central condensations in the range ( Then, they might contain no dark matter at all and The galaxy is approximately 4.3 million light years distant from Earth. For reference, the full moon is 30' (arc minutes) or 0.5° in size. However, dwarf spheroidals do not possess gas in the peripherical effects which are now completely absent. A dwarf spheroidal galaxy (dSph) is a term in astronomy applied to low luminosity galaxies that are companions to the Milky Way and to the similar systems that are companions to the Andromeda Galaxy (M31). The Carina Dwarf Spheroidal Galaxy is a dwarf galaxy in the Carina constellation.It was discovered in 1977 with the UK Schmidt Telescope by Cannon et al. Search for other works by this author on: Oxford Academic. It was discovered in 1997 by Mike Irwin, Alan Whiting, and George Hau in a survey of the northern sky. near the satellite. The color-magnitude diagram of the dwarf spheroidal galaxy in Draco is redetermined, and forms the basis of a discussion of the evolutionary status of stars in this distant, metal-poor satellite of the Galaxy. 1992). spheroidals irrespective of their luminosity. A dwarf spheroidal galaxy (dSph) is a very common type of galaxy that is very low-luminosity, has very little dust, and has an older star population. Natural units are used throughout. It’s about 50,000 light-years away from the Milky Way center. SHELL STRUCTURE IN THE FORNAX DWARF SPHEROIDAL GALAXY Matthew Coleman and G. S. Da Costa Research School of Astronomy and Astrophysics, Institute of Advanced Studies, Australian National University, Cotter Road, Weston Creek, ACT 2611, Australia; [email protected], [email protected] Joss Bland-Hawthorn The Carina Dwarf Spheroidal Galaxy is a dwarf galaxy in the Carina constellation.It was discovered in 1977 with the UK Schmidt Telescope by Cannon et al. drop to zero. galaxies to conclude that their dark matter halos must be truncated at As the dwarf is not a rigid The dwarf spheroidal galaxy Donatiello I, discovered by amateur astronomer Giuseppe Donatiello, sits in the middle of this composite image. a large DM content, maybe 10 times SHELL STRUCTURE IN THE FORNAX DWARF SPHEROIDAL GALAXY Matthew Coleman and G. S. Da Costa Research School of Astronomy and Astrophysics, Institute of Advanced Studies, Australian National University, Cotter Road, Weston Creek, ACT 2611, Australia; Summarizing, dwarf spheroidal galaxies could contain (Mateo et al. Fornax Dwarf Spheroidal Fornax Dwarf Spheroidal annotated Date: August 2018 Location: Bateleur Nature Reserve, South Africa CCD: QHY22 Scope: Takahashi Epsilon 130D @F3.3 Mount: Takahashi EM10 FS2 GoTo Dwarf spheroidal galaxies (dSph) are among the most peculiar types of galaxies in the Universe. The tidal If the ancient Galactic globular clusters, like M92, formed concurrently with the early formation of the Milky Way galaxy itself, then the Ursa Minor dwarf spheroidal is probably as old as the Milky Way. Explanation of dwarf spheroidal galaxy which the dwarf satellites are partially disrupted in perigalactic galaxies: a) Tidal radii.- The dwarf spheroidal satellites of the Milky Way, for Pictured on the lower left is one of the many dwarf ellipticals: NGC 205. velocity dispersions are low, of the order of 10 kms-1, dwarf spheroidal galaxy are ancient. - GMmr/r3. The Sagittarius Dwarf Spheroidal Galaxy was discovered in 1994, and at the time astronomers thought it was the closest dwarf galaxy to the Milky Way. dwarf-spheroidal-galaxy definition: Noun (plural dwarf spheroidal galaxies) 1. The Sagittarius Dwarf Spheroidal Galaxy (Sgr dSph), also known as the Sagittarius Dwarf Elliptical Galaxy (Sgr dE or Sag DEG), is an elliptical loop-shaped satellite galaxy of the Milky Way. More precise calculations (e.g. Find out information about dwarf spheroidal galaxy. Ashman, 1983). A dwarf spheroidal galaxy is a low-luminosity (less than absolute visual magnitude -14), low-surface-brightness dwarf elliptical galaxy of near-spherical shape that lacks a nucleus. https://astronomical.fandom.com/wiki/Dwarf_spheroidal_galaxy?oldid=8103. used the observation of stars being removed in the tidal tails of dwarf Kuhn and Miller (1989), Sector Borealis Deep Space ℗ Sector Borealis Released on: 2019-09-24 Auto-generated by YouTube. passages; orbiting condensations can be identified after this event The galaxy shows signs of … dwarf spheroidal galaxy ( plural dwarf spheroidal galaxies ) ( astronomy) A faint galaxy, devoid of gas, having a higher than normal proportion of dark matter; especially those that orbit the Milky Way and Andromeda quotations . The dSph galaxies are character- ized by small size, low luminosity, low surface brightness, and old- to intermediate-age stellar populations. would attract one another with a force of the order of Suppose the dwarf divided into two halves. Salucci and Persic (1997) However, dwarf spheroidals do not possess gas in … Let us determine the radius of the satellite necessary for periphery, as do bright Modern instrumentation unlocks the possibility of scrutinizing Google Scholar. spheroidal galaxies were not considered as virialized systems. Dwarf Spheroidal Galaxies ENCYCLOPEDIA OF ASTRONOMY AND STROPHYSICS Dwarf Spheroidal Galaxies Our galaxy, the MILKY WAY, is surrounded by a swarm of dwarf galaxies each composed of 100000 to billions of stars.
2020 dwarf spheroidal galaxy
|
2021-01-27 10:26:20
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6098534464836121, "perplexity": 1979.2828575788449}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704821381.83/warc/CC-MAIN-20210127090152-20210127120152-00309.warc.gz"}
|
http://www.ams.org/mathscinet-getitem?mr=MR0487472
|
MathSciNet bibliographic data MR487472 (58 #7102) 46J99 (17C99) Wright, J. D. Maitland; Youngson, M. A. A Russo-Dye theorem for Jordan \$C\sp*\$$C\sp*$-algebras. Functional analysis: surveys and recent results (Proc. Conf., Paderborn, 1976), pp. 279–282. North-Holland Math. Studies, Vol. 27; Notas de Mat., No. 63, North-Holland, Amsterdam, 1977. Links to the journal or article are not yet available
For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.
|
2015-11-28 06:47:42
|
{"extraction_info": {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9647230505943298, "perplexity": 7219.044242162957}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398451648.66/warc/CC-MAIN-20151124205411-00040-ip-10-71-132-137.ec2.internal.warc.gz"}
|
https://data-analytics.netlify.app/machine-learning/2018-02_logistic_regression/
|
# Logistic Regression
Don’t get confused by its name! It is a classification not a regression algorithm. It is used to estimate discrete values (binary values like 0/1, yes/no, true/false ) based on given set of independent variable(s). In simple words, it predicts the probability of occurrence of an event by fitting data to a logit function. Hence, it is also known as logit regression. Since, it predicts the probability, its output values lies between 0 and 1 (as expected).
Again, let us try and understand this through a simple example.
Let’s say your friend gives you a puzzle to solve. There are only two outcome scenarios – either you solve it or you don’t. Now imagine, that you are being given wide range of puzzles / quizzes in an attempt to understand which subjects you are good at. The outcome to this study would be something like this – if you are given a trignometry based tenth grade problem, you are 70% likely to solve it. On the other hand, if it is grade fifth history question, the probability of getting an answer is only 30%. This is what Logistic Regression provides you.
Coming to the math, the log odds of the outcome is modeled as a linear combination of the predictor variables.
odds = p/ (1-p) = probability of event occurrence / probability of not event occurrence ln(odds) = ln(p/(1-p)) logit(p) = ln(p/(1-p)) = b0+b1X1+b2X2+b3X3….+bkXk
Above, p is the probability of presence of the characteristic of interest. It chooses parameters that maximize the likelihood of observing the sample values rather than that minimize the sum of squared errors (like in ordinary regression).
Now, you may ask, why take a log? For the sake of simplicity, let’s just say that this is one of the best mathematical way to replicate a step function. I can go in more details, but that will beat the purpose of this article.
Logistic_RegressionPython Code
#Import Library
from sklearn.linear_model import LogisticRegression
#Assumed you have, X (predictor) and Y (target) for training data set and x_test(predictor) of test_dataset
# Create logistic regression object
model = LogisticRegression()
# Train the model using the training sets and check score
model.fit(X, y)
model.score(X, y)
#Equation coefficient and Intercept
print('Coefficient: \n', model.coef_)
print('Intercept: \n', model.intercept_)
#Predict Output
predicted= model.predict(x_test)
R Code
x <- cbind(x_train,y_train)
# Train the model using the training sets and check score
logistic <- glm(y_train ~ ., data = x,family='binomial')
summary(logistic)
#Predict Output
predicted= predict(logistic,x_test)
Furthermore..
There are many different steps that could be tried in order to improve the model:
including interaction terms
removing features
regularization techniques
using a non-linear model
|
2020-06-02 21:25:33
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7357454299926758, "perplexity": 1043.9684785315337}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347426801.75/warc/CC-MAIN-20200602193431-20200602223431-00165.warc.gz"}
|
https://secure.sky-map.org/starview?object_type=1&object_id=1043&object_name=Anser&locale=ZH
|
SKY-MAP.ORG
首页 开始 To Survive in the Universe News@Sky 天文图片 收集 论坛 Blog New! 常见问题 新闻 登录
# α Vul (Anser)
### 图像
DSS Images Other Images
### 相关文章
CHARM2: An updated Catalog of High Angular Resolution MeasurementsWe present an update of the Catalog of High Angular ResolutionMeasurements (CHARM, Richichi & Percheron \cite{CHARM}, A&A,386, 492), which includes results available until July 2004. CHARM2 is acompilation of direct measurements by high angular resolution methods,as well as indirect estimates of stellar diameters. Its main goal is toprovide a reference list of sources which can be used for calibrationand verification observations with long-baseline optical and near-IRinterferometers. Single and binary stars are included, as are complexobjects from circumstellar shells to extragalactic sources. The presentupdate provides an increase of almost a factor of two over the previousedition. Additionally, it includes several corrections and improvements,as well as a cross-check with the valuable public release observationsof the ESO Very Large Telescope Interferometer (VLTI). A total of 8231entries for 3238 unique sources are now present in CHARM2. Thisrepresents an increase of a factor of 3.4 and 2.0, respectively, overthe contents of the previous version of CHARM.The catalog is only available in electronic form at the CDS viaanonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or via http://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/431/773 Local kinematics of K and M giants from CORAVEL/Hipparcos/Tycho-2 data. Revisiting the concept of superclustersThe availability of the Hipparcos Catalogue has triggered many kinematicand dynamical studies of the solar neighbourhood. Nevertheless, thosestudies generally lacked the third component of the space velocities,i.e., the radial velocities. This work presents the kinematic analysisof 5952 K and 739 M giants in the solar neighbourhood which includes forthe first time radial velocity data from a large survey performed withthe CORAVEL spectrovelocimeter. It also uses proper motions from theTycho-2 catalogue, which are expected to be more accurate than theHipparcos ones. An important by-product of this study is the observedfraction of only 5.7% of spectroscopic binaries among M giants ascompared to 13.7% for K giants. After excluding the binaries for whichno center-of-mass velocity could be estimated, 5311 K and 719 M giantsremain in the final sample. The UV-plane constructed from these datafor the stars with precise parallaxes (σπ/π≤20%) reveals a rich small-scale structure, with several clumpscorresponding to the Hercules stream, the Sirius moving group, and theHyades and Pleiades superclusters. A maximum-likelihood method, based ona Bayesian approach, has been applied to the data, in order to make fulluse of all the available stars (not only those with precise parallaxes)and to derive the kinematic properties of these subgroups. Isochrones inthe Hertzsprung-Russell diagram reveal a very wide range of ages forstars belonging to these groups. These groups are most probably relatedto the dynamical perturbation by transient spiral waves (as recentlymodelled by De Simone et al. \cite{Simone2004}) rather than to clusterremnants. A possible explanation for the presence of younggroup/clusters in the same area of the UV-plane is that they have beenput there by the spiral wave associated with their formation, while thekinematics of the older stars of our sample has also been disturbed bythe same wave. The emerging picture is thus one of dynamical streamspervading the solar neighbourhood and travelling in the Galaxy withsimilar space velocities. The term dynamical stream is more appropriatethan the traditional term supercluster since it involves stars ofdifferent ages, not born at the same place nor at the same time. Theposition of those streams in the UV-plane is responsible for the vertexdeviation of 16.2o ± 5.6o for the wholesample. Our study suggests that the vertex deviation for youngerpopulations could have the same dynamical origin. The underlyingvelocity ellipsoid, extracted by the maximum-likelihood method afterremoval of the streams, is not centered on the value commonly acceptedfor the radial antisolar motion: it is centered on < U > =-2.78±1.07 km s-1. However, the full data set(including the various streams) does yield the usual value for theradial solar motion, when properly accounting for the biases inherent tothis kind of analysis (namely, < U > = -10.25±0.15 kms-1). This discrepancy clearly raises the essential questionof how to derive the solar motion in the presence of dynamicalperturbations altering the kinematics of the solar neighbourhood: doesthere exist in the solar neighbourhood a subset of stars having no netradial motion which can be used as a reference against which to measurethe solar motion?Based on observations performed at the Swiss 1m-telescope at OHP,France, and on data from the ESA Hipparcos astrometry satellite.Full Table \ref{taba1} is only available in electronic form at the CDSvia anonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/430/165} Ca II K Emission-Line Asymmetries Among Red GiantsMeasurements of the asymmetry of the K2 emission line of CaII have been made for a sample of bright field giants with B-V>1.15observed with the Cassegrain echelle spectrometer on the McDonaldObservatory 2.1 m telescope. The asymmetry of the Ca II K2line is quantified through measurement of a parameter V/R, which isdefined as the ratio between the maximum counts recorded in the violetand red components of the double-peaked emission profile. Red-maximumasymmetry (V/R<1.0) is found in our sample of 35 giants only amongstars with B-V>1.35, a trend that is still maintained (with oneexception) with the inclusion of an additional sample of giantspreviously observed by us with the same spectrograph. Althoughexceptional stars can be found in the literature, the data support anearlier finding by R. Stencel that among luminosity class III fieldgiants the occurrence of V/R<1.0 is generally restricted to effectivetemperatures cooler than 4320 K. This limit may coincide with the onsetof pulsation. Improved Baade-Wesselink surface brightness relationsRecent, and older accurate, data on (limb-darkened) angular diameters iscompiled for 221 stars, as well as BVRIJK[12][25] magnitudes for thoseobjects, when available. Nine stars (all M-giants or supergiants)showing excess in the [12-25] colour are excluded from the analysis asthis may indicate the presence of dust influencing the optical andnear-infrared colours as well. Based on this large sample,Baade-Wesselink surface brightness (SB) relations are presented fordwarfs, giants, supergiants and dwarfs in the optical and near-infrared.M-giants are found to follow different SB relations from non-M-giants,in particular in V versus V-R. The preferred relation for non-M-giantsis compared to the earlier relation by Fouqué and Gieren (basedon 10 stars) and Nordgren et al. (based on 57 stars). Increasing thesample size does not lead to a lower rms value. It is shown that theresiduals do not correlate with metallicity at a significant level. Thefinally adopted observed angular diameters are compared to thosepredicted by Cohen et al. for 45 stars in common, and there isreasonable overall, and good agreement when θ < 6 mas.Finally, I comment on the common practice in the literature to average,and then fix, the zero-point of the V versus V-K, V versus V-R and Kversus J-K relations, and then rederive the slopes. Such a commonzero-point at zero colour is not expected from model atmospheres for theV-R colour and depends on gravity. Relations derived in this way may bebiased. Unveiling Mira stars behind the molecules. Confirmation of the molecular layer model with narrow band near-infrared interferometryWe have observed Mira stars with the FLUOR beamcombiner on the IOTAinterferometer in narrow bands around 2.2 μm wavelength. We findsystematically larger diameters in bands contaminated by water vapor andCO. The visibility measurements can be interpreted with a modelcomprising a photosphere surrounded by a thin spherical molecular layer.The high quality of the fits we obtain demonstrates that this simplemodel accounts for most of the star's spatial structure. For each starand each period we were able to derive the radius and temperature of thestar and of the molecular layer as well as the optical depth of thelayer in absorption and continuum bands. The typical radius of themolecular layer is 2.2 R* with a temperature ranging between1500 and 2100 K. The photospheric temperatures we find are in agreementwith spectral types of Mira stars. Our photospheric diameters are foundsmaller than in previous studies by several tens of percent. We believeprevious diameters were biased by the use of unsuited geometrical modelsto explain visibilities. The conclusions of this work are various.First, we offer a consistent view of Mira stars over a wide range ofwavelengths. Second, the parameters of the molecular layer we find areconsistent with spectroscopic studies. Third, from our diametermeasurements we deduce that all Mira stars are fundamental modepulsators and that previous studies leading to the conclusion of thefirst-overtone mode were biased by too large diameter estimates.Based on observations collected at the IOTA interferometer, WhippleObservatory, Mount Hopkins, Arizona.Table 3 is only available in electronic form athttp://www.edpsciences.org Angular Diameters of Stars from the Mark III Optical InterferometerObservations of 85 stars were obtained at wavelengths between 451 and800 nm with the Mark III Stellar Interferometer on Mount Wilson, nearPasadena, California. Angular diameters were determined by fitting auniform-disk model to the visibility amplitude versus projected baselinelength. Half the angular diameters determined at 800 nm have formalerrors smaller than 1%. Limb-darkened angular diameters, effectivetemperatures, and surface brightnesses were determined for these stars,and relationships between these parameters are presented. Scatter inthese relationships is larger than would be expected from themeasurement uncertainties. We argue that this scatter is not due to anunderestimate of the angular diameter errors; whether it is due tophotometric errors or is intrinsic to the relationship is unresolved.The agreement with other observations of the same stars at the samewavelengths is good; the width of the difference distribution iscomparable to that estimated from the error bars, but the wings of thedistribution are larger than Gaussian. Comparison with infraredmeasurements is more problematic; in disagreement with models, coolerstars appear systematically smaller in the near-infrared than expected,warmer stars larger. Hipparcos red stars in the HpV_T2 and V I_C systemsFor Hipparcos M, S, and C spectral type stars, we provide calibratedinstantaneous (epoch) Cousins V - I color indices using newly derivedHpV_T2 photometry. Three new sets of ground-based Cousins V I data havebeen obtained for more than 170 carbon and red M giants. These datasetsin combination with the published sources of V I photometry served toobtain the calibration curves linking Hipparcos/Tycho Hp-V_T2 with theCousins V - I index. In total, 321 carbon stars and 4464 M- and S-typestars have new V - I indices. The standard error of the mean V - I isabout 0.1 mag or better down to Hp~9 although it deteriorates rapidly atfainter magnitudes. These V - I indices can be used to verify thepublished Hipparcos V - I color indices. Thus, we have identified ahandful of new cases where, instead of the real target, a random fieldstar has been observed. A considerable fraction of the DMSA/C and DMSA/Vsolutions for red stars appear not to be warranted. Most likely suchspurious solutions may originate from usage of a heavily biased color inthe astrometric processing.Based on observations from the Hipparcos astrometric satellite operatedby the European Space Agency (ESA 1997).}\fnmsep\thanks{Table 7 is onlyavailable in electronic form at the CDS via anonymous ftp tocdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/397/997 A catalogue of calibrator stars for long baseline stellar interferometryLong baseline stellar interferometry shares with other techniques theneed for calibrator stars in order to correct for instrumental andatmospheric effects. We present a catalogue of 374 stars carefullyselected to be used for that purpose in the near infrared. Owing toseveral convergent criteria with the work of Cohen et al.(\cite{cohen99}), this catalogue is in essence a subset of theirself-consistent all-sky network of spectro-photometric calibrator stars.For every star, we provide the angular limb-darkened diameter, uniformdisc angular diameters in the J, H and K bands, the Johnson photometryand other useful parameters. Most stars are type III giants withspectral types K or M0, magnitudes V=3-7 and K=0-3. Their angularlimb-darkened diameters range from 1 to 3 mas with a median uncertaintyas low as 1.2%. The median distance from a given point on the sky to theclosest reference is 5.2degr , whereas this distance never exceeds16.4degr for any celestial location. The catalogue is only available inelectronic form at the CDS via anonymous ftp to cdsarc.u-strasbg.fr(130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/393/183 CHARM: A Catalog of High Angular Resolution MeasurementsThe Catalog of High Angular Resolution Measurements (CHARM) includesmost of the measurements obtained by the techniques of lunaroccultations and long-baseline interferometry at visual and infraredwavelengths, which have appeared in the literature or have otherwisebeen made public until mid-2001. A total of 2432 measurements of 1625sources are included, along with extensive auxiliary information. Inparticular, visual and infrared photometry is included for almost allthe sources. This has been partly extracted from currently availablecatalogs, and partly obtained specifically for CHARM. The main aim is toprovide a compilation of sources which could be used as calibrators orfor science verification purposes by the new generation of largeground-based facilities such as the ESO Very Large Interferometer andthe Keck Interferometer. The Catalog is available in electronic form atthe CDS via anonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/386/492, and from theauthors on CD-Rom. Spectrophotometry and structural analysis of 5 cometsWe discuss the morphology and spectrophotometry of 5 comets visible inAugust, 2001. We decompose comae into coma profiles and azimuthallyrenormalized images, in which general and local features arequantitatively comparable. Comet 19P/Borrelly showed a strong gas fantoward the solar direction, but no detectable gas in the tail. Dust inits inner coma was collimated toward the antisolar direction and thetail, with no dust in the outer coma. The contribution of spatialvariations structure was moderate, about 35%. Comet29P/Schwassmann-Wachmann 1 was observed in outburst: we detectedspinning'' jet structures. A high level of dust production resulted inan unusually high Afρ=16 600 cm. The spatial variations reached-77%, at the minimum, due in part to a jet and a ring-like structure in1 arcmin distance from the nucleus. In comet C/2001 A2, we detected astrong post-perihelion increase of dust and gas activity, in which theC2 profile became one magnitude brighter over a 3-day period.For comets C/2000 SV74 and C/2000 WM1, we present detailedpre-perihelion spectrophotometry and morphological information. CometC/2000 SV74 showed high dust production (Afρ=1479 cm). Its comasuggests a steady-state outflow of material, while the low contributionof spatial variations support high level activity. The coma of C/2000WM1 is dominated by solar effects, and CO+ forms the bulk of its gasactivity. Despite its large heliocentric distance, we observed a nicetail. Based on observations taken at the German-Spanish AstronomicalCentre, Calar Alto, operated by the Max-Planck-Institute for Astronomy,Heidelberg, jointly with the Spanish National Commission for Astronomy. Catalogue of Apparent Diameters and Absolute Radii of Stars (CADARS) - Third edition - Comments and statisticsThe Catalogue, available at the Centre de Données Stellaires deStrasbourg, consists of 13 573 records concerning the results obtainedfrom different methods for 7778 stars, reported in the literature. Thefollowing data are listed for each star: identifications, apparentmagnitude, spectral type, apparent diameter in arcsec, absolute radiusin solar units, method of determination, reference, remarks. Commentsand statistics obtained from CADARS are given. The Catalogue isavailable in electronic form at the CDS via anonymous ftp tocdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcar?J/A+A/367/521 The proper motions of fundamental stars. I. 1535 stars from the Basic FK5A direct combination of the positions given in the HIPPARCOS cataloguewith astrometric ground-based catalogues having epochs later than 1939allows us to obtain new proper motions for the 1535 stars of the BasicFK5. The results are presented as the catalogue Proper Motions ofFundamental Stars (PMFS), Part I. The median precision of the propermotions is 0.5 mas/year for mu alpha cos delta and 0.7mas/year for mu delta . The non-linear motions of thephotocentres of a few hundred astrometric binaries are separated intotheir linear and elliptic motions. Since the PMFS proper motions do notinclude the information given by the proper motions from othercatalogues (HIPPARCOS, FK5, FK6, etc.) this catalogue can be used as anindependent source of the proper motions of the fundamental stars.Catalogue (Table 3) is only available at the CDS via anonymous ftp tocdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strastg.fr/cgi-bin/qcat?J/A+A/365/222 Radiometric Validation of the Midcourse Space Experiment's (MSX) Point Source Catalogs and the MSX Properties of Normal StarsWe describe our methods to validate the absolute calibration of theinfrared (IR) radiometry of the Midcourse Space Experiment (MSX) PointSource Catalog, version 1.2 (PSC1.2). These are based upon stars drawnfrom our current all-sky network of absolute calibrators and upon otherstars that also support the calibration of ESA's Infrared SpaceObservatory and of several other satellites. Based on the mean ratios of(observed/predicted) irradiances for this ensemble of stars, all four ofMSX's mid-IR bands are consistent with this network; i.e., there are nomean ratios that deviate more than 3 σ from unity. However, thetwo narrowest bands (near 4.3 μm) are significantly discrepant by afew percent. This paper probes the radiometry of stars fainter thanthose validated by an independent study of the primary and secondary MSXstandards by Cohen et al. To provide the user of the PSC1.2 with a basisfor comparison, we derive the IR properties of normal stars frommeasurements by MSX's IR focal planes. Sixth Catalogue of Fundamental Stars (FK6). Part I. Basic fundamental stars with direct solutionsThe FK6 is a suitable combination of the results of the HIPPARCOSastrometry satellite with ground-based data, measured over more than twocenturies and summarized in the FK5. Part I of the FK6 (abbreviatedFK6(I)) contains 878 basic fundamental stars with direct solutions. Suchdirect solutions are appropriate for single stars or for objects whichcan be treated like single stars. From the 878 stars in Part I, we haveselected 340 objects as "astrometrically excellent stars", since theirinstantaneous proper motions and mean (time-averaged) ones do not differsignificantly. Hence most of the astrometrically excellent stars arewell-behaving "single-star candidates" with good astrometric data. Thesestars are most suited for high-precision astrometry. On the other hand,199 of the stars in Part I are Δμ binaries in the sense ofWielen et al. (1999). Many of them are newly discovered probablebinaries with no other hitherto known indication of binarity. The FK6gives, besides the classical "single-star mode" solutions (SI mode),other solutions which take into account the fact that hidden astrometricbinaries among "apparently single-stars" introduce sizable "cosmicerrors" into the quasi-instantaneously measured HIPPARCOS proper motionsand positions. The FK6 gives in addition to the SI mode the "long-termprediction (LTP) mode" and the "short-term prediction (STP) mode". TheseLTP and STP modes are on average the most precise solutions forapparently single stars, depending on the epoch difference with respectto the HIPPARCOS epoch of about 1991. The typical mean error of anFK6(I) proper motion in the single-star mode is 0.35 mas/year. This isabout a factor of two better than the typical HIPPARCOS errors for thesestars of 0.67 mas/year. In the long-term prediction mode, in whichcosmic errors are taken into account, the FK6(I) proper motions have atypical mean error of 0.50 mas/year, which is by a factor of more than 4better than the corresponding error for the HIPPARCOS values of 2.21mas/year (cosmic errors included). Spectral Irradiance Calibration in the Infrared. X. A Self-Consistent Radiometric All-Sky Network of Absolutely Calibrated Stellar SpectraWe start from our six absolutely calibrated continuous stellar spectrafrom 1.2 to 35 μm for K0, K1.5, K3, K5, and M0 giants. These wereconstructed as far as possible from actual observed spectral fragmentstaken from the ground, the Kuiper Airborne Observatory, and the IRAS LowResolution Spectrometer, and all have a common calibration pedigree.From these we spawn 422 calibrated spectral templates'' for stars withspectral types in the ranges G9.5-K3.5 III and K4.5-M0.5 III. Wenormalize each template by photometry for the individual stars usingpublished and/or newly secured near- and mid-infrared photometryobtained through fully characterized, absolutely calibrated,combinations of filter passband, detector radiance response, and meanterrestrial atmospheric transmission. These templates continue ourongoing effort to provide an all-sky network of absolutely calibrated,spectrally continuous, stellar standards for general infrared usage, allwith a common, traceable calibration heritage. The wavelength coverageis ideal for calibration of many existing and proposed ground-based,airborne, and satellite sensors, particularly low- tomoderate-resolution spectrometers. We analyze the statistics of probableuncertainties, in the normalization of these templates to actualphotometry, that quantify the confidence with which we can assert thatthese templates truly represent the individual stars. Each calibratedtemplate provides an angular diameter for that star. These radiometricangular diameters compare very favorably with those directly observedacross the range from 1.6 to 21 mas. Stellar radii of M giantsWe determine the stellar radii of the M giant stars in the Hipparcoscatalogue that have a parallax measured to better than 20% accuracy.This is done with the help of a relation between a visual surfacebrightness parameter and the Cousins (V - I) colour index, which wecalibrate with M giants with published angular diameters.The radii of(non-Mira) M giants increase from a median value of 50 R_Sun at spectraltype M0 III to 170 R_Sun at M7/8 III. Typical intermediate giant radiiare 65 R_Sun for M1/M2, 90 R_Sun for M3, 100 R_Sun for M4, 120 R_Sun forM5 and 150 R_Sun for M6. There is a large intrinsic spread for a givenspectral type. This variance in stellar radius increases with latertypes but in relative terms, it remains constant.We determineluminosities and, from evolutionary tracks, stellar masses for oursample stars. The M giants in the solar neighbourhood have masses in therange 0.8-4 M_Sun. For a given spectral type, there is a close relationbetween stellar radius and stellar mass. We also find a linear relationbetween the mass and radius of non-variable M giants. With increasingamplitude of variability we have larger stellar radii for a given mass. Averaged energy distributions in the stellar spectra.Not Available Measurements of Starspot Parameters on Active Stars using Molecular Bands in Echelle SpectraWe present results from a study of starspot areas (f_S) and temperatures(T_S), primarily on active, single-lined spectroscopic binaries,determined using molecular absorption bands. Expanding upon our previousstudies, we have analyzed multiorder echelle spectra of eight systems tosimultaneously measure several different molecular bands andchromospheric emission lines. We determined starspot parameters byfitting the molecular bands of interest, using spectra of inactive G andK stars as proxies for the nonspotted photosphere of the active stars,and using spectra of M stars as proxies for the spots. At least twobands with different T_eff sensitivities are required. We found thatfitting bands other than the TiO 7055 and 8860 Å features does notgreatly extend the temperature range or sensitivity of our technique.The 8860 Å band is particularly important because of its sharplydifferent temperature sensitivity. We did not find any substantialdepartures from f_S or T_S that we have measured previously based onsingle-order spectra. We refined our derived spot parameters usingcontemporaneous photometry where available. We found that using M giantsas spot proxies for subgiant active stars often underestimates f_Sneeded to fit the photometry; this is presumably due to the increase instrength of the TiO bands with decreasing gravity. We also investigatedcorrelations between f_S and chromospheric emission, and we developed asimple method to measure nonspot temperature (T_Q) solely from ourechelle spectra. The Tokyo PMC catalog 90-93: Catalog of positions of 6649 stars observed in 1990 through 1993 with Tokyo photoelectric meridian circleThe sixth annual catalog of the Tokyo Photoelectric Meridian Circle(PMC) is presented for 6649 stars which were observed at least two timesin January 1990 through March 1993. The mean positions of the starsobserved are given in the catalog at the corresponding mean epochs ofobservations of individual stars. The coordinates of the catalog arebased on the FK5 system, and referred to the equinox and equator ofJ2000.0. The mean local deviations of the observed positions from theFK5 catalog positions are constructed for the basic FK5 stars to comparewith those of the Tokyo PMC Catalog 89 and preliminary Hipparcos resultsof H30. Standard Stars for CCD Photometry in the Vilnius SystemThe results of seven-color photometry of 73 stars of magnitudes 9--13are given in 11 areas. The areas are of 10times 10 arcmin size at lowgalactic latitudes in the transparent parts of the Milky Way in theconstellations of Serpens, Aquila, Vulpecula, Lyra and Cygnus. Thesestars can be used as zero-point standards for future CCD photometry. Classification of Population II Stars in the Vilnius Photometric System. II. ResultsThe results of photometric classification of 848 true and suspectedPopulation II stars, some of which were found to belong to Population I,are presented. The stars were classified using a new calibrationdescribed in Paper I (Bartkevicius & Lazauskaite 1996). We combinethese results with our results from Paper I and discuss in greaterdetail the following groups of stars: UU Herculis-type stars and otherhigh-galactic-latitude supergiants, field red horizontal-branch stars,metal-deficient visual binaries, metal-deficient subgiants, stars fromthe Catalogue of Metal-deficient F--M Stars Classified Photometrically(MDPH; Bartkevicius 1993) and stars from one of the HIPPARCOS programs(Bartkevicius 1994a). It is confirmed that high galactic latitudesupergiants from the Bartaya (1979) catalog are giants or even dwarfs.Some stars, identified by Rose (1985) and Tautvaisiene (1996a) as fieldRHB stars, appear to be ordinary giants according to our classification.Some of the visual binaries studied can be considered as physical pairs.Quite a large fraction of stars from the MDPH catalog are found to havesolar metallicity. A number of new possible UU Herculis-type stars, RHBstars and metal-deficient subgiants are identified. Classification and Identification of IRAS Sources with Low-Resolution SpectraIRAS low-resolution spectra were extracted for 11,224 IRAS sources.These spectra were classified into astrophysical classes, based on thepresence of emission and absorption features and on the shape of thecontinuum. Counterparts of these IRAS sources in existing optical andinfrared catalogs are identified, and their optical spectral types arelisted if they are known. The correlations between thephotospheric/optical and circumstellar/infrared classification arediscussed. Systematic Errors in the FK5 Catalog as Derived from CCD Observations in the Extragalactic Reference Frame.Abstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1997AJ....114..850S&db_key=AST Luminosity and Temperature from Near-Infrared Spectra of Late-Type Giant StarsWe present moderate resolution (lambda / Delta lambda ~ 1380 and lambda/ Delta lambda ~ 4830) spectra of 43 K0 to M6 III stars covering 2.19 -2.34 mu m and measure equivalent widths of the strongest absorptionfeatures - Na I, Ca I, and (12) CO(2,0) - present on the spectra. Thehigh resolution Wallace & Hinkle (1996) spectral atlas shows thatour moderate resolution measurements of the atomic features havesignificant contributions from other species, such as Sc, S, Fe, Ti, Si,and V. We also find that our measured equivalent widths are affected byCN absorption present in the continuum bands. In spite of this, theequivalent widths of Na I and Ca I features at moderate resolution stillshow a strong dependence on effective temperature. The CO equivalentwidth at moderate resolution is less affected by other lines andcontinuum placement than the atomic features, because of its relativelygreater strength. We compare our data to similar data taken forlate-type dwarf stars (Ali et al. 1995) and find that a two dimensionalspectral classification can be constructed based on the near-IR spectra.The quantity log[EW(CO)/(EW(Na)+EW(Ca))] is a strong luminosityindicator independent of effective temperature, while the equivalentwidth of (12) CO(2,0) has a well-defined dependence on effectivetemperature for a given luminosity. This two dimensional spectralclassification is ideal for cool stars obscured by dust in, for example,the central part of the Galactic bulge and regions of star formation. A catalogue of [Fe/H] determinations: 1996 editionA fifth Edition of the Catalogue of [Fe/H] determinations is presentedherewith. It contains 5946 determinations for 3247 stars, including 751stars in 84 associations, clusters or galaxies. The literature iscomplete up to December 1995. The 700 bibliographical referencescorrespond to [Fe/H] determinations obtained from high resolutionspectroscopic observations and detailed analyses, most of them carriedout with the help of model-atmospheres. The Catalogue is made up ofthree formatted files: File 1: field stars, File 2: stars in galacticassociations and clusters, and stars in SMC, LMC, M33, File 3: numberedlist of bibliographical references The three files are only available inelectronic form at the Centre de Donnees Stellaires in Strasbourg, viaanonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5), or viahttp://cdsweb.u-strasbg.fr/Abstract.html High resolution infrared spectroscopy of CN and NH lines: nitrogen abundance in oxygen-rich giants through K to late MThe analyses of high resolution infrared spectra have been done for CNlines in oxygen-rich cool evolved stars including 2 K giants, 20 Mgiants and 1 S-type star. Since CN lines analyzed in the present workare weak and resolved well, they are appropriate for quantitativeanalyses. CN lines of Delta v=-2 and -1 sequences (red system) which arein the K- and the H-window regions, respectively, give the consistentnitrogen abundance for each star. The analyses of NH lines in theL-window region have been done for 5 late M giants for which CN lineshave been also analyzed. Although the triplet structure of NH linescannot be fully resolved, they are preferable because determination ofnitrogen abundance is almost independent of other elemental abundanceswhile nitrogen abundance based on CN depends on carbon abundance. Thenitrogen abundances derived from NH for late M giants agree well withthose from CN for which we adopt 7.75eV as the dissociation energy inthe analysis. The results show that the nitrogen abundances in late Mgiants are larger than those in early M giants while decrease of thecarbon abundance was found in late M giants by our previous work (Tsuji1991). These variations of abundances can not be explained by the firstdredge-up model but require additional processing by the CN cycle andmixing after the first dredge-up. However, there is no obvious evidenceof other processes such as the 3alpha -process and subsequent hot bottomburning in our program stars. Such variation of the carbon and nitrogenabundances is not well understood by the present evolutionary models oflow-mass and intermediate-mass stars. Table~10 is only available inelectronic form at the CDS (Strasbourg) via anonymous ftp tocdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/Abstract.html The Pulkovo Spectrophotometric Catalog of Bright Stars in the Range from 320 TO 1080 NMA spectrophotometric catalog is presented, combining results of numerousobservations made by Pulkovo astronomers at different observing sites.The catalog consists of three parts: the first contains the data for 602stars in the spectral range of 320--735 nm with a resolution of 5 nm,the second one contains 285 stars in the spectral range of 500--1080 nmwith a resolution of 10 nm and the third one contains 278 stars combinedfrom the preceding catalogs in the spectral range of 320--1080 nm with aresolution of 10 nm. The data are presented in absolute energy unitsW/m(2) m, with a step of 2.5 nm and with an accuracy not lower than1.5--2.0%. Obscured Active Galactic Nuclei in Luminous Infrared GalaxiesWe examine the nature of the central power source in very luminousinfrared galaxies. The infrared properties of the galaxies, includingtheir far-infrared and 2.2 micron fluxes, CO indices, and Brackett linefluxes are compared to models of starburst stellar populations. Amongseven galaxies we found two dominated by emission from young stars, twodominated by emission from an AGN, and three transition cases. Ourresults are consistent with evidence for active nuclei in the samegalaxies at other wavelengths. Nuclear mass measurements obtained forthe galaxies indicate an initial mass function biased toward high-massstars in two galaxies. After demonstrating our methods in well-studiedgalaxies, we define complete samples of high luminosity andultraluminous galaxies. We find that the space density of embedded andunembedded quasars in the local universe is similar for objects ofsimilar luminosity. If quasars evolve from embedded sources to opticallyprominent objects, it appears that the lifetime of a quasar is no morethan about 108 yr. Vitesses radiales. Catalogue WEB: Wilson Evans Batten. Subtittle: Radial velocities: The Wilson-Evans-Batten catalogue.We give a common version of the two catalogues of Mean Radial Velocitiesby Wilson (1963) and Evans (1978) to which we have added the catalogueof spectroscopic binary systems (Batten et al. 1989). For each star,when possible, we give: 1) an acronym to enter SIMBAD (Set ofIdentifications Measurements and Bibliography for Astronomical Data) ofthe CDS (Centre de Donnees Astronomiques de Strasbourg). 2) the numberHIC of the HIPPARCOS catalogue (Turon 1992). 3) the CCDM number(Catalogue des Composantes des etoiles Doubles et Multiples) byDommanget & Nys (1994). For the cluster stars, a precise study hasbeen done, on the identificator numbers. Numerous remarks point out theproblems we have had to deal with. SANTIAGO 91, a right ascension catalogue of 3387 stars (equinox J2000).The positions in right ascension of 3387 stars belonging to the Santiago67 Catalogue, observed with the Repsold Meridian Circle at Cerro Calan,National Astronomical Observatory, during the period 1989 to 1994, aregiven. The average mean square error of a position, for the wholeCatalogue, is +/-0.009 s. The mean epoch of the catalogue is 1991.84.
• - 没有找到链接 -
### 下列团体成员
#### 观测天体数据
星座: 狐狸座 右阿森松: 19h28m42.30s 赤纬: +24°39'54.0" 视星: 4.44 距离: 90.909 天文距离 右阿森松适当运动: -127.5 赤纬适当运动: -106.1 B-T magnitude: 6.382 V-T magnitude: 4.61
适当名称 Anser (Edit) Bayer α Vul Flamsteed 6 Vul HD 1989 HD 183439 TYCHO-2 2000 TYC 2129-2772-1 USNO-A2.0 USNO-A2 1125-12776004 BSC 1991 HR 7405 HIP HIP 95771 → 要求更多目录从vizier
|
2019-10-24 02:11:54
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6421398520469666, "perplexity": 7120.5340700244815}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987838289.72/warc/CC-MAIN-20191024012613-20191024040113-00038.warc.gz"}
|
https://www.tennisforum.com/threads/pick-a-winner-paw-223-canberra-2006.211127/page-15
|
281 - 289 of 289 Posts
tks
·
Registered
Joined
·
1,199 Posts
Discussion Starter
Hi
League Table updated
The winner of 2006 PAW Canberra is: jrm :worship: :worship: :worship: congrats
I hope you enjoyed.
Thanks on the compliments, pleasure is all mine.
GOOD LUCK you all next week in the AO.
Bye, :wavey:
tks
Mateo Mathieu
·
Senior Member
Joined
·
63,617 Posts
Congratulations, jrm! :yeah:
WTA Handicapper
·
Registered
Joined
·
1,781 Posts
WTA Handicapper said:
Who will Win Paw Canberra?
Betting Odds [Compiled before Sunday matches ]
9-1 Favourite:JRM
10-1:Canuck
11-1:Speedster,Daniela86
12-1:Kirilenko1Fan
14-1:Guycan
16-1:MyPapaje
18-1:Bus
20-1:Alexsydney
22-1:Cory
25-1:James
28-1:Ghosts
40-1:Mariano
50-1 Bar
Good Luck All
Betting Summary
Bookmakers were hit for 6 with the Favourite here and for the Australian Open Jrm taking the title.
Big let down was the normally reliable Canuck who never figured.
Hayato
·
Registered
Joined
·
34,578 Posts
Thanks tks! Well done jrm! :banana:
Just a little thing, in the league table:
05. . . . . 13. . . . Benjiboy (wc). . . . . . .8. . . . . . . .5. . . . . . . . .76
05 . . . . .13. . . . guycan. . . . . . . . . . . 7. . . . . . . 6. . . . . . . . .76
We shouldn't be equal because I had one extra correct PAW. I wouldn't mind if it didn't affect how many points I lose for being equal with someone.
tks
·
Registered
Joined
·
1,199 Posts
Discussion Starter
Benjiboy said:
Thanks tks! Well done jrm! :banana:
Just a little thing, in the league table:
05. . . . . 13. . . . Benjiboy (wc). . . . . . .8. . . . . . . .5. . . . . . . . .76
05 . . . . .13. . . . guycan. . . . . . . . . . . 7. . . . . . . 6. . . . . . . . .76
We shouldn't be equal because I had one extra correct PAW. I wouldn't mind if it didn't affect how many points I lose for being equal with someone.
corrected
andrew_uk
·
Registered
Joined
·
5,933 Posts
Congratulations on your second title Valentina :hug: :wavey:
Big thank you to the organiser :hug: :wavey:
speedster_
·
Registered
Joined
·
1,007 Posts
congratulations jrm :worship:
thank you tks, you run the game very well, I enjoyed to play in here :yeah:
LUXXXAS
·
Team WTAworld, Senior Member
Joined
·
15,686 Posts
congrats jrm:yeah::bounce:
Kitakaze
·
Registered
Joined
·
13,087 Posts
Congrats, Tina!
281 - 289 of 289 Posts
|
2020-05-31 14:15:59
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9044027328491211, "perplexity": 3389.8133150619256}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347413406.70/warc/CC-MAIN-20200531120339-20200531150339-00220.warc.gz"}
|
http://www.sciencemadness.org/talk/viewthread.php?tid=153237
|
Sciencemadness Discussion Board » Special topics » Biochemistry » People With Diabetes Making Their Own Insulin Select A Forum Fundamentals » Chemistry in General » Organic Chemistry » Reagents and Apparatus Acquisition » Beginnings » Responsible Practices » Miscellaneous » The Wiki Special topics » Technochemistry » Energetic Materials » Biochemistry » Radiochemistry » Computational Models and Techniques » Prepublication Non-chemistry » Forum Matters » Legal and Societal Issues » Detritus » Test Forum
Author: Subject: People With Diabetes Making Their Own Insulin
Pyro_cat
Hazard to Others
Posts: 205
Registered: 30-4-2018
Member Is Offline
Mood: No Mood
People With Diabetes Making Their Own Insulin
Biohackers With Diabetes Are Making Their Own Insulin
Diabetes is a punishingly expensive disease. In an Oakland warehouse, scientists are going DIY.
https://elemental.medium.com/biohackers-with-diabetes-are-ma...
In 1940, during World War II, she and her husband, Victor Saxl, fled to Shanghai, China. In Shanghai, a year later, Saxl was diagnosed with type 1 diabetes. When the Japanese attacked Pearl Harbor in 1941 the Japanese occupation of China was tightened, and soon all the pharmacies in Shanghai were closed. Saxl had no legal access to insulin. It was possible to buy insulin on the black market using one-ounce gold bars for payment. But that was not the safest option; one of Eva's friends died from using the black market insulin.
Eventually, Victor and Eva decided to get insulin another—highly unconventional—way: make it themselves. The book "Beckman's Internal Medicine"[Note 1] described the methods that Dr. Frederick Banting and Charles Best first used to extract insulin from the pancreases of dogs, calves, and cows in 1921.
https://en.wikipedia.org/wiki/Eva_Saxl
Metacelsus
International Hazard
Posts: 2445
Registered: 26-12-2012
Location: Boston, MA
Member Is Offline
Mood: Double, double, toil and trouble
I can make my own insulin, and I don't even have diabetes
On a more serious note, I worry about safety for this. Injectable medications need to be sterile and free of bacterial toxins. And with insulin, the dosage must be carefully controlled (an overdose is fatal). If the alternative is death, it's worth it. But I don't think these people are following CGMP.
As below, so above.
Ubya
International Hazard
Posts: 1151
Registered: 23-11-2017
Location: Rome-Italy
Member Is Offline
how are they going to measure the dosage? every batch is going to be different, as metacelsus pointed out, overdosing is quite easy, you need very little insulin. i really doubt they can produce cheaper insulin in a domestic setting, and of good enough quality. each batch needs to be tested, tests can be expensive if not in house (if they want to test by themselves good luck paying for the equipment).
surely it can be done, but not for 20$a vial, you would need economy of scales, and a small production can't lower the costs --------------------------------------------------------------------- feel free to correct my grammar, or any mistakes i make If you are looking for chemicals check this out: [For Sale]300 chemicals (rare & unusual) --------------------------------------------------------------------- Pyro_cat Hazard to Others Posts: 205 Registered: 30-4-2018 Member Is Offline Mood: No Mood 2018 - Alec Raeshawn Smith was 23 when diagnosed with Type 1 diabetes and 26 when he died. He couldn't afford$1300 per month for his insulin ... Approximately 7.4 million Americans with diabetes use one or more formulations of insulin.
My idea is for diabetics to get together and start making it instead. Essentially starting small pharmaceutical companies.
People are already building their own closed loop systems. #T1D #WeAreNotWaiting #Loop
DIY is where its at. Plenty of smart people.
How difficult is this process of making insulin ?
[Edited on 22-8-2019 by Pyro_cat]
Ubya
International Hazard
Posts: 1151
Registered: 23-11-2017
Location: Rome-Italy
Member Is Offline
Synthesizing human insulin is a multi-step biochemical process that depends on basic recombinant DNA techniques and an understanding of the insulin gene. DNA carries the instructions for how the body works and one small segment of the DNA, the insulin gene, codes for the protein insulin. Manufacturers manipulate the biological precursor to insulin so that it grows inside simple bacteria. While manufacturers each have their own variations, there are two basic methods to manufacture human insulin.
Working with human insulin
1 The insulin gene is a protein consisting of two separate chains of amino acids, an A above a B chain, that are held together with bonds. Amino acids are the basic units that build all proteins. The insulin A chain consists of 21 amino acids and the B chain has 30.
2 Before becoming an active insulin protein, insulin is first produced as preproinsulin. This is one single long protein chain with the A and B chains not yet separated, a section in the middle linking the chains together and a signal sequence at one end telling the protein when to start secreting outside the cell. After preproinsulin, the chain evolves into proinsulin, still a single chain but without the signaling sequence. Then comes the active protein insulin, the protein without the section linking the A and B chains. At each step, the protein needs specific enzymes (proteins that carry out chemical reactions) to produce the next form of insulin.
STARTING WITH A AND B
3 One method of manufacturing insulin is to grow the two insulin chains separately. This will avoid manufacturing each of the specific enzymes needed. Manufacturers need the two mini-genes: one that produces the A chain and one for the B chain. Since the exact DNA sequence of each chain is known, they synthesize each mini-gene's DNA in an amino acid sequencing machine.
4 These two DNA molecules are then inserted into plasmids, small circular pieces of DNA that are more readily taken up by the host's DNA.
5 Manufacturers first insert the plasmids into a non-harmful type of the bacterium E. coli. They insert it next to the lacZ gene. LacZ encodes for 8-galactosidase, a gene widely used in recombinant DNA procedures because it is easy to find and cut, allowing the insulin to be readily removed so that it does not get lost in the bacterium's DNA. Next to this gene is the amino acid methionine, which starts the protein formation.
6 The recombinant, newly formed, plasmids are mixed up with the bacterial cells. Plasmids enter the bacteria in a process called transfection. Manufacturers can add to the cells DNA ligase, an enzyme that acts like glue to help the plasmid stick to the bacterium's DNA.
7 The bacteria synthesizing the insulin then undergo a fermentation process. They are grown at optimal temperatures in large tanks in manufacturing plants. The millions of bacteria replicate roughly every 20 minutes through cell mitosis, and each expresses the insulin gene.
8 After multiplying, the cells are taken out of the tanks and broken open to extract the DNA. One common way this is done is by first adding a mixture of lysozome that digest the outer layer of the cell wall, then adding a detergent mixture that separates the fatty cell wall membrane. The bacterium's DNA is then treated with cyanogen bromide, a reagent that splits protein chains at the methionine residues. This separates the insulin chains from the rest of the DNA.
9 The two chains are then mixed together and joined by disulfide bonds through the reduction-reoxidation reaction. An oxidizing agent (a material that causes oxidization or the transfer of an electron) is added. The batch is then placed in a centrifuge, a mechanical device that spins quickly to separate cell components by size and density.
10 The DNA mixture is then purified so that only the insulin chains remain. Manufacturers can purify the mixture through several chromatography, or separation, techniques that exploit differences in the molecule's charge, size, and affinity to water. Procedures used include an ion-exchange column, reverse-phase high performance liquid chromatography, and a gel filtration chromatography column. Manufacturers can test insulin batches to ensure none of the bacteria's E. coli proteins are mixed in with the insulin. They use a marker protein that lets them detect E. coli DNA. They can then determine that the purification process removes the E. coli bacteria.
PROINSULIN PROCESS
11 Starting in 1986, manufacturers began to use another method to synthesize human insulin. They started with the direct precursor to the insulin gene, proinsulin. Many of the steps are the same as when producing insulin with the A and B chains, except in this method the amino acid machine synthesizes the proinsulin gene.
12 The sequence that codes for proinsulin is inserted into the non-pathogenic E. coli bacteria. The bacteria go through the fermentation process where it reproduces and produces proinsulin. Then the connecting sequence between the A and B chains is spliced away with an enzyme and the resulting insulin is purified.
13 At the end of the manufacturing process ingredients are added to insulin to prevent bacteria and help maintain a neutral balance between acids and bases. Ingredients are also added to intermediate and long-acting insulin to produce the desired duration type of insulin. This is the traditional method of producing longer-acting insulin. Manufacturers add ingredients to the purified insulin that prolong their actions, such as zinc oxide. These additives delay absorption in the body. Additives vary among different brands of the same type of insulin.
Quality Control
After synthesizing the human insulin, the structure and purity of the insulin batches are tested through several different methods. High performance liquid chromatography is used to determine if there are any impurities in the insulin. Other separation techniques, such as X-ray crystallography, gel filtration, and amino acid sequencing, are also performed. Manufacturers also test the vial's packaging to ensure it is sealed properly.
Manufacturing for human insulin must comply with National Institutes of Health procedures for large-scale operations. The United States Food and Drug Administration must approve all manufactured insulin.
not that easy
---------------------------------------------------------------------
feel free to correct my grammar, or any mistakes i make
If you are looking for chemicals check this out: [For Sale]300 chemicals (rare & unusual)
---------------------------------------------------------------------
Felis Corax
Harmless
Posts: 35
Registered: 26-7-2019
Location: These Uninted States of Murica
Member Is Offline
Mood: Pre-project euphoria
Sorry Ubya, but that article was written by someone who doesn't know the difference between proteins (specifically folded molecular machines made of one or more polypeptide) and DNA (chains of paired nucleotides). DNA codes for protein synthesis, along other things, but it's not itself a protein. Think of it as the difference between a blueprint (DNA) and a machine built from that blueprint (a protein).
There are other mistakes too, but most could be attributed to oversimplification. Describing DNA as being made of amino acids is unambiguously and fundamentally wrong.
I'm not saying diy insulin synthesis is safe and easy, I haven't researched it, but the article referenced is not a source for anything besides dubious infotainment for those without any knowledge of molecular biology. Well, that and as a bracing reminder of the importance of using reliable sources.
Nothing is pure, nothing is perfect, nothing is clean.
Pyro_cat
Hazard to Others
Posts: 205
Registered: 30-4-2018
Member Is Offline
Mood: No Mood
I just found this https://openinsulin.org/
So they call this biohacking.
I find so much more interesting search results when I click on images instead of "all". I was looking at some stuff and decides to search: Skids & Units for Pharma Biotech Products "insulin" then clicked images to see what this stuff looks like and that is where I found the above link. The regular search results are just generic mainstream news and websites.
I don't know why they have to call independent research "hacking" like they want to make it sound dirty or illegal or something.
Sulaiman
International Hazard
Posts: 3010
Registered: 8-2-2015
Location: UK ... on extended Holiday in Malaysia
Member Is Offline
Maybe it would be safer and cheaper to go from the USA to Canada (or Cuba) ?
CAUTION : Hobby Chemist, not Professional or even Amateur
Tsjerk
International Hazard
Posts: 2355
Registered: 20-4-2005
Location: Netherlands
Member Is Online
Mood: Mood
Or come to the Netherlands.
Price of Fiasp10 ml 100 u/ml : 20 euros
USA price: 310 Dollars
We do pay the first 385 euros a year our self (in total for all care, not per medicine or treatment) and health insurance is mandatory, the cheapest being around 120 euros a month. For those with a low income there is some free financing plan to cover this 120 euros though, which is maximally 99 euros a month for those earning less than 20500 euros a year, this goes to 0 about linearly until you earn about 30.000.
At least I now understand even better why there is huge trafficking between the USA and it's neighbors...
Ubya
International Hazard
Posts: 1151
Registered: 23-11-2017
Location: Rome-Italy
Member Is Offline
Quote: Originally posted by Felis Corax Sorry Ubya, but that article was written by someone who doesn't know the difference between proteins (specifically folded molecular machines made of one or more polypeptide) and DNA (chains of paired nucleotides). DNA codes for protein synthesis, along other things, but it's not itself a protein. Think of it as the difference between a blueprint (DNA) and a machine built from that blueprint (a protein). There are other mistakes too, but most could be attributed to oversimplification. Describing DNA as being made of amino acids is unambiguously and fundamentally wrong. I'm not saying diy insulin synthesis is safe and easy, I haven't researched it, but the article referenced is not a source for anything besides dubious infotainment for those without any knowledge of molecular biology. Well, that and as a bracing reminder of the importance of using reliable sources.
yep i know, i read it before posting, there are many errors but it was good enough to give an idea of the process, not a detailed procedure
---------------------------------------------------------------------
feel free to correct my grammar, or any mistakes i make
If you are looking for chemicals check this out: [For Sale]300 chemicals (rare & unusual)
---------------------------------------------------------------------
Metacelsus
International Hazard
Posts: 2445
Registered: 26-12-2012
Location: Boston, MA
Member Is Offline
Mood: Double, double, toil and trouble
Making insulin would be quite easy for anyone with basic molecular biology skills. As I mentioned before, the issue is purification and formulation. This would require expensive equipment as well as technical expertise.
The openinsulin website doesn't have many details, so I can't really judge whether they're likely to be successful with this. But then again, absence of important details might be a bad sign.
[Edited on 2019-8-23 by Metacelsus]
As below, so above.
Pyro_cat
Hazard to Others
Posts: 205
Registered: 30-4-2018
Member Is Offline
Mood: No Mood
I am new diabetic. Always been a mad scientist.
Don't underestimate our resentments towards this system that uses our sickness to make us cash cows.
Not only cash cows but when its time to buy needles you are an addict till you prove otherwise. I just order online now.
Its actually quite sad. Why would anyone do all the school required to be a pharmacist just to be part of the state pill police ?? Take pills from big bottle put in small bottle then paperwork bullshit. Its not like the old days, before my time, when they were actual chemists. I have no respect for them.
This is international forum. As American I am ashamed. I am older and it wasn't always like this. Wasn't always a big brother rip you off police state.
No more politics, exploring science is how I chill so I should leave that alone. This is a great idea micro breweries for insulin just like with beer. I hope it catches on.
Anyway I posted this here to attract good "biohackers" to our plight.
[Edited on 25-9-2019 by Pyro_cat]
rockyit98
Hazard to Others
Posts: 222
Registered: 12-4-2019
Location: The Known Universe
Member Is Offline
Mood: no mood is a good mood
What I've Learned 1.15M subscribers watch this channel first
|
2021-02-28 01:13:16
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2867327034473419, "perplexity": 4576.556817369237}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178359624.36/warc/CC-MAIN-20210227234501-20210228024501-00038.warc.gz"}
|
https://physics.stackexchange.com/questions/94560/hilbert-g%C3%B6del-and-god-equations-a-19th-century-lesson-for-21st-century-phy
|
# Hilbert, Gödel, and "God equations" - a 19th century lesson for 21st century physicists?
It seems there are a lot of respected physicists appearing on pop-sci programs (discovery channel, science channel, etc.) these days spreading the gospel of "we can know, we must know."
Three examples, quickly: 1) Many programs feature Michio Kaku saying that he is on a quest to find an equation, "perhaps just one inch long," which will "describe the whole universe." 2) Max Tegmark has come out with a new book in which he expresses the gut feeling that "nothing is off-limits" to science. The subtitle of this book is My Quest for the Ultimate Nature of Reality. 3) In the series Through the Wormhole there is talk about a search for the "God equation."
(A good counter-example would be Feynman, but his self-described "non-axiomatic" or "Babylonian" approach does not seem popular with physicists today.)
Is there any sense among physicists that it might be impossible to articulate the "ultimate nature of reality" in equations and formal logic? It seems to me that physicists are following in the footsteps of the 19th century mathematicians (led by Hilbert) who were on a similar quest which was put to rest by Gödel's incompleteness theorems in 1931. Is there any appreciation for how the Incompleteness Theorems might apply to physics?
Has any progress been made on Hilbert's 6th problem for the 20th century? Shouldn't this be addressed before getting all worked up about a "God equation?"
• Jan 21, 2014 at 1:28
• Imho, you can safely discount $99.99 \%$ of that as just chatter. Taking advantage of the fecund atmosphere for science popularization, many people (some expert and some not so) are trying their hand at drumming up some excitement.
– Siva
Jan 21, 2014 at 2:01
• @Hunter Removed philosophy tag. meta.physics.stackexchange.com/q/80 Jan 21, 2014 at 7:21
• Interesting link with the opposite trend: What scientific idea is ready for retirement?: We'll Never Hit Barriers To Scientific Understanding. Martin Rees, Edge, 2014. Jan 21, 2014 at 11:36
• @Siva Yes, but these are respected scientists holding high positions at respected institutions.
– ben
Feb 4, 2014 at 5:30
First regarding: Is there any appreciation for how the Incompleteness Theorems might apply to physics?
To put this in perspective, imagine Newton said "Oh, looks like my $F = m a$ is pretty much a theory of everything. So now I could know everything about nature if only it were guaranteed that every sufficiently strong consistent formal system is complete."
And then later Lagrange: "Oh, looks like my $\delta L = 0$ is pretty much a theory of everything. So now I could know everything about nature if only it were guaranteed that every sufficiently strong consistent formal system is complete."
And then later Schrödinger: "Oh, looks like my $i \hbar \partial_t \psi = H \psi$ is pretty much a theory of everything. So now I could know everything about nature if only it were guaranteed that every sufficiently strong consistent formal system is complete."
And so forth.
The point being, that what prevented physicists 300 years ago, 200 years ago, 100 years ago from in principle knowing everything about physics was never any incompleteness theorem, but were always two things:
1. they didn't actually have a fundamental theory yet;
2. they didn't even have the mathematics yet to formulate what later was understood to be the more fundamental theory.
Gödel's incompleteness theorem is much like "$E = m c^2$" in the pop culture: people like to allude to it with a vague feeling of deep importance, without really knowing what the impact is. Gödel incompleteness is a statement about the relation between metalanguage and "object language" (it's the metalanguage that allows one to know that a given statement "is true", after all, even if if cannot be proven in the object language!). To even appreciate this distinction one has to delve a bit deeper into formal logic than I typically see people do who wonder about its relevance to physics.
And the above history suggests: it is in any case premature to worry about the fine detail of formal logic as long as the candidate formalization of physics that we actually have is glaringly insufficient, and in particular as long as it seems plausible that in 100 years form now fundamental physics will be phrased in new mathematics compared to which present tools of mathematical physics look as outdated as those from a 100 years back do to us now. Just open a theoretical physics textbook from the turn of the 19th to the 20th century to see that with our knowledge about physics it would have been laughable for the people back then to worry about incompleteness. They had to worry about learning linear algebra and differential geometry.
second: Has any progress been made on Hilbert's 6th problem for the 20th century?
I had recently been giving some talks which started out with considering this question, see the links on my site at Synthetic quantum field theory.
One answer is: there has been considerable progress (see the table right at the beginning of the slides or also in this talk note). Lots of core aspects of modern physics have a very clean mathematical formulation. For instance gauge theory is firmely captured by differential cohomology and Chern-Weil theory, local TQFT by higher monoidal category theory, and so forth.
But two things are remarkable here: first, the maths that formalizes aspects of modern fundamental physics involves the crown jewels of modern mathematics, so something deep might be going on, but, second, these insights remain piecemeal. There is a field of mathematics here, another there. One could get the idea that somehow all this wants to be put together into one coherent formal story, only that maybe the kind of maths used these days is not quite sufficient for doing so.
This is a point of view that, more or less implicitly, has driven the life work of William Lawvere. He is famous among pure mathematicians as being the founder of categorical logic, of topos theory in formal logic, of structural foundations of mathematics. What is for some weird reason almost unknown, however, is that all this work of his has been inspired by the desire to produce a working formal foundations for physics. (See on the nLab at William Lawvere -- Motivation from foundations of physics).
I think anyone who is genuinely interested in the formal mathematical foundations of phyiscs and questions as to whether a fundamental formalization is possible and, more importantly, whether it can be useful, should try to learn about what Lawvere has to say.
Of course reading Lawvere is not easy. (Just like reading a modern lecture on QFT would not be easy for a physicist form the 19th century had he been catapulted into our age...) That's how it goes when you dig deeply into foundations, if you are really making progress, then you won't be able to come back and explain it in five minutes on the Dicovery Channel. (As in Feynman's: If I could tell you in five minutes what gained me the Nobel, then it wouldn't have.)
You might start with the note on the nLab: "Higher toposes of laws of motion" for an idea of what Lawverian foundations of physics is about.
A little later this month I'll be giving various talks on this issue of formally founding modern physics (local Lagrangian gauge quantum field theory) in foundational mathematics in a useful way. The notes for this are titled Homotopy-type semantics for quantization.
I can only speak from my personal experience (which seems fair enough since this question is subjective). Most physicists I know, including myself, are much more humble on what physics knows now and will know in the future compared to the "celebrity physicists" you mentioned.
It's fairly easy to see from history of the field that whenever we think we become close to explaining it all, something new is observed or some small inconsistency turns out to open up an entire new branch of physics. From these experiences, it seems highly doubtful to me that we would ever become close to pushing the ability of one theory to the point where we have to worry about Goedel's Theorems (that is worry about completeness - after all it could easily be the case that the truth statements which our theory cannot predict are not relevant to our universe i.e. experiments). Furthermore, I've yet to hear a good definition for what we mean by "one theory". After all, QFT is much more of a framework and the Standard Model is just one of many possible applications of that framework. We fit the Standard Model to conform to our observed universe. So what exactly do those physicists mean then by a "god equation"? Do they mean one framework from which multiple equations can arise?
I guess I'm answering questions with questions, but it is only to make the point that these "dream theories" can become idealized to the point of myth.
It seems to me that in the future what will most likely occur is some framework or language that can be used to describe gravity and dark energy in addition the other forces. This framework will be applied to some Standard Model version 2 that incorporates dark matter and other observed matter. That does not mean one equation. It just means one unified way of thinking about things. It will likely lead to many equations with a good number of assumptions that are assumed only because they give they accurately predict experiment.
The question Is there any sense among physicists that it might be impossible to articulate the "ultimate nature of reality" in equations and formal logic? is about a belief (like a faith of a religion) that most physicists may or may not have. Just like physicists have many different faith, I think physicists have different believes on this issue. So it is hard to answer yes or no, since physicists do not have a common opinion.
However, I think many physicists share the opinion of Dao-De-Jing 道德经 on a related issue:
2500 years ago, Dao-De-Jing 道德经 expressed the following point of view: (English translation)
The Dao that can be stated cannot be the eternal Dao.
The Name that can be given cannot be the eternal Name.
The nameless nonbeing is the origin of universe;
The named being is the mother of all observed things.
Within nonbeing, we enjoy the mystery of the universe.
Among being, we observe the richness of the world.
Nonbeing and being are two aspects of the same mystery.
From nonbeing to being and from being to nonbeing is the gateway to all understanding.
Here DAO ~ "ultimate nature of reality". So the point of view is that "ultimate nature of reality" exists. But any (or current) concrete description of "ultimate nature of reality" in terms of equations and formal logic is not a faithful description of the "ultimate nature of reality".
Physicists have been trying to approximately approach "ultimate nature of reality" (or DAO). My attempt is a unification of (quantum) information and matter: http://blog.sciencenet.cn/blog-1116346-736093.html
"Matrix" is a story of two worlds: A real material world and a virtual information world (inside computers). The real material world is formed by elementary particles. The virtual information world is formed by bits. (My point of view) in fact, the real material world is not real, the virtual information world is more real. The material world and the information world is actually the same world. To be more precise, our world is a quantum information world:
• Space = a collection of many many qubits.
• Vacuum = the ground state of the qubits.
• Elementary particles = collective excitations of the qubits.
In other words, all matter are formed by the excitations of the qubits.
We live inside a quantum computer.
"ultimate reality" = qubits, "God equation" = Shreodinger equation
-- this is AN approximate approach to "ultimate nature of reality" (or DAO).
• Is that a yes or a no? Jan 21, 2014 at 11:02
• I updated my answer for your point. Jan 22, 2014 at 4:10
• In your link it mentions that gravity is not unified with the others through String-nets. However, I remember that in your paper with Levin you mention that LQG may be a string-net. I believe the program to join the two was called quantum graphity at one point? I know that there was a paper published on it in Phys. Rev. D. Separately Michael Freeman has played with some String-net like ideas "off lattice" using a "quantum gravity" Hamiltonian. Do you know if there is a good reference for how this program has continued? Jan 22, 2014 at 4:47
• I feel that string-net or LQG is a good description of gauge theory. But I still do not see (understand) if string-net or LQG is a good description of gravity or not. Gu and I have a paper on emergent (linearized) gravity, but that is not based on string-net nor LQG (see arxiv.org/abs/0907.1203 ). Jan 22, 2014 at 5:50
• If space is a collection of many many qubits. and elementary particles are collective excitations of the space-forming qubits, an observable consequence is that the U(1)xSU(2)xSU(3) standard model is incomplete. There must be additional Z_2 gauge theory which will lead to new cosmic strings (Z_2 flux-lines). See arxiv.org/abs/1210.1281 (section IV D) Feb 5, 2014 at 14:34
The history of physics shows that each generation of physicists, theoretical and experimentally biased ones, believes at some point that they have found the holy grail or are very close to finding it. Certainly this was true in the nineteenth century when mathematics reigned and theories were so complete and beautiful they thought that all that was left was applications of known theories . That is a type of hubris.
What changed the game was newer and better experiments that showed up inconsistencies in the predictions of their Theory Of Everything (TOE) .
It is fair to suppose that the goal will always be a TOE, and hypothesize that newer and better experimental data will open up again and again the scope of what the TOE describes. Because this has to be said: at the limits of the experimental domains of their applications, newer theories and older theories blend, usually older ones are shown to emerge from the newer ones ( as for example thermodynamics from statistical mechanics ). There is consistency in our theories.
Now as for Godel and his theorem, which I remember from my mathematics course in the form "the set of all sets is open" , as applied to a TOE is not inconsistent with the above view. What may happen though, we will reach the limits of our possible experimental verifications and the openness will be a moot point, going towards metaphysics.
• I don't think that "the set of all sets is open" is one of Gödel's incompleteness theorems. Jan 21, 2014 at 8:40
• well it was in a set theory course back in 1960 so I may be paraphrasing, the professor might have proven this using G theorem. Jan 21, 2014 at 9:25
• Anna, I think you may be thinking of the Russell Antinomy that shows it cannot be logically consistent to think of the set of all sets as a whole - this leads to the idea of a proper class and type theory. What's really interesting historically is the quote by Cantor that I give here mathoverflow.net/a/66187/14510. I find it fascinating that Cantor seems to have grasped that there are concepts that are off limits to set theory: he was well aware of the Russell Antinomy and seemed to take it in his stride: he just takes it as a given that sometimes set concepts are not well founded, ... Jan 22, 2014 at 4:37
• ... and seems to think that's OK - implicitly I think he was saying to Dedekind that the onus was on the mathematician to check that his or her sets were not, as he called them "infinite or inconsistent multiplicities". What a deft sidestep: "I define my theory to be sound whenever it is sound!": although sounding like a bit of a con, is really quite a stroke of genius. It is a shame that he never published his ideas on "infinite or inconsistent multiplicities", probably because Kronecker and others influential in mathematics publishing were dead against him. Jan 22, 2014 at 4:42
• BTW - there are definite likenesses between the construction of Gödel unprovable sentences and the set of all sets that do not contain themselves as members in the Russell Antinomy as well as the construction of uncomputable numbers (with respect to a given language)- they are a generalised version of the Cantor Slash argument (I must say, the English name of this procedure is the most apposite of all - most languages render it "Cantor Diagonalisation", but Cantor Slash conveys the mathematical violence the argument does!) Jan 22, 2014 at 4:46
|
2022-06-26 09:19:42
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5564842224121094, "perplexity": 866.8457757170746}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103037649.11/warc/CC-MAIN-20220626071255-20220626101255-00718.warc.gz"}
|
https://support.bioconductor.org/p/101178/
|
Counting exonic reads with DEXSeq and GenomicAlignments
0
0
Entering edit mode
igor ▴ 40
@igor
Last seen 3 months ago
United States
The DEXSeq vignette describes several protocols for obtaining exon counts. One option is to use the dexseq_count.py script (powered by htseq) that comes with the package. Another option is to use summarizeOverlaps directly in R. I tried both methods with unstranded parameters and the results were identical (based on a few genes). However, stranded settings yield very different results. Some exons are off by a few reads, but some can be as low as half. I used the settings as specified in the vignette, which I assumed would be most appropriate. I also tried adjusting parameters, but it didn't help.
The command for dexseq_count.py:
python /path/DEXSeq/python_scripts/dexseq_count.py \
--format bam \
--order pos \
--paired yes \
--stranded reverse \
$ref_gff \$bam \
\$counts_txt
The command for GenomicAlignments:
txdb = makeTxDbFromGFF(file = genes_gtf, format = "gtf")
exons = disjointExons(txdb, aggregateGenes = FALSE)
SE = summarizeOverlaps(features = exons, reads = bam_list, singleEnd = FALSE,
ignore.strand = FALSE, inter.feature = FALSE, fragments = TRUE)
Is there a mistake with one of the methods?
dexseq genomicalignments exon usage • 716 views
|
2023-02-05 17:58:50
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.400131493806839, "perplexity": 10713.30639590805}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500273.30/warc/CC-MAIN-20230205161658-20230205191658-00325.warc.gz"}
|
https://itprospt.com/num/17618831/if-a-reaction-responsible-for-spoiling-a-foodstuff-has-an
|
5
# If a reaction responsible for spoiling a foodstuff has an activation energy of 54 Kj / mol, by what factor is it slowed down in a refrigerator at...
## Question
###### If a reaction responsible for spoiling a foodstuff has an activation energy of 54 Kj / mol, by what factor is it slowed down in a refrigerator at 5 ° C and in a freezer at 18 ° C , compared to the temperature of 25 ° C? A catalyst keeps this foodstuff at 20 ° C the same time as if it were in a refrigerator at 5 ° C. What is the activation energy of the catalyzed reaction?
If a reaction responsible for spoiling a foodstuff has an activation energy of 54 Kj / mol, by what factor is it slowed down in a refrigerator at 5 ° C and in a freezer at 18 ° C , compared to the temperature of 25 ° C? A catalyst keeps this foodstuff at 20 ° C the same time as if it were in a refrigerator at 5 ° C. What is the activation energy of the catalyzed reaction?
#### Similar Solved Questions
##### Fill in the organic products complete with stereochemistry where applicable, In the blank boxes for each of the following reactions. Inorganic side products need not be shown (12 points) H,c- Ich,ch, CH,CH_OH ho CH,MgBe HjoTCi,KCNNolTecl doniolmt paro roono Jullonyl chlorido "Py" rolorepyrIdineHjc:TaciPyridino8. Clrcle one cholce in each case. Give brief but clear reasoning within the space provided_ points) a, Which is more reactive In an El reaction?Reason:
Fill in the organic products complete with stereochemistry where applicable, In the blank boxes for each of the following reactions. Inorganic side products need not be shown (12 points) H,c- Ich,ch, CH,CH_OH ho CH,MgBe Hjo TCi, KCN NolTecl doniolmt paro roono Jullonyl chlorido "Py" rolore...
##### @WCSTIAWna Yalo mellona CubIc Ieel pe aCE1 Melano Iniinny Roind Your anaurar cne Decimalnlaze 68 miton 0* /acre Wneeemaon Ir{acre 17.9 milion #Pcre 9.3 mibion 0? acre 44 Milon ZrcnWmeetMuneteMeaalneo ni0amIrhnnaFonmsFretnedn [email protected] Inouindelinic utrgrArar Jti "66695
@WCSTIA Wna Yalo mellona CubIc Ieel pe aCE1 Melano Iniinny Roind Your anaurar cne Decimalnlaze 68 miton 0* /acre Wneeemaon Ir{acre 17.9 milion #Pcre 9.3 mibion 0? acre 44 Milon Zrcn Wmeet Munete Meaalneo ni0am IrhnnaFonms Fretnedn Rrle @Doroaches AmesticN Find Inouindelinic utrgr Arar J ti "666...
##### Point) Let f be the linear function (in blue and let g be the parabolic function (in red) below:Evaluate the following:(fo 9)(2)=(g 0 f)(2)= (f o f)(2)=(9 0 9) (2)=(f + 9)(4)=(f/9)(2)=
point) Let f be the linear function (in blue and let g be the parabolic function (in red) below: Evaluate the following: (fo 9)(2)= (g 0 f)(2)= (f o f)(2)= (9 0 9) (2)= (f + 9)(4)= (f/9)(2)=...
##### 23 V -+ MV)=3 TV
23 V -+ MV)=3 TV...
##### Sontiv Pushinc] RLliny C 0' mu: 3014 0O an mncnd Npt 0a 1 snum 1x dJ tbu Coc(6,ret c fln Kinlc Lchin 4u 0usc and 4n skp O-lo ond 4nl Ccc(6ttnl 0i sta hc Gnctn ( M;a.) cnu yh cuk shd Ly YLe Yas_kxutn Wba ( Icvec nul Yw poxn Apely Paalu ( Slcp: Aw Skee allblu Yhu Cult Cconncy den cnltetn' Suecel Slcingfatcc pe sn Yal [ Aeply mmnimutm Jbt Ccle Para paralli( 4 skee Allnj dcr J Jkcec sldi )furu Ycu peakn Cua Paly Jal maxurulm s Cru* T Kall QaruLLL Mthel Causiay) SL?2 #~kale El Erak Yea) On
Sontiv Pushinc] RLliny C 0' mu: 3014 0O an mncnd Npt 0a 1 snum 1x dJ tbu Coc(6,ret c fln Kinlc Lchin 4u 0usc and 4n skp O-lo ond 4nl Ccc(6ttnl 0i sta hc Gnctn ( M; a.) cnu yh cuk shd Ly YLe Yas_kxutn Wba ( Icvec nul Yw poxn Apely Paalu ( Slcp: Aw Skee allblu Yhu Cult Cconncy den cnltetn' S...
##### A point source of $P=12 mathrm{~W}$ is located on the axis of a circular mirror plate of radius $R=3 mathrm{~cm} .$ If distance of source from the plate is $a=39 mathrm{~cm}$ and reflection coefficient of mirror plate is $a=0.7$, calculate force exerted by light rays on the plate.
A point source of $P=12 mathrm{~W}$ is located on the axis of a circular mirror plate of radius $R=3 mathrm{~cm} .$ If distance of source from the plate is $a=39 mathrm{~cm}$ and reflection coefficient of mirror plate is $a=0.7$, calculate force exerted by light rays on the plate....
##### 19) Water is being poured into a circular cone of "height 15 m and radius 5 m at & rate of 10 m3 /min_At what rate is the water level rising when the level is 3 m high? FormulasVolume of a sphere: V = ;ur3 Surface area of a sphere: A = 4nr2 Volume of a trapezoidal prism: V =4( b)hw Volume of a cone: V = trzh
19) Water is being poured into a circular cone of "height 15 m and radius 5 m at & rate of 10 m3 /min_ At what rate is the water level rising when the level is 3 m high? Formulas Volume of a sphere: V = ;ur3 Surface area of a sphere: A = 4nr2 Volume of a trapezoidal prism: V =4( b)hw Volum...
##### Find each derivative and simplify.$$rac{d}{d x} 2^{x^{3}-x^{2}+4 x+1}$$
Find each derivative and simplify. $$\frac{d}{d x} 2^{x^{3}-x^{2}+4 x+1}$$...
##### If the foci of an ellipse are $(-4,4)$ and $(6,4),$ then the coordinates of the center of the ellipse are _____.
If the foci of an ellipse are $(-4,4)$ and $(6,4),$ then the coordinates of the center of the ellipse are _____....
##### A steel boat is foating - motionless in the watcr_ The density of water is 1000 kg/m" When floating; the boat displaces 100 m? of water. What is the magnitude of the buoyant force on the boat_ in Newtons? (6) The boat springs leak and sinks, initially becoming completely submerged under the water leaving no air pockets. The density of the steel comprising the boat is 7800 kg/m" . The steel that comprises the boat occupies & volume of 12 m" What is the buoyant force On the subm
A steel boat is foating - motionless in the watcr_ The density of water is 1000 kg/m" When floating; the boat displaces 100 m? of water. What is the magnitude of the buoyant force on the boat_ in Newtons? (6) The boat springs leak and sinks, initially becoming completely submerged under the wat...
##### Jottewuk0.5 Poisson Probability Distributions (With Geomet scont Clo CCqale 5.3.12-TMw scor; 73".B or Hpl Drban - 1#Lneni 0n3nlca Fednn Dmee Mu#neucrnoie DURDEIA Atnd 'ndaulJit EleldESnee ndnnelnca I blonna umtrtal tei CtnentmtIauanetnd by ueininumicretaech cottte Kilal rauna
Jottewuk0.5 Poisson Probability Distributions (With Geomet scont Clo CCqale 5.3.12-T Mw scor; 73".B or Hpl Drban - 1#Lneni 0n3nlca Fednn Dmee Mu#neucr noie DURDEIA Atnd 'ndaulJit EleldESnee ndnnelnca I blonna umtrtal tei Ctnentmt Iauanetnd by ueini numicretaech cottte Kilal rauna...
##### An enzyme-catalysed reaction in the presence of a competitive inhibitor:has the same reaction rate at all concentrations of inhibitor. has the same velocity as the uninhibited reaction at sub-saturating substrate concentrations_ has an unaltered Km, as the shape of the enzyme is unchanged by the inhibitor may have either substrate or inhibitor bound (non-simultaneously) at the same site
An enzyme-catalysed reaction in the presence of a competitive inhibitor: has the same reaction rate at all concentrations of inhibitor. has the same velocity as the uninhibited reaction at sub-saturating substrate concentrations_ has an unaltered Km, as the shape of the enzyme is unchanged by the in...
##### (I) How fast does water flow from a hole at the bottom of a verywide, 5.3 -m-deep storage tank filled with water? Ignore viscosity.
(I) How fast does water flow from a hole at the bottom of a very wide, 5.3 -m-deep storage tank filled with water? Ignore viscosity....
##### 1. (a) Solve y''' - 6y'' + 11y'- 6y = ex by Variation of Parametermethod (b) Define UCfunction (c) Briefly explain thegeometric interpretation of linear first order differentialequationPlease solve all the questions
1. (a) Solve y''' - 6y'' + 11y' - 6y = ex by Variation of Parameter method (b) Define UC function (c) Briefly explain the geometric interpretation of linear first order differential equation Please solve all the questions...
##### A 123 mL of Pb(NO3)2 at 1.5 x 10-2 M is mixed with 123 mL of 1.5 x 10-3 M NaF. Will a precipitate form? Ksp = 4 x 10-8
A 123 mL of Pb(NO3)2 at 1.5 x 10-2 M is mixed with 123 mL of 1.5 x 10-3 M NaF. Will a precipitate form? Ksp = 4 x 10-8...
##### Suppose that [-1 0] A=[-2 -3 1] [1 ~2] dovour work by hand,show allofyour work a. Find the characteristic polvnomial of A b. Find the eigenvalues and eigenvectors of A Find the algebraic and geometric multiplicity of A d. Prove that A is defective;
Suppose that [-1 0] A=[-2 -3 1] [1 ~2] dovour work by hand,show allofyour work a. Find the characteristic polvnomial of A b. Find the eigenvalues and eigenvectors of A Find the algebraic and geometric multiplicity of A d. Prove that A is defective;...
##### Duarin i.v3ias - Lket IJC rIT 2 #e ldlloxice: {) ?,-7ttti4+2 {4) Qi,=7+23
Duarin i.v3ias - Lket IJC rIT 2 #e ldlloxice: {) ?,-7ttti 4+2 {4) Qi,=7+23...
##### Suppose that a researcher; using wage data on randomly selected workers, estimates the OLS regression:Wage : 11.32+2.05 College + 3.84 White 2.25 Collegex White + uwhere Wage is measured in dollars per hour; College is a binary variable that is 1if the person has a college degree and 0 if the person does not have a college degree, and White is a binary variable that is 1 if the person is white and 0 if the person is non-white. Collegex White is an interaction term between the two binary variable
Suppose that a researcher; using wage data on randomly selected workers, estimates the OLS regression: Wage : 11.32+2.05 College + 3.84 White 2.25 Collegex White + u where Wage is measured in dollars per hour; College is a binary variable that is 1if the person has a college degree and 0 if the pers...
##### A sample of 40 male aerobics instructors in a certain city had a mean of 173 pounds (lbs ) with standard deviation of 6 Ibs. We wish to test the claim that the mean weight is at most 175 Ibs Express the claim and the null and alternative hypotheses in symbolic form for this claim, then answer the questions below:(1)Perameter: Choose the parameter symbol to be used in the claim and the hypotheses. xbarmu(2)Claim:(3) Ho(4)H;(5)Tails : this test will be ~tailed_0 leftright [Wo
A sample of 40 male aerobics instructors in a certain city had a mean of 173 pounds (lbs ) with standard deviation of 6 Ibs. We wish to test the claim that the mean weight is at most 175 Ibs Express the claim and the null and alternative hypotheses in symbolic form for this claim, then answer the qu...
-- 0.023329--
|
2022-08-11 15:05:35
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6312629580497742, "perplexity": 7958.624447979831}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571472.69/warc/CC-MAIN-20220811133823-20220811163823-00368.warc.gz"}
|
http://en.wikipedia.org/wiki/Simple_shear
|
# Simple shear
Simple shear
In fluid mechanics, simple shear is a special case of deformation where only one component of velocity vectors has a non-zero value:
$\ V_x=f(x,y)$
$\ V_y=V_z=0$
And the gradient of velocity is constant and perpendicular to the velocity itself:
$\frac {\partial V_x} {\partial y} = \dot \gamma$,
where $\dot \gamma$ is the shear rate and:
$\frac {\partial V_x} {\partial x} = \frac {\partial V_x} {\partial z} = 0$
The deformation gradient tensor $\Gamma$ for this deformation has only one non-zero term:
$\Gamma = \begin{bmatrix} 0 & {\dot \gamma} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
Simple shear with the rate $\dot \gamma$ is the combination of pure shear strain with the rate of $\dot \gamma \over 2$ and rotation with the rate of $\dot \gamma \over 2$:
$\Gamma = \begin{matrix} \underbrace \begin{bmatrix} 0 & {\dot \gamma} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ \mbox{simple shear}\end{matrix} = \begin{matrix} \underbrace \begin{bmatrix} 0 & {\dot \gamma \over 2} & 0 \\ {\dot \gamma \over 2} & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ \mbox{pure shear} \end{matrix} + \begin{matrix} \underbrace \begin{bmatrix} 0 & {\dot \gamma \over 2} & 0 \\ {- { \dot \gamma \over 2}} & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ \mbox{solid rotation} \end{matrix}$
Important examples of simple shear include laminar flow through long channels of constant cross-section (Poiseuille flow), and elastomeric bearing pads in base isolation systems to allow critical buildings to survive earthquakes undamaged.
## Simple shear in solid mechanics
In solid mechanics, a simple shear deformation is defined as an isochoric plane deformation in which there are a set of line elements with a given reference orientation that do not change length and orientation during the deformation.[1] This deformation is differentiated from a pure shear by virtue of the presence of a rigid rotation of the material.[2][3]
If $\mathbf{e}_1$ is the fixed reference orientation in which line elements do not deform during the deformation and $\mathbf{e}_1-\mathbf{e}_2$ is the plane of deformation, then the deformation gradient in simple shear can be expressed as
$\boldsymbol{F} = \begin{bmatrix} 1 & \gamma & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.$
We can also write the deformation gradient as
$\boldsymbol{F} = \boldsymbol{\mathit{1}} + \gamma\mathbf{e}_1\otimes\mathbf{e}_2.$
|
2014-04-17 23:12:04
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 15, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8519086241722107, "perplexity": 644.2969017560158}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00317-ip-10-147-4-33.ec2.internal.warc.gz"}
|
https://www.risczero.com/docs
|
Getting Started
Examples and Explanations
To get started building with RISC Zero, take a look at these examples, which feature explanations and tutorials for code available in our GitHub repositories:
• Hello World in Rust - Our starter project for the RISC Zero zkVM. If you're just getting started writing code for the zkVM, we recommend reading (and playing with) this repository.
• RISC Zero Battleship - To see a more complex use of the RISC Zero zkVM in action, take a look at this working Battleship game. Here, players use private game board states to track whether opponents have sunk their battleships.
• RISC Zero Digital Signatures - In this example, you'll see how to verifiably sign code run inside the RISC Zero zkVM. This example features a post-quantum digital signature generated using only SHA-2 as a cryptographic primitive.
• RISC Zero Password Validity Checker - In this example, you'll see Alice convince Bob's Identity Service that her password meets Bob's validity requirements. This example makes use of public shared outputs that Alice can write to the RISC Zero zkVM's journal.
Open Source Repositories
• Rust Crates - If you're a Rust user, you'll find RISC Zero crates here, ready to be included in your existing projects.
• Contribute to RISC Zero - If you're interested in how RISC Zero projects for the zkVM work, or curious about contributing to this project, come take a look at our main project repository.
|
2022-10-06 23:45:04
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2725849747657776, "perplexity": 6023.556692598839}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337889.44/warc/CC-MAIN-20221006222634-20221007012634-00449.warc.gz"}
|
https://prepinsta.com/top-100-puzzles/ant-and-triangle-puzzle/
|
# Ant and Triangle Puzzle
## Solve the Ant and Triangle Puzzle
Solve the Ant and Triangle Puzzle. Three ants are sitting at the three corners of an equilateral triangle. Each ant starts randomly picks a direction and starts to move along the edge of the triangle. What is the probability that none of the ants collide?
## Detailed Explanation
The ants can only avoid a collision if they all decide to move in the same direction (either clockwise or anti-clockwise). If the ants do not pick the same direction, there will be a collision. Each ant has the option to either move clockwise or anti-clockwise. There is a one-in-two chance that an ant decides to pick a particular direction. Using simple probability calculations, we can determine the probability of no collision.
N(No collision) = N(All ants go in a clockwise direction) + N( All ants go in an anti-clockwise direction) = $0.5 \times 0.5 \times 0.5 + 0.5 \times 0.5 \times 0.5 = 0.25$
## Get over 200+ course One Subscription
Courses like AI/ML, Cloud Computing, Ethical Hacking, C, C++, Java, Python, DSA (All Languages), Competitive Coding (All Languages), TCS, Infosys, Wipro, Amazon, DBMS, SQL and others
## Checkout list of all the video courses in PrepInsta Prime Subscription
• Crossing the Bridge Puzzle
• Death & Marbles Puzzle
• Airplane Seat Puzzle
• 2 Eggs & 100 Floors Puzzle
|
2022-12-04 02:17:51
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 1, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4293873906135559, "perplexity": 2339.6915749652544}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710953.78/warc/CC-MAIN-20221204004054-20221204034054-00084.warc.gz"}
|
https://gmatclub.com/forum/solve-time-and-work-problems-efficiently-using-efficiency-method-266202.html
|
GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 14 Nov 2019, 01:55
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Solve Time and Work Problems Efficiently using Efficiency Method!
Author Message
TAGS:
### Hide Tags
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3138
Solve Time and Work Problems Efficiently using Efficiency Method! [#permalink]
### Show Tags
Updated on: 07 Aug 2018, 06:11
10
22
Added the PDF of the article at the end of the post!
Solve Time and Work Problems Efficiently using Efficiency Method!
Objective of the article
• Are you someone who takes a long time to solve Time and Work?
• Do you get stuck in the fractional and decimal calculation related to Time and Work problems?
• Do you face problem in understanding how positive and negative efficiency works in Time and Work problems?
• The application of LCM to solve Time and Work questions easily.
• A comparative study of methods of solving Time and Work problems – by using fraction and by using LCM method, which one should be used and when?
• How efficiency is defined and in what ways it can be implemented in problem solving
In this article, we have only used official sources to help you understand the above aspects. Also, at the end of the article, we have provided some practice questions where you can apply your learning from the article.
Solving Time and Work Problem: Conventional Method
In time and work problem, where people with different constant efficiencies do a certain work, we assume the total work to be constant. Mostly we solve these questions assuming the total work to be 1 unit. Let’s look at the following example:
Example 1
It would take one machine 4 hours to complete a large production order and another machine 3 hours to complete the same order. How many hours would it take both machines, working simultaneously at their respective constant rates, to complete the order?
Solution 1A: Conventional Method
Here it is given that two machines, working independently with their different constant efficiency, complete a large production order in 4 hours and 3 hours respectively.
• As the total work done by both the machines are same, the work is constant here.
Let us assume the total work of production order is 1 unit. Therefore, we can say:
• 1st machine in 4 hours can compete 1 unit of work
• Hence, 1st machine in 1 hour can complete $$\frac{1}{4}$$ units of work
• Similarly, 2nd machine in 3 hours can complete 1 unit of work
• Hence, 2nd machine in 1 hour can complete $$\frac{1}{3}$$ units of work
• If both the machines work together simultaneously, then in 1 hour they will complete ($$\frac{1}{4} + \frac{1}{3}$$) units of work = $$\frac{(3+4)}{12}$$ units of work = $$\frac{7}{12}$$ units of work
• As the machines can complete $$\frac{7}{12}$$ units of work in 1 hour time, to complete the whole work of 1 unit, they will take = $$\frac{1}{(7/12)}$$ hours = $$\frac{12}{7}$$ hours
Hence, working together, both the machines will take $$\frac{12}{7}$$ hours to complete the production order.
Important observation
In the beginning of the solution, we have assumed the total constant work to be 1 unit.
• Important thing one can observe here is that assuming the total work to be 1 unit or x units can make life difficult for us – because, when we calculate the 1-day work of that person or machine, chances are very high that we will be getting fractional values.
Now we all know, dealing with fractional calculations can be difficult and it increases the chances of making error to a great extent. But the question is how we can avoid it?
Again, if you look carefully, the 1-day work of a person or a machine is calculated as the ratio of total work and the time taken.
• Therefore, if the assumed value of the total work is a multiple of the time taken, then the 1-day work will certainly be an integer value.
• So, when we are assuming a value for the total work, we need to consider such a value which should be a multiple of the time taken by the different parties working on that job.
• In other words, the assumed value of the total work can be taken as the least common multiple (LCM) of the given individual time taken by the persons or machines working on that job.
Let us try to solve the same question using the method that we just mentioned.
Solution 1B: Changing the total work value
Let us assume the total work of production order is 12 units. Therefore, we can say:
• In 4 hours 1st machine can complete 12 units of work
• Hence, in 1 hour 1st machine can complete 3 units of work
• Similarly, 2nd machine in 3 hours can complete 12 units of work
• Hence, in 1 hour 2nd machine can complete 4 units of work
• If both the machines work simultaneously together, in 1 hour they will complete (3 + 4) units of work = 7 units of work
• As the complete work is 12 units, to complete the order they need to do total 12 units of work together
• Combined together, the machines are completing 7 units of work in 1 hour
• Therefore, the machines will complete 12 units of work together in $$\frac{12}{7}$$ hours
Hence, working together, both the machines will take $$\frac{12}{7}$$ hours to complete the production order.
Important Takeaway
If we compare the methods explained above, fundamentally they are same, with some difference in value-wise derivation. But the most important thing to observe is that there are no fractional intermediate calculations involved in the second solution.
• As the total work is assumed to be the LCM of 3 and 4, we are avoiding the fractional calculation, and thus, saving the overall calculation time.
Let us take another example to validate this learning:
Example 2
One inlet pipe fills an empty tank in 5 hours. A second inlet pipe fills the same tank in 3 hours. If both pipes are used together, how long will it take to fill 2/3rd of the tank?
Solution 2A: Conventional Method
In this question, it is given that one inlet pipe can fill an empty tank in 5 hours. A second inlet pipe fills the same tank in 3 hours.
As the capacity of the tank remains same, let’s assume the capacity of the tank is 1 unit.
• As the 1st pipe takes 5 hours to fill 1 unit,
• In 1 hour it will fill $$\frac{1}{5}$$ units
• Similarly, the 2nd pipe takes 3 hours to fill 1 unit
• Hence, in 1 hour it will fill $$\frac{1}{3}$$ units
• If both pipes are used together, in 1 hour they will fill = ($$\frac{1}{5} + \frac{1}{3}$$) units = $$\frac{(3+5)}{15}$$ units = $$\frac{8}{15}$$ units
Hence, to fill $$\frac{2}{3}$$rd of the tank, they will take = $$\frac{2}{3}/\frac{8}{15}$$ hrs = $$\frac{2}{3} * \frac{15}{8}$$ hrs = $$\frac{5}{4}$$ hrs
Now, let’s try solving the same question using the LCM method:
Solution 2B: LCM Method
As the capacity of the tank remains same, let’s assume the capacity of the tank is 15 units.
[notice that 15 = LCM (3,5)]
• As the 1st pipe takes 5 hours to fill 15 units,
• In 1 hour it will fill 3 units
• Similarly, the 2nd pipe takes 3 hours to fill 15 units
• Hence, in 1 hour it will fill 5 units
• If both pipes are used together, in 1 hour they will fill = (3 + 5) units = 8 units
• The total work to be completed = ($$\frac{2}{3} * 15$$) = 10 units
Hence, to fill $$\frac{2}{3}$$rd of the tank, they will take = $$\frac{10}{8}$$ hrs = $$\frac{5}{4}$$ hrs
Important Takeaway
As you can see, both the problems can be solved using conventional method as well as LCM method. However, application of LCM method significantly decreases the calculation related, and therefore the time taken to solve a particular question becomes less.
Let us solve couple of more examples using the LCM method to reinforce our learning.
Example 2
Solving Time and Work Problem: LCM Method
In the following questions, we will solve the problems by applying LCM method only.
Example 3
Machines A and B always operate independently and at their respective constant rates. When working alone, Machine A can fill a production lot in 3 hours, and Machine B can fill the same lot in x hours. If A and B worked alternatively, while each work for 1 hour at a time, the total work gets completed in 4 hours. What is the value of x?
Solution 3
In this question it is given that machine A can fill a production lot in 3 hours, and machine B can fill the same production lot in x hours, when they are working alone.
However, they will take 4 hours to complete the job, when they are working alternatively for 1 hour each.
Let us assume the total work to be 3x units (LCM of 3 and x)
• As machine A takes 3 hours to fill 3x units,
• In 1 hour it will fill $$\frac{3x}{3}$$ units = x units
• Similarly, machine B takes x hours to fill 3x units,
• Hence, in 1 hour it will fill $$\frac{3x}{x}$$ units = 3 units
• It is also given that; machine A and machine B, working alternatively, take 4 hours to fill 3x units
• Therefore, in 2 hours’ time they will fill = (x + 3) units
• Hence, in 1 hour the production lot they fill = $$\frac{1}{2} * (x + 3)$$ units
• Now we can say in 4 hours,
• Units filled by A + units filled by B = total production units
Or, $$4 * \frac{1}{2} * (x + 3)$$ = 3x
Or, 2x + 6 = 3x
Or, x = 6 hours
Hence, the value of x is 6.
Important Takeaway
Unlike the previous questions, in this question we have a variable defined as x. Even in such case, we can apply the LCM method, as we did in the solved example, to get quick and accurate answer.
Example 4
Working at a constant rate, pump X pumped out half of the water in a flooded basement in 4 hours. Then pump Y was started and the two pumps, working independently at their respective constant rates, pumped out the rest of the water in 3 hours. How many hours would it have taken pump Y, operating alone at its own constant rate, to pump out all of the water that was pumped out of the basement?
Solution 4
In this question there are two pumps working separately at their own constant rate.
• Pump X cleared half of the water present in the flooded basement in 4 hours
• Therefore, to clear all the water, pump A would have taken twice the amount of time, i.e. 8 hours
• After pump X worked for 4 hours, pump Y also started working, and together they took 3 hours to clear the remaining half of the water
• As they took 3 hours to clear half the total amount of water, they would have taken double of that time, i.e. 6 hours, to clear the whole water
With this inference, we actually have a simple problem, which is very similar to the previous ones that we solved.
• Pump X takes 8 hours to clear the whole water
• Pump X and Y together take 6 hours to clear the whole water
Let us assume the total amount of water to be 24 units
• In 8 hours, pump X clears 24 units,
• Hence, in 1 hour, pump X clears $$\frac{24}{8}$$ units = 3 units
• Similarly, in 6 hours, pump X and Y clears 24 units,
• Hence, in 1 hour, pump X and Y together clear $$\frac{24}{6}$$ units = 4 units
• Now in 1 hour, out of the 4 units of water cleared by pump X and Y together, 3 units were cleared by X only
• Therefore, amount of water cleared by Y in 1 hour = (4 – 3) unit = 1 unit
As pump Y can clear 1 unit of water in 1 hour, it will take $$\frac{24}{1}$$ hours = 24 hours to clear the whole amount of water, when working alone at its constant rate.
Important Observation
Apart from the usual application of the LCM method, another distinct observation we can do from this question, which is applicable in the previous question also. Let us see that:
• X can complete the whole work alone in 8 hours
• Y can complete the whole work alone in 24 hours
• X and Y together can complete the whole work in 6 hours
It is noticeable that the time taken by them together to complete the work is much less than their individual time. This is happening because of their difference in individual efficiencies.
Application of Efficiency in Time and Work problems
As we know, efficiency essentially means the rate of work, i.e. the amount of work done in unit time.
• The more time one takes to complete a job, the less efficient the person is.
• Similarly, the less time one takes to complete a job, the more efficient the person is.
Therefore, we can say that, efficiency is inversely proportional to the time taken to complete the work.
In a given scenario, efficiency can be used to explain the work rate either of an individual or of a group consists of similar objects. For a group, it is generally assumed that everyone within that group has the same efficiency.
Let us take an example of a time and work problem, which relates to the concept of efficiency:
Example 5
At a loading dock, each worker on the night crew loaded 3/4 as many boxes as each worker on the day crew. If the night crew has 4/5 as many workers as the day crew, what fraction of all the boxes loaded by the two crews did the day crew load?
Solution 5
In this question we have two separate groups of people to consider.
• Each member within the group has the same efficiency as other members
• However, group wise the efficiencies are different from each other
• This means that everyone in the night crew has the same efficiency and everyone in the day crew has the same efficiency. However, the efficiency of day and night crew in this case is different.
Let us look at the problem in detail.
It is given that there are two groups of workers – one operates at day time and the other one operates at night time.
• Each worker on the night crew loaded $$\frac{3}{4}$$ as many boxes as each worker on the day crew
• This statement defines the difference in efficiency of each worker belongs to night crew and day crew
• If we assume that each worker in the day crew loaded 4 boxes, then each worker in the night crew must have loaded $$4 * \frac{3}{4}$$ boxes = 3 boxes
• It is also given that night crew has $$\frac{4}{5}$$ as many workers as the day crew
• If we assume that there are 5 workers in the day crew, then there will be $$5 * \frac{4}{5}$$ = 4 workers in the night crew
[notice that we have intentionally assumed the number of box loaded and number of day crew workers as 4 and 5, to avoid any fractional calculation. Both the numbers could have been assumed as 1 or any other number – the end result would have been same, but may lead to fractional calculations]
We can show the whole data in the following table:
As we can see, total boxes loaded by day crew are 20 and by night crew are 12.
Hence, the boxes loaded by day crew as the fraction of all the boxes loaded by both the crew = $$\frac{20}{(20+12)} = \frac{20}{32} = \frac{5}{8}$$
Let us take one last example to demonstrate the application of efficiency:
Example 6
Machines X and Y work at their respective constant rates. How many more hours does it take machine Y, working alone, to fill a production order of a certain size than it takes machine X, working alone?
A. Machines X and Y, working together, fill a production order of this size in two-thirds the time that machine X, working alone, does.
B. Machine Y, working alone, fills a production order of twice the size in 6 hrs.
Option choices:
A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are NOT sufficient.
Solution 6
In this question, two machines X and Y, with different efficiency, can work on a certain job.
• The question asks the number of extra hours that machine Y takes to complete the job when working alone, compared to the time taken by machine X to complete the same job, working alone.
As no other information was provided in the question stem, we will analyse the given statements.
Analysing Statement 1:
• As per the information given in statement 1, machines X and Y, working together, fill a production order of the same size in two-thirds the time that machine X, working alone, does.
• If we assume that machine X, working alone, takes t hours to complete the job, then from statement 1 we can say
• Machine X and Y, working together, complete the whole work in $$\frac{2}{3}$$t hours
• As both the machines work on their constant efficiencies throughout, we can say in $$\frac{2}{3}$$t hrs machine X can complete $$\frac{2}{3}$$rd of the total work.
• Therefore, in that $$\frac{2}{3}$$t hours, machine Y completes $$\frac{1}{3}$$rd of the total work.
• Hence, machine Y can complete the total work in $$\frac{2}{3}$$t * 3 hours = 2t hours
• But we cannot deduce the exact value of t from this statement.
Hence, statement 1 is not sufficient to answer the question.
Analysing Statement 2:
• As per the information given in statement 2, machine Y, working alone, fills a production order of twice the size in 6 hrs
• Hence, if we consider a production order of the original volume, machine Y will take 3 hours to complete it
• But from this statement, we cannot figure out any information regarding the work rate or time taken by X to complete the work
Hence, statement 2 is not sufficient to answer the question.
Combining Both Statements:
• If we consider both statements together, we can say machine Y takes 2t hours to complete the whole job, which is equal to 3 hours
• Hence, the value of t = $$\frac{3}{2}$$ hours
• We also know that machine X takes t hours to complete the whole job working alone
• As we already know t is equal to $$\frac{3}{2}$$ hours, we can say machine X takes $$\frac{3}{2}$$ hours to complete the job working alone
• Therefore, machine Y takes ($$3 – \frac{3}{2}$$) hours = $$\frac{3}{2}$$ hours more than what machine X takes to complete the work, working individually.
As information from both the statements are required to solve the question, we can say the correct answer choice is Option C.
• In Time and Work problems where we are supposed to assume the value of total work, it is always a good practice to assume the total work as the LCM of the given days (or any multiple of LCM), rather than assuming the total work as 1 unit – as it decreases the complex calculations involve with fractions.
• LCM method can still be applied when one of the data points are given in terms of variable
• Efficiency is inversely proportional to time taken.
• When working in groups, all are considered to be equally efficient within a group, whereas people have different efficiencies across groups.
What next in this series of article??
Next week we will come up with one more article in the Distance series.
In that article, we are going to discuss 3 common errors students make while solving Time and Work problems
If you have enjoyed the article, then try the practice questions given below.
Attachments
Time and work.pdf [599.89 KiB]
_________________
Originally posted by EgmatQuantExpert on 23 May 2018, 04:18.
Last edited by EgmatQuantExpert on 07 Aug 2018, 06:11, edited 6 times in total.
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3138
Solve Time and Work Problems Efficiently using Efficiency Method! [#permalink]
### Show Tags
Updated on: 13 Aug 2018, 05:25
Originally posted by EgmatQuantExpert on 23 May 2018, 04:20.
Last edited by EgmatQuantExpert on 13 Aug 2018, 05:25, edited 1 time in total.
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3138
Re: Solve Time and Work Problems Efficiently using Efficiency Method! [#permalink]
### Show Tags
27 May 2018, 05:34
1
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3138
Re: Solve Time and Work Problems Efficiently using Efficiency Method! [#permalink]
### Show Tags
27 May 2018, 23:06
1
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3138
Re: Solve Time and Work Problems Efficiently using Efficiency Method! [#permalink]
### Show Tags
13 Jun 2018, 23:54
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3138
Re: Solve Time and Work Problems Efficiently using Efficiency Method! [#permalink]
### Show Tags
25 Jun 2018, 22:59
Intern
Joined: 22 Apr 2018
Posts: 32
Re: Solve Time and Work Problems Efficiently using Efficiency Method! [#permalink]
### Show Tags
24 Jul 2018, 00:44
I didn't understand the following line in Solution 4:
Similarly, in 6 hours, pump Y clears 24 units,---- How do we know that Pump Y clears 24 units in 6 hours this is what we have to find right??
• Hence, in 1 hour, pump X and Y together clear
24
6
246
units = 4 units
Also in Solution 6:
As both the machines work on their constant efficiencies throughout, we can say in
2/3
Can anybody please explain? I'm really confused
2/3t hrs machine X can complete
2/3
2/3rd of the total work.
------------------- How do we know that Machine X will complete 2/3rd of work??
_________________
If you like my post, please don't shy away from giving Kudos, as it boosts my confidence to post more
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3138
Re: Solve Time and Work Problems Efficiently using Efficiency Method! [#permalink]
### Show Tags
24 Jul 2018, 01:13
Quote:
I didn't understand the following line in Solution 4:
Similarly, in 6 hours, pump Y clears 24 units,---- How do we know that Pump Y clears 24 units in 6 hours this is what we have to find right??
• Hence, in 1 hour, pump X and Y together clear
24
6
246
units = 4 units
It should be X and Y, not Y only - the X was missing. However, it can be understood from the next immediate line, where the 1 hour work of X and Y together is calculated.
Quote:
Also in Solution 6:
As both the machines work on their constant efficiencies throughout, we can say in
2/3
Can anybody please explain? I'm really confused
2/3t hrs machine X can complete
2/3
2/3rd of the total work.
------------------- How do we know that Machine X will complete 2/3rd of work??
It is assumed that machine X, working alone, takes t hours to complete the job.
Hence, in $$\frac{2}{3}$$t hours, it will complete $$\frac{2}{3}$$rd of the whole job.
_________________
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3138
Solve Time and Work Problems Efficiently using Efficiency Method! [#permalink]
### Show Tags
Updated on: 29 Oct 2018, 23:35
Originally posted by EgmatQuantExpert on 27 Sep 2018, 03:51.
Last edited by EgmatQuantExpert on 29 Oct 2018, 23:35, edited 2 times in total.
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3138
Re: Solve Time and Work Problems Efficiently using Efficiency Method! [#permalink]
### Show Tags
03 Oct 2018, 05:19
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3138
Re: Solve Time and Work Problems Efficiently using Efficiency Method! [#permalink]
### Show Tags
13 Feb 2019, 06:31
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3138
Re: Solve Time and Work Problems Efficiently using Efficiency Method! [#permalink]
### Show Tags
14 Feb 2019, 22:42
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3138
Re: Solve Time and Work Problems Efficiently using Efficiency Method! [#permalink]
### Show Tags
21 Feb 2019, 05:25
Re: Solve Time and Work Problems Efficiently using Efficiency Method! [#permalink] 21 Feb 2019, 05:25
Display posts from previous: Sort by
|
2019-11-14 08:54:58
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6507670283317566, "perplexity": 1265.4307047840239}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496668334.27/warc/CC-MAIN-20191114081021-20191114105021-00557.warc.gz"}
|
https://www.physicsforums.com/threads/visualising-tensors.155186/
|
# Visualising tensors
1. Feb 7, 2007
### nolanp2
Can anyone help me to unerstand what tensors are in physics, a few basic examples would probably do, and how they actually go about giving you the results. the only explanations i've been given of them are through maths which are useless to me since i still cant manage to visualise what a matrix is actually doing in many cases. need help urgently!!
2. Feb 7, 2007
### arunma
I've always had trouble with this too. I took both differential geometry and cosmology, and I still didn't get it. I only started to understand tensors when I got an easier introduction to them in special relativity.
Here's my understanding. Tensors are essentially matrices, except they can exist in more than two dimensions. For example, a tensor with one index is a vector. If you were to write it out as a matrix, it would exist in one dimension. Tensors with two dimensions are just typical square matrices. A tensor of three indices can be thought of as a three-dimensional box of numbers. And it just goes from there.
Of course, I've never quite understood the difference between covarient and contravarient tensors. Let me know if you ever find out!
3. Feb 7, 2007
### cristo
Staff Emeritus
Do you know the difference between a covariant and a contravariant vector?
4. Feb 7, 2007
5. Feb 7, 2007
### arunma
Well, I know that one has raised indices, and the other has lowered indices. And I know that raising or lowering both indices simultaneously means that that tensor becomes its inverse. So I can brainlessly do the operation. But I've never understood why this is the case.
6. Feb 7, 2007
### quasar987
The words covariant and contravariant refer to the way the components of the tensor transform as the basis is transformed (contravariant = in a contrary way and covariant = in a ...well in the same way :P).
I feel I cannot say more without rewriting the whole theory from scratch so I refer you to say Sharipov's ebook: http://uk.arxiv.org/abs/math/0403252/
7. Feb 8, 2007
### mjsd
how do you actually want to visualise it? As apples and oranges ? or as a collection of numbers/symbols/sea shells group together in some meaningful way? and that they follow particular rules when you try to juggle the content....?
I guess you may have a chance to visualise tensors if you limit yourself to tensors with very small ranks and very low dimensionality. Otherwise, it would be as difficult as to visualising a 10-dimensional hypercube for instance.
remember tensors are defined by how they transform, and certainly not all tensors naturally lead to some "everyday" visualizations.
8. Feb 14, 2007
### baryon
A tensor is a collection of values that alters a field.
Last edited: Feb 14, 2007
9. Feb 14, 2007
### CPL.Luke
from my very limited understanding of them they can work as follows
an mXm matrix will take a vector and transform it to produce another vector in the same "space" (ex. R^m)
now what if you had an array of several matrices? if you were to multiply this by a vector such that each component of the vector gave a certain "weight" to each of the component matrices in the tensor, thus producing a new matrix that could then be used to tranform a vector.
I believe that general relativity works in this way.
but most simply put a tensor is an array of matrices, and ghigher dimensional tensors are then arrays of lower dimensional tensors. similar in effect to a hypercube.
NOTE: I have no formal traiing or even informal self-study of tensors, I just gathered this from a couple of brief chapters on general relativity a while back, so what I said could easily be very wrong.
10. Feb 14, 2007
### Magister
The contravariant and convariant vectors are related by the way you get the components of the vector. If you use the paralelogram rule of the vectors you get the contravariants components, for instance if you get the components by tracing a line ortogonal to the axis you get the covariant components. Is because of this that in $R^3$ the contravariants and covariants of a vector are equal.
I suppose that the name covariant cames from the fact that when you change one axis only the components related to this axis change while when working with contravariant components all the components change.
I hope I have made my self clear.
11. Feb 14, 2007
### Hurkyl
Staff Emeritus
IMHO, one thing that would help is to try to understand what a tensor is before you try and understand what a tensor field is.
(much like it's easier to learn about vectors before you try to learn about a vector field)
If you've chosen a basis, the coordinates of a vector form an nx1 array of numbers. The coordinates of a covector form a 1xn array of numbers.
In coordinates, a tensor product is a generalization of a scalar product. For example, in three dimensions, if I tensor the covector [a b c] with something (could be anything: a scalar, a vector, a covector, a matrix, or some other kind of tensor)
$$\left[\begin{array}{ccc}a & b & c\end{array} \right] \otimes \mathbf{v} = \left[\begin{array}{ccc}a\mathbf{v} & b\mathbf{v} & c\mathbf{v}\end{array}\right]$$
If v was, say, the vector with coordinates x, y, and z, this tensor product would be the matrix:
$$\left[\begin{array}{ccc}a & b & c\end{array} \right] \otimes \left[\begin{array}{c}x \\ y \\ z\end{array} \right] = \left[\begin{array}{ccc} a \left[\begin{array}{c}x \\ y \\ z\end{array} \right] & b \left[\begin{array}{c}x \\ y \\ z\end{array} \right] & c \left[\begin{array}{c}x \\ y \\ z\end{array} \right]\end{array} \right] = \left[\begin{array}{c|c|c}ax & bx & cx \\ ay & by & cy \\ az & bz & cz \end{array} \right]$$
Actually, we would drop the partitions in this case: they're more useful for other cases, such as if v happened to be the covector [p q r]:
$$\left[\begin{array}{ccc}a & b & c\end{array} \right] \otimes \left[\begin{array}{ccc}p & q & r\end{array} \right] = \left[\begin{array}{ccc|ccc|ccc} ap & aq & ar & bp & bq & br & cp & cq & cr \end{array}\right]$$
Mind you, this is all in a coordinate representation: just like with vectors, we can do algebra with tensors without resorting to coordinates.
Last edited: Feb 14, 2007
12. Feb 15, 2007
### Mentz114
This intended to help the fellows above who want to know about covariant and contravariant components. As magister above points out, if the axes of our co-ordinate system are not orthogonal, or are curved, there are two ways of relating a point to an axis. One way is to drop a perpendicular, the other way is to go parallel to another axis. So we have 2 numbers where we had one. These are the contravariant and covariant components. We used to describe a point in 2D with 2 co-ords, x and y, now we have a choice of x (contra), x (covar), y(contra), y (covar).
Now switching notation so super- and subscripts are the dimension ( eg 0=time, 1=x, 2=y etc), if we have 2 points in 2D (X,Y) we have 8 numbers ( n=0,1)
$$x^n, x_n , y^n, y_n$$
Luckily the only quantities we can use in physics are products of contravariant and covariant components, so only two sensible choices exist to work with these numbers,
$$x_m y^m = x_0 y^0 + x_1 y^1$$
and
$$x^n y_n$$
which are the same thing. All the intricacies of tensor analysis follow from the fact that physical quantities are scalars.
I haven't mentioned the metric tensor, to keep it simpler.
Last edited: Feb 15, 2007
13. Feb 15, 2007
### CPL.Luke
hmm I'm intrigued
14. Feb 16, 2007
### Mentz114
Hi CPL - what I've posted is the first step in a journey of a thousand leagues.
B Reimann in the 1840's(roughly) investigated systems where the axes were not straight or orthogonal, realising that concepts like position coords are no longer useful, but that distances could be kept invariant.
I can recommend Stephani's book "General Relativity" as an introduction.
M.
Last edited: Feb 16, 2007
15. Feb 16, 2007
### nolanp2
ok i'm trying to focus on seeing low dimensional tensors in simple instances, such as how tensors would be useful in a force field or something similar. can anybody help me with an example of this?
16. Feb 16, 2007
### robphy
Although it has been raised earlier, it's not clear if you are looking for
a geometric visualization (akin to visualizing a vector as an arrow and vector addition via the parallelogram rule)
or
an organizational typeset visualization (akin to visualizing a vector as a column matrix and vector addition via component-wise addition).
If you are looking for a geometric visualization, did you consult the arxiv.org paper?
17. Feb 16, 2007
### Mentz114
Hi Nolanp2. I don't think visualising tensors helps to understand them.
Thinking about geometry might help. You mention force fields, so I'll give you an example of how and why tensors are used. Consider the equation of motion of a charge in an electric field,
m.a = e(E + v x B ), i.e. mass times acceleration is electric charge times E + v x B, where v is the velocity vector and E and B are electric and magnetic field vectors. x is the vector cross-product.Note that this equation is really 3 equations, one each for x, y and z.
This formula is fine but it is not relativistic. It won't work if you put it in a moving frame of reference.
To make the equation obey special relativity, we go to four dimensions,
t,x,y,z, with stipulation that the time dimension has a negative sign, and
that we multiply all times by c=velocity of light to make them distances.
We can now rewrite the equation in tensor notation,
$$ma^{\mu} = e F^{\mu\nu}v_{\nu}$$
which is four equations since $$\mu$$ can be t,x,y,or z. F is the EM field tensor ( see Wiki for instance).
But the point is that this equation is relativistically covariant, i.e it transforms properly between inertial frames. It's also very elegant.
To do a calculation from this formula, say for x ( mu=1), we expand the tensor to give,
$$ma^{1} = e F^{10}v_{0} + e F^{11}v_{1} + e F^{12}v_{2} + e F^{13}v_{3}$$
Note that all the indexes are numbers, and we can plug in the values of the components to get an algebraic differential equation. The tensors are gone.
Last edited: Feb 16, 2007
18. Feb 16, 2007
### nolanp2
as a specific example i'm looking through small oscillations in s dimensions and it gives the formula U= 1/2 SUM[K_(ik)*x_(i)*x_(k)]. is the constant K a tensor in this case and if so what is it doing? need help badly!
19. Feb 16, 2007
### Mentz114
I think this is just using the tensor summation convention. Expand it over all
values of i and k thus,
U = K(0,0)*x(0)*x(0) + K(0,1)*x(0)*x(1) + K(0,2)*x(0)*x(2) + ....
In fact this just matrix multiplication.
All you need now are the values ( components) of K and some x's.
M
Last edited: Feb 16, 2007
20. Feb 17, 2007
### nolanp2
so do the different k values represent different springs (for example) being stretched when one particle connected to them is moved or does it mean a spring is acting different when pulled in different directions? or does a second degree of freedom mean there's a second partcle and the two are connected via springs? in particular what would k_11 or k_22 represent?
|
2017-03-01 20:20:20
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7721461057662964, "perplexity": 752.9236465511079}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501174276.22/warc/CC-MAIN-20170219104614-00236-ip-10-171-10-108.ec2.internal.warc.gz"}
|
http://jaspervdg.wordpress.com/
|
What could be easier? Still, percentages in polls often add up to all sorts of numbers other than 100. For example, suppose exactly one third of the participants votes for option A, another third for option B, and another third for option C. All options have exactly one third of the votes, or 33.333…%. Rounding obviously gives 33% for all options, which adds up to 99%, rather than 100%. In a similar situation with six options, every option gets 16.666…% of the votes, rounding gives 17%, and 6 times 17 equals 102!
When presenting poll results, it might make sense to ensure that the percentages always add up to 100, but there are also some more subtle problems. Suppose there are only two options and only 4 out of 1000 participants vote for option A. Rounding would result in 0% for option A and 100% for option B. This does add up to 100%, but the ratio 4:996 is quite different from 0:100. Here we will give a method for finding the most likely assignment of percentages (or integer fractions of some other number than 100). Why? Because it’s fun :) Also, if you ever happen to be founding a democracy, the method used here could be considered to give a more representative result than the method currently in use (in the Netherlands at least). Oh, and their might also be some connections to (Huffman) entropy coding (for the connaisseurs: it’s possible to show that we’re actually minimizing the Kullback-Leibler divergence).
So how can we translate a set of votes into a partition of 100 (or any other number for that matter)? One option is to consider the most likely partition, given the counted votes. The “likelihood” of a partition, given the data, is the probability of the data given the partition. If you’re a statistician, that last sentence probably made perfect sense, if you’re a mere mortal, your head might have come off. The idea is really simple. If we wish to evaluate how likely it is that it rains on 50% of the days in the Netherlands (which is apparently reasonably accurate, according to the royal Dutch meteorological institute), then we just go outside each day for a year and see whether it rains. After a year we have a series of measurements: rain, no rain, rain, no rain, etc. Now we can compute the probability of this sequence given our model that each day you have a 50% chance of rain, this probability (of the data) is the likelihood of the model.
First, lets suppose that we have a number of options: $\mathcal{O}$={A, B, …}. The number of votes on option A is denoted by n(A), and similarly for the other options. Say that we also assign probabilities pA, pB, etc. to each option, then the probability of the data for two options A and B equals
$\displaystyle\binom{n}{n(A)}\,p_A^{n(A)}p_B^{n(B)}\text{ with }p_A+p_B=1.$
This is the well-know Binomial distribution.
We will equate the computed percentages with the probabilities in our basic model, so everything stays fixed, except for pA and pB. We can thus ignore the binomial and, in general, maximize the following (with $\mathcal{O}$ being the set of options):
$\displaystyle\prod_{A\in\mathcal{O}} p_A^{n(A)}\text{ subject to }\sum_{A\in\mathcal{O}} p_A=1.$
If desired, the logarithm can be taken to yield:
$\displaystyle\sum_{A\in\mathcal{O}} n(A)\,\log(p_A)\text{ subject to }\sum_{A\in\mathcal{O}} p_A=1.$
However, the idea is that the probabilities stem from fractions with some denominator N (typically N equals 100), so each probability has the form num/N. This makes it an integer progamming problem, and a non-convex, non-linear one at that. Perhaps contrary to expectation, integer optimization problems are typically harder than their continuous counterparts, so this is not a good thing. Luckily (in this case) there is a way out.
We can rely on what is quite possibly the most useful technique for making seemingly intractable problems tractable: dynamic programming. First of all, let’s define the following score function, using numA to denote the (integer) numerator for option A:
$\displaystyle S(\mathcal{O},M)=\underset{num}{\rm{maximize}}\prod_{A\in\mathcal{O}} (num_A/N)^{n(A)}\text{ subject to }\sum_{A\in\mathcal{O}} num_A=M.$
It should be clear that we are interested in the partition that gives us S($\mathcal{O}$,N).
We now proceed as follows:
• First create an array S with N+1 elements, fill index zero with the value 1 and fill the rest of the array with zeroes. You can think of this as stating that if there are no options, then the only possibility is to have a total of 0%.
• Now execute the following for the first option A (assuming n(A)>0):
for total=N to 0 (backwards)
S[total] = 0;
for numerator=1 to total
newS = S[total-numerator]*pow(numerator/N, n(A))
if newS > S[total] then
S[total] = newS
Now we have S[M] = S({A},M). Note that the code never overwrites a value of S that we need later1.
• Now we repeat the previous step for all options, and after having processed the last option S[N] will equal S($\mathcal{O}$,N). The trick is that at each step we do not care about how we got to the sum in S[total-numerator], as this does not affect the factor we multiply it with (pow(numerator/N, n)). So it suffices to remember only the best possible way of getting to total-numerator.
We glossed over the fact that we now just have a procedure for getting the value of S($\mathcal{O}$,N), while we are more interested in the partition that gave us this value. This is typical in dynamic programming, and has a really simple solution, we just keep track of the decisions we made. Practically speaking this means that we would have to keep an array of links around that keeps track of the configurations that gave the values in S (which we update each time we update S).
I implemented the above algorithm in a Fiddle, where you can easily try it out for yourself. The code was kept simple, but basically works fine as long as the number of classes does not exceed N. In comparison with the algorithm above, I eliminated the division by N in the score function, and used the logarithm of the score function to avoid raising numbers to insane powers. Some possible enhancements and optimizations include:
• Ensuring that there is a monotone mapping from counts to numerators (that is, a larger count always gets a larger numerator, and equal counts get equal numerators). Currently a list of counts like 1,1,1 results in the numerators 34,33,33 (although 33,34,33 is equally likely, more on that later), which feels a bit odd (although at least it adds up to 100%). Also, if the number of options exceeds N, weird things can and do happen.
• One may want to detect all partitions that give the maximal score, and then average them or output ranges for example. The idea is that this would solve problems like outputting 34,33,33 when it would have been just as valid to output 33,34,33 or 33,33,34. However, the resulting numerators would no longer be integers, so I’m not sure it’s a good idea. Also, apparently even when assigning seats in the parliament do they eventually fall back to letting fate decide. (At least, that’s what it seems to say in §P.2 of the Dutch “kieswet”.)
• Only evaluate total=100 for the last class. Similarly, the first class technically does not need to store links (they should all point to zero).
• As effectively our only problem is rounding fractions, it might make sense to enforce that we either round each numerator up or down. This can restrict the problem space enormously and speed up the algorithm quite a bit, especially for larger values of N. I have not checked whether this will ever change the end result, but if it does the changes will probably be minor.
• Various optimizations of a purely technical nature (like using typed arrays, which is often faster, although I have not checked that it is so in this case).
Finally, there are also other ways of defining “sensible” partitions. For example, one could look for a partition so that the ratios between the numerators most closely resemble the ratios between fractions. “Most closely resemble” could be defined in terms of differences between logarithms for example. For some purposes such a scheme might make more sense than the approach discussed here.
1 In the inner loop, S[total] is the first value we access, and the outer loop is going backwards (while the inner loop does not go beyond the current total), so after having updated S[total], we never access it again.
Radial gradients have long been a staple of vector graphics. They allow for all sorts of interesting effects, like highlights. However, there are some subtleties to them. This has led to differences between SVG, Canvas, CSS and PDF. And recently there has been some discussion on what the upcoming SVG 2 specification should allow, so time to take a closer look.
First of all, what is a radial gradient? The easiest way to think about it is to imagine a whole bunch of circles being drawn. Conceptually you start at t=0 with a circle with radius r0, colour c0 and centered at (x0,y0), and end at t=1 with a circle with radius r1, colour c1 and centered at (x1,y1). These values are all linearly interpolated to yield intermediate circles:
By convention we consider circles with higher values of t to be above those with lower values of t. This can give the impression of a 3D tube or cone:
Also, in many cases there is support for extending the circles for negative values of t and values of t larger than one:
In these images the first and last circle have equal radius, resulting in a kind of tube. In the second image the tube has been extended, repeating the colour gradient for values of t outside the interval 0-1. For example, the circle for t=-0.8 has the same colour as the circle for t=0.2, and that for t=1.2.
Seems pretty straightforward, doesn’t it? Well, not quite. First of all, there is an obvious choice with respect to the kinds of extensions allowed. But also, more fundamentally, how much freedom one has in specifying the two circles between which we interpolate. There are some technical issues with not enforcing the starting circle to be fully contained in the ending circle.
Here is table showing some of the different choices that have been made:
Format Start outside of end Extension
SVG (2) No. Always, using pad, reflect or repeat.
CSS No (always at the center). Always, using pad or repeat.
PDF Yes. At start, end, neither or both, using pad.
So what is the problem with a starting circle outside the ending circle? (Or vice-versa!) First of all, it is not so much about the starting circle that needs to be contained in the ending circle, it is about the point where the radius is zero: the zero-radius point. This is most easily pictured by imagining the circles building a 3D cone (with t being the 3rd dimension), if the point where the radius is zero is inside the starting and/or ending circle we only see the top-half of the (double) cone, while if the point where the radius is zero is outside the ending circle we’re looking down on the side of the double cone (for simplicity, we will just talk about being inside or outside the ending circle, but in principle any circle on the gradient will do):
In principle the above is no problem. However, if the zero-radius point is very close to the ending circle, the gradient can quickly switch from one behaviour to the next. In other words, it is unstable:
In the left-most image we see that the zero-radius point (at the intersection of the two lines) is inside the ending circle, with negative values of t giving circles inside the starting circle and t>1 giving circles outside the ending circle. In the middle image, where the zero-radius point is on the ending circle, positive radii map to the left half of the image and negative radii map to the right half. In the third image the zero-radius point is no longer contained in the ending circle, and the plane will no longer be completely filled (you will see the outlines of the double cone).
If we imagine moving the zero-radius point from within the ending circle to outside the ending circle, we see the following:
• First the plane is mostly filled by circles with large values of t, all with positive radii.
• Just before the zero-radius point touches the ending circle the right half of the plane is filled with circles that are arbitrarily closely packed (so it makes sense to render the average colour of all circles with t>1)
• As soon as the zero-radius point is on or outside the ending circle we get into a regime where positive radii form a cone on the left half of the plane, while negative radii form a cone on the right half (initially filling both half-planes entirely).
Examples of the “inside” and “outside” regimes (spread method is repeat, the unextended gradients are superimposed over the extended gradients):
SVG appears to be the only specification dealing with this problem, and it does so by specifying that the zero-radius point must be contained within the ending circle. The advantage of this approach is that it can be specified that if the zero-radius point just touches the ending circle (or is even outside it), it should simply behave as if it was very slightly within the ending circle. On the other hand, having the zero-radius point outside the ending circle is essentially just as easy: negative radii map to a cone in one half-plane, positive radii to a cone in the other.
Several possible scenario’s for allowing the zero-radius point to not be contained within the ending circle come to mind:
• Let the author specify what behaviour should be assumed (inside or outside).
• Let renderers interpolate between the two behaviours. Conceptually the renderer would draw the same gradient many times, with slightly varying radii/positions, and take a weighted average of the results.
• Allow specification of the starting and/or ending radius (in principle the problem is “symmetric”) as a percentage/fraction of the distance between the starting center and ending circle. 100% would then correspond to the starting circle touching the ending circle exactly. Essentially this puts the responsibility
With one of the above schemes, or some other creative solution, SVG 2 and other future specifications should be able to provide a maximum amount of flexibility, while avoiding the pitfalls of allowing the zero-radius point to be on the starting/ending circle. It should be emphasized that it is the zero-radius point (and not the starting circle) that should be contained in the starting and/or ending circle (if it is contained in one, it is also contained in the other), and that the problem is not with the zero-radius point being outside the circles, but rather with it being on (or very close to) them.
Finally, it should be noted that there are some further opportunities for generalizing radial gradients. For example, CSS 4 might have conic(al) gradients, which essentially vary colour over angle rather than t. It would even be possible to create hybrid “spiral” gradients. Also, there is no particular reason to restrict yourself to circles, the mathematics are virtually just as easy for ellipses (especially with axis-aligned radii).
Posted in Vector graphics | Tagged | 1 Comment
|
2014-03-11 15:27:21
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6739783883094788, "perplexity": 573.3979247674826}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394011217448/warc/CC-MAIN-20140305092017-00090-ip-10-183-142-35.ec2.internal.warc.gz"}
|
https://en.wikipedia.org/wiki/Local_class_field_theory
|
# Local class field theory
In mathematics, local class field theory, introduced by Helmut Hasse,[1] is the study of abelian extensions of local fields; here, "local field" means a field which is complete with respect to an absolute value or a discrete valuation with a finite residue field: hence every local field is isomorphic (as a topological field) to the real numbers R, the complex numbers C, a finite extension of the p-adic numbers Qp (where p is any prime number), or a finite extension of the field of formal Laurent series Fq((T)) over a finite field Fq.[2]
It is the analogue for local fields of global class field theory.
## Connection to Galois groups
Local class field theory gives a description of the Galois group G of the maximal abelian extension of a local field K via the reciprocity map which acts from the multiplicative group K×=K\{0}. For a finite abelian extension L of K the reciprocity map induces an isomorphism of the quotient group K×/N(L×) of K× by the norm group N(L×) of the extension L× to the Galois group Gal(L/K) of the extension.[3]
The absolute Galois group G of K is compact and the group K× is not compact. Taking the case where K is a finite extension of the p-adic numbers Qp or formal power series over a finite field, the group K× is the product of a compact group with an infinite cyclic group Z. The main topological operation is to replace K× by its profinite completion, which is roughly the same as replacing the factor Z by its profinite completion Z^. The profinite completion of K× is the group isomorphic with G via the local reciprocity map.
The actual isomorphism used and the existence theorem is described in the theory of the norm residue symbol. There are several different approaches to the theory, using central division algebras or Tate cohomology or an explicit description of the reciprocity map. There are also two different normalizations of the reciprocity map: in the case of an unramified extension, one of them asks that the (arithmetic) Frobenius element corresponds to the elements of "K" of valuation 1; the other one is the opposite.
## Lubin–Tate theory
Main article: Lubin–Tate theory
Lubin–Tate theory is important in explicit local class field theory. The unramified part of any abelian extension is easily constructed, Lubin–Tate finds its value in producing the ramified part. This works by defining a family of modules (indexed by the natural numbers) over the ring of integers consisting of what can be considered as roots of the power series repeatedly composed with itself. The compositum of all fields formed by adjoining such modules to the original field gives the ramified part.
A Lubin–Tate extension of a local field K is an abelian extension of K obtained by considering the p-division points of a Lubin–Tate group. If g is an Eisenstein polynomial, f(t) = t g(t) and F the Lubin–Tate formal group, let θn denote a root of gfn-1(t)=g(f(f(⋯(f(t))⋯))). Then Kn) is an abelian extension of K with Galois group isomorphic to U/1+pn where U is the unit group of the ring of integers of K and p is the maximal ideal.[4]
## Higher local class field theory
For a higher-dimensional local field $K$ there is a higher local reciprocity map which describes abelian extensions of the field in terms of open subgroups of finite index in the Milnor K-group of the field. Namely, if $K$ is an $n$-dimensional local field then one uses $\mathrm{K}^{\mathrm{M}}_n(K)$ or its separated quotient endowed with a suitable topology. When $n=1$ the theory becomes the usual local class field theory. Unlike the classical case, Milnor K-groups do not satisfy Galois module descent if $n>1$. Higher-dimensional class field theory was pioneered by A.N. Parshin in positive characteristic and K. Kato, I. Fesenko, Sh. Saito in the general case.
|
2015-09-03 16:48:14
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8934774994850159, "perplexity": 325.7114700642545}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645315643.73/warc/CC-MAIN-20150827031515-00345-ip-10-171-96-226.ec2.internal.warc.gz"}
|
http://ieeexplore.ieee.org/xpl/articleDetails.jsp?reload=true&tp=&arnumber=6462165&contentType=Books+%26+eBooks
|
This chapter contains sections titled: The Problem, The Nineteenth Century, The Twentieth Century, Conclusion
|
2017-11-24 13:37:47
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8225895762443542, "perplexity": 1735.9996642292517}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934808133.70/warc/CC-MAIN-20171124123222-20171124143222-00215.warc.gz"}
|
http://mathhelpforum.com/discrete-math/204313-wheel-graph-complete-graph.html
|
# Math Help - Wheel graph and complete graph
1. ## Wheel graph and complete graph
Hello, since the wheel graph with 3 vertices on the rim and 1 vertex in the middle is the same as the complete graph on 4 vertices, should I not be able to derive the general wheel graph chromatic polynomial for W3 from the general chromatic polynomial for a Complete graph(4)? .
I would write it up here, but it's pointless because all i am doing is expanding the elements inside of the brackets, but this is clearly not going to reach the general chromatic polynomial for a wheel graph.
Wheel graph Chromatic polynomial with 3 vertices on the rim is x(x-2)^3 +(-1)^3(x-2)
Complete graph on 4 vertices is x(x-1)(x-2)(x-3).
so should is not be the case that x(x-1)(x-2)(x-3)==x((x-2)^3 +(-1)^3(x-2)) ?
Edit: OK, i substituted x with a random number and then computed each polynomial and have gotten the same result from both polynomials with the same value for x, so they are clearly the same. I just need to show it algebraically.
Edit: OK, that was easy. I was clearly over thinking it. If i knew how to delete the thread i would.
Thanks.
|
2014-09-19 02:56:29
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8460593819618225, "perplexity": 320.9244535704932}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657129431.12/warc/CC-MAIN-20140914011209-00145-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"}
|
http://ssconlineexam.com/onlinetest/SSC-CGL-Tier-1/Quantitative-Aptitude/QA-Test-58
|
SSC CGL Tier 1 :: QA Test 58
Home » QA Test 58 » General Questions
# SSC CGL Tier 1 Quantitative Aptitude Questions and Answers Online Test 58
Get free ssc cgl tier 1 free Quantitative Aptitude questions and answers free online test mock tests quiz sample questions ssc online test ssc practice test freeonline Test & Mock Test Quiz 58
Exercise
1 .
If Ais the area of a right angled triangle and bis one of the sides containing the right angle, then what is the length of the altitude on the hypotenuse?
$2Ab \over \sqrt {b^4 + 4 A^2}$ $2A^2 b \over \sqrt {b^4 + 4 A^2}$ $2Ab^2 \over \sqrt {b^4 + 4 A^2}$ $2A^2 b^2 \over \sqrt {b^4 + 4 A^2}$
2 .
A parallelogram and a rectangle stand on the same base and on the same side of the base with the same height. If $I_ 1$ , $I_ 2$ be the perimeters of the parallelogram and the rectangle respectively, then which one of the following is correct?
$I_ 1$ < $I_ 2$ $I_ 1$ = $I_ 2$ $I_ 1$ > $I_ 2$ but $I_ 1$ $\neq$2$I_ 2$ $I_ 1$ = 2$l_ 2$
3 .
The diameter of two circles are 18 cm and 8 cm. The distance between their centres is 13 cm. What is the number of common tangents?
1 2 3 None of the above
4 .
What is the geometric means of the observations
125, 729, 1331?
495 1485 2221 None of the above
5 .
The mean of 100 values is 45. If 15 is added to each of the first forty values and 5 is subtracted from each of the remaining sixty values, the new mean becomes
45 48 51 55
6 .
A cylindrical tube open at both ends is made of metal. The internal diameter of the tube is 6 cm and length of the tube is 10 cm. If the thickness of the metal used is 1 cm, then the outer curved surface area of the tube is
140$\pi$sq cm 146.5 $\pi$sq cm 70 $\pi$sq cm None of the above
7 .
For a plot of land of 100 m $\times$ 80 m, the length to be raised by spreading the earth from stack of a rectangular base 10 m $\times$ 8 m and vertical section being a trapezium of height 2 m. The top of the stack is 8 m $\times$ 5 m. How many centimeters can the level raised?
3 cm 2.5 m 2 cm 1.5 cm
8 .
If $3^ x$ + 27($3^ {-x}$ ) = 12, then what is the value of x?
Only 1 Only 2 1 or 2 0 or 1
For which value of kdoes the pair of equations$x^ 2$ - $y^ 2$ = 0 and $(x- k)^ 2$ + $y^ 2$ = 1 yield a unique positive solution of x?
2 0 $\sqrt 2$ - $\sqrt 2$
The expression $sin^ 2$ x +$cos^ 2$ x - 1 = 0 is satisfied by how many values of x?
|
2017-08-21 17:46:36
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7598904967308044, "perplexity": 548.3809868895697}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886109470.15/warc/CC-MAIN-20170821172333-20170821192333-00204.warc.gz"}
|
https://pythagitup.com/page/5/
|
# What Do My Students Really Know? Part 2
During the second semester, I only assigned homework using Delta Math. This worked well for several reasons, with the most important one being that it gave me better insight into what my students actually understood than pencil-and-paper homework ever did.
The Delta Math homework I assigned surprised me right off the bat. The image below displays one student’s results on the first assignment of the second semester. Many students had similar results.
This assignment consisted entirely of review material. We had assessed linear functions at least once in December, and for the most part, students did fine. I knew that students would be a bit rusty after Winter Break. That’s why we completed this assignment. The surprise came from the misconceptions I saw in student responses.
This slideshow requires JavaScript.
Although I could only see their answers, I found it relatively easy to determine their misconceptions. Unsurprisingly, I saw all of the classic slope mistakes.
• Writing slope as an ordered pair instead of a ratio
• Subtraction errors, especially with negative numbers
• Putting the change in $x$ in the numerator instead of the denominator
• Mismatching $x$ and $y$
• Forgetting to simplify (not exactly a mistake, but Delta Math marks this wrong)
• Misinterpreting $\frac{4}{0}$ as either $4$ or as $0$
These aren’t new mistakes. I’ve seen them many times, and we’ve corrected them together many times. We said “vertical divided by horizontal” at least ten times each class period every day for a month. And the students got this! Like I said, students did fine when we assessed linear functions in December.
So why did so many students continue to display these misconceptions? I’m not naive. I realize that students master the material at different rates. But how could so many of them who had mastered the material make these mistakes? Why did they return to misconceptions that they had overcome a month earlier?
One explanation is that students managed to know the material well enough to pass an assessment, but they did not develop the robust understanding necessary to maintain their skills even a month later. That’s the explanation I originally subscribed to, but I think it lacks something important. Many of my students did understand slope. Throughout the fall, they worked hard to connect proportional relationships, steepness, and lines. They built a solid conceptual foundation. They thought mathematically. They solved problems. I think what they lacked – and this is on me – was a proper emphasis on procedural fluency. Sure, we had plenty of opportunities to practice and to develop that fluency, but rarely did they have the “Uh oh, I’m wrong!” moments that Delta Math gave them. My Delta Math assignments required students to get a certain number of problems correct to get credit. You just can’t fake a right answer. That’s what I like about Delta Math – it holds them accountable.
The other major advantage of Delta Math over pencil-and-paper homework? I can look through as many assignments as I want to in a relatively short time period without having to carry papers around. I have a bit of a problem with keeping papers organized, so moving homework online and avoiding paper altogether saves me some serious time and energy. The obvious drawback is that I cannot see the work that the students did. In the examples above, it’s relatively straightforward to identify misconceptions without seeing the student’s work, but that certainly won’t always be the case. I think, though, that simply being able to see that a student struggled with a problem type may be enough, especially given that some sort of intervention would need to take place anyway.
I’m not sure yet what homework will look like in my classes this year, but it seems like I’ll want to place more emphasis on procedural fluency. Perhaps such an emphasis earlier in the year will lead to better understanding all year long. I hope 2018-19 is the year I finally figure out how to make homework work for me and for my students!
# Sandra Cisneros – “One Holy Night”
Last summer, I read two books with a student. We read The Awakening by Kate Chopin and Mrs. Dalloway by Virginia Woolf. Our conversations about The Awakening proved fascinating, but we didn’t have as much success reading and discussing Mrs. Dalloway. Nevertheless, I found the experience meaningful, and I think the student did too. She sent me the following Remind message before the school year ended:
Hey Carlson, are we reading this summer?
How could I say no? We haven’t yet started, but we plan to read The Piano Lesson by August Wilson. Additionally, I asked one of my students from this past year if she would like to read a book with me, and she agreed. We’ve been reading Woman Hollering Creek and Other Stories by Sandra Cisneros. The story “One Holy Night” led to an interesting discussion. Here are the highlights and my rambling commentary. To maintain her privacy, I will refer to my student simply as C.
In “One Holy Night,” an eighth-grade girl tells of how she loses her virginity to a man who calls himself Chaq Uxmal Paloquín. He claims to be descended from Mayan kings. The people in the neighborhood call him Boy Baby, and no one seems to know much about him. We find out later – after he has left town and our narrator has become pregnant – that he is actually thirty-seven years old and that he is a serial killer. The story is only nine pages long, and it focuses more on mood and feeling than on specific details.
For whatever reason, our conversation fixated on rape. C rightly pointed out that Boy Baby committed statutory rape. She expressed some discomfort with the story and mentioned that some readers might be deeply disturbed (“triggered”) by the story. We talked about the choice Cisneros made to avoid detail and use poetic language to describe the rape:
Then something inside bit me, and I gave out a cry as if the other, the one I wouldn’t be anymore, leapt out.
C was glad that there wasn’t more detail. Because rape is so disturbing, she asserted, only a bad writer would need to include more detail to bring about the desired response from the reader. This assertion seemed to extend past “One Holy Night” to all literature (and other works of art, for the matter), so I asked if there could ever be a situation in which more detail might be necessary. She said no, again emphasizing that a good writer wouldn’t need to include the details of the rape. I told C that I didn’t disagree but that I wanted to press the point further. Could there be any value in graphic description of such a violent act? And what about other violent acts like murder? C felt that rape belonged to a category of its own, that it was even worse than murder. I wondered whether a rape victim might consider it necessary to express the horror and the violence she went through. C agreed this might be possible, but she expressed her concerns about works of art that use such violence for shock or entertainment value. She talked about the show 13 Reasons Why (which I have not seen) and how she felt like it glamorized suicide. We talked about how the narrator seemed to romanticize her own rape, describing Boy Baby’s face as “the face I am in love with” even after discovering he’s a serial killer. It was a meandering conversation, but it was a meaningful one.
When I chose Woman Hollering Creek and Other Stories, I didn’t know much about its content. I didn’t intend to pick a book with a story about a young girl’s rape, so I’m glad that C and I managed to have such an interesting discussion. I’m forced to wonder where this sort of conversation takes place. Or if it even does. It’s not easy to talk about rape, but it seems important that we do. Sometimes we don’t give our students enough credit for the depth of their insights. I look forward to learning more with C as the summer continues.
# Black History Month
I took a risk this February. My school celebrates Black History Month in a number of ways, but I always feel like I need to do more. Here’s what I tried.
I’m not sure what made me think of this, but I decided that we would read a poem by an African-American poet each morning in homeroom. I figured this could be a simple yet powerful way to celebrate African-American culture, and honestly, I just thought it would be interesting. You can find the poems I used here.
I wasn’t really sure what to expect the first day. Would the students really listen? Would anyone want to participate? Would the experience be meaningful to anyone but me?
I explained to my homeroom the plan for the month. I told them that anyone could volunteer to read a poem or even bring in a poem of their choice. As I prepared to read the first poem, I paused and thought “Why not ask for a volunteer now?” I expected dead silence and blank stares. Instead, an energetic, excitable young man – who happens to be African-American – said he wanted to read the poem. Overjoyed. I was absolutely overjoyed.
As the month continued, I kept bringing in poems, and my students kept volunteering to read. It might have only been 7 or 8 students, but when I started, I had no expectations whatsoever. And while my students sometimes struggled to read the poems, they truly committed themselves to their delivery. And the rest of the class? Quiet, respectful, attentive. Did they find the poems interesting or meaningful or enjoyable? I can’t say, but I do know that they respected my idea and made it a reality.
Students read nearly all of the poems. I had to read 1 or 2 because of time constraints, and I asked a guidance counselor to read one. Her reading of Audre Lorde’s “Hanging Fire” truly moved me. I had hoped that having a “guest reader” would be special, but I was totally blown away. I think the kids were too.
For the last day of February, I decided to talk briefly about the idea of Black History Month and close with a short selection from a poem that means something to me. I thanked my students for committing to the poem readings all month and told them that I would really miss not having a poem to read every day. Then, I attempted to tell them how I’d like us all to carry the message of Black History Month forward. That we need to spend all year trying to make our school, our community, and our country more tolerant and more just. I think I stumbled over my words a bit here. I was emotional, especially knowing what would come next. I closed with the last few lines of Amiri Baraka’s “Three Modes of History and Culture.”
I think about a time when I will be relaxed.
When flames and non-specific passion wear themselves
away. And my eyes and hands and mind can turn
and soften, and my songs will be softer
and lightly weight the air.
I’m not a poet. I’m not an English teacher. I’m not a literary scholar. Maybe this poem or any of the others mean something totally different than I think. I don’t think it matters, though. What matters is that we pushed ourselves to do something different, that we worked outside of our comfort zone, that we really tried to learn and understand.
But it’s not enough. I need to do more next year. I need to do more for my students. To let them know that their history and their culture matter. To let them know that they matter. To help us all learn to be better, more tolerant, more understanding, more generous in spirit.
This was a risk. I don’t know if I did a good thing. I don’t know if I made a mistake. I badly want feedback, but I’m also terrified that I sent a message I didn’t intend to send. It’s uncomfortable sometimes – teaching – but it’s worth it for those moments. Those powerful moments when twenty-five thirteen- and fourteen-year-old students devote their attention to listening to a classmate read a poem. I hope that I made a difference.
# Reflections on The Classroom Chef
Take risks. More than anything else, that’s the message of The Classroom Chef by John Stevens and Matt Vaudrey. It’s 2018. We owe it to our students to do better than just teach the same lesson we’ve been using. We owe them better than life in a textbook universe. We owe them more than just lectures and practice. We owe our students a meaningful classroom experience that will help them develop understanding and not just procedural fluency. We owe them more.
Take risks. I told this to my students when we returned from winter break. I told them that I planned to take risks this semester to make class more interesting, to help them find meaning in what we’re doing, and to allow them to learn and refine academic and life skills.
Take risks. Grades don’t matter. Test scores don’t matter. Coverage doesn’t matter. Standard algorithms don’t matter. Compliance doesn’t matter. The book doesn’t matter. The pacing guide doesn’t matter. Standards don’t matter.
Take risks. What does matter? The students in front of you right now. Their thoughts and ideas and energy and interests and passions and enthusiasm and suggestions and questions and feelings and understandings and beliefs and knowledge and motivation and … What matters? Their future. What matters? They do.
Take risks. We owe our students the best education we can possibly give them. John and Matt understand this. They understand that we need to change if we really want our students to grow. They understand that it’s not enough to do the same old thing – even if it has been effective in the past. They understand that we need to keep pushing forward lest we end up going backward. They understand that education in 2018 can’t look like education in 1950 or 1990 or 2005 or even in 2017. They understand that students need us to value their engagement, their thinking, and their future.
Take risks. The Classroom Chef offers a ton of great ideas and useful advice. But beneath all of the stories and suggestions lies one simple message – take risks.
|
2021-04-18 11:37:33
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2799115478992462, "perplexity": 1426.4057231990826}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038476606.60/warc/CC-MAIN-20210418103545-20210418133545-00535.warc.gz"}
|
https://experts.mcmaster.ca/display/publication1409979
|
# Beurling densities and frames of exponentials on the union of small balls Academic Article
•
• Overview
•
• Research
•
• View All
•
### abstract
• If $x_1,\dots,x_m$ are finitely many points in $\mathbb{R}^d$, let $E_\epsilon=\cup_{i=1}^m\,x_i+Q_\epsilon$, where $Q_\epsilon=\{x\in \mathbb{R}^d,\,\,|x_i|\le \epsilon/2, \, i=1,...,d\}$ and let $\hat f$ denote the Fourier transform of $f$. Given a positive Borel measure $\mu$ on $\mathbb{R}^d$, we provide a necessary and sufficient condition for the frame inequalities $$A\,\|f\|^2_2\le \int_{\mathbb{R}^d}\,|\hat f(\xi)|^2\,d\mu(\xi)\le B\,\|f\|^2_2,\quad f\in L^2(E_\epsilon),$$ to hold for some $A,B>0$ and for some $\epsilon>0$ sufficiently small. If $m=1$, we show that the limits of the optimal lower and upper frame bounds as $\epsilon\rightarrow 0$ are equal, respectively, to the lower and upper Beurling density of $\mu$. When $m>1$, we extend this result by defining a matrix version of Beurling density. Given a (possibly dense) subgroup $G$ of $\mathbb{R}$, we then consider the problem of characterizing those measures $\mu$ for which the inequalities above hold whenever $x_1,\dots,x_m$ are finitely many points in $G$ (with $\epsilon$ depending on those points, but not $A$ or $B$). We point out an interesting connection between this problem and the notion of well-distributed sequence when $G=a\,\mathbb{Z}$ for some $a>0$. Finally, we show the existence of a discrete set $\Lambda$ such that the measure $\mu=\sum_{\lambda}\,\delta_\lambda$ satisfy the property above for the whole group $\mathbb{R}$.
|
2019-04-26 00:30:57
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9768068194389343, "perplexity": 82.73358710085444}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578743307.87/warc/CC-MAIN-20190425233736-20190426015736-00176.warc.gz"}
|
https://physicscomputingblog.com/category/computing/page/2/
|
## Numerical solution of PDE:s, Part 6: Adiabatic approximation for quantum dynamics
Having solved the time-dependent Schrödinger equation both in real and imaginary time, we can move forward to investigate systems where the potential energy function $V$ has an explicit time dependence in it:
In this kind of systems, the expectation value of the Hamiltonian operator doesn’t have to stay constant.
Time-dependent perturbation theory is one method for finding approximate solutions for this kind of problems, but here I will handle a simpler example, which is called adiabatic approximation.
Suppose that the potential energy function $V(x,t)$ is known. Now, let’s say that we also know the solutions of the time-independent Schrödinger equation
for any value of $t$. I denote the solutions as $\psi_n (x;t)$, where it is understood that $x$ is a variable and $t$ is a parameter. Now, if the function $V(x,t)$ changes very slowly as a function of time, i.e. its partial derivative with respect to $t$ is small at all points of the domain, we can use the adiabatic approximation, which says that if the initial state $\Psi (x,0)$ is the ground state for the potential $V(x,0)$, then the state at time $t$ is the ground state for the potential $V(x,t)$.
So, we can change a ground state of one potential $V_1 (x)$ into the ground state of another potential $V_2 (x)&bg=ffffff&fg=000000$ by making a continuous change from $V_1 (x)$ to $V_2 (x)$ slowly enough.
Let’s test this by chooosing a function $V$ as
i.e. a Hookean potential that moves to the positive x-direction with constant speed. If we set the wavefunction at $t=0$ to
which is the ground state corresponding to $V(x,0)$, the time depelopment of the wavepacket should be like
which means that is moves with the same constant speed as the bottom of the potential $V$. This can be calculated with the R-Code below:
library(graphics) #load the graphics library needed for plotting
lx <- 6.0 #length of the computational domain
lt <- 15.0 #length of the simulation time interval
nx <- 100 #number of discrete lattice points
nt <- 300 #number of timesteps
dx <- lx/nx #length of one discrete lattice cell
dt <- lt/nt #length of timestep
V = c(1:nx) #potential energies at discrete points
for(j in c(1:nx)) {
V[j] = as.complex(2*(j*dx-3)*(j*dx-3)) #harmonic potential
}
kappa1 = (1i)*dt/(2*dx*dx) #an element needed for the matrices
kappa2 <- c(1:nx) #another element
for(j in c(1:nx)) {
kappa2[j] <- as.complex(kappa1*2*dx*dx*V[j])
}
psi = as.complex(c(1:nx)) #array for the wave function values
for(j in c(1:nx)) {
psi[j] = as.complex(exp(-(j*dx-3)*(j*dx-3))) #Gaussian initial wavefunction
}
xaxis <- c(1:nx)*dx #the x values corresponding to the discrete lattice points
A = matrix(nrow=nx,ncol=nx) #matrix for forward time evolution
B = matrix(nrow=nx,ncol=nx) #matrix for backward time evolution
for(j in c(1:nx)) {
for(k in c(1:nx)) {
A[j,k]=0
B[j,k]=0
if(j==k) {
A[j,k] = 1 + 2*kappa1 + kappa2[j]
B[j,k] = 1 - 2*kappa1 - kappa2[j]
}
if((j==k+1) || (j==k-1)) {
A[j,k] = -kappa1 #off-diagonal elements
B[j,k] = kappa1
}
}
}
for (k in c(1:nt)) { #main time stepping loop
for(j in c(1:nx)) {
V[j] = as.complex(2*(j*dx-3-k*dt*0.05)*(j*dx-3-k*dt*0.05)) #time dependent potential
}
for(j in c(1:nx)) {
kappa2[j] <- as.complex(kappa1*2*dx*dx*V[j])
}
for(l in c(1:nx)) {
for(m in c(1:nx)) {
A[l,m]=0
B[l,m]=0
if(l==m) {
A[l,m] = 1 + 2*kappa1 + kappa2[m]
B[l,m] = 1 - 2*kappa1 - kappa2[m]
}
if((l==m+1) || (l==m-1)) {
A[l,m] = -kappa1
B[l,m] = kappa1
}
}
}
sol <- solve(A,B%*%psi) #solve the system of equations
for (l in c(1:nx)) {
psi[l] <- sol[l]
}
if(k %% 3 == 1) { #make plots of psi(x) on every third timestep
jpeg(file = paste("plot_",k,".jpg",sep=""))
plot(xaxis,abs(psi)^2,xlab="position (x)", ylab="Abs(Psi)^2",ylim=c(0,2))
title(paste("|psi(x,t)|^2 at t =",k*dt))
lines(xaxis, abs(psi)^2)
lines(xaxis, V)
dev.off()
}
}
and in the following sequences of images you see that the approximation is quite good for $v = 0.05$
As an animation, this process looks like shown below:
By doing the same calculation again, but this time with $v = 3$, the image sequence looks like this:
where it is obvious that the approximation doesn’t work anymore.
## Numerical solution of PDE:s, Part 5: Schrödinger equation in imaginary time
In the last post, I mentioned that the solution of the time dependent 1D Schrödinger equation
can be written by expanding the initial state $\Psi (x,0)$ in the basis of the solutions of time-independent Schrödinger equation
and multiplying each term in the expansion by a complex-valued time dependent phase factor:
Now, assuming that the energies $E_n$ are all positive and are ordered so that $E_0$ is the smallest of them, we get an interesting result by replacing the time variable $t$ with an imaginary time variable $s = it$. The function $\Psi (x,s)$ is then
which is a sum of exponentially decaying terms, of which the first one, $c_0 \exp(-E_0 s)\psi_0 (x)$ decays slowest. So, in the limit of large $s$, the wave function is
i.e. after normalizing it, it’s approximately the same as the ground state $\psi_0$. Also, if $s$ is a large number, we have
or
So, here we have a way to use the TDSE to numerically calculate the ground state wavefunction and corresponding energy eigenvalue for any potential energy function $V(x)$. This is very useful, as the ground state can’t usually be solved analytically, except for some very simple potential energy functions such as the harmonic oscillator potential.
To test this imaginary time method, I will approximately calculate the ground state of an anharmonic oscillator, described by a Schrödinger equation
as an initial state, I will choose a Gaussian function
and the computational domain is defined by 0 < x < 6, 0 < t < 3, $\Delta x = 0.05$, $\Delta t = 0.01$.
An R-Code that performs this calculation is shown below:
library(graphics) #load the graphics library needed for plotting
lx <- 6.0 #length of the computational domain
lt <- 3.0 #length of the simulation time interval
nx <- 120 #number of discrete lattice points
nt <- 300 #number of timesteps
dx <- lx/nx #length of one discrete lattice cell
dt <- (-1i)*lt/nt #length of imaginary timestep
V = c(1:nx) #potential energies at discrete points
for(j in c(1:nx)) {
V[j] = as.complex(0.25*(j*dx-3)*(j*dx-3)+0.15*(j*dx-3)*(j*dx-3)*(j*dx-3)*(j*dx-3)) #anharmonic potential
}
kappa1 = (1i)*dt/(2*dx*dx) #an element needed for the matrices
kappa2 <- c(1:nx) #another element
for(j in c(1:nx)) {
kappa2[j] <- as.complex(kappa1*2*dx*dx*V[j])
}
psi = as.complex(c(1:nx)) #array for the wave function values
for(j in c(1:nx)) {
psi[j] = as.complex(exp(-2*(j*dx-3)*(j*dx-3))) #Gasussian initial wavefunction
}
xaxis <- c(1:nx)*dx #the x values corresponding to the discrete lattice points
A = matrix(nrow=nx,ncol=nx) #matrix for forward time evolution
B = matrix(nrow=nx,ncol=nx) #matrix for backward time evolution
for(j in c(1:nx)) {
for(k in c(1:nx)) {
A[j,k]=0
B[j,k]=0
if(j==k) {
A[j,k] = 1 + 2*kappa1 + kappa2[j]
B[j,k] = 1 - 2*kappa1 - kappa2[j]
}
if((j==k+1) || (j==k-1)) {
A[j,k] = -kappa1
B[j,k] = kappa1
}
}
}
for (k in c(1:nt)) { #main time stepping loop
sol <- solve(A,B%*%psi) #solve the system of equations
for (l in c(1:nx)) {
psi[l] <- sol[l]
}
nrm = 0
for (l in c(1:nx)) nrm <- nrm + dx*abs(psi[l])*abs(psi[l])
if(k %% 3 == 1) { #make plots of psi(x) on every third timestep
jpeg(file = paste("plot_",k,".jpg",sep=""))
plot(xaxis, abs(psi)^2,xlab="position (x)", ylab="Abs(Psi)^2")
title(paste("|psi(x,t)|^2 at t =",k*dt))
lines(xaxis, abs(psi)^2)
lines(xaxis, V*max(abs(psi)^2))
dev.off()
}
}
And a set of plots of $|\Psi (x,t)|^2$ for several values of $t$ look like this:
Here the shape of the anharmonic potential has been plotted to the same images.
The problem with this method for finding a ground state is that if the system has more degrees of freedom than a single coordinate x, the matrices in the linear systems needed in the time stepping quickly become very large and the calculation becomes prohibitively slow. To make this worse, the Crank-Nicolson method of time stepping can’t be parallelized for multiple processors. However, there is another way to compute the evolution of a wave function in imaginary time, which is called Diffusion Monte Carlo, and that is easily parallelizable. DMC is one practical way for calculating ground state wave functions for multi-particle systems such as a helium or a lithium atom.
## Numerical solution of PDE:s, Part 4: Schrödinger equation
In the earlier posts, I showed how to numerically solve a 1D or 2D diffusion or heat conduction problem using either explicit or implicit finite differencing. In the 1D example, the relevant equation for diffusion was
and an important property of the solution was the conservation of mass,
i.e. the integral of the concentration field over whole domain stays constant.
Next, I will show how to integrate the 1D time-dependent Schrödinger equation, which in a nondimensional form where we set $\hbar = 1$ and $m = 1$ reads:
Here $i$ is the imaginary unit and $V(x)$ is the potential energy as a function of $x$. The solutions of this equation must obey a conservation law similar to the mass conservation in the diffusion equation, the conservation of norm:
where the quantity $|\Psi (x,t)|$ is the modulus of the complex-valued function $\Psi (x,t)$ . This property of the solution is also called unitarity of the time evolution.
Apart from the TDSE, another way to represent the time development of this system is to find the normalized solutions $\psi_0 (x)$, $\psi_1 (x)$, $\psi_2 (x) \dots$ of the time-independent Schrödinger equation
and write the initial state $\Psi (x,0)$ as a linear combination of those basis functions:
This is possible because the solutions of the time-independent equation form a basis for the set of acceptable wave functions $\psi (x)$. Then, every term in that eigenfunction expansion is multiplied by a time dependent phase factor $\exp(-iE_n t)$:
The numbers $E_n$ are the eigenvalues corresponding to the solutions $\psi_n (x)$ and the function $\psi_0 (x)$ is called the ground state corresponding to potential $V(x)$, while the functions $\psi_1 (x)$ is the first excited state and so on.
The Schrödinger equation can’t be discretized by using either the explicit or implicit method that we used when solving the diffusion equation. The method is either numerically unstable or doesn’t conserve the normalization of the wave function (or both) if you try to do that. The correct way to discretize the Schrödinger equation is to replace the wave function with a discrete equivalent
and the potential energy function $V(x)$ with $V_{i;j}$ (or $V_i$ in the case of time-independent potential), and write an equation that basically tells that propagating the state $\Psi_{i;j}$ forward by half a time step gives the same result as propagating the state $\Psi_{i;j+1}$ backwards by half a time step:
Here we have
and
This kind of discretization is called the Crank-Nicolson method. As boundary conditions, we usually set that at the boundaries of the computational domain the wavefunction stays at value zero: $\Psi (0,t) = \Psi (L,t) = 0$ for any value of $t$. In the diffusion problem, this kind of a BC corresponded to infinite sinks at the boundaries, that annihilated anything that diffused through them. In the Schrödinger equation problem, which is a complex diffusion equation, the equivalent condition makes the boundaries impenetrable walls that deflect elastically anything that collides with them.
An R-Code that calculates the time evolution of a Gaussian initial wavefunction
in an area of zero potential:
for a domain 0 < x < 6, a lattice spacing $\Delta x = 0.05$, time interval 0 < t < 2 and time step $\Delta t = 0.01$, is given below:
library(graphics) #load the graphics library needed for plotting
lx <- 6.0 #length of the computational domain
lt <- 2.0 #length of the simulation time interval
nx <- 120 #number of discrete lattice points
nt <- 200 #number of timesteps
dx <- lx/nx #length of one discrete lattice cell
dt <- lt/nt #length of timestep
V = c(1:nx) #potential energies at discrete points
for(j in c(1:nx)) {
V[j] = 0 #zero potential
}
kappa1 = (1i)*dt/(2*dx*dx) #an element needed for the matrices
kappa2 <- c(1:nx) #another element
for(j in c(1:nx)) {
kappa2[j] <- as.complex(kappa1*2*dx*dx*V[j])
}
psi = as.complex(c(1:nx)) #array for the wave function values
for(j in c(1:nx)) {
psi[j] = as.complex(exp(-2*(j*dx-3)*(j*dx-3))) #Gaussian initial wavefunction
}
xaxis <- c(1:nx)*dx #the x values corresponding to the discrete lattice points
A = matrix(nrow=nx,ncol=nx) #matrix for forward time evolution
B = matrix(nrow=nx,ncol=nx) #matrix for backward time evolution
for(j in c(1:nx)) {
for(k in c(1:nx)) {
A[j,k]=0
B[j,k]=0
if(j==k) {
A[j,k] = 1 + 2*kappa1 + kappa2[j]
B[j,k] = 1 - 2*kappa1 - kappa2[j]
}
if((j==k+1) || (j==k-1)) {
A[j,k] = -kappa1
B[j,k] = kappa1
}
}
}
for (k in c(1:nt)) { #main time stepping loop
sol <- solve(A,B%*%psi) #solve the system of equations
for (l in c(1:nx)) {
psi[l] <- sol[l]
}
if(k %% 3 == 1) { #make plots of |psi(x)|^2 on every third timestep
jpeg(file = paste("plot_",k,".jpg",sep=""))
plot(xaxis,abs(psi)^2,xlab="position (x)", ylab="Abs(Psi)^2",ylim=c(0,2))
title(paste("|psi(x,t)|^2 at t =",k*dt))
lines(xaxis,abs(psi)^2)
dev.off()
}
}
The output files are plots of the absolute squares of the wavefunction, and a few of them are shown below.
In the next simulation, I set the domain and discrete step sizes the same as above, but the initial state is:
Which is a Gaussian wave packet that has a nonzero momentum in the positive x-direction. This is done by changing the line
for(j in c(1:nx)) {
psi[j] = as.complex(exp(-2*(j*dx-3)*(j*dx-3))) #Gaussian initial wavefunction
}+(1i)*j*dx
into
for(j in c(1:nx)) {
psi[j] = as.complex(exp(-2*(j*dx-3)*(j*dx-3)+(1i)*j*dx)) #Gaussian initial wavefunction
}
The plots of $|\Psi (x,t)|^2$ for several values of $t$ are shown below
and there you can see how the wave packet collides with the right boundary of the domain and bounces back.
In the last simulation, I will set the potential function to be
which is a harmonic oscillator potential, and with the nondimensional mass $m =1$ and Planck constant $\hbar = 1$ the ground state $\psi _0 (x)$ of this system is
If I’d set the initial state to be $\Psi (x,0) = \psi_0 (x)$, or any other solution of the time-independent SE, the modulus of the wavefunction would not change at all. To get something interesting to happen, I instead set an initial state that is a displaced version of the ground state:
The solution can be obtained with the code shown below:
library(graphics) #load the graphics library needed for plotting
lx <- 6.0 #length of the computational domain
lt <- 3.0 #length of the simulation time interval
nx <- 360 #number of discrete lattice points
nt <- 300 #number of timesteps
dx <- lx/nx #length of one discrete lattice cell
dt <- lt/nt #length of timestep
V = c(1:nx) #potential energies at discrete points
for(j in c(1:nx)) {
V[j] = as.complex(2*(j*dx-3)*(j*dx-3)) #Harmonic oscillator potential with k=4
}
kappa1 = (1i)*dt/(2*dx*dx) #an element needed for the matrices
kappa2 <- c(1:nx) #another element
for(j in c(1:nx)) {
kappa2[j] <- as.complex(kappa1*2*dx*dx*V[j])
}
psi = as.complex(c(1:nx)) #array for the wave function values
for(j in c(1:nx)) {
psi[j] = as.complex(exp(-(j*dx-2)*(j*dx-2))) #Gaussian initial wavefunction, displaced from equilibrium
}
xaxis <- c(1:nx)*dx #the x values corresponding to the discrete lattice points
A = matrix(nrow=nx,ncol=nx) #matrix for forward time evolution
B = matrix(nrow=nx,ncol=nx) #matrix for backward time evolution
for(j in c(1:nx)) {
for(k in c(1:nx)) {
A[j,k]=0
B[j,k]=0
if(j==k) {
A[j,k] = 1 + 2*kappa1 + kappa2[j]
B[j,k] = 1 - 2*kappa1 - kappa2[j]
}
if((j==k+1) || (j==k-1)) {
A[j,k] = -kappa1
B[j,k] = kappa1
}
}
}
for (k in c(1:nt)) { #main time stepping loop
sol <- solve(A,B%*%psi) #solve the system of equations
for (l in c(1:nx)) {
psi[l] <- sol[l]
}
if(k %% 3 == 1) { #make plots of Abs(psi(x))^2 on every third timestep
jpeg(file = paste("plot_",k,".jpg",sep=""))
plot(xaxis,abs(psi)^2, xlab="position (x)", ylab="Abs(Psi)^2",ylim=c(0,2))
title(paste("|psi(x,t)|^2 at t =",k*dt))
lines(xaxis,abs(psi)^2)
lines(xaxis,V)
dev.off()
}
}
and the solution at different values of $t$ look like this (images and video):
Here the shape of the Hookean potential energy is plotted in the same images. So, here you see how the center of the Gaussian wavefunction oscillates around the point $x = 3$, just like a classical mechanical harmonic oscillator does when set free from a position that is displaced from equilibrium.
By changing the code that produces the output images, we can also get a sequence of plots of the imaginary part of the wavefunction:
if(k %% 3 == 1) { #make plots of Im(psi(x)) on every third timestep
jpeg(file = paste("plot_",k,".jpg",sep=""))
plot(xaxis,Im(psi), xlab="position (x)", ylab="Im(Psi)",ylim=c(-1.5,1.5))
title(paste("Im(psi(x,t)) at t =",k*dt))
lines(xaxis,Im(psi))
lines(xaxis,V)
dev.off()
}
and the resulting plots look like this:
## Numerical solution of PDE:s, Part 3: 2D diffusion problem
In the earlier posts related to PDE numerical integration, I showed how to discretize 1-dimensional diffusion or heat conduction equations either by explicit or implicit methods. The 1d model can work well in some situations where the symmetry of a physical system makes the concentration or temperature field practically depends on only one cartesian coordinate and is independent of the position along two other orthogonal coordinate axes.
The 2-dimensional version of the diffusion/heat equation is
assuming that the diffusion is isotropic, i.e. its rate does not depend on direction.
In a discretized description, the function C(x,y,t) would be replaced by a three-index object
assuming that one of the corners of the domain is at the origin. Practically the same can also be done with two indices, as in
where $N_x$ is the number of discrete points in x-direction.
Using the three-index version of the 2D diffusion equation and doing an explicit discretization, we get this kind of a discrete difference equation:
which can be simplified a bit if we have $\Delta x = \Delta y$ .
Below, I have written a sample R-Code program that calculates the evolution of a Gaussian concentration distribution by diffusion, in a domain where the x-interval is 0 < x < 6 and y-interval is 0 < y < 6. The boundary condition is that nothing diffuses through the boundaries of the domain, i.e. the two-dimensional integral of C(x,y,t) over the area $[0,6]\times [0,6]$ does not depend on t.
library(graphics) #load the graphics library needed for plotting
lx <- 6.0 #length of the computational domain in x-direction
ly <- 6.0 #length of the computational domain in y-direction
lt <- 6 #length of the simulation time interval
nx <- 30 #number of discrete lattice points in x-direction
ny <- 30 #number of discrete lattice points in y-direction
nt <- 600 #number of timesteps
dx <- lx/nx #length of one discrete lattice cell in x-direction
dy <- ly/ny #length of one discrete lattice cell in y-direction
dt <- lt/nt #length of timestep
D <- 1.0 #diffusion constant (assumed isotropic)
Conc2d = matrix(nrow=ny,ncol=nx)
DConc2d <- matrix(nrow=ny, ncol=nx) #a vector for the changes in concentration during a timestep
xaxis <- c(0:(nx-1))*dx #the x values corresponding to the discrete lattice points
yaxis <- c(0:(ny-1))*dy #the y values corresponding to the discrete lattice points
kappax <- D*dt/(dx*dx) #a parameter needed in the discretization
kappay <- D*dt/(dy*dy) #a parameter needed in the discretization
for (i in c(1:ny)) {
for (j in c(1:nx)) {
Conc2d[i,j] = exp(-(i*dy-3)*(i*dy-3)-(j*dx-3)*(j*dx-3)) #2D Gaussian initial concentration distribution
DConc2d[i,j] <- 0 #all initial values in DConc vector zeroed
}
}
for (j in c(1:nt)) { #main time stepping loop
for(k in c(1:nx))
{
Conc2d[1,k] <- Conc2d[k,2] #fluxes through the boundaries of the domain are forced to stay zero
Conc2d[nx,k] <- Conc2d[nx-1,k]
}
for(k in c(1:ny)) {
Conc2d[k,1] <- Conc2d[k,2]
Conc2d[k,nx] <- Conc2d[k,nx-1]
}
for (k in c(2:(ny-1))) {
for (l in c(2:(nx-1))) {
DConc2d[k,l] <- kappax*(Conc2d[k,l-1]-2*Conc2d[k,l]+Conc2d[k,l+1]) + kappay*(Conc2d[k-1,l]-2*Conc2d[k,l]+Conc2d[k+1,l]) #time stepping
}
}
for (k in c(2:(ny-1))) {
for (l in c(2:(nx-1))) {
Conc2d[k,l] <- Conc2d[k,l]+DConc2d[k,l] #add the changes to the vector Conc
}
}
k <- 0
l <- 0
if(j %% 3 == 1) { #make plots of C(x,y) on every third timestep
jpeg(file = paste("plot_",j,".jpg",sep=""))
persp(yaxis,xaxis,Conc2d,zlim=c(0,1))
title(paste("C(x,y) at t =",j*dt))
dev.off()
}
}
To plot the distribution C(x,y) with a color map instead of a 3D surface, you can change the line
persp(yaxis,xaxis,Conc2d,zlim=c(0,1))
to
image(yaxis,xaxis,Conc2d,zlim=c(0,0.3))
The two kinds of graphs are shown below for three different values of t.
Figure 1. Surface plots of the time development of a Gaussian mass or temperature distribution spreading by diffusion.
Figure 2. Time development of a Gaussian mass or temperature distribution spreading by diffusion, plotted with a red-orange-yellow color map.
To obtain an implicit differencing scheme, we need to write the discretized equation as a matrix-vector equation where $C_{i;j;k}$ is obtained from $C_{i;j;k+1}$ by backward time stepping. In the square matrix, there is one row (column) for every lattice point of the discrete coordinate system. Therefore, if there are N points in x-direction and N points in y-direction, then the matrix is an $N^2 \times N^2$ – matrix. From this it’s quite obvious that the computation time increases very quickly when the spatial resolution is increased.
Initially, one may think that the equation corresponding to diffusion in a really coarse $3 \times 3$ grid is the following one:
Where I’ve used the denotation $k_x = \frac{D\Delta t}{(\Delta x)^2}$ and $k_y = \frac{D\Delta t}{(\Delta x)^2}$ The problem with this is that there are unnecessary terms $-k_x$ in here, creating a cyclic boundary condition (mass that is diffusing through the boundary described by line x = L reappears from the boundary on the other side, x = 0. The correct algorithm for assigning the nonzero elements $A_{ij}$ of the matrix A is
1. $A_{ij} = 1 + 2k_x + 2k_y$ , when $i = j$
2. $A_{ij} = -k_x$ , when $j=i-1$ AND $i\neq 1$ (modulo $N_x$)
3. $A_{ij} = -k_x$ , when $j=i+1$ AND $i\neq 0$ (modulo $N_x$)
4. $A_{ij} = -k_y$ , when $j=i+N_x$ OR $j=i-N_x$
when you want to get a boundary condition which ensures that anything that diffuses through the boundaries is lost forever. Then the matrix-vector equation in the case of $3 \times 3$ lattice is
Unlike the linear system in the 1D diffusion time stepping, this is not a tridiagonal problem and consequently is slower to solve. An R-Code that produces a series of images of the diffusion process for a Gaussian concentration distribution in a $15 \times 15$ discrete lattice is given below.
library(graphics) #load the graphics library needed for plotting
lx <- 6.0 #length of the computational domain in x-direction
ly <- 6.0 #length of the computational domain in y-direction
lt <- 6 #length of the simulation time interval
nx <- 15 #number of discrete lattice points in x-direction
ny <- 15 #number of discrete lattice points in y-direction
nt <- 180 #number of timesteps
dx <- lx/nx #length of one discrete lattice cell in x-direction
dy <- ly/ny #length of one discrete lattice cell in y-direction
dt <- lt/nt #length of timestep
D <- 1.0 #diffusion constant
C = c(1:(nx*ny))
Cu = c(1:(nx*ny))
Conc2d = matrix(nrow=ny,ncol=nx)
xaxis <- c(0:(nx-1))*dx #the x values corresponding to the discrete lattice points
yaxis <- c(0:(ny-1))*dy #the y values corresponding to the discrete lattice points
kappax <- D*dt/(dx*dx) #a parameter needed in the discretization
kappay <- D*dt/(dy*dy) #a parameter needed in the discretization
A = matrix(nrow=(nx*ny),ncol=(nx*ny))
for(i in c(1:(nx*ny))) {
for(j in c(1:(nx*ny))) {
A[i,j] <- 0
if(i==j) A[i,j] <- 1+2*kappax+2*kappay
if(j==i+1 && (i%%nx != 0)) A[i,j] <- -kappax
if(j==i-1 && (i%%nx != 1)) A[i,j] <- -kappax
if(j==i+nx) A[i,j] <- -kappay
if(j==i-nx) A[i,j] <- -kappay
}
}
for (i in c(1:ny)) {
for (j in c(1:nx)) {
Conc2d[i,j] <- exp(-2*(i*dy-3)*(i*dy-3)-2*(j*dx-3)*(j*dx-3)) #2D Gaussian initial concentration distribution
}
}
for (j in c(1:nt)) { #main time stepping loop
for(k in c(1:ny)) {
for(l in c(1:nx)) {
C[(k-1)*nx+l] <- Conc2d[k,l]
}
}
Cu <- solve(A,C)
for(k in c(1:ny)) {
for(l in c(1:nx)) {
Conc2d[k,l] <- Cu[(k-1)*nx+l]
}
}
k <- 0
l <- 0
if(j %% 3 == 1) { #make plots of C(x,y) on every third timestep
jpeg(file = paste("plot_",j,".jpg",sep=""))
persp(y=xaxis,x=yaxis,z=Conc2d,zlim=c(0,1))
title(paste("C(x) at t =",j*dt))
dev.off()
}
}
Note that the function C(x,y) approaches a constant zero function as t increases, try modifying the code yourself to make a boundary condition that doesn’t let anything diffuse through the boundaries!
## Numerical solution of PDE:s, Part 2: Implicit method
In the previous blog post, I showed how to solve the diffusion equation
using the explicit method, where the equation is converted to a discrete one
which can be simplified by using the notation
The problem with this approach is that if we want to have a high resolution, i.e. $\Delta x$ is very small, the timestep $\Delta t$ has to be made even much smaller to keep the procedure numerically stable. The stability of this kind of calculations can be investigated with Von Neumann stability analysis, and there you will find out that the instability acts by amplifying the short-wavelength Fourier components of the numerical discretization error.
The specific stability condition for the explicit solution of diffusion equation is
In the better method to solve the diffusion equation, the implicit method, we will not solve the numbers $f_{i;j+1}$ in a straightforward way from the numbers $f_{i;j}$. Instead, we use backward time stepping to write the $f_{i;j}$ as a function of the numbers $f_{i-1;j+1}$, $f_{i;j+1}$ and $f_{i+1;j+1}$, as in the equation below:
which represents a linear system of equations. More specifically, it is a tridiagonal system, and in matrix-vector form it reads
for the case of a relatively small mesh $n_x = 7$. So, now we have to solve this tridiagonal system on each timestep, and this take more computation time than the explicit method but has the advantage of making the calculation stable even when $\Delta t$ is not necessarily made much smaller than $\Delta x$.
A code for solving this kind of a linear system can be found from the book “Numerical Recipes for C” or from the corresponding book written for FORTRAN.
Now, let’s use the implicit method to solve a diffusion problem where the x-domain is
$x \in [0,6]$,
and the step sizes are
$\Delta x = 0.01$ and $\Delta t = 0.05$ .
The initial concentration profile is chosen to be
(don’t be confused by the fact that the function is now called C instead of f) and we use a boundary condition that forces the value of $C(x,t)$ at the left endpoint to be $C(0,t)=1$ and that at the right endpoint to be $C(6,t) = 0$. This means that the left boundary is an infinite source of concentration (or heat in the case of conduction problem) and the right boundary is an infinite “sink”. With physical intuition, it is easy to deduce that this kind of system evolves toward a steady state where the value of $C(x)$ decreases linearly from 1 to 0 on the interval $[0,6]$
A C++ code for doing this calculation is shown below.
// This program calculates the time development of a diffusion or heat conduction
// system with implicit finite differencing. The system described is the development of a concentration or temperature field
// between boundary points where it is constrained to stay at different constant values.
// Teemu Isojärvi, Feb 2017
#include
#include
using namespace std;
#define LX 6. // Length of spatial domain
#define NX 600 // Number of lattice points
#define LT 3. // Length of time interval
#define NT 60 // Number of timesteps
#define D 1. // Diffusion coefficient
int main(void)
{
double dx = (double)LX/(double)NX; // Lattice spacing
double dt = (double)LT/(double)NT; // Length of timestep
double c[NX]; // Concentration values at lattice points
double x; // Auxiliary position variable
double kappa = D*dt/(dx*dx); // Auxiliary variable for representing the linear system
double g[NX];
double b;
double q;
double u[NX]; // Vector for storing the solution on each timestep
for(int m = 0; m<NX; m++)
{
x = (double)m*dx;
c[m]=exp(-3*x*x); // Gaussian initial concentration
}
for(int n = 0; n<NT; n++)
{
c[0]=1;
c[NX-1]=0;
u[0]=(c[0]+kappa)/(2*kappa+1);
q = 2*kappa + 1;
for(int m = 1; m < NX; m++) { // First loop for solving the tridiagonal system g[m]=-kappa/q; q = (2*kappa + 1) + kappa*g[m]; u[m]=(c[m]+kappa*u[m-1])/q; } for(int m=(NX-2); m>=0; m--) u[m] -= g[m+1]*u[m+1]; // Second loop
for(int m=0; m<NX; m++) c[m] = u[m]; // Updating the concentration or temperature field
}
for(int m = 0; m<NX; m++)
{
x = (double)m*dx;
cout << x << " " << c[m] << "\n"; // Output with the results at time t = LT
}
return 0;
}
Running this program three times, with domain lenght parameters $L=0.5$, $L=1.0$ and $L=3.0$ and keeping the time step constant, we get data points that can be plotted in the same coordinate system with a graphing program, like below:
Figure 1. Time evolution of a concentration field C(x,t) in a system where the concentration is forced to stay at constant values at the endpoints of the domain.
The simulation seems to proceed as expected, approaching a linearly decreasing function $C(x)$
An equivalent code for FORTRAN is in the next box:
! Calculates the time development of a concentration distribution C(x,t) with implicit
! finite-differencing of a diffusion equation. The boundary condition is that the left boundary is an infinite source
! of solute/heat and the value of C(x) at x=0 stays at constant value 1. The value of C at the right boundary stays zero.
! Therefore, the function C(x) evolves towards a linearly decreasing function.
! Teemu Isojärvi, Feb 2017
PROGRAM MAIN
real :: DX, DT, LX, LT, D, KAPPA,B,Q ! Real variables for discretization and the diffusion constant
integer :: NT,NX ! Number of time and position steps
REAL :: C(600) ! Concentration field as an array of data points, dimension same as value of NX
REAL :: G(600)
REAL :: U(600)
REAL :: X ! Auxiliary variable
INTEGER :: M ! Integer looping variables
INTEGER :: N
LX = 6. ! Length of x-interval
LT = 3.0 ! Length of t-interval
NX = 600 ! Number of lattice points
NT = 300 ! Number of time steps
DX = LX/NX ! Distance between neighboring lattice points
DT = LT/NT ! Length of time step
D = 1. ! Diffusion/heat conduction coefficient
KAPPA = D*DT/(DX*DX)
do M = 1, NX ! Initial values of concentration
X = M*DX
C(M) = EXP(-3*X*X) ! Gaussian initial concentration distribution
end do
do N = 1, NT ! Time stepping loop
C(1)=1
C(NX)=0
U(1)=(C(1) + 2 * KAPPA)/(2 * KAPPA + 1)
Q = 2 * KAPPA + 1
do M = 2, NX ! First loop for solving the tridiagonal system
G(M) = -KAPPA/Q
Q = (2 * KAPPA + 1) + KAPPA * G(M)
U(M) = (C(M) + KAPPA * U(M-1))/Q
end do
do M = (NX-1), 1
U(M) = U(M) - G(M+1) * U(M+1) ! Second loop
end do
do M = 1, NX-1
C(M) = U(M) ! Updating the concentration or temperature field
end do
end do
do M = 1, NX
X = M*DX
print *,X,C(M) ! Print the x and concentration values at data points
end do
END
To test the implicit method with R-Code, let’s solve a problem where the length of the x-domain is 20, the time interval has length 3 and the initial distribution is
to ensure that we can set the boundary conditions $C(0,t)=C(L,t)=0$ (the Gaussian distribution doesn’t have time to spread all the way to the boundaries in a time interval of 3 units). The code for this calculation is shown below:
library(graphics) #load the graphics library needed for plotting
library(limSolve)
lx <- 20.0 #length of the computational domain
lt <- 3. #length of the simulation time interval
nx <- 4000 #number of discrete lattice points
nt <- 300 #number of timesteps
dx <- lx/nx #length of one discrete lattice cell
dt <- lt/nt #length of timestep
D <- 1.0 #diffusion constant
Conc <- c(1:nx) #define a vector to put the x- and t-dependent concentration values in
xaxis <- c(1:nx) #the x values corresponding to the discrete lattice points
kappa <- D*dt/(dx*dx) #a parameter needed in the discretization
offdiagonal <- rep(-kappa, times = nx-1)
ondiagonal <- rep(1+2*kappa, times = nx)
for (i in c(1:nx)) {
Conc[i] <- exp(-2*(i*dx-10)*(i*dx-10)) #a Gaussian initial concentration field
xaxis[i] <- i*dx #calculate the x coordinates of lattice points
}
for (j in c(1:nt)) { #main time stepping loop
sol <- Solve.tridiag(offdiagonal, ondiagonal, offdiagonal, Conc)
for (k in c(1:nx)) {
Conc[k] <- sol[k]
}
if(j %% 3 == 1) { #make plots of C(x) on every third timestep
jpeg(file = paste("plot_",j,".jpg",sep=""))
plot(xaxis,Conc,xlab="position (x)", ylab="concentration",ylim=c(0,2))
title(paste("C(x) at t =",j*dt))
lines(xaxis,Conc)
dev.off()
}
}
An animation of the results can be viewed in this link. If you test the code yourself, remember to install the limSolve package first, by writing the command
install.packages(“limSolve”)
in the R console. If you don’t want to load the video, here are the plots for 3 different values of t:
When solving PDE:s other than the ordinary diffusion equation, the implicit method is often even more necessary that it was in the examples here. For example, when solving the time-development of a quantum wave packet from the time-dependent Schroedinger equation, the solution function doesn’t stay normalized if one tries to do a simple explicit time stepping. The solution of Schroedinger equations with implicit method will be shown in the next post. The TDSE is a complex diffusion equation of the form
where the equation has been simplified by setting the Planck constant to value $2\pi$ and the particle mass to $m=1$.
The 1D diffusion and Schroedinger equations are simple in the sense that the linear system to be solved on each timestep is tridiagonal, with makes the solution considerably faster than is the case with a general linear system. Systems that are not tridiagonal will appear when one wants to solve equations with more than one space coordinate (i.e. x and y instead of just x), or when the equation contains higher than 2nd order derivatives with respect to the position coordinate (this can be demonstrated with the thin-film equation, which is fourth-order with respect to x).
## Numerical solution of PDE:s, Part 1: Explicit method
In this post and a few ones following it, I will show how to write C++, Fortran or R-Code programs that solve partial differential equations like diffusion equation or Schroedinger equation numerically, either by explicit or implicit method. An explicit method is a very simple direct calculation, while the implicit (Crank-Nicolson) method involves solving a linear system of equations on every discrete timestep.
A simple example of a PDE is the Poisson equation
,
which describes things like the electric potential field produced by a charge distribution which is the function $\rho$ multiplied by a constant, and can be solved if the function $\rho$ and the boundary values of the function $f$ on some closed surface are known.
In the case of these posts, the equations that I will handle are time-evolution equations, where we are solving a function $f(x,t)$, where the variables are 1D position and time, and the boundary conditions are the function $f(x,0)$ (function $f$ at an initial moment $t=0$) and some conditions for $f(0,t)$ and $f(L,t)$, where $x=0$ and $x=L$ are the left and right boundaries of a computational domain.
The equation that describes both diffusion (spreading of a dissolved substance in solution, or spreading of an unevenly distributed component in gas phase) and heat conduction in a 1-dimensional domain is of the form
.
Here $f(x,t)$ is a function that gives solute concentration or temperature at point $x$ at time $t$, and $D$ is a constant that tells how quickly the diffusion or conduction takes place in medium.
To solve this kind of an equation numerically, we have to choose some domains in time and space
,
and define the object $f_{i;j}$, where the $i$ and $j$ are discrete indices and there is an approximate correspondence
.
Here $\Delta x$ is the spatial discretization step and $\Delta t$ is the timestep. The initial condition means that we know the numbers $f_{i;0}$ for all values of $i$. This is equivalent to knowing the function $f(x,0)$.
The most simple way of converting the PDE into a difference equation for the object $f_{i;j}$ is to use straightforward finite differencing for the derivatives in $x$ and $t$:
With these substitutions and some rearrangement, the PDE becomes
So, now we know what kind of numbers to add to the values $f_{i;j}$ for $i = 0,1,2,\dots ,n_x - 1$ at timestep $j$ to get the numbers $f_{i;j+1}$ for $i = 0,1,2,\dots ,n_x - 1$. Note that I now call the total number of discrete x-axis points $n_x$.
Next, as an example I will solve this discretized equation with the domains and step sizes
$L = 6$
$n_x = 30$
$t_{end} = 6$
$n_t = 600$
$D = 1$
and an initial condition
i.e. a Gaussian temperature/concentration profile. I will also set a boundary condition for the position derivative of $f(x,t)$ at the domain boundaries:
which basically means that the endpoints are impenetrable walls that don’t let anything to diffuse/conduct through them. In other words, the system can be described as “closed” or “adiabatic” depending on whether it’s a diffusion or conduction system.
An example C++ code to do this numerical calculation and to print the values $f_{i;j}$ at the final time $j = n_t$ is shown below:
// This program calculates the time development of a diffusion or heat conduction
// system with explicit finite differencing.
// Teemu Isojärvi, Feb 2017
#include <math.h>;
#include <iostream>;
using namespace std;
#define LX 6. // Length of spatial domain
#define NX 30 // Number of lattice points
#define LT 6 // Length of time interval
#define NT 600 // Number of timesteps
#define D 1. // Diffusion coefficient
int main(void)
{
double dx = (double)LX/(double)NX; // Lattice spacing
double dt = (double)LT/(double)NT; // Length of timestep
double c[NX]; // Concentration values at lattice points
double dc[NX]; // Change of concentration during timestep
double x; // Auxiliary position variable
for(int m = 0; m<NX; m++)
{
x = ((double)m+0.5)*dx;
c[m]=exp(-2*(x-3)*(x-3)); // Gaussian initial concentration
}
for(int n = 0; n<NT; n++)
{
c[0]=c[1]; // Derivative zeroed at left boundary point
c[NX-1] = c[NX-2]; // Derivative zeroed at right boundary point
for(int m = 0; m<NX; m++)
{
dc[m] = (D*dt/(dx*dx))*(c[m+1]-2*c[m]+c[m-1]); // Main finite differencing
}
for(int m = 1; m<NX-1; m++) c[m] += dc[m]; // Updating the concentration
}
for(int m = 0; m<NX; m++)
{
x = (double)m*dx;
cout << x << " " << c[m] << "\n"; // Output with the results at time t = LT
}
return 0;
}
The code can easily be edited to use different initial or boundary conditions, or different discretization step sizes or domain intervals. Doing the calculation three times with time interval lenghths L=0.5, L=1, and L=6, and plotting the output data in a single graph with Grace or similar free graphing program, the result is Fig 1.
Figure 1. Diffusive time evolution of a Gaussian temperature/concentration distribution
To make this kind of graphs, you need to run the program from command line and direct the standard output into a file – in Linux this is done with a command like
./diffusion > out.txt
and in Windows command line it’s done as
diffusion.exe > out.txt
The same code is also easy to write with FORTRAN. This code is shown below.
! Calculates the time development of an initially uneven concentration distribution C(x,t) with explicit
! finite-differencing of a diffusion equation. The boundary condition is that no solute diffuses through the boundaries
! Teemu Isojärvi, Feb 2017
PROGRAM MAIN
real :: DX, DT, LX, LT, D ! Real variables for discretization and the diffusion constant
integer :: NT,NX ! Number of time and position steps
REAL :: C(30) ! Concentration field as an array of data points, dimension same as value of NX
REAL :: DC(30) ! Change of the above quantity during a time step
REAL :: X ! Auxiliary variable
INTEGER :: M ! Integer looping variables
INTEGER :: N
LX = 6. ! Length of x-interval
LT = 6. ! Length of t-interval
NX = 30 ! Number of lattice points
NT = 600 ! Number of time steps
DX = LX/NX ! Distance between neighboring lattice points
DT = LT/NT ! Length of time step
D = 1 ! Diffusion/heat conduction coefficient
do M = 1, NX ! Initial values of concentration
X = M*DX
C(M) = EXP(-2*(X-3)*(X-3)) ! Gaussian initial concentration distribution
end do
do N = 1, NT ! Time stepping loop
C(1)=C(2)
C(NX)=C(NX-1)
do M = 2, NX-1
DC(M) = (D*DT/(DX*DX))*(C(M+1)-2*C(M)+C(M-1)) ! Main time stepping calculation
end do
do M = 1, NX
C(M) = C(M)+DC(M) ! Update the data points
end do
end do
do M = 1, NX
X = M*DX
print *,X,C(M) ! Print the x and concentration values at data points
end do
END
Finally, I will show a way to write, with R-Code programming language, a program that will do the same calculation and automatically produce output files with plots of the function $f(x,t)$ on every one timestep out of 3:
library(graphics) #load the graphics library needed for plotting
lx <- 6.0 #length of the computational domain
lt <- 6 #length of the simulation time interval
nx <- 30 #number of discrete lattice points
nt <- 600 #number of timesteps
dx <- lx/nx #length of one discrete lattice cell
dt <- lt/nt #length of timestep
D <- 1.0 #diffusion constant
Conc <- c(1:nx) #define a vector to put the x- and t-dependent concentration values in
DConc <- c(1:nx) #a vector for the changes in concentration during a timestep
xaxis <- c(1:nx) #the x values corresponding to the discrete lattice points
kappa <- D*dt/(dx*dx) #a parameter needed in the discretization
for (i in c(1:nx)) {
Conc[i] = exp(-2*(i*dx-3)*(i*dx-3)) #Gaussian initial concentration distribution
DConc[i] <- 0 #all initial values in DConc vector zeroed
xaxis[i] <- i*dx #calculate the x coordinates of lattice points
}
for (j in c(1:nt)) { #main time stepping loop
Conc[1] <- Conc[2] #derivatives of C(x) at both ends of domain forced to stay zero
Conc[nx] <- Conc[nx-1]
for (k in c(2:(nx-1))) {
DConc[k] <- kappa*(Conc[k-1]-2*Conc[k]+Conc[k+1]) #calculate the changes in concentration
}
for (l in c(1:nx)) {
Conc[l] <- Conc[l]+DConc[l] #add the changes to the vector Conc
}
k <- 0
l <- 0
if(j %% 3 == 0) { #make plots of C(x) on every third timestep
jpeg(file = paste("plot_",j,".jpg",sep=""))
plot(xaxis,Conc,xlab="position (x)", ylab="concentration",ylim=c(0:1))
title(paste("C(x) at t =",j*dt))
lines(xaxis,Conc)
dev.off()
}
So, the output image files are named as plot_1.jpg, plot_2.jpg, and so on. Now we can use the free ImageJ program (or something else with the same functionality) to import this sequence of images and to save it as an AVI video file. The video can be viewed here.
As an another example, I will give an R-Code that calculates the diffusion or conduction through a layer of intermediate material, given that the function $f(x,t)$ is set to have a constant nonzero value at $x=0$ and the boundary condition that its derivative with respect to x is zero at the domain endpoints:
library(graphics) #load the graphics library needed for plotting
lx <- 6.0 #length of the computational domain
lt <- 6 #length of the simulation time interval
nx <- 30 #number of discrete lattice points
nt <- 600 #number of timesteps
dx <- lx/nx #length of one discrete lattice cell
dt <- lt/nt #length of timestep
D <- 1.0 #diffusion constant
Conc <- c(1:nx) #define a vector to put the x- and t-dependent concentration values in
DConc <- c(1:nx) #a vector for the changes in concentration during a timestep
xaxis <- c(1:nx) #the x values corresponding to the discrete lattice points
kappa <- D*dt/(dx*dx) #a parameter needed in the discretization
for (i in c(1:nx)) {
Conc[i] <- 0.5*(1+tanh(5-5*i*dx)) #a hyperbolic tangent initial concentration field
DConc[i] <- 0 #all elements zeroed initially
xaxis[i] <- i*dx #calculate the x coordinates of lattice points
}
for (j in c(1:nt)) { #main time stepping loop
Conc[1] <- 1 #force the C to stay at value 1 at left boundary
Conc[2] <- 1 #force the derivative of C to zero at left boundary
Conc[nx] <- Conc[nx-1] #force the derivative of C to zero at right boundary
for (k in c(2:(nx-1))) {
DConc[k] <- kappa*(Conc[k-1]-2*Conc[k]+Conc[k+1]) #calculate the changes in concentration
}
for (l in c(2:(nx-1))) {
Conc[l] <- Conc[l]+DConc[l] #add the changes to the vector Conc
}
if(j %% 3 == 1) { #make plots of C(x) on every third timestep
jpeg(file = paste("plot_",j,".jpg",sep=""))
plot(xaxis,Conc,xlab="position (x)", ylab="concentration",ylim=c(0,2))
title(paste("C(x) at t =",j*dt))
lines(xaxis,Conc)
dev.off()
}
}
The animation made of this simulation can be viewed here
So, here were the first examples of PDE solving with discretization. The problem with the explicit finite differencing used here, is that if one decides to use a very small spatial step size $\Delta x$, the timestep $\Delta t$ may have to be made prohibitively small to keep the simulation numerically stable. Suppose we do the simulation with parameters $L = 6$, $n_x = 300$, $t_{end} = 1$, $n_t = 100$. Now the end result plotted with Grace looks like this:
Figure 2. Unstable behavior of explicit method discretization of diffusion equation.
So obviously the numerical errors done in the discretization accumulate exponentially and “blow up” when trying to use too short a $\Delta x$ compared to $\Delta t$. In later blog posts I will show how to use these programming languages to write an implicit PDE solver that corrects this problem for the diffusion/conduction equation, as well as makes it possible to solve Schroedinger equation (a complex-valued version of diffusion equation, which can have wave-like solutions with reflection, interference, etc.). Also, I will extend the method to problems with more than one spatial dimension, and show how to solve a nonlinear PDE called thin-film equation.
## Some things about iterative sequences
How would one, using only pen and paper, construct a sequence of rational numbers that approaches the number $\sqrt{2}$ ? This is not a simple question for many, as nowadays most people haven’t had to calculate approximations for square roots without a calculator. A simple way to solve this problem is by noting that, if x is any positive rational number, then the number $\sqrt{2}$ is always between the numbers x and 2/x. So, we can make the sequence by starting with $a_0 = 1$ and then calculating the number $a_n$ as the average value of $a_{n-1}$ and $2/a_{n-1}$:
$a_1 = \frac{1}{2}(1 + \frac{2}{1}) = 3/2 = 1.5$
$a_2 = \frac{1}{2}(3/2 + 2/(3/2)) = 17/12 \approx 1.416666$
$a_3 = \frac{1}{2}(17/12 + 2/(17/12)) = 577/408 \approx 1.414216$
This sequence approaches the number $\sqrt{2} = 1.414213562$ rather quickly. The square root of some other integer m could be calculated with the same method by using the iteration rule $a_n = \frac{1}{2}(a_{n-1} + m/a_{n-1})$. This method also produces error bars for the approximate results, as we always have $a_n \leq \sqrt{m} \leq m/a_n$ or $m/a_n \leq \sqrt{m} \leq a_n$
More generally, if one wants to find an iterative sequence that approaches a number x that has some desired property described by the equation $f(x) = 0$, the iteration rules to try are of the form $a_n = a_{n-1} + cf(a_{n-1})$, where c is some positive or negative constant. Note that the number x is a “fixed point” of this iteration, which means that if for some n, the number $a_n$ is already exactly $a_n = x$, then all the numbers $a_k$ with k>n are equal to x.
Depending on the choice of the initial guess $a_0$ and the factor c, these iterations can converge towards the desired result either quickly (like the iteration for $\sqrt{2}$ above), slowly or not at all. Suppose we want to find a solution for the cubic equation $P(x) = x^3 + 3x^2 -3x - 9 = 0$ by iteration. If we try to set $a_0 = 1$ and $a_n = a_{n-1}+P(a_{n-1})$, we get the quickly diverging sequence
$a_1 = -7$
$a_2 = -191$
$a_3 = -6858055$
and so on. If we choose $a_n = a_{n-1} + 0.1P(a_{n-1})$ instead, the sequence converges towards the solution $x = -\sqrt{3} \approx -1.732050808$, but rather slowly:
$a_0 = 1$
$a_1 = 0.2$
$a_2 = -0.7472$
$a_3 = -1.29726$
$a_8 = -1.71421$
Sometimes this kind of iterations can also get “stuck in a loop”, bouncing between two values and not converging towards any number, like the sequence $a_0 = 1$, $a_1 = 2$, $a_2 = 1$, $a_3 = 2$, $a_4 = 1$, …
Sometimes the property that defines the number that we’re seeking is not defined by a polynomial equation. For example, the Lambert W function W(x) is defined by
$x = W(x)e^{W(x)}$.
Obviously, if we want to find an approximation of W(x) for some x, we could try an iteration with rule $a_n = a_{n-1} + c(x-a_{n-1}e^{a_{n-1}})$ where c is some positive or negative number between -1 and 1. But this, of course doesn’t produce rational number approximations like the iterations in the examples above did. Other things that have to be calculated iteratively, are the solutions of transcendental equations like $e^x = 3x$, which appear in the quantum theory of finite potential wells and the analysis of nonlinear electronic circuits (those that have diodes or other nonlinear components in them).
So, if you have to keep a presentation about convergence and iterations in some maths course, playing with this kind of calculations is a good way to construct multiple example cases of slowly or quickly converging sequences!
|
2019-10-24 00:25:39
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 215, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7500398755073547, "perplexity": 1726.166728727612}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987836368.96/warc/CC-MAIN-20191023225038-20191024012538-00398.warc.gz"}
|
https://www.physicsforums.com/threads/leaning-ladder.59870/
|
1. Jan 15, 2005
laminatedevildoll
I am having problems with the following problem.
A person is standing on a leaning ladder rested on a wall, where m1 is the mass of the person and m2 is the mass of the ladder. He is up at the distance d on the ladder, L (the length). There is also an angle at the bottom.
The first part of the question asks to find the minimum coefficient friction, so that the ladder does not slip, which I found was 0.5*L*m_2*g+m_1*g*d)/(L*tan(theta)*(m_2*g+m_1*g))
Then, part B of the question is to find the magnitude of the force of friction that the floor applies to the ladder. The coefficient of the static friction force, mu_s is equal to (3/2)*mu_min which is given. In this case, I know that f doesn't equal mu_s*N. but it is less than mu_s*F_normal .
In the end, I got (3/2)*(d/L*m_1*g+0.5*m_2*g)*tan(theta) , which is completely wrong.
Any help is appreciated.
Thank you
Last edited: Jan 16, 2005
2. Jan 15, 2005
Q_Goest
These kind of problems use summing of moments around a point and summing forces in various orthogonal directions.
Start by summing forces in the verticle (Y) direction and horizontal (X) direction. You should find verticle forces don't really matter here, so look at forces in the horizontal direction. In this case, the force against the wall, N, is equal to the frictional force. Note there are no other forces in the X direction, only these two, so the force against the wall is equal to the frictional force.
Now find the force against the wall, N, by summing moments around a point. If the point you select is the point where the ladder rests on the ground, you should find moments created by the verticle forces due to weight for the ladder and the person are countered by the normal force on the wall, N times the moment arm. This solves for N, and since N = frictional force, you also have the frictional force.
3. Jan 15, 2005
Gokul43201
Staff Emeritus
laminated,
Rather than just writing down what you got, if you tell us how you got that, it would be possible for us to show you where you made your mistake. Now we really don't know how you got what you got.
4. Jan 16, 2005
laminatedevildoll
I started summing forces from the x, y direction
Fx = fs+n1=0
Fy=n2+(-m1g)+(-m2g)=0
Then also the torque
TB=n1Lsin(theta)-0.5(m2g)(Lcos(theta)-m1g(Dcos(theta))+n2(0)+fs(0)=0
fs=mu_min*n2, plugging in all the values and solving for mu_min...
This yielded to 0.5*L*m_2*g+m_1*g*d)/(L*tan(theta)*(m_2*g+m_1*g)) which is CORRECT.
But, how do I find the force of friction that the floor applies to the ladder? Ignoring the slipping...
5. Jan 16, 2005
laminatedevildoll
So this means f=mu_min*N right? N = (m2g+m1g)
Fx = fs+n1=0
Fy=n2+(-m1g)+(-m2g)=0
Kind of confused...
6. Jan 16, 2005
Gokul43201
Staff Emeritus
Should one assume that there is no friction from the wall ?
7. Jan 16, 2005
Gokul43201
Staff Emeritus
Assuming this, I get, for the first part :
$$\mu _{min} = \frac {m1(d/L) + (m2/2)}{(m1+m2)tan \theta }$$
You have the same result for this. So far, so good.
For the second part, the only importance of the additional data is to tell you that the coefficient is large enough for stable equilibrium.
In this case, you simply redo the calculations that you did for (A), but instead of $\mu _{min} N$, you just call the frictional force F, and find F. Balancing horizontal forces will tell you that F = R, the reaction force from the wall. So, replacing R with F in the torque equation gives :
$$F = \frac {g[m1(d/L) + (m2/2)]}{tan \theta}$$
Is this what the answer is supposed to be ?
8. Jan 16, 2005
laminatedevildoll
Yes, that is the correct answer! I have been pondering about that second part for days. I thought that I had to integrate 3/2mu_min into the equation somehow. But, I am still a rookie at this stuff.
In any case, thank you for your help. I really appreciate your time.
|
2017-03-24 00:36:37
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8051159977912903, "perplexity": 407.3531791080377}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218187227.84/warc/CC-MAIN-20170322212947-00127-ip-10-233-31-227.ec2.internal.warc.gz"}
|
http://annals.math.princeton.edu/2010/171
|
Volume 171
## The classification of Kleinian surface groups, I: models and bounds
Pages 1-107 by Yair Minsky | From volume 171-1
## Uniqueness for the signature of a path of bounded variation and the reduced path group
Pages 109-167 by Ben Hambly, Terry Lyons | From volume 171-1
## Transformations of elliptic hypergeometric integrals
Pages 169-243 by Eric M. Rains | From volume 171-1
## Mirković-Vilonen cycles and polytopes
Pages 245-294 by Joel Kamnitzer | From volume 171-1
## Noise stability of functions with low influences: Invariance and optimality
Pages 295-341 by Elchanan Mossel, Ryan O’Donnell, Krzysztof Oleszkiewicz | From volume 171-1
## Nilpotency, almost nonnegative curvature, and the gradient flow on Alexandrov spaces
Pages 343-373 by Vitali Kapovitch, Anton Petrunin, Wilderich Tuschmann | From volume 171-1
## On the classification of finite-dimensional pointed Hopf algebras
Pages 375-417 by Nicolás Andruskiewitsch , Hans-Jürgen Schneider | From volume 171-1
## Dyson’s ranks and Maass forms
Pages 419-449 by Kathrin Bringmann, Ken Ono | From volume 171-1
## On the ergodicity of partially hyperbolic systems
Pages 451-489 by Keith Burns, Amie Wilkinson | From volume 171-1
## The Smith-Toda complex $V((p+1)/2)$ does not exist
Pages 491-509 by Lee S. Nave | From volume 171-1
## The Brauer-Manin obstruction for subvarieties of abelian varieties over function fields
Pages 511-532 by Bjorn Poonen, José Felipe Voloch | From volume 171-1
## Essential dimension, spinor groups, and quadratic forms
Pages 533-544 by Patrick Brosnan, Zinovy Reichstein, Angelo Vistoli | From volume 171-1
## Pentagon and hexagon equations
Pages 545-556 by Hidekazu Furusho | From volume 171-1
## A nonhomogeneous orbit closure of a diagonal subgroup
Pages 557-570 by François Maucourant | From volume 171-1
## Tight closure does not commute with localization
Pages 571-588 by Holger Brenner, Paul Monsky | From volume 171-1
## Characterization of Lee-Yang polynomials
Pages 589-603 by David Ruelle | From volume 171-1
## Ergodic billiards that are not quantum unique ergodic (with an appendix by Andrew Hassell and Luc Hillairet)
Pages 605-618 by Andrew Hassell, Luc Hillairet | From volume 171-1
## Quantitative noise sensitivity and exceptional times for percolation
Pages 619-672 by Oded Schramm, Jeffrey E. Steif | From volume 171-2
## Free boundaries in optimal transport and Monge-Ampère obstacle problems
Pages 673-730 by Luis Caffarelli, Robert J. McCann | From volume 171-2
## The cohomology ring of the real locus of the moduli space of stable curves of genus $0$ with marked points
Pages 731-777 by Pavel Etingof, André Henriques, Joel Kamnitzer, Eric M. Rains | From volume 171-2
## A family of Calabi-Yau varieties and potential automorphy
Pages 779-813 by Michael Harris, Nick Shepherd-Barron, Richard Taylor | From volume 171-2
## Arithmetic quantum unique ergodicity for symplectic linear maps of the multidimensional torus
Pages 815-879 by Dubi Kelmer | From volume 171-2
## A group-theoretic approach to a family of $2$-local finite groups constructed by Levi and Oliver
Pages 881-978 by Michael Aschbacher, Andrew Chermak | From volume 171-2
## Polynomial parametrization for the solutions of Diophantine equations and arithmetic groups
Pages 979-1009 by Leonid Vaserstein | From volume 171-2
## Constructible exponential functions, motivic Fourier transform and transfer principle
Pages 1011-1065 by Raf Cluckers, François Loeser | From volume 171-2
## Global solutions of shock reflection by large-angle wedges for potential flow
Pages 1067-1182 by Gui-Qiang Chen, Mikhail Feldman | From volume 171-2
## Boundary rigidity and filling volume minimality of metrics close to a flat one
Pages 1183-1211 by Dmitri Burago, Sergei Ivanov | From volume 171-2
## An algorithm for computing some Heegaard Floer homologies
Pages 1213-1236 by Sucharit Sarkar, Jiajun Wang | From volume 171-2
## Order of current variance and diffusivity in the asymmetric simple exclusion process
Pages 1237-1265 by Márton Balázs , Timo Seppäläinen | From volume 171-2
## Quantization of coboundary Lie bialgebras
Pages 1267-1345 by Benjamin Enriquez, Gilles Halbout | From volume 171-2
## Differentiating maps into $L^1$, and the geometry of $\rm BV$ functions
Pages 1347-1385 by Jeff Cheeger, Bruce Kleiner | From volume 171-2
Pages 1387-1400 by Lewis Phylip Bowen | From volume 171-2
## The global stability of Minkowski space-time in harmonic gauge
Pages 1401-1477 by Hans Lindblad, Igor Rodnianski | From volume 171-3
## Divergent square averages
Pages 1479-1530 by Zoltán Buczolich , R. Daniel Mauldin | From volume 171-3
## Un indice qui affine l’indice de Poincaré-Lefschetz pour les homeomorphismes de surfaces
Pages 1531-1589 by Frédéric Le Roux | From volume 171-3
## Sur un problème de Gelfond : la somme des chiffres des nombres premiers
Pages 1591-1646 by Christian Mauduit, Joël Rivat | From volume 171-3
## The Atiyah-Singer index formula for subelliptic operators on contact manifolds. Part I
Pages 1647-1681 by Erik van Erp | From volume 171-3
## The Atiyah-Singer index formula for subelliptic operators on contact manifolds. Part II
Pages 1683-1706 by Erik van Erp | From volume 171-3
## Cones and gauges in complex spaces: Spectral gaps and complex Perron-Frobenius theory
Pages 1707-1752 by Hans Henrik Rugh | From volume 171-3
## Linear equations in primes
Pages 1753-1850 by Ben Green, Terence Tao | From volume 171-3
## Measure equivalence rigidity of the mapping class group
Pages 1851-1901 by Yoshikata Kida | From volume 171-3
## Drift diffusion equations with fractional diffusion and the quasi-geostrophic equation
Pages 1903-1930 by Luis Caffarelli, Alexis Vasseur | From volume 171-3
## Perturbations of orthogonal polynomials with periodic recursion coefficients
Pages 1931-2010 by David Damanik, Rowan Killip, Barry Simon | From volume 171-3
## A characterization of the entropies of multidimensional shifts of finite type
Pages 2011-2038 by Michael Hochman, Tom Meyerovitch | From volume 171-3
## Vacant set of random interlacements and percolation
Pages 2039-2087 by Alain-Sol Sznitman | From volume 171-3
## The perverse filtration and the Lefschetz hyperplane theorem
Pages 2089-2113 by Mark Andrea A. de Cataldo, Luca Migliorini | From volume 171-3
## Densities for rough differential equations under Hörmander’s condition
Pages 2115-2141 by Thomas Cass, Peter Friz | From volume 171-3
## Some adjoints in homotopy categories
Pages 2143-2155 by Amnon Neeman | From volume 171-3
## Insufficiency of the Brauer-Manin obstruction applied to étale covers
Pages 2157-2169 by Bjorn Poonen | From volume 171-3
## Cappell-Shaneson homotopy spheres are standard
Pages 2171-2175 by Selman Akbulut | From volume 171-3
|
2015-09-02 21:41:47
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5695759057998657, "perplexity": 10182.112050531829}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645293619.80/warc/CC-MAIN-20150827031453-00216-ip-10-171-96-226.ec2.internal.warc.gz"}
|
https://linearalgebras.com/tag/transitivity
|
If you find any mistakes, please make a comment! Thank you.
Basic properties of blocks of a group action
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.1 Exercise 4.1.7 Let $G \leq S_A$ act transitively on the set $A$. A block is a nonempty subset…
Every doubly transitive group action is primitive
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.1 Exercise 4.1.8 A transitive permutation group $G \leq S_A$ acting on $A$ is called doubly transitive if for…
Transitive group actions induce transitive actions on the orbits of the action of a subgroup
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.1 Exercise 4.1.9 Suppose $G \leq S_A$ acts transitively on $A$ and let $H \leq G$ be normal. Let…
An abelian group has the same cardinality as any sets on which it acts transitively
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.1 Exercise 4.1.3 Suppose $G \leq S_A$ is an abelian and transitive subgroup. Show that $\sigma(a) \neq a$ for…
Stabilizer commutes with conjugation II
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.1 Exercise 4.1.2 Let $G$ be a permutation group on the set $A$ (i.e. $G \leq S_A$), let \$\sigma…
|
2022-12-02 02:23:56
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8896107077598572, "perplexity": 139.54727905778728}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710890.97/warc/CC-MAIN-20221202014312-20221202044312-00287.warc.gz"}
|
http://gkkg.solidalweb.it/implementation-of-dfs-using-adjacency-matrix.html
|
# Implementation Of Dfs Using Adjacency Matrix
In algorithmic problems we use 3 ways to represent the graph. 1 - Adjacency Matrix. Of these two the adjacency matrix is the simplest, as long as you don't mind having a (possibly huge) n * n array, where n is the number of vertices. DFS (Depth First Search) BFS (Breadth First Search) DFS (Depth First Search) DFS traversal of a graph produces a spanning tree as final result. BUGLIFE (SPOJ): Solution Using BFS. An implementation. Representation. –Call DFS(G) to get finishing times u. Set of edges representation. DFS Implementation in C++. The idea is also simple - imagine an n by n grid, where each row and each column represents a vertex. To represent the traversal of a graph using the AbstractGraph. If such edge doesn't exist, we store zero. C program to implement Breadth First Search(BFS). Graph traversal Algorithms: Breadth first search in java Depth first search in java Breadth first search is graph traversal algorithm. A graph having n vertices, will have a dimension n x n. •If E = O(V) (sparse graph), adjacency lists are more space efficient. The advantage of DFS is it requires less memory compare to Breadth …. DFS Using Adjacency Matrix. Representation. 1 - Adjacency Matrix. My code does not read the row if all the numbers are all 0 and doesn't show the components visited. java implements the same API using the adjacency-matrix representation. asked Apr 20 in UTU B. The size of the matrix is VxV where V is the number of vertices in the graph and the value of an entry Aij is either 1 or 0 depending on whether there is an edge from vertex i to vertex j. Here is a simple code to implement Breadth First Search (Using Queue Data Structure ) and Depth First Search (Using Stack Data Structure )algorithm using adjacency matrix implemented in C Using Stack Data Structure: Code: [code]#include i. In the last post, we discussed how to represent a graph in a programmatic way using adjacency matrix and adjacency list representation. adjacency list representation. 5 (127 ratings) Course Ratings are calculated from individual students’ ratings and a variety of other signals, like age of rating and reliability, to ensure that they reflect course quality fairly and accurately. I am implementing a graph via adjacency matrix but cannot get the insertEdge method to function properly. To ask us a question or send us a comment, write us at. A graph is a collection of nodes and edges. In this video, we have discussed how we can know about the vertices that is creating cycles in our graph by the simplest method. This is because the number of entries in adjacency list is 2 X M. The Adjacency Matrix implementation initially outperforms the Adjacency List implementation, but the situation changes quickly, and the Adjacency Matrix implementation progresses twice as slowly. Let's get our hands dirty and use backtracking to solve N-Queens problem. A 10 minute video with very good explanation BFS implementation in java using Adjacency Matrix 2014 (6) December (4) November (2) About Me. IPseudocode for Depth-first-search of graph G=(V,E) dfs(v) count := count + 1 mark v with count for each vertex w adjacent to v do if w is marked with 0 dfs(w) DFS(G) count :=0 mark each vertex with 0 (unvisited) for each vertex v∈ V do if v is marked with 0 dfs(v) Design and Analysis of Algorithms - Chapter 5 26 Example – undirected graph. Dijkstra algorithm is a greedy algorithm. Other graph representations are possible. The algorithm to determine whether a graph is bipartite or not uses the concept of graph colouring and BFS and finds it in O(V+E) time complexity on using an adjacency list and O(V^2) on using adjacency matrix. Show the state of your auxiliary data structures (e. We will discuss two of them: adjacency matrix and adjacency list. We define two private variable i. Notice that if the graph is undirected, the adjacency matrix will be symmetric across its diagonal (from the top left to the bottom right corners). A graph can be represented using an adjacency matrix or an adjacency list. For dense graphs (represented with an adjacency matrix), as noted in Section 19. For dense graphs (represented with an adjacency matrix), as noted in Section 19. Adjacency List. – Example: it takes O(1) constant time to check whether two vertices are connected using an adjacency matrix – it takes linear time O(n) to print all edges in a graph using adjacency lists, where n is the number of edges. Adjacency matrix for undirected graph is always symmetric. An adjacency list, also called an edge list, is one of the most basic and frequently used representations of a network. DFS can be implemented in two ways. 1 Undirected Graphs. undirected. 3: Let parent, pre- and post-order be global arrays. Depth-first search. Use The Adjacency Matrix To Implement The Graph. the DFS traversal makes use of an stack. There are several possible ways to represent a graph inside the computer. The disadvantage of BFS is it requires more memory compare to Depth First Search(DFS). I'm still new to this, sorry if my mistake is too obvious. We have already seen about breadth first search in level order traversal of binary tree. The idea is also simple - imagine an n by n grid, where each row and each column represents a vertex. NET Library. from CSE 2315. STEP 3: Using the adjacency matrix of the graph find all the unvisited adjacency node to search node. … Read More ». I am implementing a graph via adjacency matrix but cannot get the insertEdge method to function properly. - Adjacency matrix implementation - Adjacency lists implementation. It involves exhaustive searches of all the nodes by going ahead, if possible, else by backtracking. HA-B A-G A-C L-M J-M J-L J-K E-D F-D H-I F-E A-F G-E A G E B C F D H M J K L I A G E B C D H M K J L I 10 Adjacency Matrix Representation Adjacency matrix representation. That's why in most implementation we would use an adjacency list rather than the matrix. The graphs we'll be working with are simple enough that it doesn't matter which implementation we opt for. We introduce two classic algorithms for searching a graph—depth-first search and breadth-first search. It indicates direct edge from vertex i to vertex j. One starts at the root (selecting some arbitrary node as the root in the case of a graph) and explores as far as possible along each branch before backtracking. The advantage of DFS is it requires less memory compare to Breadth First Search(BFS). Using Warshall algorithm we can modify adjacency matrix of graph to generate transistive closure of graph using which we can know what all vertices are reachable from a particular vertex. Topological Sort (DFS) Small Graph Adjacency Matrix Representation: Animation Speed: w: h:. It finds a shortest path tree for a weighted undirected graph. Sparse graph: very few edges. You initialize G[0] to NULL and then begin inserting all the edges before you finish initializing the rest of G[]. I am implementing a graph via adjacency matrix but cannot get the insertEdge method to function properly. Use MathJax to format equations. Dijkstra algorithm is a greedy algorithm. Data have been retrieved using the scholar package, the pipeline is describe in this github repository. Rewrite the function sssp() for finding the shortest path, such that the digraph is represented by its adjacency lists instead of the adjacency matrix. C Program to insert and delete nodes in graph using adjacency matrix. Space for adjacency matrix (AMAT) is n^2. // Use a stack for the iterative DFS version public static void dfs_iterative(ArrayList> adj, int s){ boolean[] visited = new boolean[adj. save Save Graphs-1. An 'x' means that that vertex does not exist (deleted). Depth-first search is a classic recursive method for systematically examining each of the vertices and edges in a graph. An adjacency list representation for a graph associates each vertex in the graph with the collection of its neighboring vertices or edges. If you use an adjacency matrix, you'd have to scan all the way through a row of the matrix, even if the vertex you're interested in is adjacent to only a few other vertices. Adjacency matrix. Below friends, please find the code for Implementation of BFS using Adjacency Matrix /** * * @author krishna kumar * * Below code shows basic implementation of DFS in a graph. NetworkX is the Python library that we are going to use to create entities on a graph (nodes) and then allow us to connect them together (edges). Space for adjacency matrix (AMAT) is n^2. Implementation of Floyd Warshall algorithm is very simple which is its main advantage. When visiting a vertex v and its adjacency vertices, if there is a back edge (w, v) which directs to v then that graph has a cycle Learn more. The idea is also simple - imagine an n by n grid, where each row and each column represents a vertex. The number of weakly connected components is. We use the following steps to implement DFS. here is my code package newtestgraph;. Adjacency matrix. b) In the given graph, simulate traversal of all vertices starting at node A using breadth first search (BFS). Graph Implementations Graph API as a design problem, two graph representations, and the implications of these representations on algorithms. A graph can also be represented in an adjacency matrix form which we have discussed during Djikstra algorithm implementation. As the graph is. Edge List, Adjacency List, Adjacency Matrix is the 3 common ways to represent ways to represent the graph. Tags adjacency list representation of graph adjacency matrix java code algorithm for dijkstra algorithm of dijkstra c program to find shortest path using dijkstra's algorithm data structures and algorithms in java tutorial dfs traversal dijkstra dijkstra algorithm c++ code dijkstra algorithm example directed graph dijkstra algorithm example in. me/9198774. •Adjacency lists •More compact than adjacency matrices if graph has few edges •Requires more time to find if an edge exists •Adjacency. Let Your Program Be Menu Driven Represent A Vertex By Using The Index Value Of The Array. The value that is stored in the cell at the intersection of row $$v$$ and column $$w$$ indicates if there is an edge from vertex $$v$$ to vertex. Dijkstra algorithm is a greedy algorithm. Description. DFS can be thought of as what you’d get it you replaced the queue from BFS with a stack. One representation is an adjacency matrix. C Program #include #include int […]. Suppose I have a predecessor array that tracks down the minimum edges i Have found so far. Adjacency lists are the right data structure for most applications of graphs. Edge Sets. Each cell represents an edge between the row vertex and the column vertex with a value of true if the edge exists, or false if it does not exist. Row i has "neighbor" information about vertex i. java from the text. If the graph has some edges from i to j vertices, then in the adjacency matrix at i th row and j th column it will be 1 (or some non-zero value for weighted graph), otherwise that place will hold 0. In this matrix in each side V vertices are marked. predecessor[u]=v {this is also the final MST} Now I want to modify the current A[i][j] matrix and change the weights to 1. Show the state of your auxiliary data structures (e. Use The Adjacency Matrix To Implement The Graph. And if the coordinates we want to visit next is not in this list, we are going to visit that position. Adjacency matrix representation: In adjacency matrix representation of a graph, the matrix mat [] [] of size n*n (where n is the number of vertices) will represent the edges of the graph where mat [i] [j] = 1 represents that there is an edge between the vertices. java * Execution: java AdjMatrixGraph V E * Dependencies: StdOut. If the graph is undirected (i. Breadth-first search (BFS) is an algorithm for traversing or searching tree or graph data structures. C Program to implement Graph Coloring using Backtracking C Program to implement Double Ended Queue (Deque) C Program to implement Hashing using Linear and Quadratic Probing. I am representing this graph in code using an adjacency matrix via a Python Dictionary. Space for adjacency matrix (AMAT) is n^2. Apart from the data structures used, there is also a factor of whether the graph is densely populated or sparsely populated. Serial code. Two common representations for graphs. We can either use a hashmap or an array or a list or a set to implement graph using adjacency list. java implements the same API using the adjacency-matrix representation. I'm basically wanting to see how something is inserted into graph (vertices/edges) and printing it out. If there are still unvisited nodes, repeat from step 1. The adjacency matrix, sometimes also called the connection matrix, of a simple labeled graph is a matrix with rows and columns labeled by graph vertices, with a 1 or 0 in position according to whether and are adjacent or not. Graphs in Python - DFS be represented using an adjacency list, an adjacency matrix or an incidence matrix. To be more specific, it’s vectors inside a vector. The advantages of representing the edges using adjacency matrix are: Simplicity in implementation as you need a 2-dimensional array; Creating edges/removing edges is also easy as you need to update the Booleans; The drawbacks of using the adjacency matrix are: Increased memory as you need to declare N*N matrix where N is the total number of nodes. me/9198774. Graph representations/storage implementations – adjacency matrix, adjacency list, adjacency multi-list 2: 21: Graph traversal algorithm: Breadth First Search (BFS) and Depth First Search (DFS) – Classification of edges - tree, forward, back and cross edges – complexity and comparison 1: 22. We use an array of jVjalong with the list of edges incident to each vertex. ALGORITHM:- STEP 1: START STEP 2: Create node in the program marked any one node on visited. In a weighted graph. Here is the source code of the Java program to perform the dfs traversal. Runs Depth First Search (DFS) algorithm on the graph starting from vertex vertexId. 23 BILLION users, so if we were to use an adjacency matrix it would take far far too much memory. What's a good rule of thumb for picking the implementation?. Im trying to do a cycle detection using adjacency matrix. I am implementing a graph via adjacency matrix but cannot get the insertEdge method to function properly. specifies the start node, end node and. For MultiGraph/MultiDiGraph with parallel edges the weights are summed. In my examples, the numbers represent this natural order. Each algorithm has its own characteristics, features, and side-effects that we will explore in this visualization. That’s why in most implementation we would use an adjacency list rather than the matrix. Leiserson, S. Prim's Algorithm code is well commented, indented and highlighted. For any particular vertex, we create a. Adjacency Matrix; Incidence Matrix; Adjacency List; Adjacency Matrix. me/9198774. Runs Depth First Search (DFS) algorithm on the graph starting from vertex vertexId. I am representing this graph in code using an adjacency matrix via a Python Dictionary. Dfs In 2d Array. Adjacency matrix representation is below. If such edge doesn't exist, we store zero. Adjacency Lists vs. hence you have a adjacency matrix, then you could use algorithm that Imran mentioned in comment, you just need to compute An, for n. Here, there is different a bit between the implementation of Depth First Search and Breadth First Search algorithm. (1) Adjacency matrix. … Read more Graph Depth First Search in Java, easy in 5 minutes. I'm working on a program that can take in an adjacency matrix from a mile, then output a few things: the input graph, the order that vertices are first encountered, displaying their count rather than the actual vertex numbers, the order that vertices become dead ends, and trees and back edges. For the vertex 1, we only store 2, 4, 5 in our adjacency list, and skip 1,3,6 (no edges to them from 1). We can either use a hashmap or an array or a list or a set to implement graph using adjacency list. The matrix has one row and column for every node in the graph, and the element at row u column v is set to one if there is an edge from u to v. ¥Vertex = intersection. There are two most common ways to implement graph: Adjacency Matrix; Adjacency List; 2. In this video, we have discussed the DFS Traversal of a graph that is being represented using adjacency matrix. This C program generates graph using Adjacency Matrix Method. depth_first_search(adj_matrix, source_index, end_index) We have to pass in adj_matrix , which is the adjacency matrix representation of the graph implemented as an array of arrays in Ruby. If an adjacency matrix is used, they will take O(N^2) time (N^2 is the maximum number of edges that can be present). java implements the same API using the adjacency-matrix representation. C program to implement Depth First Search(DFS). We are still implementing it using the adjacency list but instead of an object (map), we'll store the vertices in an array. If you know how to implement DFS iteratively, BFS is easy to implement and analogous. depth-first search. Below is the syntax highlighted version of AdjMatrixGraph. Sequential organization. It prints the adjacency matrix in a more readable format. Dijkstra algorithm is a greedy algorithm. In previous post, we have seen breadth-first search(bfs). Implementation strategies 1. Dijkstra algorithm is a greedy algorithm. Represent a graph using adjacency list and perform DFS and BFS OUTPUT: 1)Create a graph 2)BFS 3)DFS 4)Quit Enter Your Choice : 1 Enter no. The elements of the matrix typically have values '0' or '1'. If the number is negative, the nth element from the end is ret. Here, I give you the code for implementing the Adjacency List using C++ STL where each vertex is a string instead of and integer. Adjacency Matrix. Depth First Search is a graph traversal technique. Graph Representation Using Adjacency Matrix. Consider the DFS routine for the adjacency list representation. An adjacency matrix is used to represent the graph to be. Depth-First Search and Breadth-First Search in Python 05 Mar 2014. Adjacency matrix for a graph with n vertices numbered 0, 1, …, n – 1. Adjacency Matrix If a graph has n vertices, we use n x n matrix to represent the graph. Java program to describe the representation of graph using adjacency matrix. We define two private variable i. Detect Cycle in a Directed Graph. I know I need to set it up in a way, which I can call dfs() again. In this article we will implement Djkstra's - Shortest Path Algorithm (SPT) using Adjacency Matrix. I'm basically wanting to see how something is inserted into graph (vertices/edges) and printing it out. /* C program to implement BFS(breadth-first search) and DFS(depth-first search) algorithm */ #include int q[20],top=-1,f Red Black Tree (RB-Tree) Using C++ A red-black tree is a special type of binary tree, used in computer science to organize pieces of comparable data, such as text fragments o. Implement BFS and DFS for a given directed graph as adjacency matrix and show output USING GRAPHICS. Solution: Use a Boolean visited array. Well, it makes no sense if the algorithm is using STL if the input graph isn't built by STL. Given an undirected or a directed graph, implement the graph data structure without using any container provided by any programming language library (e. I am implementing a graph via adjacency matrix but cannot get the insertEdge method to function properly. The drawback to this approach lies in that we want to add vertices. In a weighted graph. The simplest way to store a graph is probably the adjacency matrix. Dfs In 2d Array. please type the code out and show that it works. edges (Why?) Not in this chapter, but required!. Space for adjacency linked lists (ALL) is n + 3*2e = n + 6e. Let Your Program Be Menu Driven Represent A Vertex By Using The Index Value Of The Array. adjacency matrix for cycle detection Hi. What is the runtime of DFS for an adjacency matrix vs an adjacency list?. 4 : Represent a graph using adjacency list and perform DFS and BFS. Algorithm for DFS in Python. Apply the DFS-based algorithm to solve the topological sorting problem for the following digraphs: d g a c f b e (a) b a f c d e g (b). •If E = O(V) (sparse graph), adjacency lists are more space efficient. You can use booleans or bits in the matrix to save memory. That’s pretty obvious. NET Library. Adjacency matrix — Use hash table to determine integer i associated with Boston — Use hash table to determine integer j associated with Providence — Edge exists if edge[i][j] > O Adjacency lists — Use hash table to determine integer i associated with Boston — Search the linked list vertices[i] to see if a node exists whose. Using a graph class to abstract the graph, as well as a std::stack to hold vertices. Let The Maximum Number Of Vertices Be 100. Adjacency list. Traversals and Searching Graphs and Digraphs- Depth First Search (DFS) and DF Spanning Tree - Breadth First Search (BFS) and BF- ST. Implement for both weighted and unweighted graphs using Adjacency List representation. I'm working on a program that can take in an adjacency matrix from a mile, then output a few things: the input graph, the order that vertices are first encountered, displaying their count rather than the actual vertex numbers, the order that vertices become dead ends, and trees and back edges. Adjacency matrix representation: In adjacency matrix representation of a graph, the matrix mat[][] of size n*n (where n is the number of vertices) will represent the edges of the graph where mat[i][j] = 1. An adjacency matrix is a matrix of size n x n where n is the number of vertices in the graph. ・ Algorithms based on iterating over vertices adjacent to v. Just like in breadth first search, if a vertex has several neighbors it would be equally correct to go through them in any order. DetailsInput to your program consists of an integer n representing the number of vertices followed an adjacency matrix of n rows, each row with n entries. As it is, it finds some of the connected edges when I increment v also, but it misses edge 3,1 and 4,0 which are edges with initial node having a higher index than the one it is mapped to. Each node has a list of all the nodes connected to it. This algorithm is a recursive algorithm which follows the concept of backtracking and implemented using stack data structure. 2-2Suppose that we represent the graph G= (V;E) as an adjacency matrix. … Read more Graph Depth First Search in Java, easy in 5 minutes. In this tutorial, we'll explore the Depth-first search in Java. Let the 2D array be adj[][], a slot adj[i][j] = 1 indicates that there is an edge from vertex i to vertex j. An undirected graph and its adjacency matrix representation is shown in the following figure. b) The graph has 10,000 vertices and 20,000,000 edges, and it is important to use as little space as possible. E is a set of pairs of vertices,these pairs are called as edges V(G) and E(G) will represent the sets of vertices and edges of graph G. The adjacency matrix takes ( n) operations to enumerate the neighbors of a vertex v since it must iterate across an entire row of the matrix. Detect Cycle in a Directed Graph. 1 - Adjacency Matrix. I do not know how to implement a depth first search tree to address unreachable node. What is the runtime of DFS for an adjacency matrix vs an adjacency list?. Adjacency Matrix C Program Data Structure This Program is for Adjacency Matrix in C , and is a part of Mumbai University MCA Colleges Data Structures C program MCA Sem 2 #includeThis is because using. The Depth First Search Algorithm. Graph Implementations Graph API as a design problem, two graph representations, and the implications of these representations on algorithms. the DFS traversal makes use of an stack. CSC 321: Data Structures Fall 2015 Graphs & search ! graphs, isomorphism ! simple vs. To visit a node, mark it explored. If the graph has some edges from i to j vertices, then in the adjacency matrix at i th row and j th column it will be 1 (or some non-zero value for weighted graph), otherwise that place will hold 0. There are two popular options for representing a graph, the first being an adjacency matrix (effective with dense graphs) and second an adjacency list (effective with sparse graphs). I am implementing a graph via adjacency matrix but cannot get the insertEdge method to function properly. Adjacency Matrices With an adjacency matrix, a graph with N nodes is stored using an N X N matrix. Part of the processing is pushing any adjacent nodes. It prints the adjacency list in a more readable format. hello; hope all is great. Pattern matching algorithm- The Knuth-Morris-Pratt algorithm. You can read more about storing graphs here. matrices, and adjacency lists (§28. Dijkstra algorithm is a greedy algorithm. We would recommend to read the theory part of Graph Representation - Adjacency Matrix and Adjacency List before continue reading this article. Let’s see how depth first search works with respect to the following graph:. Use The Adjacency Matrix To Implement The Graph. Description. • This scheme can be used when the vertices are represented using an array. Assuming names of vertices and pointers use 2 bytes each, adjacency list requires 2n+4mbytes of space (2n+8mfor undirected graphs), adjacency matrix n2=8. Adjacency matrix. Each cell represents an edge between the row vertex and the column vertex with a value of true if the edge exists, or false if it does not exist. Depth-first search is a classic recursive method for systematically examining each of the vertices and edges in a graph. For the adjacency-matrix representation, the order of edge insertion does not affect search dynamics, but we use the parallel term standard adjacency-matrix DFS to refer to the process of inserting a sequence of edges into a graph ADT implemented with an adjacency-matrix representation (Program 17. In the case of an undirected graph the adjacency matrix is symmetrical. 1 - Adjacency Matrix. The (i,j)th entry of the graph is set to 1 if and only if there exists an edge between vertices v i and v j. Here is the source code of the Java program to perform the dfs traversal. Implementing Breadth first search using Learn more about clustering, image analysis, bfs, connectivity matrix, graph theory MATLAB. Another matrix. The last disadvantage, we want to draw you attention to, is that adjacency matrix requires huge efforts for adding/removing a vertex. All trademarks and registered trademarks are the property of their respective owners 200+ pages. Graphs out in the wild usually don't have too many connections and this is the major reason why adjacency lists are the better choice for most tasks. The Adjacency Matrix implementation initially outperforms the Adjacency List implementation, but the situation changes quickly, and the Adjacency Matrix implementation progresses twice as slowly. The overall depth first search algorithm then simply initializes a set of markers so we can tell which vertices are visited, chooses a starting vertex x, initializes tree T to x, and calls dfs(x). Adjacency Matrix for Flight Connections. (j) T F [3 points] An undirected graph is said to be Hamiltonian if it has a cycle con-taining all the vertices. here is my code package newtestgraph;. n-1} can be represented using two dimensional integer array of size n x n. Adjacency list. Exercise 1: Write a method that outputs all the edges of a graph given using an. Adjacency matrix c d b f e a Figure 1: Graph, adjacency matrix and CSC sparse storage format Algorithm 1 Sequential DFS (Recursive) 1: Let G= (V;E) be a graph with a root node r. Adjacency Lists vs. java implements the same API using the adjacency-matrix representation. Let The Maximum Number Of Vertices Be 100. Topological Sort (DFS) Algorithm Visualizations. Prove that the topological sorting problem has a solution for a digraph if and only if it is a dag. But a good object oriented design strictly prohibits the use of global variables or methods, since they are against the fundamental principles of object. computer programming , hacking news , hacking tricks , hacking tutorials , c++ programming , java programming , how to , engineering Tutorials point ,. In this article we will implement Djkstra's – Shortest Path Algorithm (SPT) using Adjacency List and Priority queue. Two drawing represent same graph. please help me with this assigment, read the instruction really carefully. No idea about this question please help me out. In the special case of a finite simple graph, the adjacency matrix is a (0,1)-matrix with zeros on its diagonal. BUGLIFE (SPOJ): Solution Using BFS. Similarly, all the other non-zero values are changed to their respective weights. 1 Undirected Graphs. The idea is also simple - imagine an n by n grid, where each row and each column represents a vertex. In this article we will implement Djkstra's – Shortest Path Algorithm (SPT) using Adjacency List and Priority queue. DFS on a graph G = (V, E) in adjacency list representation: Search(graph G = (V, E), vertex s ∈ V, integer k) 1 mark vertex s as number k 2 set k ← k + 1 3 let L be the linked list of neighbors for s 4 repeat until all entries in L are marked with an X 5 mark the first un-marked entry y in L with an X (going from left to right in the list) 6 let v be the vertex named in entry y 7 if v is. The graphs we'll be working with are simple enough that it doesn't matter which implementation we opt for. Adjacency Matrix is also used to represent weighted graphs. Iterator edgeIterator() Returns iterator object for the edges. ALGORITHM:- STEP 1: START STEP 2: Create node in the program marked any one node on visited. And the new piece, the new field, that we're going to define for objects that are of type graph adjacency matrix are these adjacency matrix that are going to be 2D arrays of integers. To traverse in trees we have traversal algorithms like inorder, preorder, postorder. Depth First Search is a graph traversal technique. Sample Questions: Give pseudocode (or C++) code for recursive function DFS. Description. Use adjacency matrix representation of the graph and find runtime of the function /* C++ Program to Implement Graph using Adjacency List, Adjacency matrix, Traverse using BFS nd DFS */. For example I thought about this DS last time I had to implement a graph for DFS: struct Edge {int start; int end; struct Edge* nextEdge;}. Question: COMP 280 Programing Assignment 5 (Graphus) Write A Program To Implement An Undirected Graph. Adding vertices would require either making the 2 arrays (vertex and adjacency array) some large maximum size OR reallocating new arrays and copying the contents from the old to the new. Seidel adjacency matrix — a matrix similar to the usual adjacency matrix but with 1. [code] #include void DFS(int); int G[10][10],visited[10],n; //n is no of vertices and graph is sorted in array G[10][10] void main() { int i,j; printf("Enter. In the context of computer science, a matrix representation is used to represent a graph, called an adjacency matrix. adjacency matrix V2 1 V adjacency list E + V degree(v) degree(v) adjacency set E + V log (degree(v)) degree(v) huge number of vertices, small average vertex degree 20!graph API!maze exploration!depth-first search!breadth-first search!connected components!challenges 21 Maze exploration Maze graphs. Computing a spanning forest of graph. 1 Depth First Search Using a Stack All DFS algorithms, as far as I know, use a stack. Depth-First Search and Breadth-First Search in Python 05 Mar 2014. The adjacency matrix is a good way to represent a weighted graph. Adjacency Matrix is a 2D array of size V x V where V is the number of vertices in a graph. please help me with this assigment, read the instruction really carefully. NetworkX is the Python library that we are going to use to create entities on a graph (nodes) and then allow us to connect them together (edges). In such situation, we need to extract all the relevant detail and transform the problem into graph problem. Unlike BFS, DFS goes in depth and from there it backtracks the. • Sparse graph: very few edges. This algorithm is a recursive algorithm which follows the concept of backtracking and implemented using stack data structure. Serial code. In DFS, You start with an un-visited node and start picking an adjacent node, until you have no choice, then you backtrack until you have another choice to pick a node, if not, you select another un-visited node. Adding a Vertex. If such edge doesn't exist, we store zero. This C program generates graph using Adjacency Matrix Method. adjacency matrix V2 1 V adjacency list E + V degree(v) degree(v) adjacency set E + V log (degree(v)) degree(v) huge number of vertices, small average vertex degree 20!graph API!maze exploration!depth-first search!breadth-first search!connected components!challenges 21 Maze exploration Maze graphs. We define an undirected graph API and consider the adjacency-matrix and adjacency-lists representations. INTRODUCTION Many of the problems that fall within the purview of artificial intelligence are too complex to be solved by direct techniques rather they must be attacked by appropriate search methods armed with what ever direct techniques. Adjacency-List implementation: 0:0023V+ 0:0012E Adjacency-Matrix implementation: 0:0005V2 Complete Graph Case : Adjacency-List implementation: 0:0011V2 + 8E 05V+ 0:928 Notice that the time bounds in the case of adjacency-matrix is quadratic. Graph Algorithms - Min Cost Spanning Tree- Shortest Path Spanning Tree. Sparse graph: very few edges. 5: routine. One starts at the root (selecting some node as the root in the graph case) and explores as far as possible along each branch before backtracking. Since dfs examines each node in the adjacency lists at most once, the time to complete the search is O(e). if visited[i] = 0 then call dfs(i) procedure dfs(v) visited[v] 1; for each node w such that (v;w) 2E do if visited[w] = 0 then call dfs(w) Questions { How to implement the for-loop (i) if an adjacency matrix is used to represent the graph and (ii) if adjacency lists are used? { How many times is dfs called in all?. Adjacency matrix for undirected graph is always symmetric. Up to O(v2) edges if fully connected. Previous Next If you want to practice data structure and algorithm programs, you can go through data structure and algorithm interview questions. But for algorithms like DFS, BFS (and those that use it, like Edmonds-Karp), Priority-first search (Dijkstra, Prim, A*), etc. A graph G,consists of two sets V and E. In such situation, we need to extract all the relevant detail and transform the problem into graph problem. For this algorithm, the adjacency list representation is better. Graph Representation Methods, Adjacency Matrix. The Adjacency Matrix implementation initially outperforms the Adjacency List implementation, but the situation changes quickly, and the Adjacency Matrix implementation progresses twice as slowly. My code does not read the row if all the numbers are all 0 and doesn't show the components visited. Two drawing represent same graph. Your Class Is To Implement The Following Functions - DFS - BFS - DFS Spanning Tree - BFS Spanning Tree Each Function Should Take The Node To Start At As Its Parameter Your Program Will Take In 2 Files. And the new piece, the new field, that we're going to define for objects that are of type graph adjacency matrix are these adjacency matrix that are going to be 2D arrays of integers. Therefore, DFS complexity is O(V + E). There are two types of popular implementation for graph. In this algorithm, lets. Adjacency Matrix An easy way to store connectivity information – Checking if two nodes are directly connected: O(1) time Make an n ×n matrix A – aij = 1 if there is an edge from i to j – aij = 0 otherwise Uses Θ(n2) memory – Only use when n is less than a few thousands, – and when the graph is dense Adjacency Matrix and Adjacency List 7. By: Ankush Singla Online course insight for Competitive Programming Course. Apart from the data structures used, there is also a factor of whether the graph is densely populated or sparsely populated. Let Your Program Be Menu Driven Represent A Vertex By Using The Index Value Of The Array. Concepts: vertex, edge, path, cycle. C Program To Implement DFS Algorithm using Recursion and Adjacency Matrix. And it consumes only that much space that is required. Though it’s called a matrix, it’s actually a NxN form. The value that is stored in the cell at the intersection of row $$v$$ and column $$w$$ indicates if there is an edge from vertex $$v$$ to vertex $$w$$. If you can get from vertex A to vertex B by travelling over a sequence of edges, then we say that there is a path between them. specifies the start node, end node and. Each cell represents an edge between the row vertex and the column vertex with a value of true if the edge exists, or false if it does not exist. Let The Maximum Number Of Vertices Be 100. The above implementation of BFS runs in O(m + n) time if the graph is given by its adjacency representation. Thus, BFS will take O(V2) time using an adjacency matrix. The simplest way to store a graph is probably the adjacency matrix. hello; hope all is great. In mathematics and computer science, an adjacency matrix is a means of representing which vertices (or nodes) of a graph are adjacent to which other vertices. This code for Depth First Search in C Programming makes use of Adjacency Matrix and Stack. This C program generates graph using Adjacency Matrix Method. In C++, it is possible to declare a global object, which can be used anywhere inside the program. Use comma "," as separator. Adjacency List A type of graph representation wherein each vertex holds a list of every other vertex adjacent to it. The VxV space requirement of the adjacency matrix makes it a memory hog. An adjacency matrix is used to represent the graph to be. DFS search starts from root node then traversal into left child node and continues, if item found it stops other wise it continues. An adjacency matrix is a square matrix used to represent a finite graph. In above case 3 times DFS will run 1st for Node-1, 2nd for Node 8 and 3rd for Node-9. My code does not read the row if all the numbers are all 0 and doesn't show the components visited. Implementation of Floyd Warshall algorithm is very simple which is its main advantage. Can you suggest such DS(main structs or classes and what will be in them). an adjacency matrix has size Θ(n2) to calculate all out-degrees for a digraph represented as an adjacency matrix: for v ←1 to n do {d[v] ←0; for w ←1 to n do d[v] ←d[v] +A[v,w]; } this algorithm takes time Θ(n2) Conclusion. Adjacency Matrix Adjacency List In this post, we start with the first method Edges and Vertices list to represent a graph. We add visited node to the stack in the process of exploring the depth until we reach the leaf node (end of a branch). Graph Implementation: Use of Array Adjacency Matrix Depth First Search (DFS) DFS (v) DFS : Implementation. ! This is a special extension for my discussion on Graph Theory Basics. Dfs Python - czly. A lot of problems in real life are modeled as graphs, and we need to be able to represent those graphs in our code. Depth First Search is a graph traversal technique. Scenarios where the from and to parameters refer to the same vertex or when multiple edges. To represent the traversal of a graph using the AbstractGraph. The drawback is that it consumes large amount of space if the number of vertices increases. Adjacency Matrix C Program Data Structure This Program is for Adjacency Matrix in C , and is a part of Mumbai University MCA Colleges Data Structures C program MCA Sem 2 #includeThis is because using. We introduce two classic algorithms for searching a graph—depth-first search and breadth-first search. This is because the number of entries in adjacency list is 2 X M. Here, I give you the Adjacency List Implementation in C Sharp (C#) using the. Implementation of Floyd Warshall algorithm is very simple which is its main advantage. the first being an adjacency matrix (effective with dense graphs) and second an adjacency list (effective with sparse graphs). Consider the DFS routine for the adjacency list representation. The V is the number of vertices of the graph G. Previous Next If you want to practice data structure and algorithm programs, you can go through data structure and algorithm interview questions. Use adjacency lists in increasing numerical order. Let the 2D array be matrix[][], a slot matrix[i][j] = 1 indicates that there is an edge from vertex i to vertex j. Adjacency matrix. 1 - Adjacency Matrix. C Program to implement Graph Coloring using Backtracking C Program to implement Double Ended Queue (Deque) C Program to implement Hashing using Linear and Quadratic Probing. In this video, we have discussed the DFS Traversal of a graph that is being represented using adjacency matrix. Adjacency Matrix is a 2D array of size V x V where V is the number of vertices in a graph. The igraph package includes a convenient function for finding the shortest paths between every dyad in a network. We introduce two classic algorithms for searching a graph—depth-first search and breadth-first search. Note: This C Program for Depth First Search Algorithm using Recursion and Adjacency Matrix for Traversing a Graph has been compiled with GNU GCC Compiler and developed using gEdit Editor in Linux Ubuntu Operating System. Breadth First Search (BFS) has been discussed in this article which uses adjacency list for the graph representation. The idea is also simple - imagine an n by n grid, where each row and each column represents a vertex. I had put hope on the “Falcon” plugin, but it only seems to deal with 4x4 matrices. for sparse graphs, adjacency list representations are preferable to adjacency matrices: they use less space. PROBLEM FINDING THE SHORTEST PATH BY USING AN ADJACENCY LIST. adjacency matrix for cycle detection Hi. You can read more about storing graphs here. Using the Code. Let the 2D array be adj[][], a slot adj[i][j] = 1 indicates that there is an edge from vertex i to vertex j. Depth First Search - Graph example In this blog post we will have a look at the Depth First Search (DFS) algorithm in Java. An Adjacency Matrix¶ One of the easiest ways to implement a graph is to use a two-dimensional matrix. Depth First Search Analysis. 1 on macOS, and I am coding with VS Code. For the vertex 1, we only store 2, 4, 5 in our adjacency list, and skip 1,3,6 (no edges to them from 1). The overall depth first search algorithm then simply initializes a set of markers so we can tell which vertices are visited, chooses a starting vertex x, initializes tree T to x, and calls dfs(x). Show the state of your auxiliary data structures (e. In this representation we have an array of lists The array size is V. Unknown View my complete profile. Graph representations/storage implementations – adjacency matrix, adjacency list, adjacency multi-list 2: 21: Graph traversal algorithm: Breadth First Search (BFS) and Depth First Search (DFS) – Classification of edges - tree, forward, back and cross edges – complexity and comparison 1: 22. Adjacency Matrix. Use The Adjacency Matrix To Implement The Graph. 2-2Suppose that we represent the graph G= (V;E) as an adjacency matrix. Adding a Vertex. Contact Info: WhatsApp: https://wa. Using an adjacency list also leads to lesser number of iterations in later operation (eg. Incidence Matrix: graph4. Because it uses the stack, we can implement it recursively without too much trouble. Other graphs examples: searching a maze, Dijkstra’s algorithm, graph coloring. If the number is negative, the nth element from the end is ret. printMat: this function can be used to aid with debugging. Adjacency Matrix Adjacency List In this post, we start with the first method Edges and Vertices list to represent a graph. A depth first traversal takes O(N*E) time for adjacency list representation and O(N2) for matrix representation. Adjacency list can be implemented using a linked list OR dynamic array. Other graph representations are possible. Implementation of DFS in Graph. All I need to store in the matrix are boolean v. The advantage of DFS is it requires less memory compare to Breadth First Search(BFS). save Save Graphs-1. We can solve several graph problems using these two traversals. Let the 2D array be adj[][], a slot adj[i][j] = 1 indicates that there is an edge from vertex i to vertex j. Implement DFS for an adjacency list. The idea is also simple - imagine an n by n grid, where each row and each column represents a vertex. Since I read it is not safe to use arrays of the form matrix[x][y] because they don't check for range, I decided to use the vector template class of the stl. Adjacency matrix for undirected graph is always symmetric. If such edge doesn't exist, we store zero. Implementing Breadth first search using Learn more about clustering, image analysis, bfs, connectivity matrix, graph theory MATLAB. Code Related Java Topics beta. Use The Adjacency Matrix To Implement The Graph. I would greatly appreciate any help on what I'm overlooking. So I'm not trying to scare you my reader but I won't fail to point out the prerequisites so you don't say. Minimum Spanning Tree : Kruskal's Algorithm. C Program to Implement Adjacency Matrix. Matrix is incorrect. adjMaxtrix [i] [j] = 1 when there is edge between Vertex i and Vertex j, else 0. Below is the source code for C Program to implement Topological Sorting Algorithm Example which is successfully compiled and run on Windows System to produce desired output as shown below :. Input: Output: Algorithm add_edge(adj_list, u, v) Input: The u and v of an edge {u,v}, and the adjacency list. In this video, we have discussed the DFS Traversal of a graph that is being represented using adjacency matrix. Use adjacency-lists representation. Depth First Search AIM:-To write a ‘c’ program for implementation of Depth First Search. Use The Adjacency Matrix To Implement The Graph. The space it takes it O(E+V), much less than adjacency matrix implementation. The advantage of DFS is it requires less memory compare to Breadth First Search(BFS). ・Algorithms based on iterating over vertices adjacent to v. Depth First Search (DFS) The aim of DFS algorithm is to traverse the graph in such a way that it tries to go far from the root node. Here are some additional notes on the following topics: expression trees. Graphs Digraphs Minimum Spanning Trees Minimum Spanning Tree Substructure Prim's Algoritm Adjacency List Undirected Graphs Paths Strongly Connected Graphs Depth-First Search Our Philosophy TeachingTree is an open platform that lets anybody organize educational content. Show the edgeTo array resulting from this dfs. In this article we will implement Djkstra's – Shortest Path Algorithm (SPT) using Adjacency List and Priority queue. Adjacency matrix. Designate this as the new search node and mark it as visited. Data have been retrieved using the scholar package, the pipeline is describe in this github repository. Representing the digraph as an adjacency matrix makes the retrieval of the cost of an edge O(1). • Dense graph: lots of edges. Give a simple implementation of Prims algorithm for this case that runs in O(V2) time. But for algorithms like DFS, BFS (and those that use it, like Edmonds-Karp), Priority-first search (Dijkstra, Prim, A*), etc. Depth first search (DFS) is an algorithm for traversing or searching tree or graph data structures. Representing the digraph as an adjacency matrix makes the retrieval of the cost of an edge O(1). 1 - Adjacency Matrix. It finds a subset of the edges that forms a tree that includes every vertex, where the total weight of all the edges in the tree is minimized. Graph traversal Algorithms: Breadth first search in java Depth first search in java Breadth first search is graph traversal algorithm. me/9198774. The number of connected components is. This Tuple stores two values, the destination vertex, (V 2 in an edge V 1 → V 2) and the weight of the edge. For this algorithm, the adjacency list representation is better. A common way to implement a graph using an adjacency list is to use either a hashtable with an array as values or use a hashtable with linked lists as a value. Graph Algorithms - Min Cost Spanning Tree- Shortest Path Spanning Tree. DFS search starts from root node then traversal into left child node and continues, if item found it stops other wise it continues. Start Search from the peak V in G, count to count1 Count1=N? Search from the peak V in G’, count to count2. Depth-First Search. E is a set of pairs of vertices,these pairs are called as edges V(G) and E(G) will represent the sets of vertices and edges of graph G. asked Apr 20 in UTU B. What is the runtime of DFS for an adjacency matrix vs an adjacency list?. In a weighted graph. 1 Kevin Lin, with thanks to many others. The DFS traversal of the graph using stack 40 20 50 70 60 30 10 The DFS traversal of the graph using recursion 40 10 30 60 70 20 50. Adjacency list. But i do not know how to do this. DFS Implementation in C++. Depth first traversal or Depth first Search is a recursive algorithm for searching all the vertices of a graph or tree data structure. Depth First Search is an algorithm used to search the Tree or Graph. Let The Maximum Number Of Vertices Be 100. Adjacency Matrix A graph G = (V, E) where v= {0, 1, 2,. A common way to implement a graph using an adjacency list is to use either a hashtable with an array as values or use a hashtable with linked lists as a value. In this tutorial you will learn about implementation of Depth First Search in Java with example. The processes will be modeled as vertices starting with the 0th process. •Adjacency lists •More compact than adjacency matrices if graph has few edges •Requires more time to find if an edge exists •Adjacency. V is a finite non-empty set of vertices. Adjacency matrix for undirected graph is always symmetric. We have already learnt about graphs and their representation in Adjacency List and Adjacency Matrix as well we explored Breadth First Search (BFS) in our previous article. As the graph is. Adjacency matrix of a directed graph is never symmetric, adj[i][j] = 1 indicates a directed edge from vertex i to vertex j. 1 - Adjacency Matrix. The advantages of representing the edges using adjacency matrix are: Simplicity in implementation as you need a 2-dimensional array; Creating edges/removing edges is also easy as you need to update the Booleans; The drawbacks of using the adjacency matrix are: Increased memory as you need to declare N*N matrix where N is the total number of nodes.
|
2020-08-04 17:30:19
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24034464359283447, "perplexity": 768.2350510927573}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439735881.90/warc/CC-MAIN-20200804161521-20200804191521-00202.warc.gz"}
|
https://kusemanohar.wordpress.com/2019/07/08/generating-music-from-basics/
|
In this post, I am discussing on music notes, frequencies and the science behind it. Along the way I shall live demo you what I am talking about. For the demos, I am using an arduino nano to generate PWM signals (square-wave) and a cheap buzzer speaker (Piezoelectric). It is possible to use a headphone or a speaker, but that will require more elaborate amplifier circuits, may be in a future post I will attempt that. For now lets just stick to basics.
## Sound Waves
As you probably know sound is just vibrations in the air. As a membrane vibrate at a particular rate it produces a wave. For example, to produce a way of 450 hz, the membrane need to vibrate 450 times a seconds (yes, that is really fast). We sapiens can sense sound waves from 20 hz to 20,000 hz with our ears. Using a tuning fork, it is possible to produce vibrations at a particular frequency. A real source, however, rarely produces a single sine wave, it is usually a sum of several sinusoidal. Different musical instruments produces different combinations of sinusoidal which give it its own distinct sounds. The reason a guitar sounds very different than a violin is because the combinations of sine-waves each produces.
#### Demo-1 : Sa Re Ga Ma (Indian Music Notation System) under a Sound Spectrum Analyzer
Let me demonstrate this generation of sine waves as I sing the musical octaves, in the Indian music system, we use the notation of Sa Re Ga Ma. This is almost exactly like the western music system of c,d,e,f,g,…etc. In a later section I go over it a bit more details. I would like to mention that when I was a kid (probably 12-15 years old), I used to learn Indian classical music. Since a microphone captures the sound intensity as time passes it is oftentimes hard to see the sinusoidal. However, using Fourier transform of this time series data it is possible to know which frequencies exist in it, this is referred to as Frequency domain or Spectral domain representation. For the reader who are unaware of the details of Fourier transform, it suffices to know that by applying Fourier transform to the sound data, we can get the constituent frequencies in the signal. On an android phone, there are several apps which can do the fast-fourier-transform (FFT) of the sound data coming in from the phones microphone.
I use the Spectroid app , in the above video. Notice my Sa is about 130 hz and the Sa in the next octave (ie. Sa after Re, Ga, Ma, Pa, Dha, Ni) is at about 260 hz. When I say 130hz, it is the dominant frequency. No sapiens can generate a pure sine waves, notice the other higher frequency noise that make my voice less melodious. The third thing to note in this context is that my Sa is 130 hz, different person can have their Sa at different frequencies, the ratios of the frequencies of Sa:Re:Ga etc make them sound them like the octaves, the absolute frequencies so not matter. When singers sing in a concert they have a reference Sa-Pa frequency, which they often listen to with their headphones.
## Speakers
There are several types of speakers that are available in the market. However all of them have one thing in common, there is a material that vibrates at the requested frequencies which in turn produces sounds. The difference between expensive speakers and cheap speakers is precision with which it can produce the requested frequencies and the exact mechanism of sound generation. I am going to breifly go over the following types: a) Electrodynamic speakers b) Piezo-electric speakers c) Rare earth speakers. The most common form of speakers the reader might be familiar are the Electrodynamic speakers. They involve a permanent magnet attached to the base and an electro-magnet attached to the vibrating diaphragm which turn on and off at the requested frequencies which vibrate the diaphragm.
There are several materials found in nature called piezo electric materials. They have a special property that when pressure is applied to these material (with a hammer for example) they produce a voltage, conversely, when a voltage is applied to these materials they vibrate. Quartz is such a material. Speakers can be made out of such materials, called the Piezospeakers. Such speakers are in popular use in buzzers. Commercial buzzers can be bought as low as 10 US cents – 50 US cents. A commercial package is available in 3 pin or a 2 pin configuration. I have a 3 pin package. +VCC, GND and Signal. The way to drive this speaker is to send a PWM signal to the signal pin. For example, if we want to produce 130hz sound on peizospeaker we need to send PWM to the speaker at 130hz (ie. with a period of about 7692 microseconds).
I am going to use a piezo speaker for the experiments in this post, because it needs a low current to drive (easily be driven directly by Arduino’s PWM). To drive a electrodynamic speaker one needs an amplifier circuit which I shall explore in a future blog post.
A rare earth element called Dysprosium has the property that it naturally resonates on application of an external electric field. This is used to construct the so called Rare-earth Speaker. These are fairly expensive and a relatively new technology. See the video below by youtuber-Thoisoi on it.
## Indian Music Theory and Notation System
Let us understand a bit on how to derive the frequencies of Sa Re Ga etc. Once we know these frequencies we know the ABCD of music, with which we can build up songs. With this end in mind lets dive in……
There are two major Indian music system. The Hindustani system and the Carnatic System. For this discussion lets stick to the Hindustani notation system. There are 12 sounds (svaras) in an octave. 7 pure sounds (Shudha): Sa, Re, Ga Ma, Pa, Dha, Ni; 4 komal (half) sounds: komal Re, komal Ga, Komal Dha, komal Ni and 1 Tivra sound: Tivra Ma. Typically in this system we talk about 3 octaves (saptak), ie. three repetitions of Sa Re, Ga,…Ni. The three saptaks are referred to as a) Mandra sapta, Madya saptak and Taar Saptak. This is so because, human’s sound will usually fall in 3 octaves only. Note that the frequency of Sa in taar saptak is two times the frequency of Sa in madya sapta. Similarly frequency of Pa in madhya saptak will be 2 times frequency of Pa in taar saptak. The absolute values of the frequency do not matter. All the svaras are defined relative to a reference. The table below summaries the information presented in this paragraph.
For this calculation, lets set the reference as 400hz for Sa of madya sapta. The band 400hz-800hz is harmonically divided into 12 parts. By harmonically we mean the next sound’s frequency will be a constant factor times the previous. In this case if we set the constant factor g as $\sqrt[12](2)$ (the 12th root of 2). The svaras will be $400$ hz, $400 \times g$ hz, $400 \times g^2$ hz, … , $400 \times g^{11}$. If we use say the reference of 270hz for Sa, the octave will sound exactly the same to a human listener but with a difference pitch, in this case the band 270hz-540hz will be harmonically divided into 12 parts. This following python snippet will help you get the frequencies (and wave periods) of each of the sounds.
ref_madya_sa = r = 400 #hz, change this as needed
g = 2**(1.0/12.0)
Sa = r
Re_k = r * g
Re = r * g**2
Ga_k = r * g**3
Ga = r * g**4
Ma = r * g**5
Ma_t = r * g**6
Pa = r * g**7
Dha_k =r * g**8
Dha = r * g**9
Ni_k = r * g**10
Ni = r * g**11
svara_list = ['Sa', 'Re_k', 'Re', 'Ga_k', 'Ga', 'Ma', 'Ma_t', 'Pa', 'Dha_k', 'Dha', 'Ni_k', 'Ni' ]
svara_freq = [Sa, Re_k, Re, Ga_k, Ga, Ma, Ma_t, Pa, Dha_k, Dha, Ni_k, Ni ]
for i in range( len(svara_list) ):
print( 'svara=%-5s\tfreq=%4.4f\tperiod(microsecs)=%4.4f' %(svara_list[i], svara_freq[i], 1.0E6/svara_freq[i] ) )
# print( '#define %s %d //hz=%f' %( svara_list[i], 1.0E6/float(svara_freq[i]), svara_freq[i] ) )
# Result:
# svara=Sa freq=400.0000 period(microsecs)=2500.0000
# svara=Re_k freq=423.7852 period(microsecs)=2359.6858
# svara=Re freq=448.9848 period(microsecs)=2227.2468
# svara=Ga_k freq=475.6828 period(microsecs)=2102.2410
# svara=Ga freq=503.9684 period(microsecs)=1984.2513
# svara=Ma freq=533.9359 period(microsecs)=1872.8838
# svara=Ma_t freq=565.6854 period(microsecs)=1767.7670
# svara=Pa freq=599.3228 period(microsecs)=1668.5498
# svara=Dha_k freq=634.9604 period(microsecs)=1574.9013
# svara=Dha freq=672.7171 period(microsecs)=1486.5089
# svara=Ni_k freq=712.7190 period(microsecs)=1403.0776
# svara=Ni freq=755.0995 period(microsecs)=1324.3289
Now if you send PWMs to the Piezo speaker at these frequencies you could get the whole sequence played by the speaker similar to my human voice. What is really happening is with the PWMs the quartz crystal vibrates at those frequencies thus producing the sounds. The following Arduino program can do that.
#define Sa 2500 //hz=400.000000
#define Re_k 2359 //hz=423.785238
#define Re 2227 //hz=448.984819
#define Ga_k 2102 //hz=475.682846
#define Ga 1984 //hz=503.968420
#define Ma 1872 //hz=533.935942
#define Ma_t 1767 //hz=565.685425
#define Pa 1668 //hz=599.322831
#define Dha_k 1574 //hz=634.960421
#define Dha 1486 //hz=672.717132
#define Ni_k 1403 //hz=712.718975
#define Ni 1324 //hz=755.099450
// Define a special note, 'R', to represent a rest
#define R 0
#define eof 321
#define speaker 4
int list[100] = {Sa, Re, Ga, Ma, Pa, Dha, Ni, 0.5*Sa, eof };
//int duration[100] = [100,212
void setup() {
pinMode(speaker, OUTPUT);
Serial.begin(9600);
int k=0;
while( list[k] != eof ) // loop over the array
{
Serial.print( list[k] );
for( int del=0 ; del<800 ; del++ )
{
// +ve for half the time and -ve for half the time
digitalWrite(speaker, HIGH);
delayMicroseconds(list[k]/2);
digitalWrite(speaker, LOW);
delayMicroseconds(list[k]/2);
}
k++;
}
}
// the loop function runs over and over again forever
void loop() {
}
While the method I mentioned is a mathematically sound method, there are alternate ways to get to the frequencies of the octave. There are thumb rules like, the ratios of Sa:Ga:Pa = 4:5:6 etc. I got to know about this through this link or [PDF].
## Demo-3: Roja Song Tune
Now that we can generate the constituents of music, we are all set to create an whole song. Although the sound quality with the piezo speaker is of very poor quality, this is a fully working proto-type. You could get the music notations of your favorite song online, just try doing a good search. My favorite song is ‘Dil Hai Chotasa’ from the movie Roja, the music director was the great AR Rahman. I got notation for this song from here.
## Future Improvements
Although this is a functional prototype, there is a lot to learn in getting the melodies correctly. For now, although the pitches are correct, the song is totally monotone. Different pitches need to play for different time duration for the song to make sense to a human. Also getting the sound to play on a electrodynamic speaker is a worthwhile endevour. Reading and playing sounds encoded as .wav or .midi can help you build your own ipod. I would also like to mention about Teensy Audio Shield (I am not promoting), available here. Guys at teensy have put up a tutorial to work with sounds with the Teensy board (A better Arduino).
|
2019-08-24 04:34:50
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.48401108384132385, "perplexity": 3066.3021366655375}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027319724.97/warc/CC-MAIN-20190824041053-20190824063053-00299.warc.gz"}
|
https://yetanothermathprogrammingconsultant.blogspot.com/2008/05/
|
## Saturday, May 31, 2008
### GAMS/IDE bug (minor)
In case of a syntax error, the GAMS/IDE will usually place the cursor on the correct place. This makes it easy to fix errors quickly without even consulting the listing file. Here is a case where the GAMS/IDE gets confused due to substitution of a macro:
In this case the cursor should go to symbol a which has not been declared. For the first error the macro %x% causes the cursor to point to parameterb.
### More cutting stock
>I decided to have a go at the challenge set in this paper - hope you don't mind.
>
We receive bars of length 1200 and must cut them into the following
> [length, quantity] pairs:[100,100],[200,2],[300,2],[493,250],[590,2],
> [630,2],[780,2],[930,2]
....
>So the solution equates to:12 * [100,100,100,100,100,100,100,493]2 *
> [100,100,100,100,100,630]2 * [100,100,930]2 * [100,300,780]2 *
>[100,493,590]118 * [200,493,493]
>
>Total bars used = 138
>
>Theoretical min:
>(100*100+200*2+300*2+493*250+590*2+630*2+780*2+930*2)/1200=116.76
>
>...and, finally, the numbers of each length of bar cut:100: 12 * 7 + 2 * 5 + 2 * 2 +
>2 * 1 + 2 * 1 = 102200: 118 * 1 = 118300: 2 * 1 = 2493: 118 * 2 + 2 * 1 + 12 * 1 =
>250590: 2 * 1 = 2630: 2 * 1 = 2780: 2 * 1 = 2930: 2 * 1 = 2
I can find a better solution using the column generation model. The following input was used:
*-----------------------------------------------------* Data*----------------------------------------------------- scalar r 'raw width' /1200/; table demanddata(i,*) width demand width1 100 100 width2 200 2 width3 300 2 width4 493 250 width5 590 2 width6 630 2 width7 780 2 width8 930 2;
The result is a solution (not proven optimal) of 131:
---- 174 PARAMETER pat pattern usage p4 p5 p9 p10 p11 p15 Totalwidth1 1 2 100width2 1 2width3 1 2width4 2 1 2 250width5 2 2width6 1 2width7 1 2width8 1 2Count 75 1 2 2 2 49 131
### GAMS/Mosek Bug
Mosek does not return the best integer solution found sofar when interrupted. Below we interrupted a MIP run, while integer solutions were found by the solver (the model was no longer intermediate integer infeasible).
**** SOLVER STATUS 8 USER INTERRUPT **** MODEL STATUS 14 NO SOLUTION RETURNED **** OBJECTIVE VALUE 3885.0000 RESOURCE USAGE, LIMIT 6.618 3600.000 ITERATION COUNT, LIMIT 0 10000000 MOSEK Link May 1, 2008 22.7.1 WIN 3927.4700 VIS x86/MS Windows M O S E K version 5.0.0.79 (Build date: Jan 29 2008 13:41:45) Copyright (C) MOSEK ApS, Fruebjergvej 3, Box 16 DK-2100 Copenhagen, Denmark http://www.mosek.com No solution returned
I don't know if this is caused by a misbehavior of the link or by Mosek itself. Work around: when stopped on a RESLIM limit, GAMS/Mosek will report the best integer solution. Don't press the Interrupt button if you don't want to loose your solution.
Note also that the iteration count is reported as zero, which is another bug.
## Friday, May 30, 2008
### GAMS smin bug
The expression smin(s, xxx) over an empty set s may or may not return -INF. The following fragment from a real world scheduling model with a problem in its input data illustrates inconsistencies in how GAMS evaluates this.
set si /i1,i2/;
parameter duedateadj(si) /i1 10, i2 10/;
set k /k1,k2/;
parameter duration(si,k) / i1.k1 1 /;
parameters
minduration(si)
mintime(si) "calculate in one step"
mintime2(si) "calculate in two steps"
;
mintime(si) = duedateadj(si)-smin(k$duration(si,k),duration(si,k)); minduration(si) = smin(k$duration(si,k),duration(si,k));
display mintime, mintime2;
This should give the same result, however when we run it we see:
---- 17 PARAMETER mintime calculate in one step
i1 9.000, i2 -INF
---- 17 PARAMETER mintime2 calculate in two steps
i1 9.000, i2 10.000
This is of course highly undesirable. In general I would try to prevent such situations by not using SMIN or SMAX over an empty set. As GAMS does not flag these occurences you will need to add explicit tests for this.
#### Update
This problem has been fixed in GAMS release 24,3,1 (july 2014).
### Date bug in GAMS/IDE
The date format is coded as the number of days since jan 1, 1900. However, it is one off compared to Excel, Access and compared to the date format in Delphi. It is also one off compared to the IDE Charting facility:
set j /x,y/ i /i1,i2,i3/; parameter data(i,j); * may 30, may 31, june 1data(i,'x') = jdate(2008,5,30)+ord(i)-1;data(i,'y') = ord(i); * check datesparameter check(i,*);check(i,'year') = gyear(data(i,'x'));check(i,'month') = gmonth(data(i,'x'));check(i,'day') = gday(data(i,'x')); option check:0:1:1; display data,check; execute_unload 'data',data;
The result display is:
---- 20 PARAMETER data x y i1 39597.000 1.000i2 39598.000 2.000i3 39599.000 3.000 ---- 20 PARAMETER check year month day i1 2008 5 30i2 2008 5 31i3 2008 6 1
The chart is:
The dates on the x-axis are wrong (click to enlarge). The work around is to add one to each date before exporting (and subtract one after importing).
## Thursday, May 29, 2008
### Color code an Excel spreadsheet
I received a very large spreadsheet to be converted to GAMS. To quickly get a feeling how many formulas there are and where they are located, this code colors all cells with formulas. The color depends on the type of the cell (number, string, logical or error).
Sub color() ActiveSheet.UsedRange.Style = "Normal" Call colorsub(xlNumbers, "Accent1") Call colorsub(xlTextValues, "Accent2") Call colorsub(xlErrors, "Accent3") Call colorsub(xlLogical, "Accent4")End Sub Sub colorsub(SpecialCells As Integer, styletype As String) Dim r As Range On Error GoTo err Set r = ActiveSheet.UsedRange.SpecialCells(xlCellTypeFormulas, SpecialCells) r.Style = styletype err: End Sub
An earlier version used HasFormula on each individual cell. As the spreadsheet I am working on is very large, that approach was very slow. This version returns complete ranges and is much faster.
### Missing info in GAMS listing file
I would highly prefer that the listing file is self-contained. In many cases solvers stop without issuing proper messages to the listing file. Here is an example with PATHNLP:
S O L V E S U M M A R Y MODEL TIME OBJECTIVE LL TYPE NLP DIRECTION MAXIMIZE SOLVER PATHNLP FROM LINE 160 **** SOLVER STATUS 4 TERMINATED BY SOLVER **** MODEL STATUS 7 INTERMEDIATE NONOPTIMAL **** OBJECTIVE VALUE 47.2365 RESOURCE USAGE, LIMIT 35.697 1000.000 ITERATION COUNT, LIMIT 0 100000 EVALUATION ERRORS 0 0 PATH-NLP May 1, 2008 22.7.2 WEX 3906.4799 WEI x86_64/MS Windows NLP size: 829 rows, 881 cols, 10145 non-zeros, 1.39% dense.MCP size: 1708 rows/cols, 26626 non-zeros, 0.91% dense.
There is no indication why PATHNLP stopped.
### Difficult MIP
When asked for an example of a small difficult MIP, the following model may be of interest:
http://www.amsterdamoptimization.com/models/etc/ran14x18.gms. Some notes on this model from a posting on sci.op-research:
The story of the instance ran14x18
We tried CPLEX 5.0 on this instance, obtaining a solution with objective value 3712 but
without proving optimality. The same solution was also found in some runs of the genetic
algorithm from reference 1.
In response to my posting on sci.op-research, some people tried exact solvers on this
instance (fctp, mps), but without success. I also contacted CPLEX, but they also failed
to prove optimality.
On Friday, 13 November 1998, I announced this web page on sci.op-research. Some
hours later I received good news: Jeff proved optimality for ran14x18. Some time later,
a parallel CPLEX version also solved this problem to optimality. I thank the people cited
below for their efforts and comments concerning this problem!
To get an impression of the difficulty of this instance, have a look at the discussion
about the problem, which has remained unsolved for more than 10 months.
Jeff Linderoth (
[email protected])
[Tue, 18 Aug 1998] Probably (repeat probably) a MIP solver that uses flow cover
inequalities would be able to solve the instances. (At least it should do better than
CPLEX for problems of this class). If you don’t have too many instances, then you can
send me the MPS files, and I would be happy to try the solver MINTO out on them for you.
[Fri, 28 Aug 1998] For the problem ran14x18.mps, I ran MINTO until it ran out of memory,
and was only able to improve the lower bound to 3617.994472. When I get the opportunity,
I will perform some longer runs and also try out PARINO (which is like a parallel
version of MINTO) on the problem. I also tried XPRESS and CPLEX and MINTO was much
better than either of these packages (due to flow cover inequalities). How did bc-opt
do? I’ll email more results later. (If you don’t hear from me, send me an email to
remind me). Anyway, sorry I wasn’t able to prove optimality for you, but hopefully the
results are somewhat useful.
[Fri, 13 Nov 1998] I proved the optimality of the solution 3712 for the ran14x18 FCTP
instance. In order to do this I ran my PARINO parallel MIP code for about 52 hours on
32 Pentium II (300MHz) processors. 7624677 nodes of the branch and bound tree were
evaluated in this time. PARINO adds "flow cover inequalities" which are usually quite
useful in helping to improve the LP bounds for these types of problems. Also, I used a
parallel pseudocost-type branching rule which is often better than the CPLEX default or
CPLEX’s strong branching.
Mark Wiley ([email protected])
[Tue, 18 Aug 1998] I saw your recent posting regarding solving randomly generated fixed
charge transportation models. As Jeff Linderoth mentioned in a subsequent posting, flow
cover cuts should help significantly on this type of problem. Our LINGO and What’sBest
packages include this enhancement. If you have some examples in MPS format we could try
them out.
[Fri, 21 Aug 1998] Thanks for sending the model. Unfortunately, flow cover cuts (or at
least our implementation) don’t do as much for your models as I would have liked. Almost
immediately LINGO finds a feasible integer solution of 4151 and a bound of 3359.81.
However, after that it makes virtually no progress. I’d be curious to know how MINTO
performs. I wish I could have given you better news. We will keep your model in our IP
test set. Perhaps a future version will be able to solve it quickly. It is an
interesting problem, small but tough.
Good luck.
[Fri, 23 Oct 1998] We tinkered with some cut options and let it run a bit longer and
found a better solution but failed to solve to optimality. Our best solution was 3764
with a bound of 3522.
Ed Klotz ([email protected])
[Mon, 2 Feb 1998] I tried a few runs with your problem, without success. I tried this on
a very fast machine, and it did not help. I think the fundamental difficulty with this
problem is that is very easy for the optimizer to move fractional solutions in the LP
subproblems from one variable to the next with little or no degradation in the objective
value. Adding cuts, or removing the symmetry of the problem, can help.
[Tue, 3 Feb 1998] Thinking about it [the FCTP model], it’s not very surprising that this
is difficult for the branch and bound to solve. When you relax the integrality
requirement on the y_ij, the problem essentially reduces to a regular transportion
problem (except that the cost coefficients are no longer cij). So the convex hull of the
LP relaxation is quite different from the convex hull of the integer feasible solutions.
This characteristic tends to lead to difficult MIPs. And, I am not the only one who
498 of "Integer and Combinatorial Optimization." "Unfortunately, the bounds obtained
from these relaxations are frequently very poor primarily because they do not accurately
represent the fixed costs..." I do not think we will be able to get CPLEX 5.0 to solve
this problem quickly just by changing parameters. It may be possible to come up with a
priority order file, but I think the best strategy will be to tighten the formulation
[Mon, 9 Nov 1998] While CPLEX 6.0 did not perform significantly better on this problem,
we have a development code that established a bound of 3578.6418 in two hours on a fast
machine. The associated integer solution of 3774 is therefore within 5.46 percent of
optimal. We obtained these results with default settings except for setting the
heuristicfreq parameter to 5 (i.e. we applied a rounding heuristic on every 5th node).
We also did a run with variableselect 3 (strong branching), heuristicfreq 5 that we let
go for several days. It eventually found a solution of 3714 and a bound of 3655. So,
CPLEX found a solution within 2 percent of optimal, although it took a long time to do
so. I suspect that with additional experimentation we can get the development version
of CPLEX to find a 2 percent solution in a more reasonable amount of time.
Irv Lustig ([email protected])
[Wed, 25 Nov 1998] We are in the process of developing a new version of CPLEX, so we
decided to try the ran14x18 problem with our new version on a large number of parallel
processors. We used an SGI Origin system with 64 processors. Our parallel development
code solved it with the following output:
Integer optimal solution (0.0001/0): Objective = 3.7120000000e+03
Current MIP best bound = 3.7116289184e+03 (gap = 0.371082, 0.010%)
Solution time = 11517.98 sec.
Iterations = 500288158 Nodes = 8982253 (101132)
This corresponds to 3 hours, 12 minutes wall clock time.
By default, we solve the problem to a .01% gap, but we could decrease this and the time
would only go up a little bit.
Just ran the model on a (cheap) quad core PC with Cplex 11 using default settings (we used threads 4):
S O L V E S U M M A R Y MODEL m OBJECTIVE cost TYPE MIP DIRECTION MINIMIZE SOLVER CPLEX FROM LINE 168 **** SOLVER STATUS 1 NORMAL COMPLETION **** MODEL STATUS 1 OPTIMAL **** OBJECTIVE VALUE 3712.0000 RESOURCE USAGE, LIMIT 630.895 3600.000 ITERATION COUNT, LIMIT 48023779 10000000
So we need 630 seconds on this machine to prove optimality (optcr=0).
### Excel coding
Usually I do Excel programming in VBA, or Delphi and VBScript via Excel Com Automation. The Delphi tools from http://www.axolot.com/components/xlsrwii20.htm are reputed to be very fast, bypassing Excel altogether. Unfortunately it cannot read the spreadsheets I am interested in. We'll try to get the spreadsheet to them pending clients permission.
## Tuesday, May 27, 2008
### Undocumented GAMS/CPLEX options
The following options are useful to make pictures of MIP performance:
• miptrace filename : write trace file to the indicated file
• miptracenode n : write a trace record each n nodes (default: n=100)
• miptracetime r : write a trace record each r seconds (default: r=1)
Example output file:
* miptrace file trace.csv: ID = Cplex 11.0.1* fields are lineNum, seriesID, node, seconds, bestFound, bestBound1, S, 0, 0.075, na, 4053.062, N, 0, 0.084, na, 4053.063, N, 0, 0.09, 1740.37, 3852.174, N, 0, 0.095, 1740.37, 3804.845, N, 0, 0.101, 1740.37, 3798.376, N, 0, 0.107, 1740.37, 3781.777, N, 0, 0.113, 1740.37, 3781.538, N, 0, 0.149, 1826.58, 3779.849, N, 100, 0.229, 2769.62, 3327.7710, N, 200, 0.304, 2891.26, 3135.0211, N, 300, 0.391, 2891.26, 2910.5712, E, 320, 0.397, 2891.26, 2891.26* miptrace file trace.csv closed
It is easy to make a plot of this in Excel:
These options are actually available in all or most MIP solvers under GAMS.
## Monday, May 26, 2008
### Stochastic Dynamic Programming in GAMS
Can this be done? This question comes up regularly.
There are a few interesting papers by Richard Howitt of UC Davis in this respect:
### GAMS/Cplex wish list
Here is my list of features I would like to see added or improved to GAMS/CPLEX:
• Support for user cuts (redundant constraints that can help the solver)
• Support for lazy constraints (constraints that are only included when violated)
• Better language support for piecewise linear expressions
• Proper language support for indicator constraints
• Sensitivity analysis results to GDX file
• Sensitivity analysis for bounds
• Performance improvement for QP/QCP related models
• Provide unbounded ray
• Fixing a number of reported bugs (some of them reported on in this blog)
### Mosek bug
Cplex needed a long time to solve the root node of an MIQCP model. I tried Mosek, but that gave
This is with version:
MOSEK Link May 1, 2008 22.7.2 WEX 3927.4799 WEI x86_64/MS WindowsM O S E K version 5.0.0.79 (Build date: Jan 29 2008 13:47:58)
I sent it to GAMS for further investigation. This problem seems only to happen with the 64 bit version. The work around would be to use the 32 bit version.
## Sunday, May 25, 2008
### Importing large CSV files into GAMS
The tool SQL2GMS can be used to import large CSV files into GAMS. This is especially useful if the CSV file does not fit in Excel. Excel before Office 2007 could only handle up to 65,536 rows (in Office 2007 this is increased to 1,048,576). Examples:
### GAMS/Convert bug
To export GAMS models with semi-continuous variables to MPS files the tool GAMS/Convert can be used. However note that semi-continuous variables with lower bound = upper bound are exported incorrectly. Consider the small example:
semicont variable x;variable y;equation e;y.lo=0;x.fx=1;e.. y =e= x;model m /e/;solve m minimizing y using mip;
The solution is x=0.
Convert generates the following MPS file:
ROWS N obj E c1COLUMNS sc1 c1 -1 x2 obj 1 x2 c1 1RHSBOUNDS FX bnd sc1 1ENDATA
In this MPS file column sc1 corresponds to GAMS variable x. However, this representation excludes sc1=0 as sc1 is fixed to one using FX. The correct MPS representation should look like:
ROWS N obj E c1COLUMNS sc1 c1 -1 x2 obj 1 x2 c1 1RHSBOUNDS LO bnd sc1 1 SC bnd sc1 1ENDATA
Work around: use the Cplex option writemps to create an MPS file. This will generate correct MPS files with fixed semi-continuous variables.
## Saturday, May 24, 2008
### GAMS/Cplex bug: iteration count reported as zero
In some cases GAMS/Cplex reports erroneously that zero iterations have been performed. Example: run THAI model from the model library with iterlim=5:
S O L V E S U M M A R Y MODEL thainavy OBJECTIVE obj TYPE MIP DIRECTION MINIMIZE SOLVER CPLEX FROM LINE 84**** SOLVER STATUS 4 TERMINATED BY SOLVER **** MODEL STATUS 14 NO SOLUTION RETURNED **** OBJECTIVE VALUE 0.0000 RESOURCE USAGE, LIMIT 0.037 1000.000 ITERATION COUNT, LIMIT 0 5
The SOLVER STATUS is also wrong, this should be 2 ITERATION INTERRUPT. Turns out I have reported this already in january 2006, but it does not seem to get fixed.
Correct values for these settings are often important when running Cplex from a more complex algorithm formulated in GAMS.
## Friday, May 23, 2008
### Modeling question
>
> Suppose I want to maximise, say:
> a1 + a2 + a3 + b1 + b2 + b3 + c1 + c2 + c3 < 30
>
> ...but with the extra condition that I must have two items from each
> of the following sets (at least two members of each set must be
> greater than zero): {a1, a2, a3}, {b1, b2, b3}, {c1, c2, c3}
>
> Is there an easy way to ask lp_solve for this condition, please?
>
One way of achieving this is to add the constraints:
a(i) ≥ 0.001·p(i)b(i) ≥ 0.001·q(i)c(i) ≥ 0.001·r(i)sum(i,p(i)) ≥ 2sum(i,q(i)) ≥ 2sum(i,r(i)) ≥ 2p(i),q(i),r(i) binary variables
Note that we do not need to model the additional implications
p(i)=0 => a(i)=0
because we exploit the "at least" part of the condition. If we want exactly two nonzeroes for each group, this implication would need to be added.
## Thursday, May 22, 2008
### Modeling question
> Hi,
>
> Can anybody help me to formulate this ? just not able to figure this out
>
> x is a continuous variable. Can take any positive value including 0.
> w is a 0/1 binary variable
>
> if x > 0, I want w = 0
> if x = 0, I want w = 1
This should work:
x ≤ (1-w)·M where M=x.up
x ≥ 0.001·(1-w)
### GAMS Update Button and Vista
The update button (File|Options|Execute|Update) does not always work very well with Vista (access denied or some other failure). This has to do with missing administrator rights. Workaround: start cmd from start menu using Shift-Ctrl-Enter, cd to gams system directory and run GAMSINST.
## Wednesday, May 21, 2008
### Timings on Cholesky
It is quite amazing how fast these three alternatives for Cholesky are. Here is a model to time the performance and accuracy. Some results are listed below.
MethodTimeAccuracyTimeAccuracyTimeAccuracy
n=50n=100n=200
Nonlinear Equations0.1489.6e-80.9171.2e-48.9910.621
External Solver0.0241.6e-130.0251.9e-100.0561.4e-5
GAMS Algorithm0.0341.9e-130.4826.3e-117.5717.7e-6
I am surprised that the GAMS algorithm is competitive. Notice the accuracy issues when using nonlinear equations due to ill-conditioning of the system of non-linear equations (small changes in a(i,j) lead to large changes in L(i,j)).
Here is some info about the external solver:
c:\>choleskyCHOLESKY: matrix decomposition A=LL^TUsage > cholesky gdxin i a gdxout Lwhere gdxin : name of gdxfile with matrix i : name of set used in matrix a : name of 2 dimensional parameter inside gdxin gdxout : name of gdxfile for results (factor L) L : name of 2 dimensional parameter inside gdxout Calculates the Cholesky decomposition A=LL^T of a symmetricpositive definite matrix A=a(i,j) where i and j are aliased sets.L will contain the Cholesky factor L(i,j) C:\>
### Cholesky Decomposition in GAMS
Three ways to do a Cholesky decomposition in GAMS:
1. Write A=LLT as nonlinear constraints. This has the advantage it can be part of an NLP model and maintains the factorization during the solve. Of course, this makes the NLP (much) more difficult. Example gams model.
2. Program a Cholesky algorithm directly in GAMS. Example gams model. This only works on parameters
3. Call an external Cholesky solver. This is probably better for large matrices. Example gams model.
## Tuesday, May 20, 2008
### Word -> LaTeX
This seems to work reasonably well: http://www.hj-gym.dk/~hj/writer2latex/index.html (called from OpenOffice by File|Export|LaTeX 2e).
### surprising GAMS models
I just provided this example:
if(1,$set name "hello"else$set name "world");display "%name%";
which may be surprising to non-expert GAMS users.
Here is one that may even stump hardcore GAMS modelers:
scalar s /NA/;s$(s>1) = 3.14;display s; Question: can you predict the result of the display statement? ## Monday, May 19, 2008 ### EPS and$ in gams
>When I run REAP I get a execution error division by EPS message and I
>can't figure out why. There must be something I'm missing. There are $>control statements to prevent this from happening and its never been a >problem before. I've tried placing the$ control conditions on either
>side of the assignment symbol and still get the error.
Yes, EPS is numerically the same as zero. As soon as you start calculating
with it, it is considered the same as zero. However when used as a $condition it is considered to be nonzero. So 1 scalar a /EPS/; 2 scalar b; 3 b$a=1/a;
4 display a,b;
gives:
**** Exec Error at line 3: division by eps
Indeed this fix is correct:
scalar a /EPS/;
a$(a=EPS) = 0; scalar b; b$a=1/a;
display a,b;
this will make a equal to zero if it was EPS and the expression b$a works again as expected. ## Sunday, May 18, 2008 ### Steady state probabilities in Markov Chains Some proof that people actually use examples from lineq.pdf. See this posting. ## Saturday, May 17, 2008 ### fitting Weibull growth curve New GAMS/NLS model: ## Friday, May 16, 2008 ### cmd wildcards I always thought that this would work under Windows: > touch f1.txt f2.txt f3.txt > ren f*.txt gg*.txt However it fails miserably: C:\projects\test\test1>touch f1.txt f2.txt f3.txtC:\projects\test\test1>ren f*.txt gg*.txtA duplicate file name exists, or the file cannot be found.A duplicate file name exists, or the file cannot be found.C:\projects\test\test1>dir Volume in drive C has no label. Volume Serial Number is 7563-3993 Directory of C:\projects\test\test105/16/2008 08:42 PM <DIR> .05/16/2008 08:42 PM <DIR> ..05/16/2008 08:42 PM 0 f2.txt05/16/2008 08:42 PM 0 f3.txt05/16/2008 08:42 PM 0 gg.txt3 File(s) 0 bytes2 Dir(s) 159,707,815,936 bytes freeC:\projects\test\test1> A ("somewhat" complicated) solution is presented here. ### GAMS put file names with date/time attached A question on the GAMS list was: >I would like to write the results of a GAMS program into a text file >whose name should depend on the date and time of the run. How could >that be done? > >Ex: output-05.15.08-09.39.txt >The above is just a sample and it does not have to be exactly the same. The suggestion to use myfile /filename_%system.date%_%system.time%/ does not work for several reasons: quotes are needed to handle embedded blanks and the ‘/’ characters in the system date cause problems when used inside a file name. Here is a different approach that actually works (I ran it before suggesting it): $onecho > createfile.awkBEGIN { print "file f / \"f_" strftime("%Y-%m-%d_%H.%M.%S") ".txt\"/" }$offecho$call "awk -f createfile.awk > ftime.inc" $include ftime.inc put f "hello"/; The generated include file will look like: file f / "f_2008-05-16_14.28.59.txt"/ For more info on AWK see gawk.html ### Cplex message Have not seen this error before. This does not bode well for this model. Trouble polishing MIP start. ## Thursday, May 15, 2008 ### GAMS/CPLEX bug For large MIQCP/QCP models there is an issue with GAMS and with GAMS/CPLEX. It is simply very slow in handling the quadratics. There are two places where there is likely a complexity problem (e.g. a quadratic term in run time vs linear in the linear nonzeroes): 1. In GAMS itself. Besides generating the model itself as usual, GAMS calls a program called QMAKER that exports the quadratic structure in the form of a GDX file. This QMAKER program is not suited for large models. E.g. for a very large MIQCP, GAMS can generate the model in 6.7 seconds, and then QMAKER needs 906.6 seconds to handle the quadratic terms. 2. In the GAMS/CPLEX link there is a similar problem. Setting up the quadratic constraints is very slow for large models. I don't have timings because they are not provided. I can only see one work around (only useful for testing purposes): use CONVERT to generate a scalar AMPL model and then use AMPL to generate the Cplex instance. ### A GAMS/Cplex bug I was hit again by this long standing bug. In some cases GAMS/Cplex will terminate with **** SOLVER STATUS 4 TERMINATED BY SOLVER **** MODEL STATUS 14 NO SOLUTION RETURNED **** OBJECTIVE VALUE 0.0000 but the log file nor the listing file will contain any hints about why Cplex stopped. It may be related to multiple threads being used and running out of memory, but this is just a guess. I suspect the link does not capture correctly the return code from the cplex library call. This can happen with many GAMS versions including 22.1 (march 2006), 22.2, 22.3, 22.4, 22.5, 22.6, and 22.7. ### Cutting Stock Problem Example On sci.op-research the following message was posted: > There is a worked example of how to solve a basic Cutting Stock Problem at > http://docs.google.com/View?docid=dfkkh8qj_64rrpz86db > It uses simple methods, and the open source program lp_solve. Briefly, > the solution shown is: > > * create a list of cut combinations (not permutations - there would be > too many of those to work with in most cases) > > * from this list, generate a set of linear programming (simplex) expressions, > and output them to a file > > * this file becomes the input to lp_solve, which calculates an optimal solution > > * most of the cut combinations in the solution will have a score of zero. > For those that have a score greater than zero, look them up on the list > created in step 1 A clever way to solve these problems is using a Delayed Column Generation approach. Such a model is not too difficult to code in GAMS. See http://www.amsterdamoptimization.com/pdf/colgen.pdf. A nice write up on the cutting stock problem is available from http://en.wikipedia.org/wiki/Cutting_stock_problem. The only change to the model to solve this problem was to update the data to reflect this problem (see below). The complete model is here. ### Multiplication of a continuous and a binary variable The expression $$z = x \cdot \delta$$ where $$x$$ is a continuous variable and $$\delta$$ is a binary variable can also be linearized fairly easily. The exact form depends on the bounds of $$x$$. Assume $$x$$ has lower bound $$x^{lo}$$ and upper bound $$x^{up}$$, then we can form the inequalities: $\boxed{\begin{split}\min\{0,x^{lo}\} &\le z \le \max\{0,x^{up}\} \\x^{lo} \cdot \delta &\le z \le x^{up} \cdot \delta \\x − x^{up} \cdot (1 − \delta) &\le z \le x − x^{lo} \cdot (1 − \delta)\end{split}}$ The last restriction is essentially a big-M constraint: $\boxed{x − M_1 \cdot (1 − \delta) \le z \le x + M_2 \cdot (1 − \delta)}$ where $$M_1$$, $$M_2$$ are chosen as tightly as possible while not excluding $$z=0$$ when $$\delta=0$$. For the special case $$x^{lo}=0$$ this reduces to: $\boxed{\begin{split}& z \in [0,x^{up}]\\ & z ≤ x^{up} · \delta\\ & z ≤ x\\ & z ≥ x − x^{up} · (1 − \delta)\end{split}}$ The construct $$z=x· \delta$$ can be used to model an OR condition: "$$z = 0$$ OR $$z = x$$". ## Wednesday, May 14, 2008 ### SciStat.org regression models Added a few regression models from the SciStat.org nonlinear regression datasets. They are solved by the linear and non-linear regression solvers GAMS/LS and GAMS/NLS. 1. Dugongs implemented as dugongs.gms This is actual a linear regression problem, so we solve with LS. The log/log fit is better, see also dugongs.png and dugongsplot.gms. What is a Dugong? See: http://en.wikipedia.org/wiki/Dugong. 2. Dolphin in gams formulated as dolphin.gms. If we add a constant term the problem converges to a singular solution. The problem may be related to an outlier visible in the plot dolphin.png (created by dolphinplot.gms). 3. brunhild is implemented as brunhilda.gms. The first non-linear fit. The seconds one is better. The third regression is a linearized version of model 2. 4. troutpcb is implemented as troutpcb.gms. The linear regression problem provides initial values for the non-linear regression problem. 5. chloride is implemented as chloride.gms. See also chloride.png, chlorideplot.gms 6. rabbit is implemented as rabbit.gms. See also rabbit.png, rabbitplot.gms 7. chlorine: chlorine.gms I used GNUPLOT to plot results, e.g.: ### Multiplication of binary variables I see often MINLP models where the only nonlinear expression is the multiplication of two binary variables. Such models can be linearized quite easily and are in most cases better solved as a linear MIP model. A good formulation for $$z=x_1 \cdot x_2$$ where $$x_1$$, $$x_2$$ are binary variables is as follows: $\boxed{\begin{split}0 &≤ z ≤ 1\\ z &≤ x_1\\ z &≤ x_2\\ z &≥ x_1 + x_2 − 1\end{split}}$ Notice that we relaxed $$z$$ to be continuous between 0 and 1: $$z$$ is automatically integer; a property we can use to reduce the number of discrete variable. Note that modern high performance solvers may reintroduce these variables as binary as they can exploit this in their algorithms. Another weaker formulation that is sometimes used is: $\boxed{\begin{split}z &\in \{0,1\}\\ z & ≤ \frac{x_1+x_2}{2}\\z &≥ x_1 + x_2 − 1\end{split}}$ This formulation has fewer equations but has two major disadvantages: • $$z$$ needs to be integer • it is not as tight as the first formulation. E.g. $$(x_1,x_2,z)=(0,1,\frac{1}{2})$$ is infeasible in the first formulation, while it is allowed in relaxations of the second formulation. An application of this reformulation is a QP (Quadratic Program) with binary variables: $\boxed{\begin{split} \min\> & x'Qx\\ & Ax=b\\ & x \in \{0,1\} \end{split}}$ This can be modeled as: variable y(i,j); set ij(i,j);ij(i,j)$( ord(i)<ord(j) ) = yes; y.lo(ij)=0;y.up(ij)=1; obj.. z =e= sum(i, q(i,i)*x(i)) + sum(ij(i,j), (q(i,j)+q(j,i))*y(i,j) );ymul1(ij(i,j)).. y(i,j) =L= x(i);ymul2(ij(i,j)).. y(i,j) =L= x(j);ymul3(ij(i,j)).. y(i,j) =G= x(i)+x(j)-1;
The more general product $$z = \prod_i x_i$$ with $$x_i\in \{0,1\}$$ can be linearized in the same way:
$\boxed{\begin{split} &z ≤ x_i\\ & z ≥ \sum_i x_i – (n-1)\\& z \in \{0,1\}\end{split}}$
where $$n$$ is the cardinality of the set $$I$$. Again if we want we can relax $$z$$ to $$z \in [0,1]$$.
## Tuesday, May 13, 2008
In my case I have added a $statement that makes sure the user can only run this code with GAMS 22.7: ## Monday, May 12, 2008 ###$ifthen does not nest (GAMS release 22.7 bug)
My first blog starts with a down note. I was working last night on this very large GAMS model with many model settings implemented as $set/$setglobal macros. In this context I want to nest some $ifthen constructs: $set stochastic 0$set erslivestockmodel 1$ifthen %stochastic%==1$ifthen %erslivestockmodel%==1 b(i,"Exports",k) = tempbx(k); b(i,"Season Average Price",k) = tempbp(k); b(i,"Feed Use",k) = tempbdf(k); b(i,"Food",k) = tempbdi(k); cfixed(i,"Exports",k) = 0; cfixed(i,"Season Average Price",k) = 0; cfixed(i,"Feed Use",k) = 0; cfixed(i,"Food",k) = 0;$ else$include grains.inc$ endif$endif This does not work as nesting is not allowed. You don't get an error message either indicating nesting is not allowed. In some cases you may get a possibly misleading error message, in the worst case it just behaves incorrectly. The documentation in the McCarl User Guide does not mention this limitation. Looks like$ifthen will be less useful in practice than I had hoped.
The workaround is to use a run-time if statement:
$set stochastic 0$set erslivestockmodel 1scalars stochastic /%stochastic%/ erslivestockmodel /%erslivestockmodel%/;if(stochastic=1, if(erslivestockmodel=1, b(i,"Exports",k) = tempbx(k); b(i,"Season Average Price",k) = tempbp(k); b(i,"Feed Use",k) = tempbdf(k); b(i,"Food",k) = tempbdi(k); cfixed(i,"Exports",k) = 0; cfixed(i,"Season Average Price",k) = 0; cfixed(i,"Feed Use",k) = 0; cfixed(i,"Food",k) = 0; else$include grains.inc );); This has additional advantages: all branches are syntax-checked even if they are not executed, and we check the type of the parameters using the scalar initializations. Of course we lose some of the compile time features using this approach. Another workaround would be to place the inner$ifthen in an include file:
$set stochastic 0$set erslivestockmodel 1$ifthen %stochastic%==1$include gamsbugworkaround.inc$endif Conclusion: GAMS users be aware when using nested$ifthen constructs. They are not implemented (correctly) and no error message is produced to warn you about this.
|
2021-05-09 14:29:14
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.617378830909729, "perplexity": 5240.123442526662}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988986.98/warc/CC-MAIN-20210509122756-20210509152756-00044.warc.gz"}
|
http://openstudy.com/updates/518dadbbe4b062a8d1da4868
|
## omgitsjc Group Title I need math help!!! one year ago one year ago
1. omgitsjc
6/7x+1/4
2. omgitsjc
|dw:1368239559175:dw|
3. sasogeek
what is the instruction to the question?
4. omgitsjc
5. sasogeek
that's all there is to the question?
6. omgitsjc
ya
7. sasogeek
i don't think it's possible, that's like putting apples and oranges together... you can't add a variable to a constant... are u supposed to simplify? is the 'x' multiplication or it's a letter x?
8. omgitsjc
in the book it says the answer is 24+7x/28x
9. sasogeek
yeah that is possible if you want to simplify, but you can't add those two fractions because they're constants and variables mixed. but yeah if u simplify, u get that answer
10. satellite73
$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$
11. satellite73
$\frac{6}{7x}+\frac{1}{4}=\frac{7x\times 1+4\times 6}{7x\times 4}$is a start
12. omgitsjc
got it !!!!! thank you both!!
13. sasogeek
you're welcome :) anytime. more credit to @satellite73
|
2014-10-26 09:48:54
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.48546576499938965, "perplexity": 5834.2536194686045}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414119662145.58/warc/CC-MAIN-20141024030102-00262-ip-10-16-133-185.ec2.internal.warc.gz"}
|
https://minhdo.org/notes/courses/real-analysis/5-induction.html
|
# Real Analysis
Induction
Let $$\mathbb{N} = \{1, 2, 3, \dots\}$$, the natural numbers.
# The Well-Ordering Property
The well-ordering property of $$\mathbb{N}$$: every non-empty subset of $$\mathbb{N}$$ has a least element. It can be taken as an axiom of $$\mathbb{N}$$.
# Principle of Induction
Let $$S \subset \mathbb{N}$$ such that
1. $$1 \in S$$, and
2. if $$k \in S$$ then $$k + 1 \in S$$.
Then $$S = \mathbb{N}$$.
The well-ordering property is equivalent to the principle of induction.
Proof that the well-ordering property implies the principle of induction (by contradiction). Suppose $$S$$ exists with the given properties but $$S \neq \mathbb{N}$$.
Let $$A := \mathbb{N} \setminus S$$, so it is non-empty. Then $$A$$ has a least element by the well-ordering property, call it $$n$$. Notice $$n > 1$$ because $$1 \in S$$ which implies $$1 \notin A$$.
Consider $$n - 1$$. It is not in $$A$$ because $$n$$ is the least element. So $$n - 1 \in S$$. By property (2), $$n - 1 + 1 \in S$$, which implies $$n \in S$$. This is a contradiction because $$n \in A$$.
Therefore, the well-ordering property implies the principle of induction. QED.
Proof that the principle of induction implies the well-ordering property (by induction on the number of elements in the subset). Let $$S$$ be a subset of $$\mathbb{N}$$ and $$n$$ be the number of elements in $$S$$.
For the base case where $$n = 1$$, the least element of any subset $$S$$ is the only element it has, so the base case holds.
For the inductive step, assume for all subsets $$S$$ of $$\mathbb{N}$$ with size $$n$$, each has a least element. Consider a subset $$S$$ of $$\mathbb{N}$$ with size $$n + 1$$ and a subset $$S'$$ of $$S$$ with size $$n$$. By the inductive hypothesis, $$S'$$ has a least element; call it $$a$$. Let $$b$$ is the only element in $$S$$ that is not in $$S'$$. Since $$\mathbb{N}$$ is ordered, either $$a < b$$, $$a = b$$, or $$a > b$$, by law of trichotomy. When $$a < b$$, $$a$$ is a least element of $$S$$. When $$a = b$$, $$a$$ is also a least element of $$S$$. When $$a > b$$, $$b$$ is less than all elements of $$S'$$, so $$b$$ is a least element of $$S$$.
Therefore, the principle of induction implies the well-ordering property. QED.
# Proof by Induction
Let $$P(n)$$ be statements indexed by the natural numbers $$\mathbb{N}$$. To show that $$P(n)$$ is true for all $$n$$, we must show
1. (base case) $$P(1)$$ is true, and
2. (inductive step) if $$P(k)$$ is true then $$P(k + 1)$$ is true.
Then, by the principle of induction, $$P(n)$$ is true for all $$n$$.
The statement $$P(k)$$ is true is called the inductive hypothesis.
# Strong Induction
Let $$P(n)$$ be statements indexed by the natural numbers $$\mathbb{N}$$. To show that $$P(n)$$ is true for all $$n$$, we must show
1. (base case) $$P(1)$$ is true, and
2. (inductive step) if $$P(1)$$ through $$P(k)$$ are true then $$P(k + 1)$$ is true.
Then, by the principle of strong induction, $$P(n)$$ is true for all $$n$$.
For example, every $$2^n \times 2^n$$ chessboard with one square removed can be tiled by L-shaped tiles.
Proof (by induction on $$n$$).
For base case, a $$2 \times 2$$ board
|
2019-03-23 05:08:27
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9540257453918457, "perplexity": 91.11242351325141}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202723.74/warc/CC-MAIN-20190323040640-20190323062640-00529.warc.gz"}
|
http://www.docstoc.com/docs/7897158/Physics-Formulas
|
# Physics Formulas
Document Sample
Physics Formulary
By ir. J.C.A. Wevers
c 1995, 2005 J.C.A. Wevers Dear reader,
Version: April 14, 2005
A This document contains a 108 page LTEX file which contains a lot equations in physics. It is written at advanced undergraduate/postgraduate level. It is intended to be a short reference for anyone who works with physics and often needs to look up equations.
This, and a Dutch version of this file, ([email protected]).
can be obtained from the author,
Johan Wevers
It can also be obtained on the WWW. See http://www.xs4all.nl/˜johanw/index.html, where also a Postscript version is available. If you find any errors or have any comments, please let me know. I am always open for suggestions and possible corrections to the physics formulary. This document is Copyright 1995, 1998 by J.C.A. Wevers. All rights are reserved. Permission to use, copy and distribute this unmodified document by any means and for any purpose except profit purposes is hereby granted. Reproducing this document by any means, included, but not limited to, printing, copying existing prints, publishing by electronic or other means, implies full agreement to the above non-profit-use clause, unless upon explicit prior written permission of the author. This document is provided by the author “as is”, with all its faults. Any express or implied warranties, including, but not limited to, any implied warranties of merchantability, accuracy, or fitness for any particular purpose, are disclaimed. If you use the information in this document, in any way, you do so at your own risk.
A A The Physics Formulary is made with teTEX and LTEX version 2.09. It can be possible that your LTEX version has problems compiling the file. The most probable source of problems would be the use of large bezier curves and/or emTEX specials in pictures. If you prefer the notation in which vectors are typefaced in boldface, uncomment the redefinition of the \vec command in the TEX file and recompile the file.
Johan Wevers
Contents
Contents Physical Constants 1 Mechanics 1.1 Point-kinetics in a fixed coordinate system . . . . . . . . 1.1.1 Definitions . . . . . . . . . . . . . . . . . . . . 1.1.2 Polar coordinates . . . . . . . . . . . . . . . . . 1.2 Relative motion . . . . . . . . . . . . . . . . . . . . . . 1.3 Point-dynamics in a fixed coordinate system . . . . . . . 1.3.1 Force, (angular)momentum and energy . . . . . 1.3.2 Conservative force fields . . . . . . . . . . . . . 1.3.3 Gravitation . . . . . . . . . . . . . . . . . . . . 1.3.4 Orbital equations . . . . . . . . . . . . . . . . . 1.3.5 The virial theorem . . . . . . . . . . . . . . . . 1.4 Point dynamics in a moving coordinate system . . . . . 1.4.1 Apparent forces . . . . . . . . . . . . . . . . . . 1.4.2 Tensor notation . . . . . . . . . . . . . . . . . . 1.5 Dynamics of masspoint collections . . . . . . . . . . . . 1.5.1 The centre of mass . . . . . . . . . . . . . . . . 1.5.2 Collisions . . . . . . . . . . . . . . . . . . . . . 1.6 Dynamics of rigid bodies . . . . . . . . . . . . . . . . . 1.6.1 Moment of Inertia . . . . . . . . . . . . . . . . 1.6.2 Principal axes . . . . . . . . . . . . . . . . . . . 1.6.3 Time dependence . . . . . . . . . . . . . . . . . 1.7 Variational Calculus, Hamilton and Lagrange mechanics 1.7.1 Variational Calculus . . . . . . . . . . . . . . . 1.7.2 Hamilton mechanics . . . . . . . . . . . . . . . 1.7.3 Motion around an equilibrium, linearization . . . 1.7.4 Phase space, Liouville’s equation . . . . . . . . 1.7.5 Generating functions . . . . . . . . . . . . . . . 2 Electricity & Magnetism 2.1 The Maxwell equations . . . . . . . . . . 2.2 Force and potential . . . . . . . . . . . . 2.3 Gauge transformations . . . . . . . . . . 2.4 Energy of the electromagnetic field . . . . 2.5 Electromagnetic waves . . . . . . . . . . 2.5.1 Electromagnetic waves in vacuum 2.5.2 Electromagnetic waves in matter . 2.6 Multipoles . . . . . . . . . . . . . . . . . 2.7 Electric currents . . . . . . . . . . . . . . 2.8 Depolarizing field . . . . . . . . . . . . . 2.9 Mixtures of materials . . . . . . . . . . . I 1 2 2 2 2 2 2 2 3 3 3 4 4 4 5 5 5 5 6 6 6 6 6 6 7 7 7 8 9 9 9 10 10 10 10 11 11 11 12 12
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
II
Physics Formulary by ir. J.C.A. Wevers
3 Relativity 3.1 Special relativity . . . . . . . . . . . . . . . . . . 3.1.1 The Lorentz transformation . . . . . . . . 3.1.2 Red and blue shift . . . . . . . . . . . . . 3.1.3 The stress-energy tensor and the field tensor 3.2 General relativity . . . . . . . . . . . . . . . . . . 3.2.1 Riemannian geometry, the Einstein tensor . 3.2.2 The line element . . . . . . . . . . . . . . 3.2.3 Planetary orbits and the perihelion shift . . 3.2.4 The trajectory of a photon . . . . . . . . . 3.2.5 Gravitational waves . . . . . . . . . . . . . 3.2.6 Cosmology . . . . . . . . . . . . . . . . . 4 Oscillations 4.1 Harmonic oscillations . . . . . . . . . 4.2 Mechanic oscillations . . . . . . . . . 4.3 Electric oscillations . . . . . . . . . . 4.4 Waves in long conductors . . . . . . . 4.5 Coupled conductors and transformers 4.6 Pendulums . . . . . . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
13 13 13 14 14 14 14 15 16 17 17 17 18 18 18 18 19 19 19 20 20 20 20 21 21 21 21 22 22 23 24 24 24 24 25 25 25 26 26 26 27 27 28 28 29 30 30 30 31 31 32
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
5 Waves 5.1 The wave equation . . . . . . . . . . . . . . 5.2 Solutions of the wave equation . . . . . . . . 5.2.1 Plane waves . . . . . . . . . . . . . . 5.2.2 Spherical waves . . . . . . . . . . . . 5.2.3 Cylindrical waves . . . . . . . . . . . 5.2.4 The general solution in one dimension 5.3 The stationary phase method . . . . . . . . . 5.4 Green functions for the initial-value problem . 5.5 Waveguides and resonating cavities . . . . . 5.6 Non-linear wave equations . . . . . . . . . . 6 Optics 6.1 The bending of light . . . . . . 6.2 Paraxial geometrical optics . . 6.2.1 Lenses . . . . . . . . 6.2.2 Mirrors . . . . . . . . 6.2.3 Principal planes . . . . 6.2.4 Magnification . . . . . 6.3 Matrix methods . . . . . . . . 6.4 Aberrations . . . . . . . . . . 6.5 Reflection and transmission . . 6.6 Polarization . . . . . . . . . . 6.7 Prisms and dispersion . . . . . 6.8 Diffraction . . . . . . . . . . . 6.9 Special optical effects . . . . . 6.10 The Fabry-Perot interferometer
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
. . . . . . . . . . . . . .
7 Statistical physics 7.1 Degrees of freedom . . . . . . . 7.2 The energy distribution function 7.3 Pressure on a wall . . . . . . . . 7.4 The equation of state . . . . . . 7.5 Collisions between molecules . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
Physics Formulary by ir. J.C.A. Wevers
III
7.6
Interaction between molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32 33 33 33 33 34 34 35 36 36 37 37 37 38 38 39 39 39 41 41 42 42 43 43 43 43 44 44 45 45 45 45 45 45 45 46 46 46 47 47 47 48 48 49 49 49 49 50 50 50 50 51
8 Thermodynamics 8.1 Mathematical introduction . . . . . . 8.2 Definitions . . . . . . . . . . . . . . . 8.3 Thermal heat capacity . . . . . . . . . 8.4 The laws of thermodynamics . . . . . 8.5 State functions and Maxwell relations 8.6 Processes . . . . . . . . . . . . . . . 8.7 Maximal work . . . . . . . . . . . . . 8.8 Phase transitions . . . . . . . . . . . 8.9 Thermodynamic potential . . . . . . . 8.10 Ideal mixtures . . . . . . . . . . . . . 8.11 Conditions for equilibrium . . . . . . 8.12 Statistical basis for thermodynamics . 8.13 Application to other systems . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
. . . . . . . . . . . . .
9 Transport phenomena 9.1 Mathematical introduction . . . . . . . . . . . . . 9.2 Conservation laws . . . . . . . . . . . . . . . . . . 9.3 Bernoulli’s equations . . . . . . . . . . . . . . . . 9.4 Characterising of flows by dimensionless numbers . 9.5 Tube flows . . . . . . . . . . . . . . . . . . . . . . 9.6 Potential theory . . . . . . . . . . . . . . . . . . . 9.7 Boundary layers . . . . . . . . . . . . . . . . . . . 9.7.1 Flow boundary layers . . . . . . . . . . . . 9.7.2 Temperature boundary layers . . . . . . . . 9.8 Heat conductance . . . . . . . . . . . . . . . . . . 9.9 Turbulence . . . . . . . . . . . . . . . . . . . . . 9.10 Self organization . . . . . . . . . . . . . . . . . . 10 Quantum physics 10.1 Introduction to quantum physics . . . . . . . 10.1.1 Black body radiation . . . . . . . . . . 10.1.2 The Compton effect . . . . . . . . . . 10.1.3 Electron diffraction . . . . . . . . . . . 10.2 Wave functions . . . . . . . . . . . . . . . . . 10.3 Operators in quantum physics . . . . . . . . . 10.4 The uncertainty principle . . . . . . . . . . . 10.5 The Schr¨ dinger equation . . . . . . . . . . . o 10.6 Parity . . . . . . . . . . . . . . . . . . . . . . 10.7 The tunnel effect . . . . . . . . . . . . . . . . 10.8 The harmonic oscillator . . . . . . . . . . . . 10.9 Angular momentum . . . . . . . . . . . . . . 10.10 Spin . . . . . . . . . . . . . . . . . . . . . . 10.11 The Dirac formalism . . . . . . . . . . . . . . 10.12 Atomic physics . . . . . . . . . . . . . . . . 10.12.1 Solutions . . . . . . . . . . . . . . . 10.12.2 Eigenvalue equations . . . . . . . . . 10.12.3 Spin-orbit interaction . . . . . . . . . 10.12.4 Selection rules . . . . . . . . . . . . . 10.13 Interaction with electromagnetic fields . . . . 10.14 Perturbation theory . . . . . . . . . . . . . . 10.14.1 Time-independent perturbation theory 10.14.2 Time-dependent perturbation theory .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . .
IV
Physics Formulary by ir. J.C.A. Wevers
10.15 N-particle systems 10.15.1 General . 10.15.2 Molecules 10.16 Quantum statistics
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
51 51 52 52 54 54 54 55 55 56 56 56 56 57 57 57 58 58 59 60 60 62 62 62 63 63 63 63 64 65 65 65 65 66 66 66 67 67 67 67 68 68 69 69 70 70 70
11 Plasma physics 11.1 Introduction . . . . . . . . . . . . . . . . . . 11.2 Transport . . . . . . . . . . . . . . . . . . . 11.3 Elastic collisions . . . . . . . . . . . . . . . 11.3.1 General . . . . . . . . . . . . . . . . 11.3.2 The Coulomb interaction . . . . . . . 11.3.3 The induced dipole interaction . . . . 11.3.4 The centre of mass system . . . . . . 11.3.5 Scattering of light . . . . . . . . . . . 11.4 Thermodynamic equilibrium and reversibility 11.5 Inelastic collisions . . . . . . . . . . . . . . 11.5.1 Types of collisions . . . . . . . . . . 11.5.2 Cross sections . . . . . . . . . . . . 11.6 Radiation . . . . . . . . . . . . . . . . . . . 11.7 The Boltzmann transport equation . . . . . . 11.8 Collision-radiative models . . . . . . . . . . 11.9 Waves in plasma’s . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
12 Solid state physics 12.1 Crystal structure . . . . . . . . . . . . . . . . . . . 12.2 Crystal binding . . . . . . . . . . . . . . . . . . . 12.3 Crystal vibrations . . . . . . . . . . . . . . . . . . 12.3.1 A lattice with one type of atoms . . . . . . 12.3.2 A lattice with two types of atoms . . . . . 12.3.3 Phonons . . . . . . . . . . . . . . . . . . . 12.3.4 Thermal heat capacity . . . . . . . . . . . 12.4 Magnetic field in the solid state . . . . . . . . . . . 12.4.1 Dielectrics . . . . . . . . . . . . . . . . . 12.4.2 Paramagnetism . . . . . . . . . . . . . . . 12.4.3 Ferromagnetism . . . . . . . . . . . . . . 12.5 Free electron Fermi gas . . . . . . . . . . . . . . . 12.5.1 Thermal heat capacity . . . . . . . . . . . 12.5.2 Electric conductance . . . . . . . . . . . . 12.5.3 The Hall-effect . . . . . . . . . . . . . . . 12.5.4 Thermal heat conductivity . . . . . . . . . 12.6 Energy bands . . . . . . . . . . . . . . . . . . . . 12.7 Semiconductors . . . . . . . . . . . . . . . . . . . 12.8 Superconductivity . . . . . . . . . . . . . . . . . . 12.8.1 Description . . . . . . . . . . . . . . . . . 12.8.2 The Josephson effect . . . . . . . . . . . . 12.8.3 Flux quantisation in a superconducting ring 12.8.4 Macroscopic quantum interference . . . . . 12.8.5 The London equation . . . . . . . . . . . . 12.8.6 The BCS model . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
Physics Formulary by ir. J.C.A. Wevers
V
13 Theory of groups 13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1.1 Definition of a group . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1.2 The Cayley table . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1.3 Conjugated elements, subgroups and classes . . . . . . . . . . . . . . 13.1.4 Isomorfism and homomorfism; representations . . . . . . . . . . . . 13.1.5 Reducible and irreducible representations . . . . . . . . . . . . . . . 13.2 The fundamental orthogonality theorem . . . . . . . . . . . . . . . . . . . . 13.2.1 Schur’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2.2 The fundamental orthogonality theorem . . . . . . . . . . . . . . . . 13.2.3 Character . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 The relation with quantum mechanics . . . . . . . . . . . . . . . . . . . . . 13.3.1 Representations, energy levels and degeneracy . . . . . . . . . . . . 13.3.2 Breaking of degeneracy by a perturbation . . . . . . . . . . . . . . . 13.3.3 The construction of a base function . . . . . . . . . . . . . . . . . . 13.3.4 The direct product of representations . . . . . . . . . . . . . . . . . 13.3.5 Clebsch-Gordan coefficients . . . . . . . . . . . . . . . . . . . . . . 13.3.6 Symmetric transformations of operators, irreducible tensor operators . 13.3.7 The Wigner-Eckart theorem . . . . . . . . . . . . . . . . . . . . . . 13.4 Continuous groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.4.1 The 3-dimensional translation group . . . . . . . . . . . . . . . . . . 13.4.2 The 3-dimensional rotation group . . . . . . . . . . . . . . . . . . . 13.4.3 Properties of continuous groups . . . . . . . . . . . . . . . . . . . . 13.5 The group SO(3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6 Applications to quantum mechanics . . . . . . . . . . . . . . . . . . . . . . 13.6.1 Vectormodel for the addition of angular momentum . . . . . . . . . . 13.6.2 Irreducible tensor operators, matrixelements and selection rules . . . 13.7 Applications to particle physics . . . . . . . . . . . . . . . . . . . . . . . . . 14 Nuclear physics 14.1 Nuclear forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 The shape of the nucleus . . . . . . . . . . . . . . . . . . . . . . . 14.3 Radioactive decay . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.4 Scattering and nuclear reactions . . . . . . . . . . . . . . . . . . . 14.4.1 Kinetic model . . . . . . . . . . . . . . . . . . . . . . . . . 14.4.2 Quantum mechanical model for n-p scattering . . . . . . . . 14.4.3 Conservation of energy and momentum in nuclear reactions 14.5 Radiation dosimetry . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Quantum field theory & Particle physics 15.1 Creation and annihilation operators . . . . . . . . . . . . . . . 15.2 Classical and quantum fields . . . . . . . . . . . . . . . . . . 15.3 The interaction picture . . . . . . . . . . . . . . . . . . . . . . 15.4 Real scalar field in the interaction picture . . . . . . . . . . . . 15.5 Charged spin-0 particles, conservation of charge . . . . . . . . 1 15.6 Field functions for spin- 2 particles . . . . . . . . . . . . . . . 1 15.7 Quantization of spin- 2 fields . . . . . . . . . . . . . . . . . . 15.8 Quantization of the electromagnetic field . . . . . . . . . . . . 15.9 Interacting fields and the S-matrix . . . . . . . . . . . . . . . . 15.10 Divergences and renormalization . . . . . . . . . . . . . . . . 15.11 Classification of elementary particles . . . . . . . . . . . . . . 15.12 P and CP-violation . . . . . . . . . . . . . . . . . . . . . . . . 15.13 The standard model . . . . . . . . . . . . . . . . . . . . . . . 15.13.1 The electroweak theory . . . . . . . . . . . . . . . . . 15.13.2 Spontaneous symmetry breaking: the Higgs mechanism
. . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . .
71 71 71 71 71 72 72 72 72 72 72 73 73 73 73 74 74 74 75 75 75 75 76 77 77 77 78 79 81 81 82 82 83 83 83 84 84 85 85 85 86 86 87 87 88 89 89 90 90 92 93 93 94
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
VI
Physics Formulary by ir. J.C.A. Wevers
15.13.3 Quantumchromodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.14 Path integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.15 Unification and quantum gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Astrophysics 16.1 Determination of distances . . . . 16.2 Brightness and magnitudes . . . . 16.3 Radiation and stellar atmospheres 16.4 Composition and evolution of stars 16.5 Energy production in stars . . . . The -operator
94 95 95 96 96 96 97 97 98 99 100
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
The SI units
Physical Constants
Name Number π Number e Euler’s constant Elementary charge Gravitational constant Fine-structure constant Speed of light in vacuum Permittivity of the vacuum Permeability of the vacuum (4πε0 )−1 Planck’s constant Dirac’s constant Bohr magneton Bohr radius Rydberg’s constant Electron Compton wavelength Proton Compton wavelength Reduced mass of the H-atom Stefan-Boltzmann’s constant Wien’s constant Molar gasconstant Avogadro’s constant Boltzmann’s constant Electron mass Proton mass Neutron mass Elementary mass unit Nuclear magneton Diameter of the Sun Mass of the Sun Rotational period of the Sun Radius of Earth Mass of Earth Rotational period of Earth Earth orbital period Astronomical unit Light year Parsec Hubble constant Symbol π e
n
Value 3.14159265358979323846 2.71828182845904523536
k=1
Unit
γ = lim
n→∞
1/k − ln(n)
= 0.5772156649 C m3 kg−1 s−2 m/s (def) F/m H/m Nm2 C−2 Js Js Am2 ˚ A eV m m kg Wm−2 K−4 mK J·mol−1 ·K−1 mol−1 J/K kg kg kg kg J/T m kg days m kg hours days m m m km·s−1 ·Mpc−1
e G, κ α = e2 /2hcε0 c ε0 µ0 h h = h/2π ¯ µB = e¯ /2me h a0 Ry λCe = h/me c λCp = h/mp c µH σ kW R NA k = R/NA me mp mn mu = µN
1.60217733 · 10−19 6.67259 · 10−11 ≈ 1/137 2.99792458 · 108 8.854187 · 10−12 4π · 10−7 8.9876 · 109 6.6260755 · 10−34 1.0545727 · 10−34 9.2741 · 10−24 0.52918 13.595 2.2463 · 10−12 1.3214 · 10−15 9.1045755 · 10−31
5.67032 · 10−8 2.8978 · 10−3 8.31441 6.0221367 · 1023 1.380658 · 10−23
1 12 12 m( 6 C)
D M T RA MA TA Tropical year AU lj pc H
1392 · 106 1.989 · 1030 25.38 6.378 · 106 5.976 · 1024 23.96 365.24219879 1.4959787066 · 1011 9.4605 · 1015 3.0857 · 1016 ≈ (75 ± 25)
9.1093897 · 10−31 1.6726231 · 10−27 1.674954 · 10−27 1.6605656 · 10−27 5.0508 · 10−27
Chapter 1
Mechanics
1.1 Point-kinetics in a fixed coordinate system
1.1.1 Definitions
The position r, the velocity v and the acceleration a are defined by: r = (x, y, z), v = (x, y, z), a = (¨, y , z ). ˙ ˙ ˙ x ¨ ¨ The following holds: s(t) = s0 + |v(t)|dt ; r(t) = r0 + v(t)dt ; v(t) = v0 + a(t)dt
1 When the acceleration is constant this gives: v(t) = v0 + at and s(t) = s0 + v0 t + 2 at2 . For the unit vectors in a direction ⊥ to the orbit et and parallel to it en holds:
et =
v v dr ˙ e˙t et = en ; en = = |v| ds ρ |e˙t | 1 det d2 r dϕ ; ρ= = 2 = ds ds ds |k|
For the curvature k and the radius of curvature ρ holds: k=
1.1.2 Polar coordinates
Polar coordinates are defined by: x = r cos(θ), y = r sin(θ). So, for the unit coordinate vectors holds: ˙ ˙ e˙r = θeθ , e˙θ = −θer
˙ ˙ ¨ The velocity and the acceleration are derived from: r = rer , v = rer + rθeθ , a = (¨ − r θ2 )er + (2r θ + rθ)eθ . ˙ r ˙˙
1.2 Relative motion
For the motion of a point D w.r.t. a point Q holds: rD = rQ + ω × vQ ˙ with QD = rD − rQ and ω = θ. ω2
¨ Further holds: α = θ. means that the quantity is defined in a moving system of coordinates. In a moving system holds: v = vQ + v + ω × r and a = aQ + a + α × r + 2ω × v + ω × (ω × r ) with ω × (ω × r ) = −ω 2 r n
1.3 Point-dynamics in a fixed coordinate system
1.3.1 Force, (angular)momentum and energy
Newton’s 2nd law connects the force on an object and the resulting acceleration of the object where the momentum is given by p = mv: F (r, v, t) = d(mv ) dv dm m=const dp = =m +v = ma dt dt dt dt
Chapter 1: Mechanics
3
Newton’s 3rd law is given by: Faction = −Freaction . ˙ For the power P holds: P = W = F · v. For the total energy W , the kinetic energy T and the potential energy ˙ ˙ U holds: W = T + U ; T = −U with T = 1 mv 2 . 2 The kick S is given by: S = ∆p = F dt
2 2
The work A, delivered by a force, is A =
1
F · ds =
F cos(α)ds
1
˙ The torque τ is related to the angular momentum L: τ = L = r × F ; and L = r × p = mv × r, |L| = mr2 ω. The following equation is valid: τ =− Hence, the conditions for a mechanical equilibrium are: ∂U ∂θ Fi = 0 and τi = 0.
The force of friction is usually proportional to the force perpendicular to the surface, except when the motion starts, when a threshold has to be overcome: Ffric = f · Fnorm · et .
1.3.2 Conservative force fields
A conservative force can be written as the gradient of a potential: Fcons = − U . From this follows that × F = 0. For such a force field also holds:
r1
F · ds = 0 ⇒ U = U0 −
r0
F · ds
So the work delivered by a conservative force field depends not on the trajectory covered but only on the starting and ending points of the motion.
1.3.3 Gravitation
The Newtonian law of gravitation is (in GRT one also uses κ instead of G): Fg = −G m1 m2 er r2
2
The gravitational potential is then given by V = −Gm/r. From Gauss law it then follows:
V = 4πG .
1.3.4 Orbital equations
If V = V (r) one can derive from the equations of Lagrange for φ the conservation of angular momentum: ∂L ∂V d = = 0 ⇒ (mr2 φ) = 0 ⇒ Lz = mr2 φ = constant ∂φ ∂φ dt For the radial position as a function of time can be found that: dr dt The angular equation is then:
r 2
=
2(W − V ) L2 − 2 2 m m r
φ − φ0 =
mr2 L
0
L2 2(W − V ) − 2 2 m m r
−1
dr
r −2 field
=
arccos 1 +
1 r 1 r0
+ km/L2 z
−
1 r0
If F = F (r): L =constant, if F is conservative: W =constant, if F ⊥ v then ∆T = 0 and U = 0.
4
Physics Formulary by ir. J.C.A. Wevers
Kepler’s orbital equations In a force field F = kr −2 , the orbits are conic sections with the origin of the force in one of the foci (Kepler’s 1st law). The equation of the orbit is: r(θ) = with 1 + ε cos(θ − θ0 ) , or: x2 + y 2 = ( − εx)2
2W L2 k L2 ; ε2 = 1 + 2 3 2 = 1 − ; a = = 2M 2 Gµ tot G µ Mtot a 1−ε 2W a is half the√ length of the long axis of the elliptical orbit in case the orbit is closed. Half the length of the short axis is b = a . ε is the excentricity of the orbit. Orbits with an equal ε are of equal shape. Now, 5 types of orbits are possible: = 1. k < 0 and ε = 0: a circle. 2. k < 0 and 0 < ε < 1: an ellipse. 3. k < 0 and ε = 1: a parabole. 4. k < 0 and ε > 1: a hyperbole, curved towards the centre of force. 5. k > 0 and ε > 1: a hyperbole, curved away from the centre of force. Other combinations are not possible: the total energy in a repulsive force field is always positive so ε > 1. If the surface between the orbit covered between t1 and t2 and the focus C around which the planet moves is A(t1 , t2 ), Kepler’s 2nd law is LC A(t1 , t2 ) = (t2 − t1 ) 2m Kepler’s 3rd law is, with T the period and Mtot the total mass of the system: 4π 2 T2 = 3 a GMtot
1.3.5 The virial theorem
The virial theorem for one particle is: mv · r = 0 ⇒ T = − 1 F · r = 2 The virial theorem for a collection of particles is: T = −1 2 Fi · r i + Fij · rij
1 2
r
dU dr
= 1 n U if U = − 2
k rn
particles
pairs
These propositions can also be written as: 2Ekin + Epot = 0.
1.4 Point dynamics in a moving coordinate system
1.4.1 Apparent forces
The total force in a moving coordinate system can be found by subtracting the apparent forces from the forces working in the reference frame: F = F − Fapp . The different apparent forces are given by: 1. Transformation of the origin: For = −maa 2. Rotation: Fα = −mα × r 3. Coriolis force: Fcor = −2mω × v 4. Centrifugal force: Fcf = mω 2 rn = −Fcp ; Fcp = − mv 2 er r
Chapter 1: Mechanics
5
1.4.2 Tensor notation
Transformation of the Newtonian equations of motion to xα = xα (x) gives: x ∂xα d¯β dxα = ; dt ∂ xβ dt ¯ The chain rule gives: d dxα d 2 xα d = = 2 dt dt dt dt so: x ∂xα d¯β β dt ∂x ¯ = ¯ d¯β d x ∂xα d2 xβ + β dt2 ∂x ¯ dt dt ∂xα ∂ xβ ¯
d ∂xα x x ∂ ∂xα d¯γ ∂ 2 xα d¯γ = = β γ ∂ xβ dt dt ∂ x ¯ ∂x ¯ ¯ ∂ xβ ∂ xγ dt ¯ ¯ ∂xα d2 xβ ∂ 2 xα d¯γ d 2 xα ¯ x = + β γ 2 β dt2 dt ∂x ¯ ∂ x ∂ x dt ¯ ¯ d 2 xα = Fα dt2 = Fα dxβ dxγ . dt dt d¯β x dt
Hence the Newtonian equation of motion m will be transformed into: m
dxβ dxγ d 2 xα + Γα βγ 2 dt dt dt
The apparent forces are taken from he origin to the effect side in the way Γ α βγ
1.5 Dynamics of masspoint collections
1.5.1 The centre of mass
˙ The velocity w.r.t. the centre of mass R is given by v − R. The coordinates of the centre of mass are given by: rm = mi r i mi
In a 2-particle system, the coordinates of the centre of mass are given by: R= m1 r 1 + m 2 r 2 m1 + m 2
˙ With r = r1 − r2 , the kinetic energy becomes: T = 1 Mtot R2 + 1 µr2 , with the reduced mass µ given by: 2 2 ˙ 1 1 1 = + µ m1 m2 The motion within and outside the centre of mass can be separated: ˙ ˙ Loutside = τoutside ; Linside = τinside p = mvm ; Fext = mam ; F12 = µu
1.5.2 Collisions
With collisions, where B are the coordinates of the collision and C an arbitrary other position, holds: p = mv m 1 2 is constant, and T = 2 mvm is constant. The changes in the relative velocities can be derived from: S = ∆p = µ(vaft − vbefore). Further holds ∆LC = CB × S, p S =constant and L w.r.t. B is constant.
6
Physics Formulary by ir. J.C.A. Wevers
1.6 Dynamics of rigid bodies
1.6.1 Moment of Inertia
The angular momentum in a moving coordinate system is given by: L = Iω + Ln where I is the moment of inertia with respect to a central axis, which is given by: I=
i
mi r i
2
1 ; T = Wrot = 2 ωIij ei ej = 1 Iω 2 2
or, in the continuous case: I= Further holds:
m V
r n dV =
2
r n dm m k xi xj
k
2
Li = I ij ωj ; Iii = Ii ; Iij = Iji = −
Steiner’s theorem is: Iw.r.t.D = Iw.r.t.C + m(DM )2 if axis C axis D. Object Cavern cylinder Disc, axis in plane disc through m Cavern sphere Bar, axis ⊥ through c.o.m. Rectangle, axis ⊥ plane thr. c.o.m. I I = mR2
1 I = 4 mR2 2 I = 3 mR2
Object Massive cylinder Halter Massive sphere Bar, axis ⊥ through end + b2 ) Rectangle, axis b thr. m
I I = 1 mR2 2 I = 1 µR2 2 I = 2 mR2 5 I = 1 ml2 3 I = ma2
I= I=
1 2 12 ml 1 2 12 m(a
1.6.2 Principal axes
Each rigid body has (at least) 3 principal axes which stand ⊥ to each other. For a principal axis holds: ∂I ∂I ∂I = = = 0 so Ln = 0 ∂ωx ∂ωy ∂ωz The following holds: ωk = −aijk ωi ωj with aijk = ˙ Ii − I j if I1 ≤ I2 ≤ I3 . Ik
1.6.3 Time dependence
For torque of force τ holds: ¨ τ = Iθ ; The torque T is defined by: T = F × d. d L =τ −ω×L dt
1.7 Variational Calculus, Hamilton and Lagrange mechanics
1.7.1 Variational Calculus
Starting with:
b
δ
a
L(q, q, t)dt = 0 with δ(a) = δ(b) = 0 and δ ˙
du dx
=
d (δu) dx
Chapter 1: Mechanics
7
the equations of Lagrange can be derived: d ∂L ∂L = dt ∂ qi ˙ ∂qi When there are additional conditions applying to the variational problem δJ(u) = 0 of the type K(u) =constant, the new problem becomes: δJ(u) − λδK(u) = 0.
1.7.2 Hamilton mechanics
The Lagrangian is given by: L = T (qi ) − V (qi ). The Hamiltonian is given by: H = ˙ 1 ˙ ˙ dimensions holds: L = T − U = 2 m(r2 + r2 φ2 ) − U (r, φ). dqi ∂H = ; dt ∂pi dpi ∂H =− dt ∂qi qi pi − L. In 2 ˙
If the used coordinates are canonical the Hamilton equations are the equations of motion for the system:
Coordinates are canonical if the following holds: {qi , qj } = 0, {pi , pj } = 0, {qi , pj } = δij where {, } is the Poisson bracket: ∂A ∂B ∂A ∂B {A, B} = − ∂qi ∂pi ∂pi ∂qi i The Hamiltonian of a Harmonic oscillator is given by H(x, p) = p2 /2m + 1 mω 2 x2 . With new coordinates 2 √ (θ, I), obtained by the canonical transformation x = 2I/mω cos(θ) and p = − 2Imω sin(θ), with inverse 1 θ = arctan(−p/mωx) and I = p2 /2mω + 2 mωx2 it follows: H(θ, I) = ωI. The Hamiltonian of a charged particle with charge q in an external electromagnetic field is given by: H= 1 p − qA 2m
2
+ qV
This Hamiltonian can be derived from the Hamiltonian of a free particle H = p 2 /2m with the transformations p → p − q A and H → H − qV . This is elegant from a relativistic point of view: this is equivalent to the transformation of the momentum 4-vector pα → pα − qAα . A gauge transformation on the potentials Aα corresponds with a canonical transformation, which make the Hamilton equations the equations of motion for the system.
1.7.3 Motion around an equilibrium, linearization
For natural systems around equilibrium the following equations are valid: ∂V ∂qi = 0 ; V (q) = V (0) + Vik qi qk with Vik =
0
∂2V ∂qi ∂qk
0
With T = 1 (Mik qi qk ) one receives the set of equations M q + V q = 0. If qi (t) = ai exp(iωt) is substituted, ˙ ˙ ¨ 2 this set of equations has solutions if det(V − ω 2 M ) = 0. This leads to the eigenfrequencies of the problem: aT V ak 2 k 2 ωk = T . If the equilibrium is stable holds: ∀k that ωk > 0. The general solution is a superposition if ak M a k eigenvibrations.
1.7.4 Phase space, Liouville’s equation
In phase space holds: =
i
∂ , ∂qi
i
∂ ∂pi
so
·v =
i
∂ ∂H ∂ ∂H − ∂qi ∂pi ∂pi ∂qi
8
Physics Formulary by ir. J.C.A. Wevers
If the equation of continuity, ∂t +
· ( v ) = 0 holds, this can be written as: { , H} + ∂ =0 ∂t
For an arbitrary quantity A holds: ∂A dA = {A, H} + dt ∂t Liouville’s theorem can than be written as: d = 0 ; or: dt pdq = constant
1.7.5 Generating functions
Starting with the coordinate transformation: Qi = Qi (qi , pi , t) Pi = Pi (qi , pi , t) one can derive the following Hamilton equations with the new Hamiltonian K: dQi ∂K = ; dt ∂Pi Now, a distinction between 4 cases can be made: 1. If pi qi − H = Pi Qi − K(Pi , Qi , t) − ˙ pi = dF1 (qi , Qi , t) , the coordinates follow from: dt dPi ∂K =− dt ∂Qi
∂F1 ∂F1 ∂F1 ; Pi = − ; K =H+ ∂qi ∂Qi ∂t
dF2 (qi , Pi , t) ˙ 2. If pi qi − H = −Pi Qi − K(Pi , Qi , t) + ˙ , the coordinates follow from: dt pi = ∂F2 ∂F2 ∂F2 ; Qi = ; K=H+ ∂qi ∂Pi ∂t dF3 (pi , Qi , t) , the coordinates follow from: dt
˙ 3. If −pi qi − H = Pi Qi − K(Pi , Qi , t) + ˙ qi = −
∂F3 ∂F3 ∂F3 ; Pi = − ; K =H+ ∂pi ∂Qi ∂t dF4 (pi , Pi , t) , the coordinates follow from: dt
4. If −pi qi − H = −Pi Qi − K(Pi , Qi , t) + ˙ qi = −
∂F4 ∂F4 ∂F4 ; Qi = ; K=H+ ∂pi ∂pi ∂t
The functions F1 , F2 , F3 and F4 are called generating functions.
Chapter 2
Electricity & Magnetism
2.1 The Maxwell equations
The classical electromagnetic field can be described by the Maxwell equations. Those can be written both as differential and integral equations: (D · n )d2 A = Qfree,included (B · n )d2 A = 0 E · ds = − dΦ dt dΨ dt (B · n )d2 A. np2 0 3ε0 kT · D = ρfree ·B =0 ×E =− ∂B ∂t ∂D ∂t
H · ds = Ifree,included + For the fluxes holds: Ψ = (D · n )d2 A, Φ =
× H = Jfree +
The electric displacement D, polarization P and electric field strength E depend on each other according to: D = ε0 E + P = ε0 εr E, P = p0 /Vol, εr = 1 + χe , with χe =
The magnetic field strength H, the magnetization M and the magnetic flux density B depend on each other according to: B = µ0 (H + M) = µ0 µr H, M = m/Vol, µr = 1 + χm , with χm = µ0 nm2 0 3kT
2.2 Force and potential
The force and the electric field between 2 point charges are given by: F12 = F Q1 Q2 er ; E = 4πε0 εr r2 Q
The Lorentzforce is the force which is felt by a charged particle that moves through a magnetic field. The origin of this force is a relativistic transformation of the Coulomb force: FL = Q(v × B ) = l(I × B ). The magnetic field in point P which results from an electric current is given by the law of Biot-Savart, also known als the law of Laplace. In here, dl I and r points from dl to P : dBP = µ0 I dl × e r 4πr2
If the current is time-dependent one has to take retardation into account: the substitution I(t) → I(t − r/c) has to be applied.
2
The potentials are given by: V12 = −
1
1 E · ds and A = 2 B × r.
10
Physics Formulary by ir. J.C.A. Wevers
Here, the freedom remains to apply a gauge transformation. The fields can be derived from the potentials as follows: ∂A E=− V − , B = ×A ∂t Further holds the relation: c2 B = v × E.
2.3 Gauge transformations
The potentials of the electromagnetic fields transform as follows when a gauge transformation is applied: A =A− f ∂f V =V + ∂t
so the fields E and B do not change. This results in a canonical transformation of the Hamiltonian. Further, the freedom remains to apply a limiting condition. Two common choices are: 1 ∂V ρ 1. Lorentz-gauge: · A + 2 = 0. This separates the differential equations for A and V : 2V = − , c ∂t ε0 2A = −µ0 J. 2. Coulomb gauge: · A = 0. If ρ = 0 and J = 0 holds V = 0 and follows A from 2A = 0.
2.4 Energy of the electromagnetic field
The energy density of the electromagnetic field is: dW = w = HdB + EdD dVol The energy density can be expressed in the potentials and currents as follows: wmag =
1 2
J · A d3 x , wel =
1 2
ρV d3 x
2.5 Electromagnetic waves
2.5.1 Electromagnetic waves in vacuum
The wave equation 2Ψ(r, t) = −f (r, t) has the general solution, with c = (ε 0 µ0 )−1/2 : Ψ(r, t) = f (r, t − |r − r |/c) 3 d r 4π|r − r |
If this is written as: J(r, t) = J(r ) exp(−iωt) and A(r, t) = A(r ) exp(−iωt) with: A(r ) = µ 4π J(r ) 1 exp(ik|r − r |) 3 d r , V (r ) = |r − r | 4πε dP k2 = dΩ 32π 2 ε0 c ρ(r ) exp(ik|r − r |) 3 d r |r − r | r:
A derivation via multipole expansion will show that for the radiated energy holds, if d, λ
2
J⊥ (r )eik·r d3 r
The energy density of the electromagnetic wave of a vibrating dipole at a large distance is: w = ε0 E 2 = p2 sin2 (θ)ω 4 0 sin2 (kr − ωt) , 16π 2 ε0 r2 c4 w
t
=
p2 sin2 (θ)ω 4 ck 4 |p |2 0 , P = 2 ε r 2 c4 32π 0 12πε0
The radiated energy can be derived from the Poynting vector S: S = E × H = cW ev . The irradiance is the time-averaged of the Poynting vector: I = |S | t . The radiation pressure ps is given by ps = (1 + R)|S |/c, where R is the coefficient of reflection.
Chapter 2: Electricity & Magnetism
11
2.5.2 Electromagnetic waves in matter
The wave equations in matter, with cmat = (εµ)−1/2 the lightspeed in matter, are:
2
− εµ
∂2 µ ∂ − ∂t2 ρ ∂t
E =0,
2
− εµ
∂2 µ ∂ − ∂t2 ρ ∂t
B=0
give, after substitution of monochromatic plane waves: E = E exp(i(k ·r −ωt)) and B = B exp(i(k ·r −ωt)) the dispersion relation: iµω k 2 = εµω 2 + ρ The first term arises from the displacement current, the second from the conductance current. If k is written in the form k := k + ik it follows that: k =ω
1 2 εµ
1+
1+
1 and k = ω (ρεω)2
1 2 εµ
−1 +
1+
1 (ρεω)2
This results in a damped wave: E = E exp(−k n · r ) exp(i(k n · r − ωt)). If the material is a good conductor, µω . the wave vanishes after approximately one wavelength, k = (1 + i) 2ρ
2.6 Multipoles
Because 1 1 = |r − r | r
∞ 0
r r
l
Pl (cos θ) the potential can be written as: V =
Q 4πε
n
kn rn
For the lowest-order terms this results in: • Monopole: l = 0, k0 = • Dipole: l = 1, k1 = • Quadrupole: l = 2, k2 = ρdV
r cos(θ)ρdV
1 2 i 2 2 (3zi − ri )
1. The electric dipole: dipole moment: p = Qle, where e goes from ⊕ to , and F = (p · W = −p · Eout . Q 3p · r Electric field: E ≈ − p . The torque is: τ = p × Eout 4πεr3 r2 √ 2. The magnetic dipole: dipole moment: if r A: µ = I × (Ae⊥ ), F = (µ · )Bout 2 mv⊥ |µ| = , W = −µ × Bout 2B −µ 3µ · r − µ . The moment is: τ = µ × Bout Magnetic field: B = 4πr3 r2
)Eext , and
2.7 Electric currents
The continuity equation for charge is: ∂ρ + ∂t I= · J = 0. The electric current is given by: dQ = dt (J · n )d2 A
For most conductors holds: J = E/ρ, where ρ is the resistivity.
12
Physics Formulary by ir. J.C.A. Wevers
dΦ If the flux enclosed by a conductor changes this results in an induced voltage V ind = −N . If the current dt flowing through a conductor changes, this results in a self-inductance which opposes the original change: dI Vselfind = −L . If a conductor encloses a flux Φ holds: Φ = LI. dt µN I The magnetic induction within a coil is approximated by: B = √ where l is the length, R the radius 2 + 4R2 l and N the number of coils. The energy contained within a coil is given by W = 1 LI 2 and L = µN 2 A/l. 2 The capacity is defined by: C = Q/V . For a capacitor holds: C = ε0 εr A/d where d is the distance between the plates and A the surface of one plate. The electric field strength between the plates is E = σ/ε 0 = Q/ε0 A where σ is the surface charge. The accumulated energy is given by W = 1 CV 2 . The current through a 2 dV . capacity is given by I = −C dt For most PTC resistors holds approximately: R = R0 (1 + αT ), where R0 = ρl/A. For a NTC holds: R(T ) = C exp(−B/T ) where B and C depend only on the material. If a current flows through two different, connecting conductors x and y, the contact area will heat up or cool down, depending on the direction of the current: the Peltier effect. The generated or removed heat is given by: W = Πxy It. This effect can be amplified with semiconductors. The thermic voltage between 2 metals is given by: V = γ(T − T0 ). For a Cu-Konstantane connection holds: γ ≈ 0.2 − 0.7 mV/K. In an electrical net with only stationary currents, Kirchhoff’s equations apply: for a knot holds: along a closed path holds: Vn = In Rn = 0. In = 0,
2.8 Depolarizing field
If a dielectric material is placed in an electric or magnetic field, the field strength within and outside the material will change because the material will be polarized or magnetized. If the medium has an ellipsoidal shape and one of the principal axes is parallel with the external field E0 or B0 then the depolarizing is field homogeneous. Edep = Emat − E0 = − NP ε0
Hdep = Hmat − H0 = −N M N is a constant depending only on the shape of the object placed in the field, with 0 ≤ N ≤ 1. For a few 1 limiting cases of an ellipsoid holds: a thin plane: N = 1, a long, thin bar: N = 0, a sphere: N = 3 .
2.9 Mixtures of materials
The average electric displacement in a material which is inhomogenious on a mesoscopic scale is given by: −1 φ2 (1 − x) where x = ε1 /ε2 . For a sphere holds: Φ = D = εE = ε∗ E where ε∗ = ε1 1 − Φ(ε∗ /ε2 ) 1 2 3 + 3 x. Further holds: φi εi
−1
i
≤ ε∗ ≤
φi ε i
i
Chapter 3
Relativity
3.1 Special relativity
3.1.1 The Lorentz transformation
The Lorentz transformation (x , t ) = (x (x, t), t (x, t)) leaves the wave equation invariant if c is invariant: ∂2 ∂2 ∂2 1 ∂2 ∂2 ∂2 ∂2 1 ∂2 + 2+ 2− 2 2 = + + − 2 2 ∂x2 ∂y ∂z c ∂t ∂x 2 ∂y 2 ∂z 2 c ∂t This transformation can also be found when ds2 = ds 2 is demanded. The general form of the Lorentz transformation is given by: x =x+ where γ= (γ − 1)(x · v )v x·v − γvt , t = γ t − 2 |v|2 c 1
2 1 − v2 c
The velocity difference v between two observers transforms according to: v = γ 1− v1 · v2 c2
−1
v2 + (γ − 1)
v1 · v2 2 v1 − γv1 v1
If the velocity is parallel to the x-axis, this becomes y = y, z = z and: x = γ(x − vt) , x = γ(x + vt ) xv xv t =γ t− 2 , t=γ t + 2 c c If v = vex holds: px = γ px −
, v =
v2 − v 1 v1 v2 1− 2 c
βW c
, W = γ(W − vpx )
With β = v/c the electric field of a moving charge is given by: E= Q (1 − β 2 )er 4πε0 r2 (1 − β 2 sin2 (θ))3/2
The electromagnetic field transforms according to: E = γ(E + v × B ) , B = γ B− v×E c2
Length, mass and time transform according to: ∆tr = γ∆t0 , mr = γm0 , lr = l0 /γ, with 0 the quantities in a co-moving reference frame and r the quantities in a frame moving with velocity v w.r.t. it. The proper time τ is defined as: dτ 2 = ds2 /c2 , so ∆τ = ∆t/γ. For energy and momentum holds: W = mr c2 = γW0 ,
14
Physics Formulary by ir. J.C.A. Wevers
W 2 = m2 c4 + p2 c2 . p = mr v = γm0 v = W v/c2 , and pc = W β where β = v/c. The force is defined by 0 F = dp/dt. 4-vectors have the property that their modulus is independent of the observer: their components can change after a coordinate transformation but not their modulus. The difference of two 4-vectors transforms also as dxα . The relation with the “common” velocity a 4-vector. The 4-vector for the velocity is given by U α = dτ i i α i u := dx /dt is: U = (γu , icγ). For particles with nonzero restmass holds: U α Uα = −c2 , for particles with zero restmass (so with v = c) holds: U α Uα = 0. The 4-vector for energy and momentum is given by: pα = m0 U α = (γpi , iW/c). So: pα pα = −m2 c2 = p2 − W 2 /c2 . 0
3.1.2 Red and blue shift
There are three causes of red and blue shifts: 1. Motion: with ev · er = cos(ϕ) follows: ∆f κM = 2. f rc f v cos(ϕ) . =γ 1− f c This can give both red- and blueshift, also ⊥ to the direction of motion.
2. Gravitational redshift:
3. Redshift because the universe expands, resulting in e.g. the cosmic background radiation: R0 λ0 = . λ1 R1
3.1.3 The stress-energy tensor and the field tensor
The stress-energy tensor is given by: Tµν = ( c2 + p)uµ uν + pgµν + The conservation laws can than be written as:
νT µν
1 α Fµα Fν + 1 gµν F αβ Fαβ 4 c2
= 0. The electromagnetic field tensor is given by:
Fαβ =
∂Aβ ∂Aα − α ∂x ∂xβ
with Aµ := (A, iV /c) and Jµ := (J, icρ). The Maxwell equations can than be written as: ∂ν F µν = µ0 J µ , ∂λ Fµν + ∂µ Fνλ + ∂ν Fλµ = 0 The equations of motion for a charged particle in an EM field become with the field tensor: dpα = qFαβ uβ dτ
3.2 General relativity
3.2.1 Riemannian geometry, the Einstein tensor
The basic principles of general relativity are: 1. The geodesic postulate: free falling particles move along geodesics of space-time with the proper time τ or arc length s as parameter. For particles with zero rest mass (photons), the use of a free parameter is required because for them holds ds = 0. From δ ds = 0 the equations of motion can be derived: d 2 xα dxβ dxγ + Γα =0 βγ 2 ds ds ds
Chapter 3: Relativity
15
2. The principle of equivalence: inertial mass ≡ gravitational mass ⇒ gravitation is equivalent with a curved space-time were particles move along geodesics. 3. By a proper choice of the coordinate system it is possible to make the metric locally flat in each point xi : gαβ (xi ) = ηαβ :=diag(−1, 1, 1, 1).
µ The Riemann tensor is defined as: Rναβ T ν := α β T µ − by j ai = ∂j ai + Γi ak and j ai = ∂j ai − Γk ak . Here, ij jk β αT µ
, where the covariant derivative is given
Γi = jk
g il 2
∂g k ∂glj ∂glk + − jl k j ∂x ∂x ∂x
, for Euclidean spaces this reduces to: Γi = jk
∂ 2 xl ∂xi ¯ , j ∂xk ∂ xl ∂x ¯
µ µ σ σ µ are the Christoffel symbols. For a second-order tensor holds: [ α , β ]Tν = Rσαβ Tν + Rναβ Tσ , k ai = j j il i l i i l l l ij ij i lj ∂k aj − Γkj al + Γkl aj , k aij = ∂k aij − Γki alj − Γkj ajl and k a = ∂k a + Γkl a + Γkl a . The following α holds: Rβµν = ∂µ Γα − ∂ν Γα + Γα Γσ − Γα Γσ . σµ βν σν βµ βν βµ µ The Ricci tensor is a contraction of the Riemann tensor: Rαβ := Rαµβ , which is symmetric: Rαβ = Rβα . The Bianchi identities are: λ Rαβµν + ν Rαβλµ + µ Rαβνλ = 0.
α The Einstein tensor is given by: Gαβ := Rαβ − 1 g αβ R, where R := Rα is the Ricci scalar, for which 2 holds: β Gαβ = 0. With the variational principle δ (L(gµν ) − Rc2 /16πκ) |g|d4 x = 0 for variations gµν → gµν + δgµν the Einstein field equations can be derived:
Gαβ =
8πκ Tαβ c2
, which can also be written as Rαβ =
8πκ µ 1 (Tαβ − 2 gαβ Tµ ) c2
For empty space this is equivalent to Rαβ = 0. The equation Rαβµν = 0 has as only solution a flat space. The Einstein equations are 10 independent equations, which are of second order in g µν . From this, the Laplace equation from Newtonian gravitation can be derived by stating: g µν = ηµν + hµν , where |h| 1. In the stationary case, this results in 2 h00 = 8πκ /c2 . The most general form of the field equations is: Rαβ − 1 gαβ R + Λgαβ = 2 8πκ Tαβ c2 where Λ is the cosmological constant. This constant plays a role in inflatory models of the universe.
3.2.2 The line element
The metric tensor in an Euclidean space is given by: gij =
k 2 µ ν
∂ xk ∂ xk ¯ ¯ . ∂xi ∂xj
In general holds: ds = gµν dx dx . In special relativity this becomes ds2 = −c2 dt2 + dx2 + dy 2 + dz 2 . This metric, ηµν :=diag(−1, 1, 1, 1), is called the Minkowski metric. The external Schwarzschild metric applies in vacuum outside a spherical mass distribution, and is given by: ds2 = −1 + 2m r c2 dt2 + 1 − 2m r
−1
dr2 + r2 dΩ2
Here, m := M κ/c2 is the geometrical mass of an object with mass M , and dΩ2 = dθ2 + sin2 θdϕ2 . This metric is singular for r = 2m = 2κM/c2 . If an object is smaller than its event horizon 2m, that implies that its escape velocity is > c, it is called a black hole. The Newtonian limit of this metric is given by: ds2 = −(1 + 2V )c2 dt2 + (1 − 2V )(dx2 + dy 2 + dz 2 ) where V = −κM/r is the Newtonian gravitation potential. In general relativity, the components of g µν are associated with the potentials and the derivatives of gµν with the field strength. The Kruskal-Szekeres coordinates are used to solve certain problems with the Schwarzschild metric near r = 2m. They are defined by:
16
Physics Formulary by ir. J.C.A. Wevers
• r > 2m:
u = u = v = v =
r r cosh − 1 exp 2m 4m r r sinh − 1 exp 2m 4m 1− 1− r r sinh exp 2m 4m r r cosh exp 2m 4m
t 4m t 4m t 4m t 4m
• r < 2m:
• r = 2m: here, the Kruskal coordinates are singular, which is necessary to eliminate the coordinate singularity there. The line element in these coordinates is given by: ds2 = − 32m3 −r/2m 2 e (dv − du2 ) + r2 dΩ2 r
The line r = 2m corresponds to u = v = 0, the limit x0 → ∞ with u = v and x0 → −∞ with u = −v. The Kruskal coordinates are only singular on the hyperbole v 2 − u2 = 1, this corresponds with r = 0. On the line dv = ±du holds dθ = dϕ = ds = 0. For the metric outside a rotating, charged spherical mass the Newman metric applies: ds2 = 1− 2mr − e2 r2 + a2 cos2 θ c2 dt2 − r2 + a2 cos2 θ r2 − 2mr + a2 − e2 sin2 θdϕ2 + dr2 − (r2 + a2 cos2 θ)dθ2 − sin2 θ(dϕ)(cdt)
r 2 + a2 +
(2mr − e2 )a2 sin2 θ r2 + a2 cos2 θ
2a(2mr − e2 ) r2 + a2 cos2 θ
where m = κM/c2 , a = L/M c and e = κQ/ε0 c2 . √ A rotating charged black hole has an event horizon with RS = m + m2 − a2 − e2 .
Near rotating black holes frame dragging occurs because gtϕ = 0. For the Kerr metric (e = 0, a = 0) then √ follows that within the surface RE = m + m2 − a2 cos2 θ (de ergosphere) no particle can be at rest.
3.2.3 Planetary orbits and the perihelion shift
To find a planetary orbit, the variational problem δ ds = 0 has to be solved. This is equivalent to the problem δ ds2 = δ gij dxi dxj = 0. Substituting the external Schwarzschild metric yields for a planetary orbit: du dϕ d2 u +u dϕ2 = du m 3mu + 2 dϕ h
where u := 1/r and h = r 2 ϕ =constant. The term 3mu is not present in the classical solution. This term can ˙ κM h2 in the classical case also be found from a potential V (r) = − 1+ 2 . r r The orbital equation gives r =constant as solution, or can, after dividing by du/dϕ, be solved with perturbation theory. In zeroth order, this results in an elliptical orbit: u0 (ϕ) = A + B cos(ϕ) with A = m/h2 and B an arbitrary constant. In first order, this becomes: u1 (ϕ) = A + B cos(ϕ − εϕ) + ε A + B2 B2 − cos(2ϕ) 2A 6A
where ε = 3m2 /h2 is small. The perihelion of a planet is the point for which r is minimal, or u maximal. This is the case if cos(ϕ − εϕ) = 0 ⇒ ϕ ≈ 2πn(1 + ε). For the perihelion shift then follows: ∆ϕ = 2πε = 6πm2 /h2 per orbit.
Chapter 3: Relativity
17
3.2.4 The trajectory of a photon
For the trajectory of a photon (and for each particle with zero restmass) holds ds 2 = 0. Substituting the external Schwarzschild metric results in the following orbital equation: du dϕ d2 u + u − 3mu dϕ2 =0
3.2.5 Gravitational waves
Starting with the approximation gµν = ηµν + hµν for weak gravitational fields and the definition hµν = 1 hµν − 2 ηµν hα it follows that 2hµν = 0 if the gauge condition ∂hµν /∂xν = 0 is satisfied. From this, it α follows that the loss of energy of a mechanical system, if the occurring velocities are c and for wavelengths the size of the system, is given by: dE G =− 5 dt 5c with Qij = d3 Qij dt3
2
i,j
(xi xj − 1 δij r2 )d3 x the mass quadrupole moment. 3
3.2.6 Cosmology
If for the universe as a whole is assumed: 1. There exists a global time coordinate which acts as x0 of a Gaussian coordinate system, 2. The 3-dimensional spaces are isotrope for a certain value of x 0 , 3. Each point is equivalent to each other point for a fixed x0 . then the Robertson-Walker metric can be derived for the line element: ds2 = −c2 dt2 + R2 (t) kr2 1− 2 4r0 (dr2 + r2 dΩ2 )
2 r0
For the scalefactor R(t) the following equations can be derived: ¨ ˙ ˙ 2R R2 + kc2 8πκp 8πκ Λ R2 + kc2 + = − 2 + Λ and = + 2 2 R R c R 3 3 where p is the pressure and parameter q: the density of the universe. If Λ = 0 can be derived for the deceleration q=− ¨ RR 4πκ = ˙2 3H 2 R
˙ where H = R/R is Hubble’s constant. This is a measure of the velocity with which galaxies far away are moving away from each other, and has the value ≈ (75 ± 25) km·s −1 ·Mpc−1 . This gives 3 possible conditions for the universe (here, W is the total amount of energy in the universe):
1 1. Parabolical universe: k = 0, W = 0, q = 2 . The expansion velocity of the universe → 0 if t → ∞. The hereto related critical density is c = 3H 2 /8πκ.
2. Hyperbolical universe: k = −1, W < 0, q < positive forever.
1 2.
The expansion velocity of the universe remains
1 3. Elliptical universe: k = 1, W > 0, q > 2 . The expansion velocity of the universe becomes negative after some time: the universe starts collapsing.
Chapter 4
Oscillations
4.1 Harmonic oscillations
ˆ ˆ The general form of a harmonic oscillation is: Ψ(t) = Ψei(ωt±ϕ) ≡ Ψ cos(ωt ± ϕ), ˆ where Ψ is the amplitude. A superposition of several harmonic oscillations with the same frequency results in another harmonic oscillation: ˆ ˆ Ψi cos(αi ± ωt) = Φ cos(β ± ωt)
i
with: tan(β) =
i
ˆ Ψi sin(αi ) ˆ Ψi cos(αi )
i
ˆ and Φ2 =
i
ˆ Ψ2 + 2 i
j>i i
ˆ ˆ Ψi Ψj cos(αi − αj )
For harmonic oscillations holds:
x(t)dt =
dn x(t) x(t) and = (iω)n x(t). iω dtn
4.2 Mechanic oscillations
For a construction with a spring with constant C parallel to a damping k which is connected to a mass M , to ˆ which a periodic force F (t) = F cos(ωt) is applied holds the equation of motion m¨ = F (t) − k x − Cx. x ˙ 2 With complex amplitudes, this becomes −mω 2 x = F − Cx − ikωx. With ω0 = C/m follows: x= where δ =
2 m(ω0
ω0 ω − . The quantity Z = F/x is called the impedance of the system. The quality of the system ˙ ω0 √ ω Cm is given by Q = . k The frequency with minimal |Z| is called velocity resonance frequency. This is equal to ω 0 . In the resonance √ curve |Z|/ Cm is plotted against ω/ω0 . The width of this curve is characterized by the points where |Z(ω)| = √ |Z(ω0 )| 2. In these points holds: R = X and δ = ±Q−1 , and the width is 2∆ωB = ω0 /Q. The stiffness of an oscillating system is given by F/x. The amplitude resonance frequency ω A is the frequency where iωZ is minimal. This is the case for ωA = ω0
1 1 − 2 Q2 .
F F , and for the velocity holds: x = √ ˙ 2 ) + ikω −ω i Cmδ + k
The damping frequency ωD is a measure for the time in which an oscillating system comes to rest. It is given 1 . A weak damped oscillation (k 2 < 4mC) dies out after TD = 2π/ωD . For a critical by ωD = ω0 1 − 4Q2 damped oscillation (k 2 = 4mC) holds ωD = 0. A strong damped oscillation (k 2 > 4mC) drops like (if k2 4mC) x(t) ≈ x0 exp(−t/τ ).
4.3 Electric oscillations
The impedance is given by: Z = R + iX. The phase angle is ϕ := arctan(X/R). The impedance of a resistor is R, of a capacitor 1/iωC and of a self inductor iωL. The quality of a coil is Q = ωL/R. The total impedance in case several elements are positioned is given by:
Chapter 4: Oscillations
19
1. Series connection: V = IZ, Ztot =
i
Zi , Ltot =
i
Li ,
1 = Ctot
i
1 Z0 , Q= , Z = R(1 + iQδ) Ci R
2. parallel connection: V = IZ, 1 = Ztot Here, Z0 = 1 1 , = Zi Ltot 1 , Ctot = Li Ci , Q =
i
i
i
R R , Z= Z0 1 + iQδ
1 L and ω0 = √ . C LC
ˆ ˆ The power given by a source is given by P (t) = V (t) · I(t), so P t = Veff Ieff cos(∆φ) 1ˆˆ 1 ˆ2 1 ˆ2 = 2 V I cos(φv − φi ) = 2 I Re(Z) = 2 V Re(1/Z), where cos(∆φ) is the work factor.
4.4 Waves in long conductors
These cables are in use for signal transfer, e.g. coax cable. For them holds: Z 0 = The transmission velocity is given by v = dx dx . dL dC dL dx . dx dC
4.5 Coupled conductors and transformers
For two coils enclosing each others flux holds: if Φ12 is the part of the flux originating from I2 through coil 2 which is enclosed by coil 1, than holds Φ12 = M12 I2 , Φ21 = M21 I1 . For the coefficients of mutual induction Mij holds: N2 Φ 2 N1 Φ 1 = ∼ N 1 N2 M12 = M21 := M = k L1 L2 = I2 I1 where 0 ≤ k ≤ 1 is the coupling factor. For a transformer is k ≈ 1. At full load holds: I2 iωM V1 = =− ≈− V2 I1 iωL2 + Rload N1 L1 =− L2 N2
4.6 Pendulums
The oscillation time T = 1/f , and for different types of pendulums is given by: • Oscillating spring: T = 2π • Physical pendulum: T = 2π • Torsion pendulum: T = 2π m/C if the spring force is given by F = C · ∆l. I/τ with τ the moment of force and I the moment of inertia. I/κ with κ = 2lm the constant of torsion and I the moment of inertia. πr4 ∆ϕ
• Mathematical pendulum: T = 2π lum.
l/g with g the acceleration of gravity and l the length of the pendu-
Chapter 5
Waves
5.1 The wave equation
The general form of the wave equation is: 2u = 0, or:
2
u−
∂2u ∂2u ∂2u 1 ∂2u 1 ∂2u = + 2 + 2 − 2 2 =0 2 ∂t2 2 v ∂x ∂y ∂z v ∂t
where u is the disturbance and v the propagation velocity. In general holds: v = f λ. By definition holds: kλ = 2π and ω = 2πf . In principle, there are two types of waves: 1. Longitudinal waves: for these holds k 2. Transversal waves: for these holds k v u.
v ⊥ u.
The phase velocity is given by vph = ω/k. The group velocity is given by: vg = dvph k dn dω = vph + k = vph 1 − dk dk n dk
where n is the refractive index of the medium. If vph does not depend on ω holds: vph = vg . In a dispersive medium it is possible that vg > vph or vg < vph , and vg · vf = c2 . If one wants to transfer information with a wave, e.g. by modulation of an EM wave, the information travels with the velocity at with a change in the electromagnetic field propagates. This velocity is often almost equal to the group velocity. For some media, the propagation velocity follows from: • Pressure waves in a liquid or gas: v = κ/ , where κ is the modulus of compression. γp/ = γRT /M. E/
• For pressure waves in a gas also holds: v =
• Pressure waves in a thin solid bar with diameter << λ: v = • waves in a string: v = Fspan l/m
• Surface waves on a liquid: v =
2πh gλ 2πγ + tanh 2π λ λ where h is the depth of the liquid and γ the surface tension. If h
λ holds: v ≈
√
gh.
5.2 Solutions of the wave equation
5.2.1 Plane waves
In n dimensions a harmonic plane wave is defined by:
n
u(x, t) = 2n u cos(ωt) ˆ
i=1
sin(ki xi )
Chapter 5: Waves
21
If waves reflect at the end of a spring this will result in a change in phase. A fixed end gives a phase change of π/2 to the reflected wave, with boundary condition u(l) = 0. A lose end gives no change in the phase of the reflected wave, with boundary condition (∂u/∂x)l = 0. If an observer is moving w.r.t. the wave with a velocity vobs , he will observe a change in frequency: the f vf − vobs Doppler effect. This is given by: = . f0 vf
The equation for a harmonic traveling plane wave is: u(x, t) = u cos( k · x ± ωt + ϕ) ˆ
5.2.2 Spherical waves
When the situation is spherical symmetric, the homogeneous wave equation is given by: 1 ∂ 2 (ru) ∂ 2 (ru) − =0 v 2 ∂t2 ∂r2 with general solution: u(r, t) = C1 f (r − vt) g(r + vt) + C2 r r
5.2.3 Cylindrical waves
When the situation has a cylindrical symmetry, the homogeneous wave equation becomes: 1 ∂2u 1 ∂ − v 2 ∂t2 r ∂r r ∂u ∂r =0
This is a Bessel equation, with solutions which can be written as Hankel functions. For sufficient large values of r these are approximated by: u ˆ u(r, t) = √ cos(k(r ± vt)) r
5.2.4 The general solution in one dimension
Starting point is the equation: ∂m ∂ 2 u(x, t) = bm m 2 ∂t ∂x m=0
N
u(x, t)
where bm ∈ I Substituting u(x, t) = Aei(kx−ωt) gives two solutions ωj = ωj (k) as dispersion relations. R. The general solution is given by:
∞
u(x, t) =
−∞
a(k)ei(kx−ω1 (k)t) + b(k)ei(kx−ω2 (k)t) dk
Because in general the frequencies ωj are non-linear in k there is dispersion and the solution cannot be written any more as a sum of functions depending only on x ± vt: the wave front transforms.
5.3 The stationary phase method
Usually the Fourier integrals of the previous section cannot be calculated exactly. If ω j (k) ∈ I the stationary R phase method can be applied. Assuming that a(k) is only a slowly varying function of k, one can state that the parts of the k-axis where the phase of kx − ω(k)t changes rapidly will give no net contribution to the integral because the exponent oscillates rapidly there. The only areas contributing significantly to the integral are areas d with a stationary phase, determined by (kx − ω(k)t) = 0. Now the following approximation is possible: dk
∞ N
−∞
a(k)ei(kx−ω(k)t) dk ≈
2π
d2 ω(ki ) 2 dki
i=1
1 exp −i 4 π + i(ki x − ω(ki )t)
22
Physics Formulary by ir. J.C.A. Wevers
5.4 Green functions for the initial-value problem
This method is preferable if the solutions deviate much from the stationary solutions, like point-like excitations. Starting with the wave equation in one dimension, with 2 = ∂ 2 /∂x2 holds: if Q(x, x , t) is the solution with ∂Q(x, x , 0) = 0, and P (x, x , t) the solution with initial values initial values Q(x, x , 0) = δ(x − x ) and ∂t ∂P (x, x , 0) P (x, x , 0) = 0 and = δ(x − x ), then the solution of the wave equation with arbitrary initial ∂t ∂u(x, 0) is given by: conditions f (x) = u(x, 0) and g(x) = ∂t
∞ ∞
u(x, t) =
−∞
f (x )Q(x, x , t)dx +
−∞
g(x )P (x, x , t)dx
P and Q are called the propagators. They are defined by: Q(x, x , t) = P (x, x , t) =
1 2 [δ(x
1 2v 0
− x − vt) + δ(x − x + vt)] if |x − x | > vt if |x − x | < vt
Further holds the relation: Q(x, x , t) =
∂P (x, x , t) ∂t
5.5 Waveguides and resonating cavities
The boundary conditions for a perfect conductor can be derived from the Maxwell equations. If n is a unit vector ⊥ the surface, pointed from 1 to 2, and K is a surface current density, than holds: n · ( D2 − D1 ) = σ n · ( B2 − B1 ) = 0 n × ( E2 − E1 ) = 0 n × ( H2 − H1 ) = K
In a waveguide holds because of the cylindrical symmetry: E(x, t) = E(x, y)ei(kz−ωt) and B(x, t) = B(x, y)ei(kz−ωt) . From this one can now deduce that, if Bz and Ez are not ≡ 0: ∂Ez i ∂Bz − εµω k εµω 2 − k 2 ∂x ∂y ∂Bz i ∂Ez + εµω Ex = k εµω 2 − k 2 ∂x ∂y Bx = Now one can distinguish between three cases: 1. Bz ≡ 0: the Transversal Magnetic modes (TM). Boundary condition: E z |surf = 0. 2. Ez ≡ 0: the Transversal Electric modes (TE). Boundary condition: ∂Bz ∂n = 0.
surf
By =
∂Ez i ∂Bz + εµω k εµω 2 − k 2 ∂y ∂x ∂Bz i ∂Ez − εµω Ey = k εµω 2 − k 2 ∂y ∂x
For the TE and TM modes this gives an eigenvalue problem for Ez resp. Bz with boundary conditions: ∂2 ∂2 + 2 ∂x2 ∂y ψ = −γ 2 ψ with eigenvalues γ 2 := εµω 2 − k 2
This gives a discrete solution ψ with eigenvalue γ 2 : k = εµω 2 − γ 2 . For ω < ω , k is imaginary and the wave is damped. Therefore, ω is called the cut-off frequency. In rectangular conductors the following expression can be found for the cut-off frequency for modes TE m,n of TMm,n : λ = 2 (m/a)2 + (n/b)2
Chapter 5: Waves
23
3. Ez and Bz are zero everywhere: the Transversal electromagnetic mode (TEM). Than holds: k = √ ±ω εµ and vf = vg , just as if here were no waveguide. Further k ∈ I so there exists no cut-off R, frequency. In a rectangular, 3 dimensional resonating cavity with edges a, b and c the possible wave numbers are given n1 π n2 π n3 π by: kx = , ky = , kz = This results in the possible frequencies f = vk/2π in the cavity: a b c f= v 2 n2 n2 n2 y z x + 2 + 2 a2 b c
For a cubic cavity, with a = b = c, the possible number of oscillating modes N L for longitudinal waves is given by: 4πa3 f 3 NL = 3v 3 Because transversal waves have two possible polarizations holds for them: N T = 2NL .
5.6 Non-linear wave equations
The Van der Pol equation is given by: d2 x dx 2 − εω0 (1 − βx2 ) + ω0 x = 0 dt2 dt βx2 can be ignored for very small values of the amplitude. Substitution of x ∼ e iωt gives: ω = 2 1−
1 2 2 ε ). 1 2 ω0 (iε 1 2 εω0 .
The lowest-order instabilities grow as
While x is growing, the 2nd term becomes larger
±
−1 and diminishes the growth. Oscillations on a time scale ∼ ω0 can exist. If x is expanded as x = x(0) + (1) 2 (2) εx + ε x + · · · and this is substituted one obtains, besides periodic, secular terms ∼ εt. If it is assumed that there exist timescales τn , 0 ≤ τ ≤ N with ∂τn /∂t = εn and if the secular terms are put 0 one obtains:
d dt
1 2
dx dt
2 1 2 + 2 ω 0 x2
= εω0 (1 − βx2 )
dx dt
2
This is an energy equation. Energy is conserved if the left-hand side is 0. If x 2 > 1/β, the right-hand side changes sign and an increase in energy changes into a decrease of energy. This mechanism limits the growth of oscillations. The Korteweg-De Vries equation is given by: ∂u ∂3u ∂u ∂u + b2 3 = 0 + − au ∂t ∂x ∂x ∂x
non−lin dispersive
with c = 1 + 1 ad and e2 = ad/(12b2 ). 3
This equation is for example a model for ion-acoustic waves in a plasma. For this equation, soliton solutions of the following form exist: −d u(x − ct) = cosh2 (e(x − ct))
Chapter 6
Optics
6.1 The bending of light
For the refraction at a surface holds: ni sin(θi ) = nt sin(θt ) where n is the refractive index of the material. Snell’s law is: λ1 v1 n2 = = n1 λ2 v2 If ∆n ≤ 1, the change in phase of the light is ∆ϕ = 0, if ∆n > 1 holds: ∆ϕ = π. The refraction of light in a material is caused by scattering from atoms. This is described by: n2 = 1 + ne e 2 ε0 m fj 2 ω0,j − ω 2 − iδω fj = 1. From this follows
j
j
where ne is the electron density and fj the oscillator strength, for which holds:
that vg = c/(1 + (ne e2 /2ε0 mω 2 )). From this the equation of Cauchy can be derived: n = a0 + a1 /λ2 . More n ak . general, it is possible to expand n as: n = λ2k
k=0
For an electromagnetic wave in general holds: n =
√
ε r µr .
The path, followed by a light ray in material can be found from Fermat’s principle:
2 2
δ
1
dt = δ
1
n(s) ds = 0 ⇒ δ c
2
n(s)ds = 0
1
6.2 Paraxial geometrical optics
6.2.1 Lenses
The Gaussian lens formula can be deduced from Fermat’s principle with the approximations cos ϕ = 1 and sin ϕ = ϕ. For the refraction at a spherical surface with radius R holds: n2 n1 − n 2 n1 − = v b R where |v| is the distance of the object and |b| the distance of the image. Applying this twice results in: 1 = (nl − 1) f 1 1 − R2 R1
where nl is the refractive index of the lens, f is the focal length and R1 and R2 are the curvature radii of both surfaces. For a double concave lens holds R1 < 0, R2 > 0, for a double convex lens holds R1 > 0 and R2 < 0. Further holds: 1 1 1 = − f v b
Chapter 6: Optics
25
D := 1/f is called the dioptric power of a lens. For a lens with thickness d and diameter D holds to a good approximation: 1/f = 8(n − 1)d/D 2 . For two lenses placed on a line with distance d holds: 1 1 d 1 = + − f f1 f2 f1 f2 In these equations the following signs are being used for refraction at a spherical surface, as is seen by an incoming light ray: Quantity R f v b + Concave surface Converging lens Real object Virtual image − Convex surface Diverging lens Virtual object Real image
6.2.2 Mirrors
For images of mirrors holds: 1 1 2 h2 1 = + = + f v b R 2 1 1 − R v
2
where h is the perpendicular distance from the point the light ray hits the mirror to the optical axis. Spherical aberration can be reduced by not using spherical mirrors. A parabolical mirror has no spherical aberration for light rays parallel with the optical axis and is therefore often used for telescopes. The used signs are: Quantity R f v b + Concave mirror Concave mirror Real object Real image − Convex mirror Convex mirror Virtual object Virtual image
6.2.3 Principal planes
The nodal points N of a lens are defined by the figure on the right. If the lens is surrounded by the same medium on both sides, the nodal points are the same as the principal points H. The plane ⊥ the optical axis through the principal points is called the principal plane. If the lens is described by a matrix mij than for the distances h1 and h2 to the boundary of the lens holds: h1 = n m22 − 1 m11 − 1 , h2 = n m12 m12
N1 r r r O N2
6.2.4 Magnification
The linear magnification is defined by: N = − b v αsyst αnone
The angular magnification is defined by: Nα = −
where αsys is the size of the retinal image with the optical system and αnone the size of the retinal image without the system. Further holds: N · Nα = 1. For a telescope holds: N = fobjective/focular. The f-number is defined by f /Dobjective.
26
Physics Formulary by ir. J.C.A. Wevers
6.3 Matrix methods
A light ray can be described by a vector (nα, y) with α the angle with the optical axis and y the distance to the optical axis. The change of a light ray interacting with an optical system can be obtained using a matrix multiplication: n1 α 1 n2 α 2 =M y1 y2 where Tr(M ) = 1. M is a product of elementary matrices. These are: 1. Transfer along length l: MR = 1 0 l/n 1 1 −D 0 1
2. Refraction at a surface with dioptric power D: MT =
6.4 Aberrations
Lenses usually do not give a perfect image. Some causes are: 1. Chromatic aberration is caused by the fact that n = n(λ). This can be partially corrected with a lens which is composed of more lenses with different functions ni (λ). Using N lenses makes it possible to obtain the same f for N wavelengths. 2. Spherical aberration is caused by second-order effects which are usually ignored; a spherical surface does not make a perfect lens. Incomming rays far from the optical axis will more bent. 3. Coma is caused by the fact that the principal planes of a lens are only flat near the principal axis. Further away of the optical axis they are curved. This curvature can be both positive or negative. 4. Astigmatism: from each point of an object not on the optical axis the image is an ellipse because the thickness of the lens is not the same everywhere. 5. Field curvature can be corrected by the human eye. 6. Distorsion gives abberations near the edges of the image. This can be corrected with a combination of positive and negative lenses.
6.5 Reflection and transmission
If an electromagnetic wave hits a transparent medium part of the wave will reflect at the same angle as the incident angle, and a part will be refracted at an angle according to Snell’s law. It makes a difference whether the E field of the wave is ⊥ or w.r.t. the surface. When the coefficients of reflection r and transmission t are defined as: E0r E0r E0t E0t r ≡ , r⊥ ≡ , t ≡ , t⊥ ≡ E0i E0i ⊥ E0i E0i ⊥ where E0r is the reflected amplitude and E0t the transmitted amplitude. Then the Fresnel equations are: r = t = tan(θi − θt ) sin(θt − θi ) , r⊥ = tan(θi + θt ) sin(θt + θi )
The following holds: t⊥ − r⊥ = 1 and t + r = 1. If the coefficient of reflection R and transmission T are defined as (with θi = θr ): Ir It cos(θt ) R≡ and T ≡ Ii Ii cos(θi )
2 sin(θt ) cos(θi ) 2 sin(θt ) cos(θi ) , t⊥ = sin(θt + θi ) cos(θt − θi ) sin(θt + θi )
Chapter 6: Optics
27
with I = |S| it follows: R + T = 1. A special case is r = 0. This happens if the angle between the reflected and transmitted rays is 90◦ . From Snell’s law it then follows: tan(θi ) = n. This angle is called Brewster’s angle. The situation with r⊥ = 0 is not possible.
6.6 Polarization
The polarization is defined as: P = Ip Imax − Imin = Ip + I u Imax + Imin
where the intensity of the polarized light is given by Ip and the intensity of the unpolarized light is given by Iu . Imax and Imin are the maximum and minimum intensities when the light passes a polarizer. If polarized light passes through a polarizer Malus law applies: I(θ) = I(0) cos2 (θ) where θ is the angle of the polarizer. The state of a light ray can be described by the Stokes-parameters: start with 4 filters which each transmits half the intensity. The first is independent of the polarization, the second and third are linear polarizers with the transmission axes horizontal and at +45◦ , while the fourth is a circular polarizer which is opaque for L-states. Then holds S1 = 2I1 , S2 = 2I2 − 2I1 , S3 = 2I3 − 2I1 and S4 = 2I4 − 2I1 . The state of a polarized light ray can also be described by the Jones vector: E= E0x eiϕx E0y eiϕy
For the √ horizontal P -state holds: E = (1, 0), for the vertical P -state E = (0, 1), the R-state is given by √ 1 1 E = 2 2(1, −i) and the L-state by E = 2 2(1, i). The change in state of a light beam after passage of optical equipment can be described as E2 = M · E1 . For some types of optical equipment the Jones matrix M is given by: Horizontal linear polarizer: Vertical linear polarizer: Linear polarizer at +45◦ Lineair polarizer at −45◦
1 4 -λ 1 4 -λ 1 2 1 2
1 0 0 0 0 0 0 1 1 1 1 1
1 −1 −1 1 1 0 0 −i 1 0 0 i 1 i −i 1 1 −i i 1
plate, fast axis vertical plate, fast axis horizontal
eiπ/4 eiπ/4
1 2 1 2
Homogene circular polarizor right Homogene circular polarizer left
6.7 Prisms and dispersion
A light ray passing through a prism is refracted twice and aquires a deviation from its original direction δ = θi + θi + α w.r.t. the incident direction, where α is the apex angle, θi is the angle between the incident angle and a line perpendicular to the surface and θi is the angle between the ray leaving the prism and a line perpendicular to the surface. When θi varies there is an angle for which δ becomes minimal. For the refractive index of the prism now holds: 1 sin( 2 (δmin + α)) n= sin( 1 α) 2
28
Physics Formulary by ir. J.C.A. Wevers
The dispersion of a prism is defined by: dδ dn dδ = dλ dn dλ where the first factor depends on the shape and the second on the composition of the prism. For the first factor follows: 2 sin( 1 α) dδ 2 = 1 dn cos( 2 (δmin + α)) D= For visible light usually holds dn/dλ < 0: shorter wavelengths are stronger bent than longer. The refractive index in this area can usually be approximated by Cauchy’s formula.
6.8 Diffraction
Fraunhofer diffraction occurs far away from the source(s). The Fraunhofer diffraction of light passing through multiple slits is described by: 2 2 sin(N v) I(θ) sin(u) · = I0 u sin(v) where u = πb sin(θ)/λ, v = πd sin(θ)/λ. N is the number of slits, b the width of a slit and d the distance between the slits. The maxima in intensity are given by d sin(θ) = kλ. The diffraction through a spherical aperture with radius a is described by: I(θ) = I0 J1 (ka sin(θ)) ka sin(θ)
2
The diffraction pattern of a rectangular aperture at distance R with length a in the x-direction and b in the y-direction is described by: 2 2 I(x, y) sin(α ) sin(β ) = I0 α β where α = kax/2R and β = kby/2R. When X rays are diffracted at a crystal holds for the position of the maxima in intensity Bragg’s relation: 2d sin(θ) = nλ where d is the distance between the crystal layers. Close at the source the Fraunhofermodel is invalid because it ignores the angle-dependence of the reflected waves. This is described by the obliquity or inclination factor, which describes the directionality of the secondary emissions: E(θ) = 1 E0 (1 + cos(θ)) where θ is the angle w.r.t. the optical axis. 2 Diffraction limits the resolution of a system. This is the minimum angle ∆θ min between two incident rays coming from points far away for which their refraction patterns can be detected separately. For a circular slit holds: ∆θmin = 1.22λ/D where D is the diameter of the slit. For a grating holds: ∆θmin = 2λ/(N a cos(θm )) where a is the distance between two peaks and N the number of peaks. The minimum difference between two wavelengths that gives a separated diffraction pattern in a multiple slit geometry is given by ∆λ/λ = nN where N is the number of lines and n the order of the pattern.
6.9 Special optical effects
• Birefringe and dichroism. D is not parallel with E if the polarizability P of a material is not equal in all directions. There are at least 3 directions, the principal axes, in which they are parallel. This results in 3 refractive indices ni which can be used to construct Fresnel’s ellipsoid. In case n2 = n3 = n1 , which happens e.g. at trigonal, hexagonal and tetragonal crystals there is one optical axis in the direction of n1 . Incident light rays can now be split up in two parts: the ordinary wave is linear polarized ⊥ the plane through the transmission direction and the optical axis. The extraordinary wave is linear polarized
Chapter 6: Optics
29
in the plane through the transmission direction and the optical axis. Dichroism is caused by a different absorption of the ordinary and extraordinary wave in some materials. Double images occur when the incident ray makes an angle with the optical axis: the extraordinary wave will refract, the ordinary will not. • Retarders: waveplates and compensators. Incident light will have a phase shift of ∆ϕ = 2πd(|n 0 − ne |)/λ0 if an uniaxial crystal is cut in such a way that the optical axis is parallel with the front and back plane. Here, λ0 is the wavelength in vacuum and n0 and ne the refractive indices for the ordinary and extraordinary wave. For a quarter-wave plate holds: ∆ϕ = π/2. • The Kerr-effect: isotropic, transparent materials can become birefringent when placed in an electric field. In that case, the optical axis is parallel to E. The difference in refractive index in the two directions is given by: ∆n = λ0 KE 2 , where K is the Kerr constant of the material. If the electrodes have an effective length and are separated by a distance d, the retardation is given by: ∆ϕ = 2πK V 2 /d2 , where V is the applied voltage. • The Pockels or linear electro-optical effect can occur in 20 (from a total of 32) crystal symmetry classes, namely those without a centre of symmetry. These crystals are also piezoelectric: their polarization changes when a pressure is applied and vice versa: P = pd + ε0 χE. The retardation in a Pockels cell is ∆ϕ = 2πn3 r63 V /λ0 where r63 is the 6-3 element of the electro-optic tensor. 0 • The Faraday effect: the polarization of light passing through material with length d and to which a magnetic field is applied in the propagation direction is rotated by an angle β = VBd where V is the Verdet constant. ˘ • Cerenkov radiation arises when a charged particle with vq > vf arrives. The radiation is emitted within a cone with an apex angle α with sin(α) = c/cmedium = c/nvq .
6.10 The Fabry-Perot interferometer
For a Fabry-Perot interferometer holds in general: T + R + A = 1 where T is the transmission factor, R the reflection factor and A the absorption factor. If F is given by F = 4R/(1 − R)2 it follows for the intensity distribution: It A = 1− Ii 1−R
2
1 1 + F sin2 (θ)
Source
q Lens 'E d
Screen Focussing lens √ √ The width of the peaks at half height is given by γ = 4/ F . The finesse F is defined as F = 1 π F . The 2 maximum resolution is then given by ∆fmin = c/2ndF.
The term [1 + F sin2 (θ)]−1 := A(θ) is called the Airy function.
Chapter 7
Statistical physics
7.1 Degrees of freedom
A molecule consisting of n atoms has s = 3n degrees of freedom. There are 3 translational degrees of freedom, a linear molecule has s = 3n − 5 vibrational degrees of freedom and a non-linear molecule s = 3n − 6. A linear molecule has 2 rotational degrees of freedom and a non-linear molecule 3. Because vibrational degrees of freedom account for both kinetic and potential energy they count double. So, for linear molecules this results in a total of s = 6n − 5. For non-linear molecules this gives s = 6n − 6. The average energy of a molecule in thermodynamic equilibrium is E tot = 1 skT . Each degree of freedom of a 2 molecule has in principle the same energy: the principle of equipartition. The rotational and vibrational energy of a molecule are: Wrot = h2 ¯ 1 h l(l + 1) = Bl(l + 1) , Wvib = (v + 2 )¯ ω0 2I
The vibrational levels are excited if kT ≈ hω, the rotational levels of a hetronuclear molecule are excited if ¯ kT ≈ 2B. For homonuclear molecules additional selection rules apply so the rotational levels are well coupled if kT ≈ 6B.
7.2 The energy distribution function
The general form of the equilibrium velocity distribution function is P (vx , vy , vz )dvx dvy dvz = P (vx )dvx · P (vy )dvy · P (vz )dvz with P (vi )dvi = 1 v2 √ exp − i2 α α π dvi
where α = 2kT /m is the most probable velocity of a particle. The average velocity is given by v = √ 3 2α/ π, and v 2 = 2 α2 . The distribution as a function of the absolute value of the velocity is given by: 4N dN mv 2 = 3 √ v 2 exp − dv α π 2kT The general form of the energy distribution function then becomes: c(s) P (E)dE = kT where c(s) is a normalization constant, given by: 1. Even s: s = 2l: c(s) = 1 (l − 1)! 2l π(2l − 1)!! E kT
1 2 s−1
exp −
E kT
dE
2. Odd s: s = 2l + 1: c(s) = √
Chapter 7: Statistical physics
31
7.3 Pressure on a wall
The number of molecules that collides with a wall with surface A within a time τ is given by:
∞ π 2π
d N=
0 0 0
3
nAvτ cos(θ)P (v, θ, ϕ)dvdθdϕ
1 From this follows for the particle flux on the wall: Φ = 4 n v . For the pressure on the wall then follows:
d3 p =
2 2mv cos(θ)d3 N , so p = n E Aτ 3
7.4 The equation of state
If intermolecular forces and the volume of the molecules can be neglected then for gases from p = and E = 3 kT can be derived: 2 1 pV = ns RT = N m v 2 3
2 3n
E
Here, ns is the number of moles particles and N is the total number of particles within volume V . If the own volume and the intermolecular forces cannot be neglected the Van der Waals equation can be derived: p+ an2 s V2 (V − bns ) = ns RT
There is an isotherme with a horizontal point of inflection. In the Van der Waals equation this corresponds with the critical temperature, pressure and volume of the gas. This is the upper limit of the area of coexistence between liquid and vapor. From dp/dV = 0 and d2 p/dV 2 = 0 follows: Tcr = a 8a , pcr = , Vcr = 3bns 27bR 27b2
3 8,
For the critical point holds: pcr Vm,cr /RTcr = general gas law.
which differs from the value of 1 which follows from the
∗ Scaled on the critical quantities, with p∗ := p/pcr , T ∗ = T /Tcr and Vm = Vm /Vm,cr with Vm := V /ns holds:
p∗ +
3
∗ (Vm )2
∗ Vm −
1 3
8 = 3T∗
Gases behave the same for equal values of the reduced quantities: the law of the corresponding states. A virial expansion is used for even more accurate views: p(T, Vm ) = RT B(T ) C(T ) 1 + + +··· 2 3 Vm Vm Vm
The Boyle temperature TB is the temperature for which the 2nd virial coefficient is 0. In a Van der Waals gas, this happens at TB = a/Rb. The inversion temperature Ti = 2TB . The equation of state for solids and liquids is given by: 1 V = 1 + γp ∆T − κT ∆p = 1 + V0 V ∂V ∂T ∆T +
p
1 V
∂V ∂p
∆p
T
32
Physics Formulary by ir. J.C.A. Wevers
7.5 Collisions between molecules
The collision probability of a particle in a gas that is translated over a distance dx is given by nσdx, where σ is v1 2 2 the cross section. The mean free path is given by = with u = v1 + v2 the relative velocity between nuσ u 1 1 m1 √ . This means the particles. If m1 m2 holds: = 1+ , so = . If m1 = m2 holds: = v1 m2 nσ nσ 2 1 . If the molecules are approximated by hard that the average time between two collisions is given by τ = nσv 1 2 2 spheres the cross section is: σ = 4 π(D1 + D2 ). The average distance between two molecules is 0.55n−1/3. Collisions between molecules and small particles in a solution result in the Brownian motion. For the average 1 motion of a particle with radius R can be derived: x2 = 3 r2 = kT t/3πηR. i A gas is called a Knudsen gas if the dimensions of the gas, something that can easily occur at low pressures. The equilibrium condition for a vessel which has a hole with surface A in it for which holds that √ √ √ √ A/π is: n1 T1 = n2 T2 . Together with the general gas law follows: p1 / T1 = p2 / T2 . Awx . d The velocity profile between the plates is in that case given by w(z) = zw x /d. It can be derived that η = 1 v where v is the thermal velocity. 3 If two plates move along each other at a distance d with velocity w x the viscosity η is given by: Fx = η dQ T2 − T 1 , which results in a temper= κA dt d 1 ature profile T (z) = T1 + z(T2 − T1 )/d. It can be derived that κ = 3 CmV n v /NA . Also holds: κ = CV η. A better expression for κ can be obtained with the Eucken correction: κ = (1 + 9R/4c mV )CV · η with an error <5%. The heat conductance in a non-moving gas is described by:
7.6 Interaction between molecules
For dipole interaction between molecules can be derived that U ∼ −1/r 6 . If the distance between two molecules approaches the molecular diameter D a repulsing force between the electron clouds appears. This force can be described by Urep ∼ exp(−γr) or Vrep = +Cs /rs with 12 ≤ s ≤ 20. This results in the Lennard-Jones potential for intermolecular forces: ULJ = 4 D r
12
−
D r
6
with a minimum at r = rm . The following holds: D ≈ 0.89rm . For the Van der Waals coefficients a and b 2 and the critical quantities holds: a = 5.275NAD3 , b = 1.3NA D3 , kTkr = 1.2 and Vm,kr = 3.9NA D3 . A more simple model for intermolecular forces assumes a potential U (r) = ∞ for r < D, U (r) = U LJ for D ≤ r ≤ 3D and U (r) = 0 for r ≥ 3D. This gives for the potential energy of one molecule: E pot =
3D
U (r)F (r)dr.
D
with F (r) the spatial distribution function in spherical coordinates, which for a homogeneous distribution is given by: F (r)dr = 4nπr 2 dr. Some useful mathematical relations are:
∞ ∞
x e
0
n −x
dx = n! ,
0
x e
2n −x2
√ (2n)! π , dx = n!22n+1
∞
x2n+1 e−x dx = 1 n! 2
0
2
Chapter 8
Thermodynamics
8.1 Mathematical introduction
If there exists a relation f (x, y, z) = 0 between 3 variables, one can write: x = x(y, z), y = y(x, z) and z = z(x, y). The total differential dz of z is than given by: dz = ∂z ∂x dx +
y
∂z ∂y
dy
x
By writing this also for dx and dy it can be obtained that ∂x ∂y Because dz is a total differential holds · ∂y ∂z · ∂z ∂x = −1
z
x
y
dz = 0.
A homogeneous function of degree m obeys: εm F (x, y, z) = F (εx, εy, εz). For such a function Euler’s theorem applies: ∂F ∂F ∂F mF (x, y, z) = x +y +z ∂x ∂y ∂z
8.2 Definitions
• The isochoric pressure coefficient: βV = • The isothermal compressibility: κT = − • The isobaric volume coefficient: γp = 1 V 1 V 1 p 1 V ∂p ∂T ∂V ∂p ∂V ∂T ∂V ∂p
V
T
p
• The adiabatic compressibility: κS = −
S
For an ideal gas follows: γp = 1/T , κT = 1/p and βV = −1/V .
8.3 Thermal heat capacity
• The specific heat at constant X is: CX = T • The specific heat at constant pressure: Cp = • The specific heat at constant volume: CV = ∂S ∂T ∂H ∂T ∂U ∂T
X
p
V
34
Physics Formulary by ir. J.C.A. Wevers
For an ideal gas holds: Cmp − CmV = R. Further, if the temperature is high enough to thermalize all internal 1 rotational and vibrational degrees of freedom, holds: CV = 2 sR. Hence Cp = 1 (s + 2)R. For their ratio now 2 follows γ = (2 + s)/s. For a lower T one needs only to consider the thermalized degrees of freedom. For a 1 Van der Waals gas holds: CmV = 2 sR + ap/RT 2. In general holds: Cp − C V = T ∂p ∂T
V
·
∂V ∂T
p
= −T
∂V ∂T
2 p
∂p ∂V
T
≥0
Because (∂p/∂V )T is always < 0, the following is always valid: Cp ≥ CV . If the coefficient of expansion is 0, Cp = CV , and also at T = 0K.
8.4 The laws of thermodynamics
The zeroth law states that heat flows from higher to lower temperatures. The first law is the conservation of energy. For a closed system holds: Q = ∆U + W , where Q is the total added heat, W the work done and ∆U the difference in the internal energy. In differential form this becomes: d Q = dU + d W , where d means that the it is not a differential of a quantity of state. For a quasi-static process holds: d W = pdV . So for a reversible process holds: d Q = dU + pdV . For an open (flowing) system the first law is: Q = ∆H + Wi + ∆Ekin + ∆Epot . One can extract an amount of work Wt from the system or add Wt = −Wi to the system. The second law states: for a closed system there exists an additive quantity S, called the entropy, the differential of which has the following property: dQ dS ≥ T If the only processes occurring are reversible holds: dS = d Qrev /T . So, the entropy difference after a reversible process is:
2
S2 − S 1 = So, for a reversible cycle holds: For an irreversible cycle holds: d Qrev = 0. T d Qirr < 0. T
d Qrev T
1
The third law of thermodynamics is (Nernst): lim ∂S ∂X =0
T
T →0
From this it can be concluded that the thermal heat capacity → 0 if T → 0, so absolute zero temperature cannot be reached by cooling through a finite number of steps.
8.5 State functions and Maxwell relations
The quantities of state and their differentials are: Internal energy: Enthalpy: Free energy: Gibbs free enthalpy: U H = U + pV F = U − TS G = H − TS dU = T dS − pdV dH = T dS + V dp dF = −SdT − pdV dG = −SdT + V dp
Chapter 8: Thermodynamics
35
From this one can derive Maxwell’s relations: ∂T ∂V =− ∂p ∂S ,
V
S
∂T ∂p
=
S
∂V ∂S
,
p
∂p ∂T
=
V
∂S ∂V
,
T
∂V ∂T
p
=−
∂S ∂p
T
From the total differential and the definitions of CV and Cp it can be derived that: T dS = CV dT + T For an ideal gas also holds: Sm = CV ln T T0 + R ln V V0 + Sm0 and Sm = Cp ln T T0 − R ln p p0 + Sm0 ∂p ∂T dV and T dS = Cp dT − T ∂V ∂T dp
p
V
Helmholtz’ equations are: ∂U ∂V =T
T
∂p ∂T
V
−p ,
∂H ∂p
T
=V −T
∂V ∂T
p
for an enlarged surface holds: d Wrev = −γdA, with γ the surface tension. From this follows: γ= ∂U ∂A =
S
∂F ∂A
T
8.6 Processes
The efficiency η of a process is given by: η = Work done Heat added Cold delivered Work added
The Cold factor ξ of a cooling down process is given by: ξ = Reversible adiabatic processes
For adiabatic processes holds: W = U1 − U2 . For reversible adiabatic processes holds Poisson’s equation: with γ = Cp /CV one gets that pV γ =constant. Also holds: T V γ−1 =constant and T γ p1−γ =constant. Adiabatics exhibit a greater steepness p-V diagram than isothermics because γ > 1. Isobaric processes Here holds: H2 − H1 = The throttle process This is also called the Joule-Kelvin effect and is an adiabatic expansion of a gas through a porous material or a small opening. Here H is a conserved quantity, and dS > 0. In general this is accompanied with a change in temperature. The quantity which is important here is the throttle coefficient: αH = ∂T ∂p =
H 2 1
Cp dT . For a reversible isobaric process holds: H2 − H1 = Qrev .
1 T Cp
∂V ∂T
p
−V
The inversion temperature is the temperature where an adiabatically expanding gas keeps the same temperature. If T > Ti the gas heats up, if T < Ti the gas cools down. Ti = 2TB , with for TB : [∂(pV )/∂p]T = 0. The throttle process is e.g. applied in refridgerators. The Carnotprocess The system undergoes a reversible cycle with 2 isothemics and 2 adiabatics: 1. Isothermic expansion at T1 . The system absorbs a heat Q1 from the reservoir. 2. Adiabatic expansion with a temperature drop to T2 .
36
Physics Formulary by ir. J.C.A. Wevers
3. Isothermic compression at T2 , removing Q2 from the system. 4. Adiabatic compression to T1 . The efficiency for Carnot’s process is: η =1− |Q2 | T2 =1− := ηC |Q1 | T1
The Carnot efficiency ηC is the maximal efficiency at which a heat machine can operate. If the process is applied in reverse order and the system performs a work −W the cold factor is given by: ξ= The Stirling process Stirling’s cycle exists of 2 isothermics and 2 isochorics. The efficiency in the ideal case is the same as for Carnot’s cycle. |Q2 | |Q2 | T2 = = W |Q1 | − |Q2 | T1 − T 2
8.7 Maximal work
Consider a system that changes from state 1 into state 2, with the temperature and pressure of the surroundings given by T0 and p0 . The maximum work which can be obtained from this change is, when all processes are reversible: 1. Closed system: Wmax = (U1 − U2 ) − T0 (S1 − S2 ) + p0 (V1 − V2 ). 2. Open system: Wmax = (H1 − H2 ) − T0 (S1 − S2 ) − ∆Ekin − ∆Epot . The minimal work needed to attain a certain state is: Wmin = −Wmax .
8.8 Phase transitions
Phase transitions are isothermic and isobaric, so dG = 0. When the phases are indicated by α, β and γ holds: Gα = Gβ and m m rβα α β ∆Sm = Sm − Sm = T0 where rβα is the transition heat of phase β to phase α and T0 is the transition temperature. The following holds: rβα = rαβ and rβα = rγα − rγβ . Further Sm = ∂Gm ∂T
p
so G has a twist in the transition point. In a two phase system Clapeyron’s equation is valid:
β dp rβα S α − Sm = = m α−Vβ α − V β )T dT Vm (Vm m m
For an ideal gas one finds for the vapor line at some distance from the critical point: p = p0 e−rβα/RT There exist also phase transitions with rβα = 0. For those there will occur only a discontinuity in the second derivates of Gm . These second-order transitions appear at organization phenomena. A phase-change of the 3rd order, so with e.g. [∂ 3 Gm /∂T 3 ]p non continuous arises e.g. when ferromagnetic iron changes to the paramagnetic state.
Chapter 8: Thermodynamics
37
8.9 Thermodynamic potential
When the number of particles within a system changes this number becomes a third quantity of state. Because addition of matter usually takes place at constant p and T , G is the relevant quantity. If a system exists of more components this becomes: dG = −SdT + V dp + µi dni
i
where µ =
∂G ∂ni
is called the thermodynamic potential. This is a partial quantity. For V holds:
p,T,nj c
V =
i=1
ni
∂V ∂ni
c
:=
nj ,p,T i=1
n i Vi
where Vi is the partial volume of component i. The following holds: Vm =
i
xi V i xi dVi
i
0 =
where xi = ni /n is the molar fraction of component i. The molar volume of a mixture of two components can be a concave line in a V -x2 diagram: the mixing contracts the volume. The thermodynamic potentials are not independent in a multiple-phase system. It can be derived that ni dµi = −SdT + V dp, this gives at constant p and T : xi dµi = 0 (Gibbs-Duhmen).
i i
Each component has as much µ’s as there are phases. The number of free parameters in a system with c components and p different phases is given by f = c + 2 − p.
8.10 Ideal mixtures
For a mixture of n components holds (the index 0 is the value for the pure component): Umixture =
i
ni Ui0 , Hmixture =
i
ni Hi0 , Smixture = n
i
0 xi Si + ∆Smix
where for ideal gases holds: ∆Smix = −nR
xi ln(xi ).
i
For the thermodynamic potentials holds: µi = µ0 + RT ln(xi ) < µ0 . A mixture of two liquids is rarely ideal: i i this is usually only the case for chemically related components or isotopes. In spite of this holds Raoult’s law for the vapour pressure holds for many binary mixtures: pi = xi p0 = yi p. Here is xi the fraction of the ith i component in liquid phase and yi the fraction of the ith component in gas phase. A solution of one component in another gives rise to an increase in the boiling point ∆T k and a decrease of the freezing point ∆Ts . For x2 1 holds: ∆Tk =
2 RTk RT 2 x2 , ∆Ts = − s x2 rβα rγβ
with rβα the evaporation heat and rγβ < 0 the melting heat. For the osmotic pressure Π of a solution holds: 0 ΠVm1 = x2 RT .
8.11 Conditions for equilibrium
When a system evolves towards equilibrium the only changes that are possible are those for which holds: (dS)U,V ≥ 0 or (dU )S,V ≤ 0 or (dH)S,p ≤ 0 or (dF )T,V ≤ 0 or (dG)T,p ≤ 0. In equilibrium for each component holds: µα = µβ = µγ . i i i
38
Physics Formulary by ir. J.C.A. Wevers
8.12 Statistical basis for thermodynamics
The number of possibilities P to distribute N particles on n possible energy levels, each with a g-fold degeneracy is called the thermodynamic probability and is given by: P = N!
i n gi i ni !
The most probable distribution, that with the maximum value for P , is the equilibrium state. When Stirling’s equation, ln(n!) ≈ n ln(n) − n is used, one finds for a discrete system the Maxwell-Boltzmann distribution. The occupation numbers in equilibrium are then given by: ni = N Wi gi exp − Z kT gi exp(−Wi /kT ). For an ideal gas holds:
i
The state sum Z is a normalization constant, given by: Z =
Z=
V (2πmkT )3/2 h3
The entropy can then be defined as: S = k ln(P ) . For a system in thermodynamic equilibrium this becomes: S= U + kN ln T Z N + kN ≈ U + k ln T ZN N!
For an ideal gas, with U = 3 kT then holds: S = 5 kN + kN ln 2 2
V (2πmkT )3/2 N h3
8.13 Application to other systems
Thermodynamics can be applied to other systems than gases and liquids. To do this the term d W = pdV has to be replaced with the correct work term, like d Wrev = −F dl for the stretching of a wire, d Wrev = −γdA for the expansion of a soap bubble or d Wrev = −BdM for a magnetic system.
A rotating, non-charged black hole has a temparature of T = hc/8πkm. It has an entropy S = Akc 3 /4¯ κ ¯ h with A the area of its event horizon. For a Schwarzschild black hole A is given by A = 16πm 2 . Hawkings area theorem states that dA/dt ≥ 0. Hence, the lifetime of a black hole ∼ m3 .
Chapter 9
Transport phenomena
9.1 Mathematical introduction
An important relation is: if X is a quantity of a volume element which travels from position r to r + dr in a time dt, the total differential dX is then given by: dX = ∂X ∂X ∂X dX ∂X ∂X ∂X ∂X ∂X dx + dy + dz + dt ⇒ = vx + vy + vz + ∂x ∂y ∂z ∂t dt ∂x ∂y ∂z ∂t dX ∂X = + (v · dt ∂t d dt )X . ∂ ∂t Xd3 V + X(v · n )d2 A operator are: rot gradφ = 0 div rotv = 0
This results in general to:
From this follows that also holds:
Xd3 V =
where the volume V is surrounded by surface A. Some properties of the div(φv ) = φdivv + gradφ · v div(u × v ) = v · (rotu ) − u · (rotv ) div gradφ = 2 φ
rot(φv ) = φrotv + (gradφ) × v rot rotv = grad divv − 2 v 2 v ≡ ( 2 v1 , 2 v2 , 2 v3 )
Here, v is an arbitrary vector field and φ an arbitrary scalar field. Some important integral theorems are: Gauss: Stokes for a scalar field: Stokes for a vector field: This results in: Ostrogradsky: (v · n )d2 A = (φ · et )ds = (v · et )ds = (divv )d3 V (n × gradφ)d2 A (rotv · n )d2 A
(rotv · n )d2 A = 0 (n × v )d2 A = (φn )d2 A = (rotv )d3 A (gradφ)d3 V ds.
Here, the orientable surface
d2 A is limited by the Jordan curve
9.2 Conservation laws
On a volume work two types of forces: 1. The force f0 on each volume element. For gravity holds: f0 = g. 2. Surface forces working only on the margins: t. For these holds: t = n T, where T is the stress tensor.
40
Physics Formulary by ir. J.C.A. Wevers
T can be split in a part pI representing the normal tensions and a part T representing the shear stresses: T = T + pI, where I is the unit tensor. When viscous aspects can be ignored holds: divT= −gradp. When the flow velocity is v at position r holds on position r + dr: v(dr ) = v(r )
translation
+
rotation, deformation, dilatation
The quantity L:=gradv can be split in a symmetric part D and an antisymmetric part W. L = D + W with Dij := 1 2 ∂vi ∂vj + ∂xj ∂xi , Wij := 1 2 ∂vi ∂vj − ∂xj ∂xi
When the rotation or vorticity ω = rotv is introduced holds: Wij = 1 εijk ωk . ω represents the local rotation 2 velocity: dr · W = 1 ω × dr. 2 For a Newtonian liquid holds: T = 2ηD. Here, η is the dynamical viscosity. This is related to the shear stress τ by: ∂vi τij = η ∂xj For compressible media can be stated: T = (η divv )I + 2ηD. From equating the thermodynamical and mechanical pressure it follows: 3η + 2η = 0. If the viscosity is constant holds: div(2D) = 2 v + grad divv. The conservation laws for mass, momentum and energy for continuous media can be written in both integral and differential form. They are:
Integral notation:
1. Conservation of mass: ∂ ∂t ∂ ∂t d3 V + (v · n )d2 A = 0 v(v · n )d2 A = f0 d3 V + n · T d2 A
2. Conservation of momentum: 3. Conservation of energy: ∂ ∂t −
vd3 V + ( 1 v 2 + e) d3 V + 2
1 ( 2 v 2 + e) (v · n )d2 A =
(q · n )d2 A +
(v · f0 )d3 V +
(v · n T)d2 A
Differential notation:
1. Conservation of mass: ∂ + div · ( v ) = 0 ∂t ∂v +( v· ∂t )v = f0 + divT = f0 − gradp + divT
2. Conservation of momentum: 3. Conservation of energy: T
ds de p d = − = −divq + T : D dt dt dt
Here, e is the internal energy per unit of mass E/m and s is the entropy per unit of mass S/m. q = −κ T is the heat flow. Further holds: ∂e ∂E ∂e ∂E =− , T = = p=− ∂V ∂1/ ∂S ∂s so ∂e ∂h CV = and Cp = ∂T V ∂T p with h = H/m the enthalpy per unit of mass.
Chapter 9: Transport phenomena
41
From this one can derive the Navier-Stokes equations for an incompressible, viscous and heat-conducting medium: divv ∂v + (v · ∂t C ∂T + C(v · ∂t )v )T = 0 = g − gradp + η
2 2
v
= κ
T + 2ηD : D
with C the thermal heat capacity. The force F on an object within a flow, when viscous effects are limited to the boundary layer, can be obtained using the momentum law. If a surface A surrounds the object outside the boundary layer holds: F =− [pn + v(v · n )]d2 A
9.3 Bernoulli’s equations
Starting with the momentum equation one can find for a non-viscous medium for stationary flows, with
1 (v · grad)v = 2 grad(v 2 ) + (rotv ) × v
and the potential equation g = −grad(gh) that:
1 2 2v
+ gh +
dp
= constant along a streamline
For compressible flows holds: 1 v 2 + gh + p/ =constant along a line of flow. If also holds rotv = 0 and 2 1 the entropy is equal on each streamline holds 2 v 2 + gh + dp/ =constant everywhere. For incompressible 1 2 flows this becomes: 2 v + gh + p/ =constant everywhere. For ideal gases with constant C p and CV holds, with γ = Cp /CV : γ p c2 1 2 1 = 2 v2 + = constant 2v + γ − 1 γ −1 With a velocity potential defined by v = gradφ holds for instationary flows: ∂φ 1 2 + 2 v + gh + ∂t dp = constant everywhere
9.4 Characterising of flows by dimensionless numbers
The advantage of dimensionless numbers is that they make model experiments possible: one has to make the dimensionless numbers which are important for the specific experiment equal for both model and the real situation. One can also deduce functional equalities without solving the differential equations. Some dimensionless numbers are given by: Strouhal: Sr = Fourier: Prandtl: ωL v a Fo = ωL2 ν Pr = a v2 gL vL P´ clet: Pe = e a Lα Nusselt: Nu = κ Froude: Fr = v c vL Reynolds: Re = ν v2 Eckert: Ec = c∆T Mach: Ma =
Here, ν = η/ is the kinematic viscosity, c is the speed of sound and L is a characteristic length of the system. α follows from the equation for heat transport κ∂y T = α∆T and a = κ/ c is the thermal diffusion coefficient. These numbers can be interpreted as follows: • Re: (stationary inertial forces)/(viscous forces)
42
Physics Formulary by ir. J.C.A. Wevers
• Sr: (non-stationary inertial forces)/(stationary inertial forces) • Fr: (stationary inertial forces)/(gravity) • Fo: (heat conductance)/(non-stationary change in enthalpy) • Pe: (convective heat transport)/(heat conductance) • Ec: (viscous dissipation)/(convective heat transport) • Ma: (velocity)/(speed of sound): objects moving faster than approximately Ma = 0,8 produce shockwaves which propagate with an angle θ with the velocity of the object. For this angle holds Ma= 1/ arctan(θ). • Pr and Nu are related to specific materials. Now, the dimensionless Navier-Stokes equation becomes, with x = x/L, v = v/V , grad = Lgrad, L2 2 and t = tω: 2 ∂v g v Sr + (v · )v = −grad p + + ∂t Fr Re
2
=
9.5 Tube flows
For tube flows holds: they are laminar if Re< 2300 with dimension of length the diameter of the tube, and turbulent if Re is larger. For an incompressible laminar flow through a straight, circular tube holds for the velocity profile: 1 dp 2 (R − r2 ) v(r) = − 4η dx
R
For the volume flow holds: ΦV =
0
v(r)2πrdr = −
π dp 4 R 8η dx
The entrance length Le is given by: 1. 500 < ReD < 2300: Le /2R = 0.056ReD √ 4R3 α π dp 3 dx 2 For flows at a small Re holds: p = η v and divv = 0. For the total force on a sphere with radius R in a 1 flow then holds: F = 6πηRv. For large Re holds for the force on a surface A: F = 2 CW A v 2 . 2. Re > 2300: Le /2R ≈ 50
For gas transport at low pressures (Knudsen-gas) holds: ΦV =
9.6 Potential theory
The circulation Γ is defined as: Γ = (v · et )ds = (rotv ) · nd2 A = (ω · n )d2 A
For non viscous media, if p = p( ) and all forces are conservative, Kelvin’s theorem can be derived: dΓ =0 dt For rotationless flows a velocity potential v = gradφ can be introduced. In the incompressible case follows from conservation of mass 2 φ = 0. For a 2-dimensional flow a flow function ψ(x, y) can be defined: with ΦAB the amount of liquid flowing through a curve s between the points A and B:
B B
ΦAB =
A
(v · n )ds =
A
(vx dy − vy dx)
Chapter 9: Transport phenomena
43
and the definitions vx = ∂ψ/∂y, vy = −∂ψ/∂x holds: ΦAB = ψ(B) − ψ(A). In general holds: ∂2ψ ∂2ψ + = −ωz 2 ∂x ∂y 2 In polar coordinates holds: ∂φ ∂ψ 1 ∂φ 1 ∂ψ = , vθ = − = r ∂θ ∂r ∂r r ∂θ Q For source flows with power Q in (x, y) = (0, 0) holds: φ = ln(r) so that vr = Q/2πr, vθ = 0. 2π vr = For a dipole of strength Q in x = a and strength −Q in x = −a follows from superposition: φ = −Qax/2πr 2 where Qa is the dipole strength. For a vortex holds: φ = Γθ/2π. If an object is surrounded by an uniform main flow with v = ve x and such a large Re that viscous effects are limited to the boundary layer holds: Fx = 0 and Fy = − Γv. The statement that Fx = 0 is d’Alembert’s paradox and originates from the neglection of viscous effects. The lift F y is also created by η because Γ = 0 due to viscous effects. Henxe rotating bodies also create a force perpendicular to their direction of motion: the Magnus effect.
9.7 Boundary layers
9.7.1 Flow boundary layers
√ If for the thickness of the boundary layer holds: δ L holds: δ ≈ L/ Re. With v∞ the velocity of the main flow it follows for the velocity vy ⊥ the surface: vy L ≈ δv∞ . Blasius’ equation for the boundary layer is, with vy /v∞ = f (y/δ): 2f + f f = 0 with boundary conditions f (0) = f (0) = 0, f (∞) = 1. From this −1/2 follows: CW = 0.664 Rex . The momentum theorem of Von Karman for the boundary layer is: d dv τ0 (ϑv 2 ) + δ ∗ v = dx dx
where the displacement thickness δ ∗ v and the momentum thickness ϑv 2 are given by:
∞ ∞ ∗
ϑv =
0
2
(v − vx )vx dy , δ v =
0
(v − vx )dy and τ0 = −η
∂vx ∂y
y=0
The boundary layer is released from the surface if
∂vx ∂y
= 0. This is equivalent with
y=0
dp 12ηv∞ = . dx δ2
9.7.2 Temperature boundary layers
If the thickness of the temperature boundary layer δT √ Pr. L holds: 1. If Pr ≤ 1: δ/δT ≈ √ 2. If Pr 1: δ/δT ≈ 3 Pr.
9.8 Heat conductance
For non-stationairy heat conductance in one dimension without flow holds: κ ∂2T ∂T = +Φ ∂t c ∂x2 where Φ is a source term. If Φ = 0 the solutions for harmonic oscillations at x = 0 are: T − T∞ x x = exp − cos ωt − Tmax − T∞ D D
44
Physics Formulary by ir. J.C.A. Wevers
with D = 2κ/ω c. At x = πD the temperature variation is in anti-phase with the surface. The onedimensional solution at Φ = 0 is 1 x2 T (x, t) = √ exp − 4at 2 πat This is mathematical equivalent to the diffusion problem: ∂n =D ∂t
2
n+P −A
where P is the production of and A the discharge of particles. The flow density J = −D n.
9.9 Turbulence
√ The time scale of turbulent velocity variations τt is of the order of: τt = τ Re/Ma2 with τ the molecular time scale. For the velocity of the particles holds: v(t) = v + v (t) with v (t) = 0. The Navier-Stokes equation now becomes: ∂ v +( v · ∂t ) v =− p +ν
2
v +
divSR
where SR ij = − vi vj is the turbulent stress tensor. Boussinesq’s assumption is: τij = − vi vj . It is stated that, analogous to Newtonian media: SR = 2 νt D . Near a boundary holds: νt = 0, far away of a boundary holds: νt ≈ νRe.
9.10 Self organization
For a (semi) two-dimensional flow holds: dω ∂ω = + J(ω, ψ) = ν 2 ω dt ∂t With J(ω, ψ) the Jacobian. So if ν = 0, ω is conserved. Further, the kinetic energy/mA and the enstrofy V are conserved: with v = × (kψ)
∞ ∞
E ∼ ( ψ) ∼
2
0
E(k, t)dk = constant , V ∼ (
2
ψ) ∼
2
0
k 2 E(k, t)dk = constant
From this follows that in a two-dimensional flow the energy flux goes towards large values of k: larger structures become larger at the expanse of smaller ones. In three-dimensional flows the situation is just the opposite.
Chapter 10
Quantum physics
10.1 Introduction to quantum physics
Planck’s law for the energy distribution for the radiation of a black body is: w(f ) = 8πhc 1 1 8πhf 3 , w(λ) = 3 5 ehc/λkT − 1 hf /kT − 1 c λ e
Stefan-Boltzmann’s law for the total power density can be derived from this: P = AσT 4 . Wien’s law for the maximum can also be derived from this: T λmax = kW .
10.1.2 The Compton effect
For the wavelength of scattered light, if light is considered to exist of particles, can be derived: λ =λ+ h (1 − cos θ) = λ + λC (1 − cos θ) mc
10.1.3 Electron diffraction
Diffraction of electrons at a crystal can be explained by assuming that particles have a wave character with wavelength λ = h/p. This wavelength is called the Broglie-wavelength.
10.2 Wave functions
The wave character of particles is described by a wavefunction ψ. This wavefunction can be described in normal or momentum space. Both definitions are each others Fourier transform: 1 Φ(k, t) = √ h 1 Ψ(x, t)e−ikx dx and Ψ(x, t) = √ h Φ(k, t)eikx dk
These waves define a particle with group velocity vg = p/m and energy E = hω. ¯ The wavefunction can be interpreted as a measure for the probability P to find a particle somewhere (Born): dP = |ψ|2 d3 V . The expectation value f of a quantity f of a system is given by: f (t) = Ψ∗ f Ψd3 V , fp (t) = Φ∗ f Φd3 Vp
This is also written as f (t) = Φ|f |Φ . The normalizing condition for wavefunctions follows from this: Φ|Φ = Ψ|Ψ = 1.
10.3 Operators in quantum physics
In quantum mechanics, classical quantities are translated into operators. These operators are hermitian because their eigenvalues must be real:
∗ ψ1 Aψ2 d3 V =
ψ2 (Aψ1 )∗ d3 V
46
Physics Formulary by ir. J.C.A. Wevers
When un is the eigenfunction of the eigenvalue equation AΨ = aΨ for eigenvalue a n , Ψ can be expanded into cn un . If this basis is taken orthonormal, then follows for the coefficients: a basis of eigenfunctions: Ψ =
n
cn = un |Ψ . If the system is in a state described by Ψ, the chance to find eigenvalue a n when measuring A is given by |cn |2 in the discrete part of the spectrum and |cn |2 da in the continuous part of the spectrum between a and a + da. The matrix element Aij is given by: Aij = ui |A|uj . Because (AB)ij = ui |AB|uj = |un un | = 1. ui |A |un un |B|uj holds:
n n
The time-dependence of an operator is given by (Heisenberg): dA ∂A [A, H] = + dt ∂t i¯ h with [A, B] ≡ AB − BA the commutator of A and B. For hermitian operators the commutator is always complex. If [A, B] = 0, the operators A and B have a common set of eigenfunctions. By applying this to p x and x follows (Ehrenfest): md2 x t /dt2 = − dU (x)/dx . The first order approximation F (x)
t
≈ F ( x ), with F = −dU/dx represents the classical equation.
Before the addition of quantummechanical operators which are a product of other operators, they should be 1 made symmetrical: a classical product AB becomes 2 (AB + BA).
10.4 The uncertainty principle
If the uncertainty ∆A in A is defined as: (∆A)2 = ψ|Aop − A |2 ψ = A2 − A
1 ∆A · ∆B ≥ 2 | ψ|[A, B]|ψ | 2
it follows:
1 1 1 From this follows: ∆E · ∆t ≥ 2 h, and because [x, px ] = i¯ holds: ∆px · ∆x ≥ 2 h, and ∆Lx · ∆Ly ≥ 2 hLz . ¯ h ¯ ¯
10.5 The Schr¨ dinger equation o
The momentum operator is given by: pop = −i¯ . The position operator is: xop = i¯ p . The energy h h operator is given by: Eop = i¯ ∂/∂t. The Hamiltonian of a particle with mass m, potential energy U and total h energy E is given by: H = p2 /2m + U . From Hψ = Eψ then follows the Schrodinger equation: ¨ − h2 ¯ 2m
2
ψ + U ψ = Eψ = i¯ h
∂ψ ∂t iEt h ¯
The linear combination of the solutions of this equation give the general solution. In one dimension it is: ψ(x, t) = The current density J is given by: J = + dE c(E)uE (x) exp −
h ¯ (ψ ∗ ψ − ψ ψ ∗ ) 2im ∂P (x, t) The following conservation law holds: = − J(x, t) ∂t
10.6 Parity
The parity operator in one dimension is given by Pψ(x) = ψ(−x). If the wavefunction is split in even and odd functions, it can be expanded into eigenfunctions of P:
1 ψ(x) = 2 (ψ(x) + ψ(−x)) + 1 (ψ(x) − ψ(−x)) 2 even:
ψ+
odd:
ψ−
1 o [P, H] = 0. The functions ψ + = 2 (1 + P)ψ(x, t) and ψ − = 1 (1 − P)ψ(x, t) both satisfy the Schr¨ dinger 2 equation. Hence, parity is a conserved quantity.
Chapter 10: Quantum physics
47
10.7 The tunnel effect
The wavefunction of a particle in an ∞ high potential step from x = 0 to x = a is given by ψ(x) = a−1/2 sin(kx). The energylevels are given by En = n2 h2 /8a2 m. If the wavefunction with energy W meets a potential well of W0 > W the wavefunction will, unlike the classical case, be non-zero within the potential well. If 1, 2 and 3 are the areas in front, within and behind the potential well, holds: ψ1 = Aeikx + Be−ikx , ψ2 = Ceik
x
+ De−ik
x
, ψ3 = A eikx
with k 2 = 2m(W − W0 )/¯ 2 and k 2 = 2mW . Using the boundary conditions requiring continuity: ψ = h continuous and ∂ψ/∂x =continuous at x = 0 and x = a gives B, C and D and A expressed in A. The amplitude T of the transmitted wave is defined by T = |A |2 /|A|2 . If W > W0 and 2a = nλ = 2πn/k holds: T = 1.
10.8 The harmonic oscillator
2 For a harmonic oscillator holds: U = 1 bx2 and ω0 = b/m. The Hamiltonian H is then given by: 2
H= with A=
1 2 mωx
p2 + 1 mω 2 x2 = 1 hω + ωA† A 2¯ 2m 2 +√ ip and A† = 2mω
1 2 mωx
−√
ip 2mω
A = A† is non hermitian. [A, A† ] = h and [A, H] = hωA. A is a so called raising ladder operator, A † a ¯ ¯ lowering ladder operator. HAuE = (E − hω)AuE . There is an eigenfunction u0 for which holds: Au0 = 0. ¯ The energy in this ground state is 1 hω: the zero point energy. For the normalized eigenfunctions follows: 2¯ 1 un = √ n! h with En = ( 1 + n)¯ ω. 2 A† √ h ¯
n
u0 with u0 =
4
mω mωx2 exp − π¯ h 2¯ h
10.9 Angular momentum
For the angular momentum operators L holds: [Lz , L2 ] = [Lz , H] = [L2 , H] = 0. However, cyclically holds: [Lx , Ly ] = i¯ Lz . Not all components of L can be known at the same time with arbitrary accuracy. For L z h holds: ∂ ∂ ∂ Lz = −i¯ h = −i¯ x h −y ∂ϕ ∂y ∂x The ladder operators L± are defined by: L± = Lx ± iLy . Now holds: L2 = L+ L− + L2 − hLz . Further, ¯ z L± = he±iϕ ± ¯ ∂ ∂ + i cot(θ) ∂θ ∂ϕ
From [L+ , Lz ] = −¯ L+ follows: Lz (L+ Ylm ) = (m + 1)¯ (L+ Ylm ). h h From [L2 , L± ] = 0 follows: L2 (L± Ylm ) = l(l + 1)¯ 2 (L± Ylm ). h From [L− , Lz ] = hL− follows: Lz (L− Ylm ) = (m − 1)¯ (L− Ylm ). ¯ h
Because Lx and Ly are hermitian (this implies L† = L ) and |L± Ylm |2 > 0 follows: l(l + 1) − m2 − m ≥ ± 0 ⇒ −l ≤ m ≤ l. Further follows that l has to be integral or half-integral. Half-odd integral values give no unique solution ψ and are therefore dismissed.
48
Physics Formulary by ir. J.C.A. Wevers
10.10 Spin
For the spin operators are defined by their commutation relations: [S x , Sy ] = i¯ Sz . Because the spin operators h do not act in the physical space (x, y, z) the uniqueness of the wavefunction is not a criterium here: also half odd-integer values are allowed for the spin. Because [L, S] = 0 spin and angular momentum operators do not have a common set of eigenfunctions. The spin operators are given by S = 1 hσ, with 2¯ σx = 0 1 1 0 , σy = 0 −i i 0 , σz = 1 0 0 −1
The eigenstates of Sz are called spinors: χ = α+ χ+ + α− χ− , where χ+ = (1, 0) represents the state with spin up (Sz = 1 h) and χ− = (0, 1) represents the state with spin down (Sz = − 1 h). Then the probability 2¯ 2¯ to find spin up after a measurement is given by |α+ |2 and the chance to find spin down is given by |α− |2 . Of course holds |α+ |2 + |α− |2 = 1. The electron will have an intrinsic magnetic dipole moment M due to its spin, given by M = −egS S/2m, with gS = 2(1 + α/2π + · · ·) the gyromagnetic ratio. In the presence of an external magnetic field this gives a potential energy U = −M · B. The Schr¨ dinger equation then becomes (because ∂χ/∂x i ≡ 0): o i¯ h egS h ¯ ∂χ(t) = σ · Bχ(t) ∂t 4m
with σ = (σ x , σ y , σ z ). If B = Bez there are two eigenvalues for this problem: χ± for E = ±egS hB/4m = ¯ ±¯ ω. So the general solution is given by χ = (ae−iωt , beiωt ). From this can be derived: Sx = 1 h cos(2ωt) h ¯ 2 1 ¯ and Sy = 2 h sin(2ωt). Thus the spin precesses about the z-axis with frequency 2ω. This causes the normal Zeeman splitting of spectral lines.
1 ¯ The potential operator for two particles with spin ± 2 h is given by:
V (r) = V1 (r) +
1 1 (S1 · S2 )V2 (r) = V1 (r) + 2 V2 (r)[S(S + 1) − 3 ] 2 h2 ¯
This makes it possible for two states to exist: S = 1 (triplet) or S = 0 (Singlet).
10.11 The Dirac formalism
If the operators for p and E are substituted in the relativistic equation E 2 = m2 c4 + p2 c2 , the Klein-Gordon 0 equation is found: 1 ∂2 m2 c 2 2 ψ(x, t) = 0 − 2 2 − 02 c ∂t h ¯ The operator 2 − m2 c2 /¯ 2 can be separated: h 0
2
−
1 ∂2 m2 c 2 − 02 = 2 ∂t2 c h ¯
γλ
∂ m2 c 2 − 02 ∂xλ h ¯
γµ
∂ m2 c 2 + 02 ∂xµ h ¯
where the Dirac matrices γ are given by: γλ γµ + γµ γλ = 2δλµ . From this it can be derived that the γ are hermitian 4 × 4 matrices given by: γk = With this, the Dirac equation becomes: γλ ∂ m2 c 2 + 02 ∂xλ h ¯ ψ(x, t) = 0 0 iσk −iσk 0 , γ4 = I 0 0 −I
where ψ(x) = (ψ1 (x), ψ2 (x), ψ3 (x), ψ4 (x)) is a spinor.
Chapter 10: Quantum physics
49
10.12 Atomic physics
10.12.1 Solutions
The solutions of the Schr¨ dinger equation in spherical coordinates if the potential energy is a function of r o alone can be written as: ψ(r, θ, ϕ) = Rnl (r)Yl,ml (θ, ϕ)χms , with Clm Ylm = √ Plm (cos θ)eimϕ 2π For an atom or ion with one electron holds: Rlm (ρ) = Clm e−ρ/2 ρl L2l+1 (ρ) n−l−1 with ρ = 2rZ/na0 with a0 = ε0 h2 /πme e2 . The Lj are the associated Laguere functions and the Plm are the i associated Legendre polynomials: Pl
|m|
(x) = (1 − x2 )m/2
d|m| (x2 − 1)l dx|m|
, Lm (x) = n
n−1 l=0
(−1)m n! −x −m dn−m −x n e x (e x ) (n − m)! dxn−m
The parity of these solutions is (−1)l . The functions are 2
(2l + 1) = 2n2 -folded degenerated.
10.12.2 Eigenvalue equations
The eigenvalue equations for an atom or ion with with one electron are: Equation Hop ψ = Eψ Lzop Ylm = Lz Ylm L2 Ylm op = L Ylm
2
Eigenvalue En = µe4 Z 2 /8ε2 h2 n2 0 Lz = m l h ¯ L = l(l + 1)¯ h Sz = m s h ¯ S 2 = s(s + 1)¯ 2 h
2 2
Range n≥1 −l ≤ ml ≤ l l<n
1 ms = ± 2 1 2
Szop χ = Sz χ
2 Sop χ = S 2 χ
s=
10.12.3 Spin-orbit interaction
The total momentum is given by J = L + M . The total magnetic dipole moment of an electron is then M = ML + MS = −(e/2me )(L + gS S) where gS = 2.0023 is the gyromagnetic ratio of the electron. Further holds: J 2 = L2 + S 2 + 2L · S = L2 + S 2 + 2Lz Sz + L+ S− + L− S+ . J has quantum numbers j 1 with possible values j = l ± 2 , with 2j + 1 possible z-components (mJ ∈ {−j, .., 0, .., j}). If the interaction energy between S and L is small it can be stated that: E = En + ESL = En + aS · L. It can then be derived that: |En |Z 2 α2 a= 2 1 h nl(l + 1)(l + 2 ) ¯ After a relativistic correction this becomes: E = En + |En |Z 2 α2 n 1 3 − 4n j +
1 2
The fine structure in atomic spectra arises from this. With gS = 2 follows for the average magnetic moment: Mav = −(e/2me )g¯ J, where g is the Land´ -factor: h e g =1+ j(j + 1) + s(s + 1) − l(l + 1) S·J =1+ J2 2j(j + 1)
For atoms with more than one electron the following limiting situations occur:
50
Physics Formulary by ir. J.C.A. Wevers
1. L − S coupling: for small atoms the electrostatic interaction is dominant and the state can be characterized by L, S, J, mJ . J ∈ {|L − S|, ..., L + S − 1, L + S} and mJ ∈ {−J, ..., J − 1, J}. The spectroscopic notation for this interaction is: 2S+1 LJ . 2S + 1 is the multiplicity of a multiplet. 2. j − j coupling: for larger atoms the electrostatic interaction is smaller than the L i · si interaction of an electron. The state is characterized by ji ...jn , J, mJ where only the ji of the not completely filled subshells are to be taken into account. The energy difference for larger atoms when placed in a magnetic field is: ∆E = gµ B mJ B where g is the Land´ factor. For a transition between two singlet states the line splits in 3 parts, for ∆m J = −1, 0 + 1. This e results in the normal Zeeman effect. At higher S the line splits up in more parts: the anomalous Zeeman effect. Interaction with the spin of the nucleus gives the hyperfine structure.
10.12.4 Selection rules
For the dipole transition matrix elements follows: p0 ∼ | l2 m2 |E · r |l1 m1 |. Conservation of angular momentum demands that for the transition of an electron holds that ∆l = ±1. For an atom where L − S coupling is dominant further holds: ∆S = 0 (but not strict), ∆L = 0, ±1, ∆J = 0, ±1 except for J = 0 → J = 0 transitions, ∆mJ = 0, ±1, but ∆mJ = 0 is forbidden if ∆J = 0. For an atom where j − j coupling is dominant further holds: for the jumping electron holds, except ∆l = ±1, also: ∆j = 0, ±1, and for all other electrons: ∆j = 0. For the total atom holds: ∆J = 0, ±1 but no J = 0 → J = 0 transitions and ∆mJ = 0, ±1, but ∆mJ = 0 is forbidden if ∆J = 0.
10.13 Interaction with electromagnetic fields
The Hamiltonian of an electron in an electromagnetic field is given by: H= 1 h2 ¯ (p + eA)2 − eV = − 2µ 2µ
2
+
e2 2 e B·L+ A − eV 2µ 2µ
where µ is the reduced mass of the system. The term ∼ A2 can usually be neglected, except for very strong fields or macroscopic motions. For B = Bez it is given by e2 B 2 (x2 + y 2 )/8µ. When a gauge transformation A = A − f , V = V + ∂f /∂t is applied to the potentials the wavefunction h is also transformed according to ψ = ψeiqef /¯ with qe the charge of the particle. Because f = f (x, t), this is called a local gauge transformation, in contrast with a global gauge transformation which can always be applied.
10.14 Perturbation theory
10.14.1 Time-independent perturbation theory
To solve the equation (H0 + λH1 )ψn = En ψn one has to find the eigenfunctions of H = H0 + λH1 . Suppose 0 that φn is a complete set of eigenfunctions of the non-perturbed Hamiltonian H 0 : H0 φn = En φn . Because φn is a complete set holds: cnk (λ)φk ψn = N (λ) φn +
k=n (1) (2)
When cnk and En are being expanded into λ: cnk = λcnk + λ2 cnk + · · · (1) (2) 0 En = En + λEn + λ2 En + · · ·
Chapter 10: Quantum physics
51
and this is put into the Schr¨ dinger equation the result is: En = φn |H1 |φn and o φm |H1 |φn c(1) = if m = n. The second-order correction of the energy is then given by: nm 0 0 En − E m φk |λH1 |φn | φk |H1 |φn |2 (2) . So to first order holds: ψn = φn + φk . En = 0 0 − E0 0 En En − E k k
k=n k=n
(1)
In case the levels are degenerated the above does not hold. In that case an orthonormal set eigenfunctions φ ni is chosen for each level n, so that φmi |φnj = δmn δij . Now ψ is expanded as: (1) cnk βi φki + · · · ψn = N (λ) αi φni + λ
i k=n i (1) i i
0 0 0 Eni = Eni + λEni is approximated by Eni := En . Substitution in the Schr¨ dinger equation and taking dot o (1) product with φni gives: αi φnj |H1 |φni = En αj . Normalization requires that |αi |2 = 1.
10.14.2 Time-dependent perturbation theory
From the Schr¨ dinger equation i¯ o h and the expansion ψ(t) =
n
∂ψ(t) = (H0 + λV (t))ψ(t) ∂t
0 −iEn t h ¯
cn (t) exp
φn with cn (t) = δnk + λcn (t) + · · · dt
(1)
follows:
c(1) (t) n
λ = i¯ h
0
t
φn |V (t )|φk exp
0 0 i(En − Ek )t h ¯
10.15 N-particle systems
10.15.1 General
Identical particles are indistinguishable. For the total wavefunction of a system of identical indistinguishable particles holds: 1. Particles with a half-odd integer spin (Fermions): ψtotal must be antisymmetric w.r.t. interchange of the coordinates (spatial and spin) of each pair of particles. The Pauli principle results from this: two Fermions cannot exist in an identical state because then ψtotal = 0. 2. Particles with an integer spin (Bosons): ψtotal must be symmetric w.r.t. interchange of the coordinates (spatial and spin) of each pair of particles. For a system of two electrons there are 2 possibilities for the spatial wavefunction. When a and b are the quantum numbers of electron 1 and 2 holds: ψS (1, 2) = ψa (1)ψb (2) + ψa (2)ψb (1) , ψA (1, 2) = ψa (1)ψb (2) − ψa (2)ψb (1) Because the particles do not approach each other closely the repulsion energy at ψ A in this state is smaller. The following spin wavefunctions are possible: √ χA = 1 2[χ+ (1)χ− (2) − χ+ (2)χ− (1)] ms = 0 2 ms = +1 χ+ (1)χ+ (2) √ 1 χS = 2[χ+ (1)χ− (2) + χ+ (2)χ− (1)] ms = 0 2 χ− (1)χ− (2) ms = −1
Because the total wavefunction must be antisymmetric it follows: ψtotal = ψS χA or ψtotal = ψA χS .
52
Physics Formulary by ir. J.C.A. Wevers
For N particles the symmetric spatial function is given by: ψS (1, ..., N ) = ψ(all permutations of 1..N )
1 The antisymmetric wavefunction is given by the determinant ψA (1, ..., N ) = √ |uEi (j)| N!
10.15.2 Molecules
The wavefunctions of atom a and b are φa and φb . If the 2 atoms approach each other there are √ possibilities: two the total wavefunction approaches the bonding function with lower total energy ψ B = 1 2(φa + φb ) or 2 √ approaches the anti-bonding function with higher energy ψ AB = 1 2(φa − φb ). If a molecular-orbital is 2 symmetric w.r.t. the connecting axis, like a combination of two s-orbitals it is called a σ-orbital, otherwise a π-orbital, like the combination of two p-orbitals along two axes. The energy of a system is: E = ψ|H|ψ . ψ|ψ
The energy calculated with this method is always higher than the real energy if ψ is only an approximation for the solutions of Hψ = Eψ. Also, if there are more functions to be chosen, the function which gives the lowest energy is the best approximation. Applying this to the function ψ = ci φi one finds: (Hij − ESij )ci = 0. This equation has only solutions if the secular determinant |Hij − ESij | = 0. Here, Hij = φi |H|φj and Sij = φi |φj . αi := Hii is the Coulomb integral and βij := Hij the exchange integral. Sii = 1 and Sij is the overlap integral. The first approximation in the molecular-orbital theory is to place both electrons of a chemical bond in the bonding orbital: ψ(1, 2) = ψB (1)ψB (2). This results in a large electron density between the nuclei and therefore a repulsion. A better approximation is: ψ(1, 2) = C1 ψB (1)ψB (2) + C2 ψAB (1)ψAB (2), with C1 = 1 and C2 ≈ 0.6. In some atoms, such as C, it is energetical more suitable to form orbitals which are a linear combination of the s, p and d states. There are three ways of hybridization in C: √ 1. SP-hybridization: ψsp = 1 2(ψ2s ± ψ2pz ). There are 2 hybrid orbitals which are placed on one line 2 under 180◦ . Further the 2px and 2py orbitals remain. √ √ √ 2. SP2 hybridization: ψsp2 = ψ2s / 3 + c1 ψ2pz + c2 ψ2py , where (c1 , c2 ) ∈ {( 2/3, 0), (−1/ 6, 1/ 2) √ √ , (−1/ 6, −1/ 2)}. The 3 SP2 orbitals lay in one plane, with symmetry axes which are at an angle of 120◦ . 3. SP3 hybridization: ψsp3 = 1 (ψ2s ± ψ2pz ± ψ2py ± ψ2px ). The 4 SP3 orbitals form a tetraheder with the 2 symmetry axes at an angle of 109◦28 .
10.16 Quantum statistics
If a system exists in a state in which one has not the disposal of the maximal amount of information about the system, it can be described by a density matrix ρ. If the probability that the system is in state ψ i is given by ai , one can write for the expectation value a of A: a = ri ψi |A|ψi .
i
If ψ is expanded into an orthonormal basis {φk } as: ψ (i) = A =
k
k
ck φk , holds:
(i)
(Aρ)kk = Tr(Aρ)
where ρlk = c∗ cl . ρ is hermitian, with Tr(ρ) = 1. Further holds ρ = ri |ψi ψi |. The probability to find k eigenvalue an when measuring A is given by ρnn if one uses a basis of eigenvectors of A for {φk }. For the time-dependence holds (in the Schr¨ dinger image operators are not explicitly time-dependent): o i¯ h dρ = [H, ρ] dt
Chapter 10: Quantum physics
53
For a macroscopic system in equilibrium holds [H, ρ] = 0. If all quantumstates with the same energy are equally probable: Pi = P (Ei ), one can obtain the distribution: Pn (E) = ρnn = e−En /kT with the state sum Z = Z e−En /kT
n
The thermodynamic quantities are related to these definitions as follows: F = −kT ln(Z), U = H = ∂ pn En = − ln(Z), S = −k Pn ln(Pn ). For a mixed state of M orthonormal quantum states with ∂kT n n probability 1/M follows: S = k ln(M ). The distribution function for the internal states for a system in thermal equilibrium is the most probable function. This function can be found by taking the maximum of the function which gives the number of states with Stirling’s equation: ln(n!) ≈ n ln(n) − n, and the conditions nk = N and nk Wk = W . For identical,
k k
indistinguishable particles which obey the Pauli exclusion principle the possible number of states is given by: P =
k
gk ! nk !(gk − nk )!
This results in the Fermi-Dirac statistics. For indistinguishable particles which do not obey the exclusion principle the possible number of states is given by: P = N!
k n gk k nk !
This results in the Bose-Einstein statistics. So the distribution functions which explain how particles are distributed over the different one-particle states k which are each g k -fold degenerate depend on the spin of the particles. They are given by: 1. Fermi-Dirac statistics: integer spin. nk ∈ {0, 1}, nk = with ln(Zg ) = gk ln[1 + exp((Ei − µ)/kT )]. N gk Zg exp((Ek − µ)/kT ) + 1 gk N Zg exp((Ek − µ)/kT ) − 1
2. Bose-Einstein statistics: half odd-integer spin. nk ∈ I , nk = N with ln(Zg ) = −
T →0
gk ln[1 − exp((Ei − µ)/kT )].
Here, Zg is the large-canonical state sum and µ the chemical potential. It is found by demanding nk = N , and for it holds: lim µ = EF , the Fermi-energy. N is the total number of particles. The Maxwell-Boltzmann distribution can be derived from this in the limit Ek − µ nk = N Ek exp − Z kT kT : gk exp − Ek kT
with Z =
k
With the Fermi-energy, the Fermi-Dirac and Bose-Einstein statistics can be written as: 1. Fermi-Dirac statistics: nk = gk . exp((Ek − EF )/kT ) + 1
2. Bose-Einstein statistics: nk =
gk . exp((Ek − EF )/kT ) − 1
Chapter 11
Plasma physics
11.1 Introduction
ne ne + n 0 where ne is the electron density and n0 the density of the neutrals. If a plasma contains also negative charged ions α is not well defined. The degree of ionization α of a plasma is defined by: α = The probability that a test particle collides with another is given by dP = nσdx where σ is the cross section. The collision frequency νc = 1/τc = nσv. The mean free path is given by λv = 1/nσ. The rate coefficient K is defined by K = σv . The number of collisions per unit of time and volume between particles of kind 1 and 2 is given by n1 n2 σv = Kn1 n2 . The potential of an electron is given by: V (r) = −e r exp − 4πε0 r λD with λD = ε0 kTe Ti ≈ e Ti + n i Te ) ε0 kTe ne e 2
e2 (n
because charge is shielded in a plasma. Here, λD is the Debye length. For distances < λD the plasma cannot be assumed to be quasi-neutral. Deviations of charge neutrality by thermic motion are compensated by oscillations with frequency ωpe = ne e 2 me ε 0
The distance of closest approximation when two equal charged particles collide for a deviation of π/2 is −1/3 2b0 = e2 /(4πε0 1 mv 2 ). A “neat” plasma is defined as a plasma for which holds: b0 < ne λD Lp . 2 Here Lp := |ne / ne | is the gradient length of the plasma.
11.2 Transport
1 Relaxation times are defined as τ = 1/νc . Starting with σm = 4πb2 ln(ΛC ) and with 2 mv 2 = kT it can be 0 found that: √ √ 8 2πε2 m(kT )3/2 4πε2 m2 v 3 0 0 = τm = 4 ne ln(ΛC ) ne4 ln(ΛC ) For momentum transfer between electrons and ions holds for a Maxwellian velocity distribution: √ √ √ √ 6π 3ε2 me (kTe )3/2 6π 3ε2 mi (kTi )3/2 0 0 τee = ≈ τei , τii = ne e4 ln(ΛC ) ni e4 ln(ΛC )
The energy relaxation times for identical particles are equal to the momentum relaxation times. Because for e-i collisions the energy transfer is only ∼ 2me /mi this is a slow process. Approximately holds: τee : τei : E τie : τie = 1 : 1 : mi /me : mi /me . The relaxation for e-o interaction is much more complicated. For T > 10 eV holds approximately: σ eo = −2/5 10−17 ve , for lower energies this can be a factor 10 lower. The resistivity η = E/J of a plasma is given by: η= √ e2 me ln(ΛC ) ne e 2 √ 2 = me νei 6π 3ε0 (kTe )3/2
Chapter 11: Plasma physics
55
The diffusion coefficient D is defined by means of the flux Γ by Γ = nvdiff = −D n. The equation of continuity is ∂t n + (nvdiff ) = 0 ⇒ ∂t n = D 2 n. One finds that D = 1 λv v. A rough estimate gives 3 τD = Lp /D = L2 τc /λ2 . For magnetized plasma’s λv must be replaced with the cyclotron radius. In electrical p v fields also holds J = neµE = e(ne µe + ni µi )E with µ = e/mνc the mobility of the particles. The Einstein ratio is: kT D = µ e Because a plasma is electrically neutral electrons and ions are strongly coupled and they don’t diffuse independent. The coefficient of ambipolar diffusion Damb is defined by Γ = Γi = Γe = −Damb ne,i . From this follows that kTe µi kTe /e − kTi /e ≈ Damb = 1/µe − 1/µi e In an external magnetic field B0 particles will move in spiral orbits with cyclotron radius ρ = mv/eB0 and with cyclotron frequency Ω = B0 e/m. The helical orbit is perturbed by collisions. A plasma is called magnetized if λv > ρe,i . So the electrons are magnetized if √ me e3 ne ln(ΛC ) ρe √ 2 = <1 λee 6π 3ε0 (kTe )3/2 B0 Magnetization of only the electrons is sufficient to confine the plasma reasonable because they are coupled to the ions by charge neutrality. In case of magnetic confinement holds: p = J × B. Combined with the two stationary Maxwell equations for the B-field these form the ideal magneto-hydrodynamic equations. For a uniform B-field holds: p = nkT = B 2 /2µ0 . If both magnetic and electric fields are present electrons and ions will move in the same direction. If E = Er er + Ez ez and B = Bz ez the E × B drift results in a velocity u = (E × B )/B 2 and the velocity in the ˙ r, ϕ plane is r(r, ϕ, t) = u + ρ(t). ˙
11.3 Elastic collisions
11.3.1 General
The scattering angle of a particle in interaction with another particle, as shown in the figure at the right is:
∞
b s d d ra b T c ϕ M χ
χ = π − 2b
dr r2 1− W (r) b2 − r2 E0
ra
Particles with an impact parameter between b and b + db, moving through a ring with dσ = 2πbdb leave the scattering area at a solid angle dΩ = 2π sin(χ)dχ. The differential cross section is then defined as: I(Ω) = dσ b ∂b = dΩ sin(χ) ∂χ
For a potential energy W (r) = kr −n follows: I(Ω, v) ∼ v −4/n . For low energies, O(1 eV), σ has a Ramsauer minimum. It arises from the interference of matter waves behind the object. I(Ω) for angles 0 < χ < λ/4 is larger than the classical value.
56
Physics Formulary by ir. J.C.A. Wevers
11.3.2 The Coulomb interaction
2 For the Coulomb interaction holds: 2b0 = q1 q2 /2πε0 mv0 , so W (r) = 2b0 /r. This gives b = b0 cot( 1 χ) and 2
I(Ω =
b ∂b b2 0 = 1 sin(χ) ∂χ 4 sin2 ( 2 χ)
Because the influence of a particle vanishes at r = λD holds: σ = π(λ2 − b2 ). Because dp = d(mv) = 0 D mv0 (1 − cos χ) a cross section related to momentum transfer σm is given by: σm = (1 − cos χ)I(Ω)dΩ = 4πb2 ln 0 1 sin( 1 χmin ) 2 = 4πb2 ln 0 λD b0 := 4πb2 ln(ΛC ) ∼ 0 ln(v 4 ) v4
where ln(ΛC ) is the Coulomb-logarithm. For this quantity holds: ΛC = λD /b0 = 9n(λD ).
11.3.3 The induced dipole interaction
The induced dipole interaction, with p = αE, gives a potential V and an energy W in a dipole field given by: V (r) = |e|p αe2 p · er , W (r) = − =− 4πε0 r2 8πε0 r2 2(4πε0 )2 r4
∞
with ba =
4
2e2 α holds: χ = π − 2b 2 (4πε0 )2 1 mv0 2
dr r2 1− b4 b2 + a4 2 r 4r
ra
If b ≥ ba the charge would hit the atom. Repulsing nuclear forces prevent this to happen. If the scattering angle is a lot times 2π it is called capture. The cross section for capture σ orb = πb2 is called the Langevin a limit, and is a lowest estimate for the total cross section.
11.3.4 The centre of mass system
If collisions of two particles with masses m1 and m2 which scatter in the centre of mass system by an angle χ are compared with the scattering under an angle θ in the laboratory system holds: tan(θ) = m2 sin(χ) m1 + m2 cos(χ)
The energy loss ∆E of the incoming particle is given by: ∆E = E
1 2 2 m2 v 2 1 2 2 m1 v 1
=
2m1 m2 (1 − cos(χ)) (m1 + m2 )2
11.3.5 Scattering of light
Scattering of light by free electrons is called Thomson scattering. The scattering is free from collective effects if kλD 1. The cross section σ = 6.65 · 10−29 m2 and 2v ∆f = sin( 1 χ) 2 f c This gives for the scattered energy Escat ∼ nλ4 /(λ2 − λ2 )2 with n the density. If λ λ0 it is called Rayleigh 0 0 scattering. Thomson sccattering is a limit of Compton scattering, which is given by λ − λ = λC (1 − cos χ) with λC = h/mc and cannot be used any more if relativistic effects become important.
Chapter 11: Plasma physics
57
11.4 Thermodynamic equilibrium and reversibility
Planck’s radiation law and the Maxwellian velocity distribution hold for a plasma in equilibrium: ρ(ν, T )dν = 8πhν 3 2πn √ 1 E E exp − dν , N (E, T )dE = 3 3/2 c exp(hν/kT ) − 1 kT (πkT ) dE
“Detailed balancing” means that the number of reactions in one direction equals the number of reactions in the opposite direction because both processes have equal probability if one corrects for the used phase space. For the reaction Xback Xforward → ←
forward back
holds in a plasma in equilibrium microscopic reversibility: ηforward = ˆ
forward back
ηback ˆ
If the velocity distribution is Maxwellian, this gives: ηx = ˆ nx h3 e−Ekin /kT gx (2πmx kT )3/2
where g is the statistical weight of the state and n/g := η. For electrons holds g = 2, for excited states usually holds g = 2j + 1 = 2n2 . With this one finds for the Boltzmann balance, Xp + e− → X1 + e− + (E1p ): ← nB gp p = exp n1 g1 Ep − E 1 kTe
And for the Saha balance, Xp + e− + (Epi ) → X+ + 2e− : ← 1 nS n+ ne h3 p 1 = + exp gp g1 ge (2πme kTe )3/2 Epi kTe
Because the number of particles on the left-hand side and right-hand side of the equation is different, a factor g/Ve remains. This factor causes the Saha-jump. From microscopic reversibility one can derive that for the rate coefficients K(p, q, T ) := σv K(q, p, T ) = gp K(p, q, T ) exp gq ∆Epq kT
pq
holds:
11.5 Inelastic collisions
11.5.1 Types of collisions
The kinetic energy can be split in a part of and a part in the centre of mass system. The energy in the centre of mass system is available for reactions. This energy is given by E= m1 m2 (v1 − v2 )2 2(m1 + m2 )
Some types of inelastic collisions important for plasma physics are: 1. Excitation: Ap + e− → Aq + e− ← 2. Decay: Aq → Ap + hf ←
58
Physics Formulary by ir. J.C.A. Wevers
3. Ionisation and 3-particles recombination: Ap + e− → A+ + 2e− ← 4. radiative recombination: A+ + e− → Ap + hf ← 5. Stimulated emission: Aq + hf → Ap + 2hf
8. Charge transfer: A+ + B → A + B+ ←
7. Penning ionisation: b.v. Ne∗ + Ar → Ar+ + Ne + e− ←
6. Associative ionisation: A∗∗ + B → AB+ + e− ←
9. Resonant charge transfer: A+ + A → A + A+ ←
11.5.2 Cross sections
Collisions between an electron and an atom can be approximated by a collision between an electron and one of the electrons of that atom. This results in πZ 2 e4 dσ = d(∆E) (4πε0 )2 E(∆E)2 Then follows for the transition p → q: σpq (E) = πZ 2 e4 ∆Eq,q+1 (4πε0 )2 E(∆E)2 pq 1 1 − Ep E ln 1.25βE Ep
For ionization from state p holds to a good approximation: σp = 4πa2 Ry 0 For resonant charge transfer holds: σex = A[1 − B ln(E)]2 1 + CE 3.3
In equilibrium holds for radiation processes: np Apq + np Bpq ρ(ν, T ) = nq Bqp ρ(ν, T )
emission stimulated emission absorption
Here, Apq is the matrix element of the transition p → q, and is given by: Apq = 8π 2 e2 ν 3 |rpq |2 with rpq = ψp |r |ψq 3¯ ε0 c3 h
q
For hydrogenic atoms holds: Ap = 1.58 · 108 Z 4 p−4.5 , with Ap = 1/τp = given by Ipq = hf Apq np /4π. The Einstein coefficients B are given by: Bpq = c3 Apq Bpq gq and = 8πhν 3 Bqp gp
Apq . The intensity I of a line is
A spectral line is broadened by several mechanisms: 1. Because the states have a finite life time. The natural life time of a state p is given by τ p = 1/
q
Apq .
From the uncertainty relation then follows: ∆(hν) · τp = 1 h, this gives 2¯ ∆ν = The natural line width is usually 1 = 4πτp
q
Apq 4π
than the broadening due to the following two mechanisms:
Chapter 11: Plasma physics
59
2. The Doppler broadening is caused by the thermal motion of the particles: ∆λ 2 = λ c 2 ln(2)kTi mi
This broadening results in a Gaussian line profile: √ kν = k0 exp(−[2 ln 2(ν − ν0 )/∆νD ]2 ), with k the coefficient of absorption or emission. 3. The Stark broadening is caused by the electric field of the electrons: ∆λ1/2 = ne C(ne , Te )
2/3
˚ with for the H-β line: C(ne , Te ) ≈ 3 · 1014 A−3/2 cm−3 . The natural broadening and the Stark broadening result in a Lorentz profile of a spectral line: 1 1 kν = 2 k0 ∆νL /[( 2 ∆νL )2 + (ν − ν0 )2 ]. The total line shape is a convolution of the Gauss- and Lorentz profile and is called a Voigt profile. The number of transitions p → q is given by np Bpq ρ and by np nhf σa c = np (ρdν/hν)σa c where dν is the line width. Then follows for the cross section of absorption processes: σ a = Bpq hν/cdν. The background radiation in a plasma originates from two processes: 1. Free-Bound radiation, originating from radiative recombination. The emission is given by: εf b = hc C 1 z i ni ne √ 1 − exp − 2 λ λkTe kTe ξf b (λ, Te )
with C1 = 1.63 · 10−43 Wm4 K1/2 sr−1 and ξ the Biberman factor. 2. Free-free radiation, originating from the acceleration of particles in the EM-field of other particles: εf f = C 1 z i ni ne hc √ exp − λ2 kTe λkTe ξf f (λ, Te )
11.7 The Boltzmann transport equation
It is assumed that there exists a distribution function F for the plasma so that F (r, v, t) = Fr (r, t) · Fv (v, t) = F1 (x, t)F2 (y, t)F3 (z, t)F4 (vx , t)F5 (vy , t)F6 (vz , t) Then the BTE is: dF ∂F = + dt ∂t
r
· (F v ) +
v
· (F a ) =
∂F ∂t
Assuming that v does not depend on r and ai does not depend on vi , holds r ·(F v ) = v· F and v ·(F a ) = a · v F . This is also true in magnetic fields because ∂ai /∂xi = 0. The velocity is separated in a thermal velocity vt and a drift velocity w. The total density is given by n = F dv and vF dv = nw. The balance equations can be derived by means of the moment method: 1. Mass balance: (BTE)dv ⇒ ∂n + ∂t · (nw) = dw + dt ∂n ∂t
cr
2. Momentum balance: 3. Energy balance:
(BTE)mvdv ⇒ mn
T + ·w+
p = mn a + R ·q =Q
(BTE)mv 2 dv ⇒
3 dp 5 + p 2 dt 2
60
Physics Formulary by ir. J.C.A. Wevers
1 Here, a = e/m(E + w × B ) is the average acceleration, q = 2 nm vt2 vt the heat flow, 2 mvt ∂F Q = dv the source term for energy production, R is a friction term and p = nkT the r ∂t cr pressure.
A thermodynamic derivation gives for the total pressure: p = nkT =
i
pi −
e2 (ne + zi ni ) 24πε0 λD wi :
For the electrical conductance in a plasma follows from the momentum balance, if w e ηJ = E − J ×B+ ene pe
In a plasma where only elastic e-a collisions are important the equilibrium energy distribution function is the Druyvesteyn distribution: N (E)dE = Cne with E0 = eEλv = eE/nσ. E E0
3/2
exp −
3me m0
E E0
2
dE
These models are first-moment equations for excited states. One assumes the Quasi-steady-state solution is valid, where ∀p>1 [(∂np /∂t = 0) ∧ ( · (np wp ) = 0)]. This results in: ∂np>1 ∂t =0 ,
cr
∂n1 + ∂t
· (n1 w1 ) =
∂n1 ∂t
,
cr
∂ni + ∂t
· (ni wi ) =
∂ni ∂t
cr
with solutions np = = Further holds for all collision-dominated levels that δbp := bp −1 = −x b0 peff with peff = Ry/Epi and 5 ≤ x ≤ 6. For systems in ESP, where only collisional (de)excitation between levels p and p ± 1 is taken into account holds x = 6. Even in plasma’s far from equilibrium the excited levels will eventually reach ESP, so from a certain level up the level densities can be calculated. To find the population densities of the lower levels in the stationary case one has to start with a macroscopic equilibrium: Number of populating processes of level p = Number of depopulating processes of level p , When this is expanded it becomes: ne
q<p coll. excit.
0 1 rp nS +rp nB p p
b p nS . p
nq Kqp + ne
q>p
nq Kqp +
q>p
nq Aqp + n2 ni K+p + ne ni αrad = e
coll. deexcit.
ne np
q<p
Kpq + ne np
q>p
Kpq + np
q<p
Apq + ne np Kp+
coll. ion.
coll. deexcit.
coll. excit.
11.9 Waves in plasma’s
Interaction of electromagnetic waves in plasma’s results in scattering and absorption of energy. For electromagnetic waves with complex wave number k = ω(n + iκ)/c in one dimension one finds: Ex = E0 e−κωx/c cos[ω(t − nx/c)]. The refractive index n is given by: n=c c k = = ω vf 1−
2 ωp ω2
Chapter 11: Plasma physics
61
ˆ For disturbances in the z-direction in a cold, homogeneous, magnetized plasma: B = B0 ez + Bei(kz−ωt) and i(kz−ωt) n = n0 + ne ˆ (external E fields are screened) follows, with the definitions α = ωp /ω and β = Ω/ω 2 2 2 and ωp = ωpi + ωpe: 1 2 1 − βs iβs 2 1 − βs 0 −iβs 2 1 − βs 1 2 1 − βs 0 0
J = σ E , with σ = iε0 ω
s
where the sum is taken over particle species s. The dielectric tensor E, with property: k · (E · E) = 0 is given by E = I − σ/iε0 ω. With the definitions S = 1 − follows:
s
α2 s
1
0
α2 s , D= 2 1 − βs
s
α 2 βs s , P =1− 2 1 − βs −iD S 0 0 0 P
α2 s
s
where n is the refractive index. From this the following solutions can be obtained: A. θ = 0: transmission in the z-direction. 1. P = 0: Ex = Ey = 0. This describes a longitudinal linear polarized wave. 2. n2 = L: a left, circular polarized wave. 3. n2 = R: a right, circular polarized wave. B. θ = π/2: transmission ⊥ the B-field.
The eigenvalues of this hermitian matrix are R = S + D, L = S − D, λ3 = P , with eigenvectors er = √ √ 1 1 2 2(1, i, 0), el = 2 2(1, −i, 0) and e3 = (0, 0, 1). er is connected with a right rotating field for which iEx /Ey = 1 and el is connected with a left rotating field for which iEx /Ey = −1. When k makes an angle θ with B one finds: P (n2 − R)(n2 − L) tan2 (θ) = S(n2 − RL/S)(n2 − P )
S E = iD 0
1. n2 = P : the ordinary mode: Ex = Ey = 0. This is a transversal linear polarized wave. 2. n2 = RL/S: the extraordinary mode: iEx /Ey = −D/S, an elliptical polarized wave. Resonance frequencies are frequencies for which n2 → ∞, so vf = 0. For these holds: tan(θ) = −P/S. For R → ∞ this gives the electron cyclotron resonance frequency ω = Ω e , for L → ∞ the ion cyclotron resonance frequency ω = Ωi and for S = 0 holds for the extraordinary mode: α2 1 − mi Ω 2 i me ω 2 = 1− m2 Ω 2 i i m2 ω 2 e 1− Ω2 i ω2
In the case that β 2
Cut-off frequencies are frequencies for which n2 = 0, so vf → ∞. For these holds: P = 0 or R = 0 or L = 0. 1 one finds Alfv´ n waves propagating parallel to the field lines. With the Alfv´ n velocity e e vA = follows: n = 1 + c/vA , and in case vA
2 ωpe
Ωe Ωi 2 2 c + ωpi
c: ω = kvA .
Chapter 12
Solid state physics
12.1 Crystal structure
A lattice is defined by the 3 translation vectors ai , so that the atomic composition looks the same from each point r and r = r + T , where T is a translation vector given by: T = u1 a1 + u2 a2 + u3 a3 with ui ∈ I . A N lattice can be constructed from primitive cells. As a primitive cell one can take a parallellepiped, with volume Vcell = |a1 · (a2 × a3 )| Because a lattice has a periodical structure the physical properties which are connected with the lattice have the same periodicity (neglecting boundary effects): ne (r + T ) = ne (r ) This periodicity is suitable to use Fourier analysis: n(r ) is expanded as: n(r ) =
G
nG exp(iG · r )
with nG =
1 Vcell
n(r ) exp(−iG · r )dV
cell
G is the reciprocal lattice vector. If G is written as G = v1 b1 + v2 b2 + v3 b3 with vi ∈ I , it follows for the N vectors bi , cyclically: ai+1 × ai+2 bi = 2π ai · (ai+1 × ai+2 ) The set of G-vectors determines the R¨ ntgen diffractions: a maximum in the reflected radiation occurs if: o ∆k = G with ∆k = k − k . So: 2k · G = G2 . From this follows for parallel lattice planes (Bragg reflection) that for the maxima holds: 2d sin(θ) = nλ. The Brillouin zone is defined as a Wigner-Seitz cell in the reciprocal lattice.
12.2 Crystal binding
A distinction can be made between 4 binding types: 1. Van der Waals bond 2. Ion bond 3. Covalent or homopolar bond 4. Metalic bond. For the ion binding of NaCl the energy per molecule is calculated by: E = cohesive energy(NaCl) – ionization energy(Na) + electron affinity(Cl) The interaction in a covalent bond depends on the relative spin orientations of the electrons constituing the bond. The potential energy for two parallel spins is higher than the potential energy for two antiparallel spins. Furthermore the potential energy for two parallel spins has sometimes no minimum. In that case binding is not possible.
Chapter 12: Solid state physics
63
12.3 Crystal vibrations
12.3.1 A lattice with one type of atoms
In this model for crystal vibrations only nearest-neighbour interactions are taken into account. The force on atom s with mass M can then be written as: Fs = M d 2 us = C(us+1 − us ) + C(us−1 − us ) dt2
Assuming that all solutions have the same time-dependence exp(−iωt) this results in: −M ω 2 us = C(us+1 + us−1 − 2us ) Further it is postulated that: us±1 = u exp(isKa) exp(±iKa). This gives: us = exp(iKsa). Substituting the later two equations in the fist results in a system of linear equations, which has only a solution if their determinant is 0. This gives: ω2 = 4C sin2 ( 1 Ka) 2 M
Only vibrations with a wavelength within the first Brillouin Zone have a physical significance. This requires that −π < Ka ≤ π. The group velocity of these vibrations is given by: vg = dω = dK Ca2 1 cos( 2 Ka) . M
and is 0 on the edge of a Brillouin Zone. Here, there is a standing wave.
12.3.2 A lattice with two types of atoms
Now the solutions are: ω2 = C 1 1 + M1 M2 ±C 1 1 + M1 M2
2
ω T − 4 sin2 (Ka) M1 M2
2C M2 2C M1
Connected with each value of K are two values of ω, as can be seen in the graph. The upper line describes the optical branch, the lower line the acoustical branch. In the optical branch, both types of ions oscillate in opposite phases, in the acoustical branch they oscillate in the same phase. This results in a much larger induced dipole moment for optical oscillations, and also a stronger emission and absorption of radiation. Furthermore each branch has 3 polarization directions, one longitudinal and two transversal.
0
E K π/a
12.3.3 Phonons
The quantum mechanical excitation of a crystal vibration with an energy hω is called a phonon. Phonons ¯ can be viewed as quasi-particles: with collisions, they behave as particles with momentum hK. Their total ¯ momentum is 0. When they collide, their momentum need not be conserved: for a normal process holds: K1 + K2 = K3 , for an umklapp process holds: K1 + K2 = K3 + G. Because phonons have no spin they behave like bosons.
64
Physics Formulary by ir. J.C.A. Wevers
12.3.4 Thermal heat capacity
The total energy of the crystal vibrations can be calculated by multiplying each mode with its energy and sum over all branches K and polarizations P : U=
K P
hω nk,p = ¯
λ
Dλ (ω)
hω ¯ dω exp(¯ ω/kT ) − 1 h
for a given polarization λ. The thermal heat capacity is then: Clattice = ∂U =k ∂T D(ω)
λ
(¯ ω/kT )2 exp(¯ ω/kT ) h h dω (exp(¯ ω/kT ) − 1)2 h
The dispersion relation in one dimension is given by: D(ω)dω = L dK L dω dω = π dω π vg
In three dimensions one applies periodic boundary conditions to a cube with N 3 primitive cells and a volume L3 : exp(i(Kx x + Ky y + Kz z)) ≡ exp(i(Kx (x + L) + Ky (y + L) + Kz (z + L))). Because exp(2πi) = 1 this is only possible if: Kx , Ky , Kz = 0; ± 4π 6π 2N π 2π ; ± ; ± ; ... ± L L L L
So there is only one allowed value of K per volume (2π/L)3 in K-space, or: L 2π
3
=
V 8π 3
allowed K-values per unit volume in K-space, for each polarization and each branch. The total number of states with a wave vector < K is: 3 L 4πK 3 N= 2π 3 for each polarization. The density of states for each polarization is, according to the Einstein model: D(ω) = dN = dω V K2 2π 2 V dK = dω 8π 3 dAω vg
The Debye model for thermal heat capacities is a low-temperature approximation which is valid up to ≈ 50K. Here, only the acoustic phonons are taken into account (3 polarizations), and one assumes that v = ωK, independent of the polarization. From this follows: D(ω) = V ω 2 /2π 2 v 3 , where v is the speed of sound. This gives:
ωD
U =3
D(ω) n hωdω = ¯
0
V ω2 hω ¯ 3V k 2 T 4 dω = 2π 2 v 3 exp(¯ ω/kT ) − 1 h 2π 2 v 3 h3 ¯
1/3
xD
0
x3 dx . ex − 1
Here, xD = hωD /kT = θD /T . θD is the Debye temperature and is defined by: ¯ θD = hv ¯ k 6π 2 N V
where N is the number of primitive cells. Because xD → ∞ for T → 0 it follows from this: U = 9N kT T θD
3 ∞
0
x3 dx 3π 4 N kT 4 12π 4 N kT 3 = ∼ T 4 and CV = ∼ T3 3 x−1 e 5θD 5θD
In the Einstein model for the thermal heat capacity one considers only phonons at one frequency, an approximation for optical phonons.
Chapter 12: Solid state physics
65
12.4 Magnetic field in the solid state
The following graph shows the magnetization versus fieldstrength for different types of magnetism: M T Msat χm ∂M = ∂H
ferro paramagnetism
0 hhhh h
diamagnetism hhhh hh h
E H
12.4.1 Dielectrics
The quantum mechanical origin of diamagnetism is the Larmorprecession of the spin of the electron. Starting with a circular electron orbit in an atom with two electrons, there is a Coulomb force F c and a magnetic force on each electron. If the magnetic part of the force is not strong enough to significantly deform the orbit holds: ω2 = Fc (r) eB eB 2 ± ω = ω0 ± (ω0 + δ) ⇒ ω = mr m m ω0 ± eB 2m
2
+ · · · ≈ ω0 ±
eB = ω0 ± ωL 2m
Here, ωL is the Larmor frequency. One electron is accelerated, the other decelerated. Hence there is a net circular current which results in a magnetic moment µ. The circular current is given by I = −Zeω L /2π, and µ = IA = Iπ ρ2 = 2 Iπ r2 . If N is the number of atoms in the crystal it follows for the susceptibility, 3 with M = µN : µ0 N Ze2 2 µ0 M =− r χ= B 6m
12.4.2 Paramagnetism
Starting with the splitting of energy levels in a weak magnetic field: ∆Um − µ · B = mJ gµB B, and with a distribution fm ∼ exp(−∆Um /kT ), one finds for the average magnetic moment µ = fm µ/ fm . After 1 linearization and because mJ = 0, J = 2J + 1 and m2 = 2 J(J + 1)(J + 2 ) it follows that: J 3 χp = This is the Curie law, χp ∼ 1/T . µ0 N µ µ0 J(J + 1)g 2 µ2 N µ0 M B = = B B 3kT
12.4.3 Ferromagnetism
A ferromagnet behaves like a paramagnet above a critical temperature T c . To describe ferromagnetism a field BE parallel with M is postulated: BE = λµ0 M . From there the treatment is analogous to the paramagnetic case: C µ0 M = χp (Ba + BE ) = χp (Ba + λµ0 M ) = µ0 1 − λ M T From this follows for a ferromagnet: χF = C µ0 M = which is Weiss-Curie’s law. Ba T − Tc
If BE is estimated this way it results in values of about 1000 T. This is clearly unrealistic and suggests another mechanism. A quantum mechanical approach from Heisenberg postulates an interaction between two neighbouring atoms: U = −2J Si · Sj ≡ −µ · BE . J is an overlap integral given by: J = 3kTc /2zS(S + 1), with z the number of neighbours. A distinction between 2 cases can now be made: 1. J > 0: Si and Sj become parallel: the material is a ferromagnet.
66
Physics Formulary by ir. J.C.A. Wevers
2. J < 0: Si and Sj become antiparallel: the material is an antiferromagnet. Heisenberg’s theory predicts quantized spin waves: magnons. Starting from a model with only nearest neighbouring atoms interacting one can write: U = −2J Sp · (Sp−1 + Sp+1 ) ≈ µp · Bp with Bp = The equation of motion for the magnons becomes: −2J (Sp−1 + Sp+1 ) gµB
dS 2J Sp × (Sp−1 + Sp+1 ) = dt h ¯
From here the treatment is analogous to phonons: postulate traveling waves of the type Sp = u exp(i(pka − ωt)). This results in a system of linear equations with solution: hω = 4JS(1 − cos(ka)) ¯
12.5 Free electron Fermi gas
12.5.1 Thermal heat capacity
The solution with period L of the one-dimensional Schr¨ dinger equation is: ψ n (x) = A sin(2πx/λn) with o nλn = 2L. From this follows h2 nπ 2 ¯ E= 2m L In a linear lattice the only important quantum numbers are n and m s . The Fermi level is the uppermost filled level in the ground state, which has the Fermi-energy EF . If nF is the quantum number of the Fermi level, it can be expressed as: 2nF = N so EF = h2 π 2 N 2 /8mL. In 3 dimensions holds: ¯ kF = 3π 2 N V
1/3
and EF = V 3π 2
h2 ¯ 2m
3π 2 N V
3/2
2/3
The number of states with energy ≤ E is then: N = and the density of states becomes: D(E) =
2mE h2 ¯ 2m h2 ¯
3/2
. √ E= 3N . 2E
V dN = dE 2π 2
The heat capacity of the electrons is approximately 0.01 times the classical expected value 3 N k. This is caused 2 by the Pauli exclusion principle and the Fermi-Dirac distribution: only electrons within an energy range ∼ kT of the Fermi level are excited thermally. There is a fraction ≈ T /TF excited thermally. The internal energy then becomes: ∂U T T and C = ≈ Nk U ≈ N kT TF ∂T TF A more accurate analysis gives: Celectrons = 1 π 2 N kT /TF ∼ T . Together with the T 3 dependence of the 2 thermal heat capacity of the phonons the total thermal heat capacity of metals is described by: C = γT +AT 3 .
12.5.2 Electric conductance
The equation of motion for the charge carriers is: F = mdv/dt = hdk/dt. The variation of k is given by ¯ δ k = k(t) − k(0) = −eEt/¯ . If τ is the characteristic collision time of the electrons, δ k remains stable if h t = τ . Then holds: v = µE, with µ = eτ /m the mobility of the electrons. The current in a conductor is given by: J = nqv = σ E = E/ρ = neµE. Because for the collision time holds: 1/τ = 1/τL + 1/τi , where τL is the collision time with the lattice phonons and τi the collision time with the impurities follows for the resistivity ρ = ρL + ρi , with lim ρL = 0.
T →0
Chapter 12: Solid state physics
67
12.5.3 The Hall-effect
If a magnetic field is applied ⊥ to the direction of the current the charge carriers will be pushed aside by the Lorentz force. This results in a magnetic field ⊥ to the flow direction of the current. If J = Jex and B = Bez than Ey /Ex = µB. The Hall coefficient is defined by: RH = Ey /Jx B, and RH = −1/ne if Jx = neµEx . The Hall voltage is given by: VH = Bvb = IB/neh where b is the width of the material and h de height.
12.5.4 Thermal heat conductivity
With = vF τ the mean free path of the electrons follows from κ = 1 C v 3 1 From this follows for the Wiedemann-Franz ratio: κ/σ = 3 (πk/e)2 T . : κelectrons = π 2 nk 2 T τ /3m.
12.6 Energy bands
In the tight-bond approximation it is assumed that ψ = eikna φ(x − na). From this follows for the energy: E = ψ|H|ψ = Eat − α − 2β cos(ka). So this gives a cosine superimposed on the atomic energy, which can often be approximated by a harmonic oscillator. If it is assumed that the electron is nearly free one can postulate: ψ = exp(ik · r ). This is a traveling wave. This wave can be decomposed into two standing waves: ψ(+) = exp(iπx/a) + exp(−iπx/a) = 2 cos(πx/a) ψ(−) = exp(iπx/a) − exp(−iπx/a) = 2i sin(πx/a) The probability density |ψ(+)|2 is high near the atoms of the lattice and low in between. The probability density |ψ(−)|2 is low near the atoms of the lattice and high in between. Hence the energy of ψ(+) is also lower than the energy of ψ)(−). Suppose that U (x) = U cos(2πx/a), than the bandgap is given by:
1
Egap =
0
U (x) |ψ(+)|2 − |ψ(−)|2 dx = U
12.7 Semiconductors
The band structures and the transitions between them of direct and indirect semiconductors are shown in the figures below. Here it is assumed that the momentum of the absorbed photon can be neglected. For an indirect semiconductor a transition from the valence- to the conduction band is also possible if the energy of the absorbed photon is smaller than the band gap: then, also a phonon is absorbed. conduction E band T • ¤ T §¥ ω ¦¤ g §¥ ¦ ◦ E T ' • Ω § T ω ¦¤ §¥ ¦ ◦
Direct transition This difference can also be observed in the absorption spectra:
Indirect transition
68
Physics Formulary by ir. J.C.A. Wevers
absorption T
absorption T ... ... . .. .. . . . . .. . . . Eg + hΩ ¯
hωg ¯
EE
E E
Direct semiconductor
Indirect semiconductor
So indirect semiconductors, like Si and Ge, cannot emit any light and are therefore not usable to fabricate lasers. When light is absorbed holds: kh = −ke , Eh (kh ) = −Ee (ke ), vh = ve and mh = −m∗ if the e conduction band and the valence band have the same structure. Instead of the normal electron mass one has to use the effective mass within a lattice. It is defined by: m∗ = F dp/dt dK = =h ¯ = h2 ¯ a dvg /dt dvg d2 E dk 2
−1
with E = hω and vg = dω/dk and p = hk. ¯ ¯ With the distribution function fe (E) ≈ exp((µ − E)/kT ) for the electrons and fh (E) = 1 − fe (E) for the holes the density of states is given by: D(E) = 1 2π 2 2m∗ h2 ¯
3/2
E − Ec
with Ec the energy at the edge of the conductance band. From this follows for the concentrations of the holes p and the electrons n:
∞
n=
Ec
De (E)fe (E)dE = 2
3
m∗ kT 2π¯ 2 h
3/2
exp
µ − Ec kT
Eg kT m∗ mh exp − e kT 2π¯ 2 h For an intrinsic (no impurities) semiconductor holds: ni = pi , for a n − type holds: n > p and in a p − type holds: n < p. For the product np follows: np = 4 An exciton is a bound electron-hole pair, rotating on each other as in positronium. The excitation energy of an exciton is smaller than the bandgap because the energy of an exciton is lower than the energy of a free electron and a free hole. This causes a peak in the absorption just under E g .
12.8 Superconductivity
12.8.1 Description
A superconductor is characterized by a zero resistivity if certain quantities are smaller than some critical values: T < Tc , I < Ic and H < Hc . The BCS-model predicts for the transition temperature Tc : Tc = 1.14ΘD exp while experiments find for Hc approximately: Hc (T ) ≈ Hc (Tc ) 1 − T2 2 Tc . −1 U D(EF )
Chapter 12: Solid state physics
69
Within a superconductor the magnetic field is 0: the Meissner effect. There are type I and type II superconductors. Because the Meissner effect implies that a superconductor is a perfect diamagnet holds in the superconducting state: H = µ0 M . This holds for a type I superconductor, for a type II superconductor this only holds to a certain value Hc1 , for higher values of H the superconductor is in a vortex state to a value Hc2 , which can be 100 times Hc1 . If H becomes larger than Hc2 the superconductor becomes a normal conductor. This is shown in the figures below. µ0 M T µ0 M T · · · · · · · · · · · · · · · Hc1 Type II
Hc Type I
EH
E H Hc2
The transition to a superconducting state is a second order thermodynamic state transition. This means that there is a twist in the T − S diagram and a discontinuity in the CX − T diagram.
12.8.2 The Josephson effect
For the Josephson effect one considers two superconductors, separated by an insulator. The electron wavefunction in one superconductor is ψ1 , in the other ψ2 . The Schr¨ dinger equations in both superconductors is o set equal: ∂ψ2 ∂ψ1 = hT ψ2 , i¯ ¯ h = hT ψ1 ¯ i¯ h ∂t ∂t hT is the effect of the coupling of the electrons, or the transfer interaction through the insulator. The electron ¯ √ √ wavefunctions are written as ψ1 = n1 exp(iθ1 ) and ψ2 = n2 exp(iθ2 ). Because a Cooper pair exist of two √ electrons holds: ψ ∼ n. From this follows, if n1 ≈ n2 : ∂θ2 ∂n2 ∂n1 ∂θ1 = and =− ∂t ∂t ∂t ∂t The Josephson effect results in a current density through the insulator depending on the phase difference as: J = J0 sin(θ2 − θ1 ) = J0 sin(δ), where J0 ∼ T . With an AC-voltage across the junction the Schr¨ dinger o equations become: ∂ψ1 ∂ψ2 i¯ h = hT ψ2 − eV ψ1 and i¯ ¯ h = hT ψ1 + eV ψ2 ¯ ∂t ∂t This gives: J = J0 sin θ2 − θ1 − 2eV t . h ¯
Hence there is an oscillation with ω = 2eV /¯ . h
12.8.3 Flux quantisation in a superconducting ring
For the current density in general holds: J = qψ ∗ vψ = nq [¯ θ − q A ] h m
From the Meissner effect, B = 0 and J = 0, follows: h θ = q A ⇒ ¯ θdl = θ2 − θ1 = 2πs with s ∈ I . N Because: Adl = (rotA, n )dσ = (B, n )dσ = Ψ follows: Ψ = 2π¯ s/q. The size of a flux quantum h follows by setting s = 1: Ψ = 2π¯ /e = 2.0678 · 10−15 Tm2 . h
70
Physics Formulary by ir. J.C.A. Wevers
12.8.4 Macroscopic quantum interference
From θ2 − θ1 = 2eΨ/¯ follows for two parallel junctions: δb − δa = h J = Ja + Jb = 2J0 sin δ0 cos eΨ h ¯ 2eΨ , so h ¯
This gives maxima if eΨ/¯ = sπ. h
12.8.5 The London equation
A current density in a superconductor proportional to the vector potential A is postulated: J= where λL = −A −B or rotJ = 2 µ 0 λL µ 0 λ2 L
2
ε0 mc2 /nq 2 . From this follows:
B = B/λ2 . L
The Meissner effect is the solution of this equation: B(x) = B0 exp(−x/λL ). Magnetic fields within a superconductor drop exponentially.
12.8.6 The BCS model
The BCS model can explain superconductivity in metals. (So far there is no explanation for high-T c superconductance). A new ground state where the electrons behave like independent fermions is postulated. Because of the interaction with the lattice these pseudo-particles exhibit a mutual attraction. This causes two electrons with opposite spin to combine to a Cooper pair. It can be proved that this ground state is perfect diamagnetic. The infinite conductivity is more difficult to explain because a ring with a persisting current is not a real equilibrium: a state with zero current has a lower energy. Flux quantization prevents transitions between these states. Flux quantization is related to the existence of a coherent many-particle wavefunction. A flux quantum is the equivalent of about 104 electrons. So if the flux has to change with one flux quantum there has to occur a transition of many electrons, which is very improbable, or the system must go through intermediary states where the flux is not quantized so they have a higher energy. This is also very improbable. Some useful mathematical relations are:
∞
π2 xdx = , ax + 1 e 12a2
∞
∞
π2 x2 dx = , x + 1)2 (e 3
∞
∞
π4 x3 dx = x+1 e 15
0 ∞
−∞
0
And, when
n=0
(−1) =
n
1 2
follows:
0
sin(px)dx =
0
cos(px)dx =
1 . p
Chapter 13
Theory of groups
13.1 Introduction
13.1.1 Definition of a group
G is a group for the operation • if: 1. ∀A,B∈G ⇒ A • B ∈ G: G is closed. 2. ∀A,B,C∈G ⇒ (A • B) • C = A • (B • C): G obeys the associative law. 3. ∃E∈G so that ∀A∈G A • E = E • A = A: G has a unit element. 4. ∀A∈G ∃A−1 ∈G so that A • A−1 = E: Each element in G has an inverse. If also holds: 5. ∀A,B∈G ⇒ A • B = B • A the group is called Abelian or commutative.
13.1.2 The Cayley table
Each element arises only once in each row and column of the Cayley- or multiplication table: because EA i = A−1 (Ak Ai ) = Ai each Ai appears once. There are h positions in each row and column when there are h k elements in the group so each element appears only once.
13.1.3 Conjugated elements, subgroups and classes
B is conjugate to A if ∃X∈G such that B = XAX −1 . Then A is also conjugate to B because B = (X −1 )A(X −1 )−1 . If B and C are conjugate to A, B is also conjugate with C. A subgroup is a subset of G which is also a group w.r.t. the same operation. A conjugacy class is the maximum collection of conjugated elements. Each group can be split up in conjugacy classes. Some theorems: • All classes are completely disjoint. • E is a class itself: for each other element in this class would hold: A = XEX −1 = E. • E is the only class which is also a subgroup because all other classes have no unit element. • In an Abelian group each element is a separate class. The physical interpretation of classes: elements of a group are usually symmetry operations which map a symmetrical object into itself. Elements of one class are then the same kind of operations. The opposite need not to be true.
72
Physics Formulary by ir. J.C.A. Wevers
13.1.4 Isomorfism and homomorfism; representations
Two groups are isomorphic if they have the same multiplication table. The mapping from group G 1 to G2 , so that the multiplication table remains the same is a homomorphic mapping. It need not be isomorphic. A representation is a homomorphic mapping of a group to a group of square matrices with the usual matrix multiplication as the combining operation. This is symbolized by Γ. The following holds: Γ(E) = I , Γ(AB) = Γ(A)Γ(B) , Γ(A−1 ) = [Γ(A)]−1 I For each group there are 3 possibilities for a representation: 1. A faithful representation: all matrices are different. 2. The representation A → det(Γ(A)). An equivalent representation is obtained by performing an unitary base transformation: Γ (A) = S −1 Γ(A)S. 3. The identical representation: A → 1.
13.1.5 Reducible and irreducible representations
If the same unitary transformation can bring all matrices of a representation Γ in the same block structure the representation is called reducible: Γ(A) = Γ(1) (A) 0 0 Γ(2) (A)
This is written as: Γ = Γ(1) ⊕ Γ(2) . If this is not possible the representation is called irreducible. The number of irreducible representations equals the number of conjugacy classes.
13.2 The fundamental orthogonality theorem
13.2.1 Schur’s lemma
Lemma: Each matrix which commutes with all matrices of an irreducible representation is a constant ×I I, where I is the unit matrix. The opposite is (of course) also true. I Lemma: If there exists a matrix M so that for two irreducible representations of group G, γ (1) (Ai ) and γ (2) (Ai ), holds: M γ (1) (Ai ) = γ (2) (Ai )M , than the representations are equivalent, or M = 0.
13.2.2 The fundamental orthogonality theorem
For a set of unequivalent, irreducible, unitary representations holds that, if h is the number of elements in the group and i is the dimension of the ith representation: ¯ Γ(i)∗ (R)Γαβ (R) = µν
R∈G (j)
h
i
δij δµα δνβ
13.2.3 Character
The character of a representation is given by the trace of the matrix and is therefore invariant for base transformations: χ(j) (R) = Tr(Γ(j) (R)) Also holds, with Nk the number of elements in a conjugacy class:
k n 2 i i=1
χ(i)∗ (Ck )χ(j) (Ck )Nk = hδij
Theorem:
=h
Chapter 13: Theory of groups
73
13.3 The relation with quantum mechanics
13.3.1 Representations, energy levels and degeneracy
Consider a set of symmetry transformations x = Rx which leave the Hamiltonian H invariant. These transformations are a group. An isomorfic operation on the wavefunction is given by: P R ψ(x ) = ψ(R−1 x ). This is considered an active rotation. These operators commute with H: P R H = HPR , and leave the volume element unchanged: d(Rx ) = dx. PR is the symmetry group of the physical system. It causes degeneracy: if ψ n is a solution of Hψn = En ψn than also holds: H(PR ψn ) = En (PR ψn ). A degeneracy which is not the result of a symmetry is called an accidental degeneracy. Assume an PR ψν
(n) n -fold
degeneracy at En : then choose an orthonormal set ψν , ν = 1, 2, . . . ,
n
(n)
n.
The function
(n) is in the same subspace: PR ψν = κ=1 (n)
(n) ψκ Γ(n) (R) κν
where Γ is an irreducible, unitary representation of the symmetry group G of the system. Each n corresponds with another energy level. One can purely mathematical derive irreducible representations of a symmetry group and label the energy levels with a quantum number this way. A fixed choice of Γ (n) (R) defines (n) the base functions ψν . This way one can also label each separate base function with a quantum number. Particle in a periodical potential: the symmetry operation is a cyclic group: note the operator describing one translation over one unit as A. Then: G = {A, A2 , A3 , . . . , Ah = E}. The group is Abelian so all irreducible representations are one-dimensional. For 0 ≤ p ≤ h − 1 follows: Γ(p) (An ) = e2πipn/h 2πp 2π mod , so: PA ψp (x) = ψp (x − a) = e2πip/h ψp (x), this gives Bloch’s ah a theorem: ψk (x) = uk (x)eikx , with uk (x ± a) = uk (x).
If one defines: k = −
13.3.2 Breaking of degeneracy by a perturbation
Suppose the unperturbed system has Hamiltonian H0 and symmetry group G0 . The perturbed system has H = H0 + V, and symmetry group G ⊂ G0 . If Γ(n) (R) is an irreducible representation of G0 , it is also a representation of G but not all elements of Γ(n) in G0 are also in G. The representation then usually becomes reducible: Γ(n) = Γ(n1 ) ⊕ Γ(n2 ) ⊕ . . .. The degeneracy is then (possibly partially) removed: see the figure below.
n1 n n2 n3
= dim(Γ(n1 ) ) = dim(Γ(n2 ) ) = dim(Γ(n3 ) )
Spectrum H0
Spectrum H
(n)
Theorem: The set of n degenerated eigenfunctions ψν irreducible representation Γ(n) of the symmetry group.
with energy En is a basis for an
n -dimensional
13.3.3 The construction of a base function
n
j
Each function F in configuration space can be decomposed into symmetry types: F =
j=1 κ=1
(j) fκ
The following operator extracts the symmetry types:
j
h
Γ(j)∗ (R)PR κκ
R∈G
(j) F = fκ
74
Physics Formulary by ir. J.C.A. Wevers
(j) This is expressed as: fκ is the part of F that transforms according to the κth row of Γ(j) . ¯
F can also be expressed in base functions ϕ: F =
ajκ
cajκ ϕκ
(aj)
. The functions fκ
(j)
are in general not
transformed into each other by elements of the group. However, this does happen if c jaκ = cja . Theorem: Two wavefunctions transforming according to non-equivalent unitary representations or according to different rows of an unitary irreducible representation are orthogonal: (i) (j) (i) (i) ϕκ |ψλ ∼ δij δκλ , and ϕκ |ψκ is independent of κ.
13.3.4 The direct product of representations
Consider a physical system existing of two subsystems. The subspace D (i) of the system transforms according (i) (1) (2) to Γ(i) . Basefunctions are ϕκ (xi ), 1 ≤ κ ≤ i . Now form all 1 × 2 products ϕκ (x1 )ϕλ (x2 ). These define a space D (1) ⊗ D(2) . These product functions transform as: PR (ϕ(1) (x1 )ϕλ (x2 )) = (PR ϕ(1) (x1 ))(PR ϕλ (x2 )) κ κ In general the space D (1) ⊗ D(2) can be split up in a number of invariant subspaces: Γ(1) ⊗ Γ(2) = ni Γ(i)
i (2) (2)
A useful tool for this reduction is that for the characters hold: χ(1) (R)χ(2) (R) =
i
ni χ(i) (R)
13.3.5 Clebsch-Gordan coefficients
With the reduction of the direct-product matrix w.r.t. the basis ϕκ ϕλ one uses a new basis ϕµ functions lie in subspaces D (ak) . The unitary base transformation is given by:
(ak) ϕµ = κλ (i) (j) (aκ)
. These base
ϕ(i) ϕλ (iκjλ|akµ) κ ϕ(aκ) (akµ|iκjλ) µ
akµ (i) (j) (ak)
(j)
and the inverse transformation by: ϕ(i) ϕλ = κ
(j)
In essence the Clebsch-Gordan coefficients are dot products: (iκjλ|akµ) := ϕ k ϕλ |ϕµ
13.3.6 Symmetric transformations of operators, irreducible tensor operators
−1 Observables (operators) transform as follows under symmetry transformations: A = PR APR . If a set of (j) operators Aκ with 0 ≤ κ ≤ j transform into each other under the transformations of G holds: −1 PR A(j) PR = κ ν
A(j) Γ(j) (R) ν νκ
(j)
If Γ(j) is irreducible they are called irreducible tensor operators A(j) with components Aκ . An operator can also be decomposed into symmetry types: A =
jk
ak , with:
(j)
a(j) = κ
j
h
−1 Γ(j)∗ (R) (PR APR ) κκ R∈G
Chapter 13: Theory of groups
75
Theorem: Matrix elements Hij of the operator H which is invariant under ∀A∈G , are 0 between states which transform according to non-equivalent irreducible unitary representations or according to different rows of such (i) (i) a representation. Further ϕκ |H|ψκ is independent of κ. For H = 1 this becomes the previous theorem. This is applied in quantum mechanics in perturbation theory and variational calculus. Here one tries to diag(i) onalize H. Solutions can be found within each category of functions ϕ κ with common i and κ: H is already diagonal in categories as a whole. Perturbation calculus can be applied independent within each category. With variational calculus the try function can be chosen within a separate category because the exact eigenfunctions transform according to a row of an irreducible representation.
13.3.7 The Wigner-Eckart theorem
Theorem: The matrix element ϕλ |Aκ |ψµ can only be = 0 if Γ(j) ⊗ Γ(k) = . . . ⊕ Γ(i) ⊕ . . .. If this is the case holds (if Γ(i) appears only once, otherwise one has to sum over a):
(k) ϕλ |A(j) |ψµ = (iλ|jκkµ) ϕ(i) A(j) ψ (k) κ (i) (i) (j) (k)
This theorem can be used to determine selection rules: the probability of a dipole transition is given by ( is the direction of polarization of the radiation): PD = 8π 2 e2 f 3 |r12 |2 with r12 = l2 m2 | · r |l1 m1 3¯ ε0 c3 h
Further it can be used to determine intensity ratios: if there is only one value of a the ratio of the matrix elements are the Clebsch-Gordan coefficients. For more a-values relations between the intensity ratios can be stated. However, the intensity ratios are also dependent on the occupation of the atomic energy levels.
13.4 Continuous groups
Continuous groups have h = ∞. However, not all groups with h = ∞ are continuous, e.g. the translation group of an spatially infinite periodic potential is not continuous but does have h = ∞.
13.4.1 The 3-dimensional translation group
For the translation of wavefunctions over a distance a holds: Pa ψ(x) = ψ(x − a). Taylor expansion near x gives: dψ(x) 1 2 d2 ψ(x) ψ(x − a) = ψ(x) − a + a −+... dx 2 dx2 h ∂ ¯ Because the momentum operator in quantum mechanics is given by: p x = , this can be written as: i ∂x
h ψ(x − a) = e−iapx /¯ ψ(x)
13.4.2 The 3-dimensional rotation group
This group is called SO(3) because a faithful representation can be constructed from orthogonal 3 × 3 matrices with a determinant of +1. For an infinitesimal rotation around the x-axis holds: Pδθx ψ(x, y, z) ≈ ψ(x, y + zδθx , z − yδθx ) ∂ ∂ − yδθx = ψ(x, y, z) + zδθx ∂y ∂z iδθx Lx = 1− ψ(x, y, z) h ¯
ψ(x, y, z)
76
Physics Formulary by ir. J.C.A. Wevers
Because the angular momentum operator is given by: Lx = So in an arbitrary direction holds: Rotations: Translations:
h ¯ i
z
∂ ∂ . −y ∂y ∂z
Pα,n = exp(−iα(n · J )/¯ ) h Pa,n = exp(−ia(n · p )/¯ ) h
Jx , Jy and Jz are called the generators of the 3-dim. rotation group, px , py and pz are called the generators of the 3-dim. translation group. The commutation rules for the generators can be derived from the properties of the group for multiplications: translations are interchangeable ↔ px py − py px = 0. Rotations are not generally interchangeable: consider a rotation around axis n in the xz-plane over an angle α. Then holds: Pα,n = P−θ,y Pα,x Pθ,y , so:
h h h h e−iα(n·J )/¯ = eiθJy /¯ e−iαJx /¯ e−iθJy /¯
If α and θ are very small and are expanded to second order, and the corresponding terms are put equal with n · J = Jx cos θ + Jz sin θ, it follows from the αθ term: Jx Jy − Jy Jx = i¯ Jz . h
13.4.3 Properties of continuous groups
The elements R(p1 , ..., pn ) depend continuously on parameters p1 , ..., pn . For the translation group this are e.g. anx , any and anz . It is demanded that the multiplication and inverse of an element R depend continuously on the parameters of R. The statement that each element arises only once in each row and column of the Cayley table holds also for continuous groups. The notion conjugacy class for continuous groups is defined equally as for discrete groups. The notion representation is fitted by demanding continuity: each matrix element depends continuously on pi (R). Summation over all group elements is for continuous groups replaced by an integration. If f (R) is a function defined on G, e.g. Γαβ (R), holds: f (R)dR :=
G p1
···
pn
f (R(p1 , ..., pn ))g(R(p1 , ..., pn ))dp1 · · · dpn
Here, g(R) is the density function. Because of the properties of the Cayley table is demanded: f (R)dR = f (SR)dR. This fixes g(R) except for a constant factor. Define new variables p by: SR(pi ) = R(pi ). If one writes: dV := dp1 · · · dpn holds: g(S) = g(E) dV dV is the Jacobian: = det dV dV ∂pi ∂pj dV dV
Here,
, and g(E) is constant.
For the translation group holds: g(a) = constant = g(0 ) because g(an )da = g(0 )da and da = da. This leads to the fundamental orthogonality theorem: Γ(i)∗ (R)Γαβ (R)dR = µν
G (j)
1
i
δij δµα δνβ
G
dR
and for the characters hold: χ(i)∗ (R)χ(j) (R)dR = δij
G G G
dR
Compact groups are groups with a finite group volume:
dR < ∞.
Chapter 13: Theory of groups
77
13.5 The group SO(3)
One can take 2 parameters for the direction of the rotational axis and one for the angle of rotation ϕ. The parameter space is a collection points ϕn within a sphere with radius π. The diametrical points on this sphere are equivalent because Rn,π = Rn,−π . Another way to define parameters is by means of Eulers angles. If α, β and γ are the 3 Euler angles, defined as: 1. The spherical angles of axis 3 w.r.t. xyz are θ, ϕ := β, α. Now a rotation around axis 3 remains possible. 2. The spherical angles of the z-axis w.r.t. 123 are θ, ϕ := β, π − γ. then the rotation of a quantum mechanical system is described by: h ¯ h h ψ → e−iαJz h e−iβJy /¯ e−iγJz /¯ ψ. So PR = e−iε(n·J )/¯ . All irreducible representations of SO(3) can be constructed from the behaviour of the spherical harmonics Ylm (θ, ϕ) with −l ≤ m ≤ l and for a fixed l: PR Ylm (θ, ϕ) =
m
Ylm (θ, ϕ)Dmm (R)
(l)
D
(l)
is an irreducible representation of dimension 2l + 1. The character of D (l) is given by:
l l
χ(l) (α) =
m=−l
eimα = 1 + 2
k=0
cos(kα) =
1 sin([l + 2 ]α) sin( 1 α) 2
In the performed derivation α is the rotational angle around the z-axis. This expression is valid for all rotations over an angle α because the classes of SO(3) are rotations around the same angle around an axis with an arbitrary orientation. Via the fundamental orthogonality theorem for characters one obtains the following expression for the density function (which is normalized so that g(0) = 1): g(α) =
1 sin2 ( 2 α) ( 1 α)2 2
With this result one can see that the given representations of SO(3) are the only ones: the character of another 1 representation χ would have to be ⊥ to the already found ones, so χ (α) sin2 ( 2 α) = 0∀α ⇒ χ (α) = 0∀α. This is contradictory because the dimension of the representation is given by χ (0).
1 Because fermions have an half-odd integer spin the states ψsms with s = 1 and ms = ± 2 constitute a 2-dim. 2 space which is invariant under rotations. A problem arises for rotations over 2π: h 1 1 ψ 1 ms → e−2πiSz /¯ ψ 1 ms = e−2πims ψ 2 ms = −ψ 2 ms 2 2
However, in SO(3) holds: Rz,2π = E. So here holds E → ±I Because observable quantities can always be I. written as φ|ψ or φ|A|ψ , and are bilinear in the states, they do not change sign if the states do. If only one state changes sign the observable quantities do change. The existence of these half-odd integer representations is connected with the topological properties of SO(3): the group is two-fold coherent through the identification R0 = R2π = E.
13.6 Applications to quantum mechanics
13.6.1 Vectormodel for the addition of angular momentum
If two subsystems have angular momentum quantum numbers j 1 and j2 the only possible values for the total angular momentum are J = j1 +j2 , j1 +j2 −1, ..., |j1 −j2 |. This can be derived from group theory as follows: from χ(j1 ) (α)χ(j2 ) (α) = nj χ(J) (α) follows:
J
D(j1 ) ⊗ D(j2 ) = D(j1 +j2 ) ⊕ D(j1 +j2 −1) ⊕ ... ⊕ D(|j1 −j2 |)
78
Physics Formulary by ir. J.C.A. Wevers
The states can be characterized by quantum numbers in two ways: with j 1 , m1 , j2 , m2 and with j1 , j2 , J, M . The Clebsch-Gordan coefficients, for SO(3) called the Wigner coefficients, can be chosen real, so: ψj1 j2 JM ψ j 1 m1 j 2 m2 =
m1 m2
ψj1 m1 j2 m2 (j1 m1 j2 m2 |JM )
=
JM
ψj1 j2 JM (j1 m1 j2 m2 |JM )
13.6.2 Irreducible tensor operators, matrixelements and selection rules
Some examples of the behaviour of operators under SO(3) 1. Suppose j = 0: this gives the identical representation with j = 1. This state is described by a (0) −1 (0) scalar operator. Because PR A0 PR = A0 this operator is invariant, e.g. the Hamiltonian of a free atom. Then holds: J M |H|JM ∼ δM M δJJ . 2. A vector operator: A = (Ax , Ay , Az ). The cartesian components of a vector operator transform equally as the cartesian components of r by definition. So for rotations around the z-axis holds: cos α − sin α 0 D(Rα,z ) = sin α cos α 0 0 0 1 The transformed operator has the same matrix elements w.r.t. P R ψ and PR φ: −1 PR ψ|PR Ax PR |PR φ = ψ|Ax |φ , and χ(Rα,z ) = 1 + 2 cos(α). According to the equation for characters this means one can choose base operators which transform like Y 1m (θ, ϕ). These turn out to be the spherical components: 1 1 (1) (1) (1) A+1 = − √ (Ax + iAy ), A0 = Az , A−1 = √ (Ax − iAy ) 2 2 3. A cartesian tensor of rank 2: Tij is a quantity which transforms under rotations like Ui Vj , where U and −1 V are vectors. So Tij transforms like PR Tij PR = Tkl Dki (R)Dlj (R), so like D(1) ⊗ D(1) =
kl
D(2) ⊕ D(1) ⊕ D(0) . The 9 components can be split in 3 invariant subspaces with dimension 1 (D (0) ), 3 (D(1) ) and 5 (D(2) ). The new base operators are: I. Tr(T ) = Txx + Tyy + Tzz . This transforms as the scalar U · V , so as D(0) . II. The 3 antisymmetric components Az = 1 (Txy − Tyx ), etc. These transform as the vector U × V , 2 so as D(1) . III. The 5 independent components of the traceless, symmetric tensor S: Sij = 1 (Tij + Tji ) − 1 δij Tr(T ). These transform as D (2) . 2 3 Selection rules for dipole transitions Dipole operators transform as D (1) : for an electric dipole transfer is the operator er, for a magnetic e( L + 2S )/2m. From the Wigner-Eckart theorem follows: J M |Aκ |JM = 0 except D (J ) is a part of D(1) ⊗ D(J) = D(J+1) ⊕ D(J) ⊕ D(|J−1|) . This means that J ∈ {J + 1, J, |J − 1|}: J = J or J = J ± 1, except J = J = 0. Land´ -equation for the anomalous Zeeman splitting e According to Land´ ’s model the interaction between a magnetic moment with an external magnetic field is e determined by the projection of M on J because L and S precede fast around J. This can also be understood
(1)
Chapter 13: Theory of groups
79
from the Wigner-Eckart theorem: from this follows that the matrix elements from all vector operators show a certain proportionality. For an arbitrary operator A follows: αjm |A|αjm = αjm|A · J |αjm j(j + 1)¯ 2 h αjm |J |αjm
13.7 Applications to particle physics
The physics of a system does not change after performing a transformation ψ = eiδ ψ where δ is a constant. This is a global gauge transformation: the phase of the wavefunction changes everywhere by the same amount. There exists some freedom in the choice of the potentials A and φ at the same E and B: gauge transformations of the potentials do not change E and B (See chapter 2 and 10). The solution ψ of the Schr¨ dinger equation o with the transformed potentials is: ψ = e−iqf (r,t) ψ. This is a local gauge transformation: the phase of the wavefunction changes different at each position. The physics of the system does not change if A and φ are also transformed. This is now stated as a guide principle: the “right of existence” of the electromagnetic field is to allow local gauge invariance. The gauge transformations of the EM-field form a group: U(1), unitary 1 × 1-matrices. The split-off of charge in the exponent is essential: it allows one gauge field for all charged particles, independent of their charge. This concept is generalized: particles have a “special charge” Q. The group elements now are PR = exp(−iQΘ). Other force fields than the electromagnetic field can also be understood this way. The weak interaction together with the electromagnetic interaction can be described by a force field that transforms according to U(1)⊗SU(2), and consists of the photon and three intermediary vector bosons. The colour force is described by SU(3), and has a gauge field that exists of 8 types of gluons. In general the group elements are given by PR = exp(−iT · Θ), where Θn are real constants and Tn operators (generators), like Q. The commutation rules are given by [Ti , Tj ] = i cijk Tk . The cijk are the structure constants of the group. For SO(3) these constants are cijk = εijk , here εijk is the complete antisymmetric tensor with ε123 = +1. These constants can be found with the help of group product elements: because G is closed holds: eiΘ·T eiΘ ·T e−iΘ·T e−iΘ ·T = e−iΘ ·T . Taylor expansion and setting equal Θn Θ m -terms results in the commutation rules. The group SU(2) has 3 free parameters: because it is unitary there are 4 real conditions over 4 complex parameters, and the determinant has to be +1, remaining 3 free parameters. Each unitary matrix U can be written as: U = e−iH . Here, H is a Hermitian matrix. Further it always holds that: det(U ) = e−iTr(H) . For each matrix of SU(2) holds that Tr(H)=0. Each Hermitian, traceless 2 × 2 matrix can be written as a linear combination of the 3 Pauli-matrices σi . So these matrices are a choice for the operators of SU(2). One can write: SU(2)={exp(− 1 iσ · Θ)}. 2 In abstraction, one can consider an isomorphic group where only the commutation rules are considered to be known regarding the operators Ti : [T1 , T2 ] = iT3 , etc. In elementary particle physics the Ti can be interpreted e.g. as the isospin operators. Elementary particles can be classified in isospin-multiplets, these are the irreducible representations of SU(2). The classification is: 1. The isospin-singlet ≡ the identical representation: e−iT ·Θ = 1 ⇒ Ti = 0 2. The isospin-doublet ≡ the faithful representation of SU(2) on 2 × 2 matrices.
k
80
Physics Formulary by ir. J.C.A. Wevers
The group SU(3) has 8 free parameters. (The group SU(N ) has N 2 − 1 free parameters). The Hermitian, traceless operators are 3 SU(2)-subgroups in the e1 e2 , e1 e3 and the e2 e3 plane. This gives 9 matrices, which are not all 9 linear independent. By taking a linear combination one gets 8 matrices. In the Lagrange density for the colour force one has to substitute ∂ D ∂ Ti A i → := − x ∂x Dx ∂x i=1
8
The terms of 3rd and 4th power in A show that the colour field interacts with itself.
Chapter 14
Nuclear physics
14.1 Nuclear forces
The mass of a nucleus is given by: Mnucl = Zmp + N mn − Ebind /c2 The binding energy per nucleon is given in the figure at the right. The top is at 56 Fe, 26 the most stable nucleus. With the constants a1 a2 a3 a4 a5 = = = = = 15.760 17.810 0.711 23.702 34.000 MeV MeV MeV MeV MeV 9 8 ↑ 7 E 6 (MeV) 5 4 3 2 1 0
0
40
80
120 160 A→
200
240
and A = Z + N , in the droplet or collective model of the nucleus the binding energy E bind is given by: Ebind (N − Z)2 Z(Z − 1) − a4 = a1 A − a2 A2/3 − a3 + a5 A−3/4 2 1/3 c A A These terms arise from: 1. a1 : Binding energy of the strong nuclear force, approximately ∼ A. 2. a2 : Surface correction: the nucleons near the surface are less bound. 3. a3 : Coulomb repulsion between the protons. 4. a4 : Asymmetry term: a surplus of protons or neutrons has a lower binding energy. 5. a5 : Pair off effect: nuclei with an even number of protons or neutrons are more stable because groups of two protons or neutrons have a lower energy. The following holds: Z even, N even: = +1, Z odd, N odd: = −1. Z even, N odd: = 0, Z odd, N even: = 0. The Yukawa potential can be derived if the nuclear force can to first approximation, be considered as an exchange of virtual pions: W0 r0 r U (r) = − exp − r r0 With ∆E · ∆t ≈ h, Eγ = m0 c2 and r0 = c∆t follows: r0 = h/m0 c. ¯ ¯ In the shell model of the nucleus one assumes that a nucleon moves in an average field of other nucleons. 1 h Further, there is a contribution of the spin-orbit coupling ∼ L · S: ∆Vls = 2 (2l + 1)¯ ω. So each level 1 1 (n, l) is split in two, with j = l ± 2 , where the state with j = l + 2 has the lowest energy. This is just the opposite for electrons, which is an indication that the L − S interaction is not electromagnetical. The energy of a 3-dimensional harmonic oscillator is E = (N + 3 )¯ ω. N = nx + ny + nz = 2(n − 1) + l 2 h 1 ¯ where n ≥ 1 is the main oscillator number. Because −l ≤ m ≤ l and ms = ± 2 h there are 2(2l + 1)
82
Physics Formulary by ir. J.C.A. Wevers
substates which exist independently for protons and neutrons. This gives rise to the so called magical numbers: nuclei where each state in the outermost level are filled are particulary stable. This is the case if N or Z ∈ {2, 8, 20, 28, 50, 82, 126}.
14.2 The shape of the nucleus
A nucleus is to first approximation spherical with a radius of R = R0 A1/3 . Here, R0 ≈ 1.4·10−15 m, constant for all nuclei. If the nuclear radius is measured including the charge distribution one obtains R 0 ≈ 1.2 · 10−15 m. The shape of oscillating nuclei can be described by spherical harmonics: R = R0 1 +
lm
alm Ylm (θ, ϕ)
l = 0 gives rise to monopole vibrations, density vibrations, which can be applied to the theory of neutron stars. √ 1 l = 1 gives dipole vibrations, l = 2 quadrupole, with a2,0 = β cos γ and a2,±2 = 2 2β sin γ where β is the deformation factor and γ the shape parameter. The multipole moment is given by µ l = Zerl Ylm (θ, ϕ). The parity of the electric moment is ΠE = (−1)l , of the magnetic moment ΠM = (−1)l+1 . e e There are 2 contributions to the magnetic moment: ML = L and MS = gS S. 2mp 2mp where gS is the spin-gyromagnetic ratio. For protons holds gS = 5.5855 and for neutrons gS = −3.8263. The z-components of the magnetic moment are given by ML,z = µN ml and MS,z = gS µN mS . The resulting magnetic moment is related to the nuclear spin I according to M = gI (e/2mp)I. The z-component is then M z = µ N g I mI .
˙ The number of nuclei decaying is proportional to the number of nuclei: N = −λN . This gives for the number of nuclei N : N (t) = N0 exp(−λt). The half life time follows from τ 1 λ = ln(2). The average life time 2 of a nucleus is τ = 1/λ. The probability that N nuclei decay within a time interval is given by a Poisson distribution: λN e−λ dt P (N )dt = N0 N! If a nucleus can decay into more final states then holds: λ = λi . So the fraction decaying into state i is λi / λi . There are 5 types of natural radioactive decay: 1. α-decay: the nucleus emits a He2+ nucleus. Because nucleons tend to order themselves in groups of 2p+2n this can be considered as a tunneling of a He2+ nucleus through a potential barrier. The tunnel probability P is P = incoming amplitude 1 = e−2G with G = outgoing amplitude h ¯ 2m [V (r) − E]dr
G is called the Gamow factor. 2. β-decay. Here a proton changes into a neutron or vice versa: p+ → n0 + W+ → n0 + e+ + νe , and n0 → p+ + W− → p+ + e− + ν e . 3. Electron capture: here, a proton in the nucleus captures an electron (usually from the K-shell). 4. Spontaneous fission: a nucleus breaks apart. 5. γ-decay: here the nucleus emits a high-energetic photon. The decay constant is given by λ= Eγ P (l) ∼ hω ¯ (¯ c)2 h Eγ R hc ¯
2l
∼ 10−4l
Chapter 14: Nuclear physics
83
where l is the quantum number for the angular momentum and P the radiated power. Usually the decay constant of electric multipole moments is larger than the one of magnetic multipole moments. 2 The energy of the photon is Eγ = Ei − Ef − TR , with TR = Eγ /2mc2 the recoil energy, which can usually be neglected. The parity of the emitted radiation is Π l = Πi · Πf . With I the quantum number of angular momentum of the nucleus, L = h I(I + 1), holds the following selection rule: ¯ |Ii − If | ≤ ∆l ≤ |Ii + If |.
14.4 Scattering and nuclear reactions
14.4.1 Kinetic model
If a beam with intensity I hits a target with density n and length x (Rutherford scattering) the number of scatterings R per unit of time is equal to R = Inxσ. From this follows that the intensity of the beam decreases as −dI = Inσdx. This results in I = I0 e−nσx = I0 e−µx . Because dR = R(θ, ϕ)dΩ/4π = Inxdσ it follows: dσ R(θ, ϕ) = dΩ 4πnxI dσ ∆N = n ∆Ω∆x N dΩ
If N particles are scattered in a material with density n then holds: For Coulomb collisions holds: dσ dΩ =
C
1 Z1 Z2 e 2 2 4 1 8πε0 µv0 sin ( 2 θ)
14.4.2 Quantum mechanical model for n-p scattering
The initial state is a beam of neutrons moving along the z-axis with wavefunction ψ init = eikz and current density Jinit = v|ψinit |2 = v. At large distances from the scattering point they have approximately a spherical wavefunction ψscat = f (θ)eikr /r where f (θ) is the scattering amplitude. The total wavefunction is then given by eikr ψ = ψin + ψscat = eikz + f (θ) r The particle flux of the scattered particles is v|ψscat |2 = v|f (θ)|2 dΩ. From this it follows that σ(θ) = |f (θ)|2 . The wavefunction of the incoming particles can be expressed as a sum of angular momentum wavefunctions: ψinit = eikz =
l
ψl
The impact parameter is related to the angular momentum with L = bp = b¯ k, so bk ≈ l. At very low energy h only particles with l = 0 are scattered, so ψ = ψ0 +
l>0
ψl and ψ0 =
sin(kr) kr
If the potential is approximately rectangular holds: ψ0 = C The cross section is then σ(θ) = sin2 (δ0 ) so σ = k2 h2 k 2 /2m ¯ W0 + W
sin(kr + δ0 ) kr 4π sin2 (δ0 ) k2
σ(θ)dΩ =
At very low energies holds: sin2 (δ0 ) =
with W0 the depth of the potential well. At higher energies holds: σ =
4π k2
sin2 (δl )
l
84
Physics Formulary by ir. J.C.A. Wevers
14.4.3 Conservation of energy and momentum in nuclear reactions
If a particle P1 collides with a particle P2 which is in rest w.r.t. the laboratory system and other particles are created, so P1 + P 2 → Pk
k>2
the total energy Q gained or required is given by Q = (m1 + m2 −
mk )c2 .
k>2
The minimal required kinetic energy T of P1 in the laboratory system to initialize the reaction is T = −Q If Q < 0 there is a threshold energy. m1 + m 2 + 2m2 mk
Radiometric quantities determine the strength of the radiation source(s). Dosimetric quantities are related to the energy transfer from radiation to matter. Parameters describing a relation between those are called interaction parameters. The intensity of a beam of particles in matter decreases according to I(s) = I 0 exp(−µs). The deceleration of a heavy particle is described by the Bethe-Bloch equation: dE q2 ∼ 2 ds v The fluention is given by Φ = dN/dA. The flux is given by φ = dΦ/dt. The energy loss is defined by Ψ = dW/dA, and the energy flux density ψ = dΨ/dt. The absorption coefficient is given by µ = (dN/N )/dx. The mass absorption coefficient is given by µ/ . The radiation dose X is the amount of charge produced by the radiation per unit of mass, with unit C/kg. An old unit is the R¨ ntgen: 1Ro= 2.58 · 10−4 C/kg. With the energy-absorption coefficient µE follows: o X= eµE dQ = Ψ dm W
where W is the energy required to disjoin an elementary charge. The absorbed dose D is given by D = dEabs /dm, with unit Gy=J/kg. An old unit is the rad: 1 rad=0.01 Gy. ˙ The dose tempo is defined as D. It can be derived that D= µE Ψ
The Kerma K is the amount of kinetic energy of secundary produced particles which is produced per mass unit of the radiated object. The equivalent dose H is a weight average of the absorbed dose per type of radiation, where for each type radiation the effects on biological material is used for the weight factor. These weight factors are called the quality factors. Their unit is Sv. H = QD. If the absorption is not equally distributed also weight factors w per organ need to be used: H = wk Hk . For some types of radiation holds: Radiation type R¨ ntgen, gamma radiation o β, electrons, mesons Thermic neutrons Fast neutrons protons α, fission products Q 1 1 3 to 5 10 to 20 10 20
Chapter 15
Quantum field theory & Particle physics
15.1 Creation and annihilation operators
A state with more particles can be described by a collection occupation numbers |n 1 n2 n3 · · · . Hence the vacuum state is given by |000 · · · . This is a complete description because the particles are indistinguishable. The states are orthonormal:
∞
n1 n2 n3 · · · |n1 n2 n3 · · · = The time-dependent state vector is given by Ψ(t) =
n1 n2 ···
δ ni ni
i=1
cn1 n2 ··· (t)|n1 n2 · · ·
The coefficients c can be interpreted as follows: |cn1 n2 ··· |2 is the probability to find n1 particles with momentum k1 , n2 particles with momentum k2 , etc., and Ψ(t)|Ψ(t) = |cni (t)|2 = 1. The expansion of the states in time is described by the Schr¨ dinger equation o i d |Ψ(t) = H|Ψ(t) dt
where H = H0 + Hint . H0 is the Hamiltonian for free particles and keeps |cni (t)|2 constant, Hint is the interaction Hamiltonian and can increase or decrease a c2 at the cost of others. All operators which can change occupation numbers can be expanded in the a and a † operators. a is the annihilation operator and a† the creation operator, and: √ ni |n1 n2 · · · ni − 1 · · · a(ki )|n1 n2 · · · ni · · · = √ † a (ki )|n1 n2 · · · ni · · · = ni + 1 |n1 n2 · · · ni + 1 · · · Because the states are normalized holds a|0 = 0 and a(ki )a† (ki )|ni = ni |ni . So aa† is an occupation number operator. The following commutation rules can be derived: [a(ki ), a(kj )] = 0 , [a† (ki ), a† (kj )] = 0 , [a(ki ), a† (kj )] = δij Hence for free spin-0 particles holds: H0 =
i
a† (ki )a(ki )¯ ωki h
15.2 Classical and quantum fields
Starting with a real field Φα (x) (complex fields can be split in a real and an imaginary part), the Lagrange density L is a function of the position x = (x, ict) through the fields: L = L(Φ α (x), ∂ν Φα (x)). The Lagrangian is given by L = L(x)d3 x. Using the variational principle δI(Ω) = 0 and with the action-integral I(Ω) = L(Φα , ∂ν Φα )d4 x the field equation can be derived: ∂L ∂ ∂L − =0 α ∂Φ ∂xν ∂(∂ν Φα ) The conjugated field is, analogous to momentum in classical mechanics, defined as: Πα (x) = ∂L ˙ ∂ Φα
86
Physics Formulary by ir. J.C.A. Wevers
˙ With this, the Hamilton density becomes H(x) = Πα Φα − L(x). Quantization of a classical field is analogous to quantization in point mass mechanics: the field functions are considered as operators obeying certain commutation rules: [Φα (x), Φβ (x )] = 0 , [Πα (x), Πβ (x )] = 0 , [Φα (x), Πβ (x )] = iδαβ (x − x )
15.3 The interaction picture
Some equivalent formulations of quantum mechanics are possible: 1. Schr¨ dinger picture: time-dependent states, time-independent operators. o 2. Heisenberg picture: time-independent states, time-dependent operators. 3. Interaction picture: time-dependent states, time-dependent operators. The interaction picture can be obtained from the Schr¨ dinger picture by an unitary transformation: o |Φ(t) = eiH0 |ΦS (t) and O(t) = eiH0 OS e−iH0 The index S denotes the Schr¨ dinger picture. From this follows: o i d d |Φ(t) = Hint (t)|Φ(t) and i O(t) = [O(t), H0 ] dt dt
S S S
15.4 Real scalar field in the interaction picture
It is easy to find that, with M := m2 c2 /¯ 2 , holds: h 0 ∂ ∂ Φ(x) = Π(x) and Π(x) = ( ∂t ∂t
2
− M 2 )Φ(x)
2 From this follows that Φ obeys the Klein-Gordon equation (2 − M 2 )Φ = 0. With the definition k0 = 2 2 2 k + M := ωk and the notation k · x − ik0 t := kx the general solution of this equation is:
1 Φ(x) = √ V
k
√
1 a(k )eikx + a† (k )e−ikx 2ωk
i , Π(x) = √ V
1 2 ωk k
−a(k )eikx + a† (k )e−ikx
In general it holds that the term with e−ikx , the positive frequency part, is the creation part, and the negative frequency part is the annihilation part. the coefficients have to be each others hermitian conjugate because Φ is hermitian. Because Φ has only one component this can be interpreted as a field describing a particle with spin zero. From this follows that the commutation rules are given by [Φ(x), Φ(x )] = i∆(x − x ) with ∆(y) = 1 (2π)3 sin(ky) 3 d k ωk
The field operators contain a volume V , which is used as normalization factor. Usually one can take the limit V → ∞.
∆(y) is an odd function which is invariant for proper Lorentz transformations (no mirroring). This is consistent with the previously found result [Φ(x, t, Φ(x , t)] = 0. In general holds that ∆(y) = 0 outside the light cone. So the equations obey the locality postulate.
1 The Lagrange density is given by: L(Φ, ∂ν Φ) = − 2 (∂ν Φ∂ν Φ + m2 Φ2 ). The energy operator is given by:
H=
H(x)d3 x =
hωk a† (k )a(k ) ¯
k
Chapter 15: Quantum field theory & Particle physics
87
15.5 Charged spin-0 particles, conservation of charge
The Lagrange density of charged spin-0 particles is given by: L = −(∂ ν Φ∂ν Φ∗ + M 2 ΦΦ∗ ). Noether’s theorem connects a continuous symmetry of L and an additive conservation law. Suppose that L ((Φα ) , ∂ν (Φα ) ) = L (Φα , ∂ν Φα ) and there exists a continuous transformation between Φα and Φα such as Φα = Φα + f α (Φ). Then holds ∂L ∂ fα = 0 ∂xν ∂(∂ν Φα ) This is a continuity equation ⇒ conservation law. Which quantity is conserved depends on the symmetry. The above Lagrange density is invariant for a change in phase Φ → Φe iθ : a global gauge transformation. The conserved quantity is the current density Jµ (x) = −ie(Φ∂µ Φ∗ − Φ∗ ∂µ Φ). Because this quantity is 0 for real fields a complex field is needed to describe charged particles. When this field is quantized the field operators are given by 1 Φ(x) = √ V √ 1 a(k )eikx + b† (k )e−ikx 2ωk 1 , Φ† (x) = √ V √ 1 a† (k )eikx + b(k )e−ikx 2ωk
k
k
Hence the energy operator is given by: H=
k
hωk a† (k )a(k ) + b† (k )b(k ) ¯
and the charge operator is given by: Q(t) = −i J4 (x)d3 x ⇒ Q = e a† (k )a(k ) − b† (k )b(k )
k
From this follows that a† a := N+ (k ) is an occupation number operator for particles with a positive charge and b† b := N− (k ) is an occupation number operator for particles with a negative charge.
15.6 Field functions for spin- 1 particles 2
Spin is defined by the behaviour of the solutions ψ of the Dirac equation. A scalar field Φ has the property ˜ that, if it obeys the Klein-Gordon equation, the rotated field Φ(x) := Φ(Λ−1 x) also obeys it. Λ denotes 4-dimensional rotations: the proper Lorentz transformations. These can be written as: ∂ ∂ ˜ Φ(x) = Φ(x)e−in·L with Lµν = −i¯ xµ h − xν ∂xν ∂xµ For µ ≤ 3, ν ≤ 3 these are rotations, for ν = 4, µ = 4 these are Lorentz transformations.
˜ ˜ A rotated field ψ obeys the Dirac equation if the following condition holds: ψ(x) = D(Λ)ψ(Λ−1 x). This 1 −1 in·S ¯ results in the condition D γλ D = Λλµ γµ . One finds: D = e with Sµν = −i 2 hγµ γν . Hence: ˜ ψ(x) = e−i(S+L) ψ(x) = e−iJ ψ(x)
Then the solutions of the Dirac equation are given by: ψ(x) = ur (p )e−i(p·x±Et) ± Here, r is an indication for the direction of the spin, and ± is the sign of the energy. With the notation v r (p ) = ur (−p ) and ur (p ) = ur (p ) one can write for the dot products of these spinors: − + ur (p )ur (p ) = + + E E δrr , ur (p )ur (p ) = δrr , ur (p )ur (p ) = 0 − − + − M M
88
Physics Formulary by ir. J.C.A. Wevers
Because of the factor E/M this is not relativistic invariant. A Lorentz-invariant dot product is defined by ab := a† γ4 b, where a := a† γ4 is a row spinor. From this follows: ur (p )ur (p ) = δrr , v r (p )v r (p ) = −δrr , ur (p )v r (p ) = 0 Combinations of the type aa give a 4 × 4 matrix:
2
u
r=1
r
(p )ur (p )
−iγλ pλ + M = , 2M
2
v r (p )v r (p ) =
r=1
−iγλ pλ − M 2M
The Lagrange density which results in the Dirac equation and having the correct energy normalization is: L(x) = −ψ(x) γµ and the current density is Jµ (x) = −ieψγµ ψ. ∂ +M ∂xµ ψ(x)
15.7 Quantization of spin- 1 fields 2
The general solution for the fieldoperators is in this case: ψ(x) = and ψ(x) =
†
M V M V
p
1 √ E 1 √ E
cr (p )ur (p )eipx + d† (p )v r (p )e−ipx r
r
c† (p )ur (p )e−ipx + dr (p )v r (p )eipx r
r
p
Here, c and c are the creation respectively annihilation operators for an electron and d † and d the creation respectively annihilation operators for a positron. The energy operator is given by
2
H=
p
Ep
r=1
c† (p )cr (p ) − dr (p )d† (p ) r r
To prevent that the energy of positrons is negative the operators must obey anti commutation rules in stead of commutation rules: [cr (p ), c† (p )]+ = [dr (p ), d† (p )]+ = δrr δpp , all other anti commutators are 0. r r The field operators obey [ψα (x), ψβ (x )] = 0 , [ψα (x), ψβ (x )] = 0 , [ψα (x), ψβ (x )]+ = −iSαβ (x − x ) ∂ − M ∆(x) ∂xλ The anti commutation rules give besides the positive-definite energy also the Pauli exclusion principle and the Fermi-Dirac statistics: because c† (p )c† (p ) = −c† (p )c† (p ) holds: {c† (p)}2 = 0. It appears to be impossible r r r r r to create two electrons with the same momentum and spin. This is the exclusion principle. Another way to see + + this is the fact that {Nr (p )}2 = Nr (p ): the occupation operators have only eigenvalues 0 and 1. with S(x) = γλ To avoid infinite vacuum contributions to the energy and charge the normal product is introduced. The expression for the current density now becomes Jµ = −ieN (ψγµ ψ). This product is obtained by: • Expand all fields into creation and annihilation operators, • Keep all terms which have no annihilation operators, or in which they are on the right of the creation operators, • In all other terms interchange the factors so that the annihilation operators go to the right. By an interchange of two fermion operators add a minus sign, by interchange of two boson operators not. Assume hereby that all commutators are zero.
Chapter 15: Quantum field theory & Particle physics
89
15.8 Quantization of the electromagnetic field
1 Starting with the Lagrange density L = − 2
∂Aν ∂Aν ∂xµ ∂xµ
it follows for the field operators A(x): 1 A(x) = √ V √ 1 2ωk
4
a m (k )
m=1
m
(k )eikx + a† (k )
m
(k )∗ e−ikx
k
The operators obey [am (k ), a† (k )] = δmm δkk . All other commutators are 0. m gives the polarization m direction of the photon: m = 1, 2 gives transversal polarized, m = 3 longitudinal polarized and m = 4 timelike polarized photons. Further holds: [Aµ (x), Aν (x )] = iδµν D(x − x ) with D(y) = ∆(y)|m=0 In spite of the fact that A4 = iV is imaginary in the classical case, A4 is still defined to be hermitian because otherwise the sign of the energy becomes incorrect. By changing the definition of the inner product in configuration space the expectation values for A1,2,3 (x) ∈ I and for A4 (x) become imaginary. R If the potentials satisfy the Lorentz gauge condition ∂µ Aµ = 0 the E and B operators derived from these potentials will satisfy the Maxwell equations. However, this gives problems with the commutation rules. It is now demanded that only those states are permitted for which holds ∂A+ µ |Φ = 0 ∂xµ This results in: ∂Aµ ∂xµ = 0.
From this follows that (a3 (k ) − a4 (k ))|Φ = 0. With a local gauge transformation one obtains N3 (k ) = 0 and N4 (k ) = 0. However, this only applies to free EM-fields: in intermediary states in interactions there can exist longitudinal and timelike photons. These photons are also responsible for the stationary Coulomb potential.
15.9 Interacting fields and the S-matrix
The S(scattering)-matrix gives a relation between the initial and final states of an interaction: |Φ(∞) = S|Φ(−∞) . If the Schr¨ dinger equation is integrated: o
t
|Φ(t) = |Φ(−∞) − i and perturbation theory is applied one finds that: S= (−i)n n! n=0
∞
Hint (t1 )|Φ(t1 ) dt1
−∞
∞
···
T {Hint (x1 ) · · · Hint (xn )} d4 x1 · · · d4 xn ≡
S (n)
n=0
Here, the T -operator means a time-ordered product: the terms in such a product must be ordered in increasing time order from the right to the left so that the earliest terms act first. The S-matrix is then given by: S ij = Φi |S|Φj = Φi |Φ(∞) . The interaction Hamilton density for the interaction between the electromagnetic and the electron-positron field is: Hint (x) = −Jµ (x)Aµ (x) = ieN (ψγµ ψAµ ) When this is expanded as: Hint = ieN (ψ + + ψ − )γµ (ψ + + ψ − )(A+ + A− ) µ µ
90
Physics Formulary by ir. J.C.A. Wevers
The expressions for S (n) contain time-ordered products of normal products. This can be written as a sum of normal products. The appearing operators describe the minimal changes necessary to change the initial state into the final state. The effects of the virtual particles are described by the (anti)commutator functions. Some time-ordened products are: T {Φ(x)Φ(y)} T ψα (x)ψβ (y)
1 = N {Φ(x)Φ(y)} + 2 ∆F (x − y) F = N ψα (x)ψβ (y) − 1 Sαβ (x − y) 2
eight terms appear. Each term corresponds with a possible process. The term ieψ + γµ ψ + A− acting on |Φ µ gives transitions where A− creates a photon, ψ + annihilates an electron and ψ + annihilates a positron. Only µ terms with the correct number of particles in the initial and final state contribute to a matrix element Φ i |S|Φj . Further the factors in Hint can create and thereafter annihilate particles: the virtual particles.
F T {Aµ (x)Aν (y)} = N {Aµ (x)Aν (y)} + 1 δµν Dµν (x − y) 2
The term 1 ∆F (x − y) is called the contraction of Φ(x) and Φ(y), and is the expectation value of the time2 ordered product in the vacuum state. Wick’s theorem gives an expression for the time-ordened product of an arbitrary number of field operators. The graphical representation of these processes are called Feynman diagrams. In the x-representation each diagram describes a number of processes. The contraction functions can also be written as: ∆F (x) = lim −2i →0 (2π)4 −2i eikx d4 k and S F (x) = lim →0 (2π)4 k 2 + m2 − i eipx iγµ pµ − M 4 d p p2 + M 2 − i
Here, S F (x) = (γµ ∂µ − M )∆F (x), DF (x) = ∆F (x)|m=0 and eikx 3 1 d k (2π)3 ωk F ∆ (x) = 1 e−ikx 3 d k 3 (2π) ωk
if x0 > 0
if x0 < 0
In the expressions for S (2) this gives rise to terms δ(p + k − p − k ). This means that energy and momentum is conserved. However, virtual particles do not obey the relation between energy and momentum.
15.10 Divergences and renormalization
It turns out that higher orders contribute infinite terms because only the sum p + k of the four-momentum of the virtual particles is fixed. An integration over one of them becomes ∞. In the x-representation this can be understood because the product of two functions containing δ-like singularities is not well defined. This is solved by discounting all divergent diagrams in a renormalization of e and M . It is assumed that an electron, if there would not be an electromagnetical field, would have a mass M 0 and a charge e0 unequal to the observed mass M and charge e. In the Hamilton and Lagrange density of the free electron-positron field appears M 0 . So this gives, with M = M0 + ∆M : Le−p (x) = −ψ(x)(γµ ∂µ + M0 )ψ(x) = −ψ(x)(γµ ∂µ + M )ψ(x) + ∆M ψ(x)ψ(x) and Hint = ieN (ψγµ ψAµ ) − i∆eN (ψγµ ψAµ ).
15.11 Classification of elementary particles
Elementary particles can be categorized as follows: 1. Hadrons: these exist of quarks and can be categorized in: I. Baryons: these exist of 3 quarks or 3 antiquarks.
Chapter 15: Quantum field theory & Particle physics
91
II. Mesons: these exist of one quark and one antiquark. 2. Leptons: e± , µ± , τ ± , νe , νµ , ντ , ν e , ν µ , ν τ . 3. Field quanta: γ, W± , Z0 , gluons, gravitons (?). An overview of particles and antiparticles is given in the following table: Particle u d s c b t e− µ− τ− νe νµ ντ γ gluon W+ Z graviton spin (¯ ) B h 1/2 1/3 1/2 1/3 1/2 1/3 1/2 1/3 1/2 1/3 1/2 1/3 1/2 0 1/2 0 1/2 0 1/2 0 1/2 0 1/2 0 1 0 1 0 1 0 1 0 2 0 L 0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 T 1/2 1/2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 T3 1/2 −1/2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 S 0 0 −1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 C 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 B∗ 0 0 0 0 −1 0 0 0 0 0 0 0 0 0 0 0 0 charge (e) +2/3 −1/3 −1/3 +2/3 −1/3 +2/3 −1 −1 −1 0 0 0 0 0 +1 0 0 m0 (MeV) 5 9 175 1350 4500 173000 0.511 105.658 1777.1 0(?) 0(?) 0(?) 0 0 80220 91187 0 antipart. u d s c b t e+ µ+ τ+ νe νµ ντ γ gluon W− Z graviton
Here B is the baryon number and L the lepton number. It is found that there are three different lepton numbers, one for e, µ and τ , which are separately conserved. T is the isospin, with T 3 the projection of the isospin on the third axis, C the charmness, S the strangeness and B∗ the bottomness. The anti particles have quantum numbers with the opposite sign except for the total isospin T. The composition of (anti)quarks of the hadrons is given in the following table, together with their mass in MeV in their ground state: π0 π+ π− K0 K0 K+ K− D+ D− D0 D0 F+ F−
1 2
√
2(uu+dd) ud du sd ds us su cd dc cu uc cs sc
134.9764 139.56995 139.56995 497.672 497.672 493.677 493.677 1869.4 1869.4 1864.6 1864.6 1969.0 1969.0
J/Ψ Υ p+ p− n0 n0 Λ Λ Σ+ Σ− Σ0 Σ0 Σ−
cc bb uud uud udd udd uds uds uus uus uds uds dds
3096.8 9460.37 938.27231 938.27231 939.56563 939.56563 1115.684 1115.684 1189.37 1189.37 1192.55 1192.55 1197.436
Σ+ Ξ0 0 Ξ − Ξ Ξ+ Ω− Ω+ Λ+ c ∆2− ∆2+ ∆+ ∆0 ∆−
dds uss uss dss dss sss sss udc uuu uuu uud udd ddd
1197.436 1314.9 1314.9 1321.32 1321.32 1672.45 1672.45 2285.1 1232.0 1232.0 1232.0 1232.0 1232.0
Each quark can exist in two spin states. So mesons are bosons with spin 0 or 1 in their ground state, while 1 baryons are fermions with spin 2 or 3 . There exist excited states with higher internal L. Neutrino’s have a 2 1 1 helicity of − 2 while antineutrino’s have only + 2 as possible value. The quantum numbers are subject to conservation laws. These can be derived from symmetries in the Lagrange density: continuous symmetries give rise to additive conservation laws, discrete symmetries result in multiplicative conservation laws. Geometrical conservation laws are invariant under Lorentz transformations and the CPT-operation. These are: 1. Mass/energy because the laws of nature are invariant for translations in time.
92
Physics Formulary by ir. J.C.A. Wevers
2. Momentum because the laws of nature are invariant for translations in space. 3. Angular momentum because the laws of nature are invariant for rotations. Dynamical conservation laws are invariant under the CPT-operation. These are: 1. Electrical charge because the Maxwell equations are invariant under gauge transformations. 2. Colour charge is conserved. 3. Isospin because QCD is invariant for rotations in T-space. 4. Baryon number and lepton number are conserved but not under a possible SU(5) symmetry of the laws of nature. 5. Quarks type is only conserved under the colour interaction. 6. Parity is conserved except for weak interactions. The elementary particles can be classified into three families: leptons 1st generation 2nd generation 3rd generation e− νe µ− νµ τ− ντ quarks d u s c b t antileptons e+ νe µ+ νµ τ+ ντ antiquarks d u s c b t
Quarks exist in three colours but because they are confined these colours cannot be seen directly. The color force does not decrease with distance. The potential energy will become high enough to create a quarkantiquark pair when it is tried to disjoin an (anti)quark from a hadron. This will result in two hadrons and not in free quarks.
15.12 P and CP-violation
It is found that the weak interaction violates P-symmetry, and even CP-symmetry is not conserved. Some processes which violate P symmetry but conserve the combination CP are: 1. µ-decay: µ− → e− + νµ + ν e . Left-handed electrons appear more than 1000× as much as right-handed ones. 2. β-decay of spin-polarized 60 Co: 60 Co →60 Ni + e− + ν e . More electrons with a spin parallel to the Co than with a spin antiparallel are created: (parallel−antiparallel)/(total)=20%. 3. There is no connection with the neutrino: the decay of the Λ particle through: Λ → p + + π − and Λ → n0 + π 0 has also these properties. The CP-symmetry was found to be violated by the decay of neutral Kaons. These are the lowest possible states with a s-quark so they can decay only weakly. The following holds: C|K 0 = η|K0 where η is a phase factor. Further holds P|K0 = −|K0 because K0 and K0 have an intrinsic parity of −1. From this follows that K0 and K0 are not eigenvalues of CP: CP|K0 = |K0 . The linear combinations √ √ 1 |K0 := 2 2(|K0 + |K0 ) and |K0 := 1 2(|K0 − |K0 ) 2 1 2 are eigenstates of CP: CP|K0 = +|K0 and CP|K0 = −|K0 . A base of K0 and K0 is practical while 1 1 2 2 1 2 describing weak interactions. For colour interactions a base of K 0 and K0 is practical because then the number u−number u is constant. The expansion postulate must be used for weak decays: |K0 = 1 ( K0 |K0 + K0 |K0 ) 1 2 2
Chapter 15: Quantum field theory & Particle physics
93
The probability to find a final state with CP= −1 is 1 0 0 2 | . 2 | K1 |K
1 2|
K0 |K0 |2 , the probability of CP=+1 decay is 2
θ1 ≡ θC is the Cabibbo angle: sin(θC ) ≈ 0.23 ± 0.01.
The relation between the mass eigenvalues of the quarks (unaccented) and the fields arising in the weak currents (accented) is (u , c , t ) = (u, c, t), and: d 1 0 0 1 0 0 cos θ1 sin θ1 0 s = 0 cos θ2 sin θ2 0 1 0 − sin θ1 cos θ1 0 b 0 − sin θ2 cos θ2 0 0 1 0 0 eiδ 1 0 0 d 0 cos θ3 sin θ3 s 0 − sin θ3 cos θ3 b
15.13 The standard model
When one wants to make the Lagrange density which describes a field invariant for local gauge transformations from a certain group, one has to perform the transformation D ∂ g ∂ → = − i Lk A k µ ∂xµ Dxµ ∂xµ h ¯ Here the Lk are the generators of the gauge group (the “charges”) and the A k are the gauge fields. g is the µ matching coupling constant. The Lagrange density for a scalar field becomes:
µν 1 a 1 L = − 2 (Dµ Φ∗ Dµ Φ + M 2 Φ∗ Φ) − 4 Fµν Fa a and the field tensors are given by: Fµν = ∂µ Aa − ∂ν Aa + gca Al Am . ν µ lm µ ν
15.13.1 The electroweak theory
The electroweak interaction arises from the necessity to keep the Lagrange density invariant for local gauge transformations of the group SU(2)⊗U(1). Right- and left-handed spin states are treated different because the weak interaction does not conserve parity. If a fifth Dirac matrix is defined by: 0 0 1 0 0 0 0 1 γ5 := γ1 γ2 γ3 γ4 = − 1 0 0 0 0 1 0 0 the left- and right- handed solutions of the Dirac equation for neutrino’s are given by: ψL = 1 (1 + γ5 )ψ and ψR = 1 (1 − γ5 )ψ 2 2 It appears that neutrino’s are always left-handed while antineutrino’s are always right-handed. The hypercharge Y , for quarks given by Y = B + S + C + B∗ + T , is defined by: Q = 1 Y + T3 2 so [Y, Tk ] = 0. The group U(1)Y ⊗SU(2)T is taken as symmetry group for the electroweak interaction because the generators of this group commute. The multiplets are classified as follows: e− R T T3 Y 0 0 −2
1 2
νeL e− L
1 2 1 2
uL dL
1 2 1 2 1 2
uR 0 0
4 3
dR 0 0 −2 3
− −1
1 3
−
94
Physics Formulary by ir. J.C.A. Wevers
Now, 1 field Bµ (x) is connected with gauge group U(1) and 3 gauge fields Aµ (x) are connected with SU(2). The total Lagrange density (minus the fieldterms) for the electron-fermion field now becomes: L0,EW g g = −(ψνe,L , ψeL )γ µ ∂µ − i Aµ · ( 1 σ) − 1 i Bµ · (−1) 2 2 h h ¯ ¯ g ψeR γ µ ∂µ − 1 i (−2)Bµ ψeR 2 h ¯ ψνe,L ψeL −
Here, 1 σ are the generators of T and −1 and −2 the generators of Y . 2
15.13.2 Spontaneous symmetry breaking: the Higgs mechanism
All leptons are massless in the equations above. Their mass is probably generated by spontaneous symmetry breaking. This means that the dynamic equations which describe the system have a symmetry which the ground state does not have. It is assumed that there exists an isospin-doublet of scalar fields Φ with electrical charges +1 and 0 and potential V (Φ) = −µ2 Φ∗ Φ + λ(Φ∗ Φ)2 . Their antiparticles have charges −1 and 0. The extra µ terms in L arising from these fields, LH = (DLµ Φ)∗ (DL Φ) − V (Φ), are globally U(1)⊗SU(2) symmetric. Hence the state with the lowest energy corresponds with the state Φ ∗ (x)Φ(x) = v = µ2 /2λ =constant. The field can be written (were ω ± and z are Nambu-Goldstone bosons which can be transformed away, and √ mφ = µ 2) as: Φ= Φ+ Φ0 = iω + √ (v + φ − iz)/ 2 and 0|Φ|0 = 0 √ v/ 2
Because this expectation value = 0 the SU(2) symmetry is broken but the U(1) symmetry is not. When the gauge fields in the resulting Lagrange density are separated one obtains: √ √ − + 1 Wµ = 1 2(A1 + iA2 ) , Wµ = 2 2(A1 − iA2 ) µ µ µ µ 2 Zµ Aµ = = gA3 − g Bµ µ g2 + g 2 g2 +g
2
≡ A3 cos(θW ) − Bµ sin(θW ) µ ≡ A3 sin(θW ) + Bµ cos(θW ) µ
g A3 + gBµ µ
where θW is called the Weinberg angle. For this angle holds: sin2 (θW ) = 0.255 ± 0.010. Relations for the masses of the field quanta can be obtained from the remaining terms: M W = 1 vg and MZ = 1 v g 2 + g 2 , 2 2 gg = g cos(θW ) = g sin(θW ) and for the elementary charge holds: e = g2 + g 2 Experimentally it is found that MW = 80.022 ± 0.26 GeV/c2 and MZ = 91.187 ± 0.007 GeV/c2 . According to the weak theory this should be: MW = 83.0 ± 0.24 GeV/c2 and MZ = 93.8 ± 2.0 GeV/c2 .
15.13.3 Quantumchromodynamics
Coloured particles interact because the Lagrange density is invariant for the transformations of the group SU(3) of the colour interaction. A distinction can be made between two types of particles: 1. “White” particles: they have no colour charge, the generator T = 0. 2. “Coloured” particles: the generators T are 8 3 × 3 matrices. There exist three colours and three anticolours. The Lagrange density for coloured particles is given by LQCD = i Ψ k γ µ Dµ Ψ k +
k k,l a µν Ψk Mkl Ψl − 1 Fµν Fa 4
The gluons remain massless because this Lagrange density does not contain spinless particles. Because leftand right- handed quarks now belong to the same multiplet a mass term can be introduced. This term can be brought in the form Mkl = mk δkl .
Chapter 15: Quantum field theory & Particle physics
95
15.14 Path integrals
The development in time of a quantum mechanical system can, besides with Schr¨ dingers equation, also be o described by a path integral (Feynman): ψ(x , t ) = F (x , t , x, t)ψ(x, t)dx
in which F (x , t , x, t) is the amplitude of probability to find a system on time t in x if it was in x on time t. Then, iS[x] F (x , t , x, t) = exp d[x] h ¯ where S[x] is an action-integral: S[x] = taken over all possible paths [x]: L(x, x, t)dt. The notation d[x] means that the integral has to be ˙
∞
1 d[x] := lim n→∞ N
dx(tn )
n
−∞
in which N is a normalization constant. To each path is assigned a probability amplitude exp(iS/¯ ). The h classical limit can be found by taking δS = 0: the average of the exponent vanishes, except where it is stationary. In quantum fieldtheory, the probability of the transition of a fieldoperator Φ(x, −∞) to Φ (x, ∞) is given by iS[Φ] F (Φ (x, ∞), Φ(x, −∞)) = exp d[Φ] h ¯ with the action-integral S[Φ] =
Ω
L(Φ, ∂ν Φ)d4 x
15.15 Unification and quantum gravity
The strength of the forces varies with energy and the reciprocal coupling constants approach each other with increasing energy. The SU(5) model predicts complete unification of the electromagnetical, weak and colour forces at 1015 GeV. It also predicts 12 extra X bosons which couple leptons and quarks and are i.g. responsible for proton decay, with dominant channel p+ → π 0 + e+ , with an average lifetime of the proton of 1031 year. This model has been experimentally falsified. Supersymmetric models assume a symmetry between bosons and fermions and predict partners for the currently known particles with a spin which differs 1 . The supersymmetric SU(5) model predicts unification at 2 1016 GeV and an average lifetime of the proton of 1033 year. The dominant decay channels in this theory are p+ → K+ + ν µ and p+ → K0 + µ+ . Quantum gravity plays only a role in particle interactions at the Planck dimensions, where λ C ≈ RS : mPl = hc/G = 3 · 1019 GeV, tPl = h/mPlc2 = hG/c5 = 10−43 sec and rPl = ctPl ≈ 10−35 m.
Chapter 16
Astrophysics
16.1 Determination of distances
The parallax is mostly used to determine distances in nearby space. The parallax is the angular difference between two measurements of the position of the object from different view-points. If the annual parallax is given by p, the distance R of the object is given by R = a/ sin(p), in which a is the radius of the Earth’s orbit. The clusterparallax is used to determine the distance of a group of stars by using their motion w.r.t. a fixed background. The tangential velocity vt and the radial velocity vr of the stars along the sky are given by vr = V cos(θ) , vt = V sin(θ) = ωR ˆ where θ is the angle between the star and the point of convergence and R the distance in pc. This results, with vt = vr tan(θ), in: R= vr tan(θ) 1 ˆ ⇒ R= ω p
-5 -4 M -3 -2 -1
0 1
Type 1
Type 2 RR-Lyrae 0,1 0,3 1 3 10 30 100
where p is the parallax in arc seconds. The parallax is then given by p= 4.74µ vr tan(θ)
P (days) →
with µ de proper motion of the star in /yr. A method to determine the distance of objects which are somewhat further away, like galaxies and star clusters, uses the period-Brightness relation for Cepheids. This relation is shown in the above figure for different types of stars.
16.2 Brightness and magnitudes
The brightness is the total radiated energy per unit of time. Earth receives s 0 = 1.374 kW/m2 from the Sun. Hence, the brightness of the Sun is given by L = 4πr2 s0 = 3.82 · 1026 W. It is also given by:
∞
L = 4πR
2 0
πFν dν
where πFν is the monochromatic radiation flux. At the position of an observer this is πf ν , with fν = (R/r)2 Fν if absorption is ignored. If Aν is the fraction of the flux which reaches Earth’s surface, the transmission factor is given by Rν and the surface of the detector is given by πa2 , then the apparent brightness b is given by:
∞
b = πa The magnitude m is defined by:
2 0
fν Aν Rν dν
b1 1 = (100) 5 (m2 −m1 ) = (2.512)m2 −m1 b2
Chapter 16: Astrophysics
97
because the human eye perceives lightintensities logaritmical. From this follows that m 2 − m1 = 2.5 ·10 log(b1 /b2 ), or: m = −2.5 ·10 log(b) + C. The apparent brightness of a star if this star would be at a distance of 10 pc is called the absolute brightness B: B/b = (ˆ/10)2 . The absolute magnitude is then given by r M = −2.5 ·10 log(B) + C, or: M = 5 + m − 5 ·10 log(ˆ). When an interstellar absorption of 10−4 /pc is taken r into account one finds: M = (m − 4 · 10−4 r ) + 5 − 5 ·10 log(ˆ) ˆ r If a detector detects all radiation emitted by a source one would measure the absolute bolometric magnitude. If the bolometric correction BC is given by BC = 2.5 ·10 log Energy flux received Energy flux detected = 2.5 ·10 log fν dν fν Aν Rν dν
holds: Mb = MV − BC where MV is the visual magnitude. Further holds Mb = −2.5 ·10 log L L + 4.72
The radiation energy passing through a surface dA is dE = Iν (θ, ϕ) cos(θ)dνdΩdAdt, where Iµ is the monochromatical intensity [Wm−2 sr−1 Hz−1 ]. When there is no absorption the quantity Iν is independent of the distance to the source. Planck’s law holds for a black body: Iν (T ) ≡ Bν (T ) = c 1 2hν 3 wν (T ) = 2 4π c exp(hν/kT ) − 1
The radiation transport through a layer can then be written as: dIν = −Iν κν + jν ds Here, jν is the coefficient of emission and κν the coefficient of absorption. ds is the thickness of the layer. The optical thickness τν of the layer is given by τν = κν ds. The layer is optically thin if τν 1, the layer is optically thick if τν 1. For a stellar atmosphere in LTE holds: jν = κν Bν (T ). Then also holds: Iν (s) = Iν (0)e−τν + Bν (T )(1 − e−τν )
16.4 Composition and evolution of stars
The structure of a star is described by the following equations: dM (r) dr dp(r) dr L(r) dr dT (r) dr rad dT (r) dr conv = 4π (r)r2 = − GM (r) (r) r2
= 4π (r)ε(r)r 2 = − = 3 L(r) κ(r) , (Eddington), or 4 4πr2 4σT 3 (r) T (r) γ − 1 dp(r) , (convective energy transport) p(r) γ dr
Further, for stars of the solar type, the composing plasma can be described as an ideal gas: p(r) = (r)kT (r) µmH
98
Physics Formulary by ir. J.C.A. Wevers
where µ is the average molecular mass, usually well approximated by: µ= nmH = 1 2X +
3 4Y 1 + 2Z
where X is the mass fraction of H, Y the mass fraction of He and Z the mass fraction of the other elements. Further holds: κ(r) = f ( (r), T (r), composition) and ε(r) = g( (r), T (r), composition) Convection will occur when the star meets the Schwartzschild criterium: dT dr <
conv
dT dr
Otherwise the energy transfer takes place by radiation. For stars in quasi-hydrostatic equilibrium hold the approximations r = 1 R, M (r) = 1 M , dM/dr = M/R, κ ∼ and ε ∼ T µ (this last assumption is only 2 2 valid for stars on the main sequence). For pp-chains holds µ ≈ 5 and for the CNO chains holds µ = 12 tot 18. 8 It can be derived that L ∼ M 3 : the mass-brightness relation. Further holds: L ∼ R 4 ∼ Teff . This results in the equation for the main sequence in the Hertzsprung-Russel diagram:
10
log(L) = 8 ·10 log(Teff ) + constant
16.5 Energy production in stars
The net reaction from which most stars gain their energy is: 41 H → 4 He + 2e+ + 2νe + γ. This reaction produces 26.72 MeV. Two reaction chains are responsible for this reaction. The slowest, speedlimiting reaction is shown in boldface. The energy between brackets is the energy carried away by the neutrino. 1. The proton-proton chain can be divided into two subchains: 1 H + p+ → 2 D + e+ + νe , and then 2 D + p → 3 He + γ. II. pp2: 3 He + α → 7 Be + γ I. pp1: 3 He +3 He → 2p+ + 4 He. There is 26.21 + (0.51) MeV released. i. 7 Be + e− → 7 Li + ν, then 7 Li + p+ → 24 He + γ. 25.92 + (0.80) MeV. ii. 7 Be + p+ → 8 B + γ, then 8 B + e+ → 24 He + ν. 19.5 + (7.2) MeV.
Both 7 Be chains become more important with raising T .
2. The CNO cycle. The first chain releases 25.03 + (1.69) MeV, the second 24.74 + (1.98) MeV. The reactions are shown below. → O + e+ → ↑ 14 N + p+ →
15 15
N+ν O+γ ←
15
−→ N + p+ → α +12 C ↓ 12 C + p+ → 13 N + γ ↓ 13 N → 13 C + e+ + ν ↓ 13 C + p+ → 14 N + γ ←−
15
15
N + p+ → 16 O + γ ↓ 16 O + p+ → 17 F + γ ↓ 17 F → 17 O + e+ + ν ↓ 17 O + p+ → α + 14 N
The
operator
99
The
-operator
∂ ∂ ∂ ex + ey + ez , gradf = ∂x ∂y ∂z ·a= ∂ay ∂az ∂ax + + , ∂x ∂y ∂z ex + ∂f ∂f ∂f ex + ey + ez ∂x ∂y ∂z ∂2f ∂2f ∂2f + 2 + 2 2 ∂x ∂y ∂z ey + ∂ay ∂ax − ∂x ∂y ez
In cartesian coordinates (x, y, z) holds: = f=
2
div a = rot a =
f=
×a=
∂az ∂ay − ∂y ∂z
∂ax ∂az − ∂z ∂x
In cylinder coordinates (r, ϕ, z) holds: = div a = rot a = 1 ∂ ∂ ∂f 1 ∂f ∂f ∂ er + eϕ + ez , gradf = er + eϕ + ez ∂r r ∂ϕ ∂z ∂r r ∂ϕ ∂z
2
ar 1 ∂aϕ ∂az ∂ar + + + , ∂r r r ∂ϕ ∂z 1 ∂az ∂aϕ − r ∂ϕ ∂z er +
f=
∂2f 1 ∂f 1 ∂2f ∂2f + + 2 + 2 2 2 ∂r r ∂r r ∂ϕ ∂z eϕ + ∂aϕ aϕ 1 ∂ar + − ∂r r r ∂ϕ ez
∂ar ∂az − ∂z ∂r
In spherical coordinates (r, θ, ϕ) holds: = gradf = ∂ 1 ∂ 1 ∂ er + eθ + eϕ ∂r r ∂θ r sin θ ∂ϕ ∂f 1 ∂f 1 ∂f er + eθ + eϕ ∂r r ∂θ r sin θ ∂ϕ ∂ar 2ar 1 ∂aθ aθ 1 ∂aϕ + + + + ∂r r r ∂θ r tan θ r sin θ ∂ϕ aθ 1 ∂aθ ∂aϕ aϕ 1 ∂aϕ 1 ∂ar + − − − er + r ∂θ r tan θ r sin θ ∂ϕ r sin θ ∂ϕ ∂r r aθ 1 ∂ar ∂aθ eϕ + − ∂r r r ∂θ ∂2f ∂f ∂2f 2 ∂f 1 ∂2f 1 1 + + 2 2 + 2 + 2 2 ∂r2 r ∂r r ∂θ r tan θ ∂θ r sin θ ∂ϕ2
div a = rot a =
eθ +
2
f
=
General orthonormal curvelinear coordinates (u, v, w) can be obtained from cartesian coordinates by the transformation x = x(u, v, w). The unit vectors are then given by: eu = 1 ∂x 1 ∂x 1 ∂x , ev = , ew = h1 ∂u h2 ∂v h3 ∂w
where the factors hi set the norm to 1. Then holds: gradf = 1 ∂f 1 ∂f 1 ∂f eu + ev + ew h1 ∂u h2 ∂v h3 ∂w ∂ ∂ ∂ 1 (h2 h3 au ) + (h3 h1 av ) + (h1 h2 aw ) h1 h2 h3 ∂u ∂v ∂w 1 ∂(h3 aw ) ∂(h2 av ) ∂(h1 au ) ∂(h3 aw ) 1 eu + − − h2 h3 ∂v ∂w h3 h1 ∂w ∂u 1 ∂(h2 av ) ∂(h1 au ) ew − h1 h2 ∂u ∂v ∂ h3 h1 ∂f ∂ 1 ∂ h2 h3 ∂f h1 h2 ∂f + + h1 h2 h3 ∂u h1 ∂u ∂v h2 ∂v ∂w h3 ∂w
div a = rot a =
ev +
2
f
=
100
The SI units
The SI units
Basic units
Quantity Length Mass Time Therm. temp. Electr. current Luminous intens. Amount of subst. Unit metre kilogram second kelvin ampere candela mol Sym. m kg s K A cd mol
Derived units with special names
Quantity Frequency Force Pressure Energy Power Charge El. Potential El. Capacitance El. Resistance El. Conductance Mag. flux Mag. flux density Inductance Luminous flux Illuminance Activity Absorbed dose Dose equivalent Unit hertz newton pascal joule watt coulomb volt farad ohm siemens weber tesla henry lumen lux bequerel gray sievert Sym. Hz N Pa J W C V F Ω S Wb T H lm lx Bq Gy Sv Derivation s−1 kg · m · s−2 N · m−2 N·m J · s−1 A·s W · A−1 C · V−1 V · A−1 A · V−1 V·s Wb · m−2 Wb · A−1 cd · sr lm · m−2 s−1 J · kg−1 J · kg−1
Extra units
Prefixes
yotta zetta exa peta tera Y Z E P T 1024 1021 1018 1015 1012 giga mega kilo hecto deca G M k h da 109 106 103 102 10 deci centi milli micro nano d c m µ n 10−1 10−2 10−3 10−6 10−9 pico femto atto zepto yocto p f a z y 10−12 10−15 10−18 10−21 10−24
DOCUMENT INFO
Shared By:
Categories:
Stats:
views: 1261 posted: 6/28/2009 language: English pages: 108
|
2013-12-18 08:11:28
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9078001379966736, "perplexity": 1565.334892321909}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1387345758214/warc/CC-MAIN-20131218054918-00006-ip-10-33-133-15.ec2.internal.warc.gz"}
|
https://intelligencemission.com/free-energy-guide-free-electricity-for-all.html
|
The high concentrations of A “push” the reaction series (A ⇌ B ⇌ C ⇌ D) to the right, while the low concentrations of D “pull” the reactions in the same direction. Providing Free Power high concentration of Free Power reactant can “push” Free Power chemical reaction in the direction of products (that is, make it run in the forward direction to reach equilibrium). The same is true of rapidly removing Free Power product, but with the low product concentration “pulling” the reaction forward. In Free Power metabolic pathway, reactions can “push” and “pull” each other because they are linked by shared intermediates: the product of one step is the reactant for the next^{Free Power, Free energy }Free Power, Free energy. “Think of Two Powerful Magnets. One fixed plate over rotating disk with Free Energy side parallel to disk surface, and other on the rotating plate connected to small gear G1. If the magnet over gear G1’s north side is parallel to that of which is over Rotating disk then they both will repel each other. Now the magnet over the left disk will try to rotate the disk below in (think) clock-wise direction. Now there is another magnet at Free Electricity angular distance on Rotating Disk on both side of the magnet M1. Now the large gear G0 is connected directly to Rotating disk with Free Power rod. So after repulsion if Rotating-Disk rotates it will rotate the gear G0 which is connected to gear G1. So the magnet over G1 rotate in the direction perpendicular to that of fixed-disk surface. Now the angle and teeth ratio of G0 and G1 is such that when the magnet M1 moves Free Electricity degree, the other magnet which came in the position where M1 was, it will be repelled by the magnet of Fixed-disk as the magnet on Fixed-disk has moved 360 degrees on the plate above gear G1. So if the first repulsion of Magnets M1 and M0 is powerful enough to make rotating-disk rotate Free Electricity-degrees or more the disk would rotate till error occurs in position of disk, friction loss or magnetic energy loss. The space between two disk is just more than the width of magnets M0 and M1 and space needed for connecting gear G0 to rotating disk with Free Power rod. Now I’ve not tested with actual objects. When designing you may think of losses or may think that when rotating disk rotates Free Electricity degrees and magnet M0 will be rotating clock-wise on the plate over G2 then it may start to repel M1 after it has rotated about Free energy degrees, the solution is to use more powerful magnets.
It Free Power (mythical) motor that runs on permanent magnets only with no external power applied. How can you miss that? It’s so obvious. Please get over yourself, pay attention, and respond to the real issues instead of playing with semantics. @Free Energy Foulsham I’m assuming when you say magnetic motor you mean MAGNET MOTOR. That’s like saying democratic when you mean democrat.. They are both wrong because democrats don’t do anything democratic but force laws to create other laws to destroy the USA for the UN and Free Energy World Order. There are thousands of magnetic motors. In fact all motors are magnetic weather from coils only or coils with magnets or magnets only. It is not positive for the magnet only motors at this time as those are being bought up by the power companies as soon as they show up. We use Free Power HZ in the USA but 50HZ in Europe is more efficient. Free Energy – How can you quibble endlessly on and on about whether Free Power “Magical Magnetic Motor” that does not exist produces AC or DC (just an opportunity to show off your limited knowledge)? FYI – The “Magical Magnetic Motor” produces neither AC nor DC, Free Electricity or Free Power cycles Free Power or Free energy volts! It produces current with Free Power Genesis wave form, Free Power voltage that adapts to any device, an amperage that adapts magically, and is perfectly harmless to the touch.
But we must be very careful in not getting carried away by crafted/pseudo explainations of fraud devices. Mr. Free Electricity, we agree. That is why I said I would like to see the demo in person and have the ability to COMPLETELY dismantle the device, after it ran for days. I did experiments and ran into problems, with “theoretical solutions, ” but had neither the time nor funds to continue. Mine too ran down. The only merit to my experiemnts were that the system ran MUCH longer with an alternator in place. Similar to what the Free Electricity Model S does. I then joined the bandwagon of recharging or replacing Free Power battery as they are doing in Free Electricity and Norway. Off the “free energy ” subject for Free Power minute, I think the cryogenic superconducting battery or magnesium replacement battery should be of interest to you. Why should I have to back up my Free Energy? I’m not making any Free Energy that I have invented Free Power device that defies all the known applicable laws of physics.
Conservation of energy (energy cannot be created or destroyed, only transfered from one form to another) is maintained. Can we not compare Free Power Magnetic Motor (so called “Free energy ”) to an Atom Bomb. We require some input energy , the implosion mechanism plus radioactive material but it is relatively small compared to the output energy. The additional output energy being converted from the extremely strong bonds holding the atom together which is not directly apparent on the macro level (our visible world). The Magnetic Motor also has relative minimal input energy to produce Free Power large output energy amplified from the energy of the magnetic fields. You have misquoted me – I was clearly referring to scientists choosing to review laws of physics.
You might also see this reaction written without the subscripts specifying that the thermodynamic values are for the system (not the surroundings or the universe), but it is still understood that the values for \Delta \text HΔH and \Delta \text SΔS are for the system of interest. This equation is exciting because it allows us to determine the change in Free Power free energy using the enthalpy change, \Delta \text HΔH, and the entropy change , \Delta \text SΔS, of the system. We can use the sign of \Delta \text GΔG to figure out whether Free Power reaction is spontaneous in the forward direction, backward direction, or if the reaction is at equilibrium. Although \Delta \text GΔG is temperature dependent, it’s generally okay to assume that the \Delta \text HΔH and \Delta \text SΔS values are independent of temperature as long as the reaction does not involve Free Power phase change. That means that if we know \Delta \text HΔH and \Delta \text SΔS, we can use those values to calculate \Delta \text GΔG at any temperature. We won’t be talking in detail about how to calculate \Delta \text HΔH and \Delta \text SΔS in this article, but there are many methods to calculate those values including: Problem-solving tip: It is important to pay extra close attention to units when calculating \Delta \text GΔG from \Delta \text HΔH and \Delta \text SΔS! Although \Delta \text HΔH is usually given in \dfrac{\text{kJ}}{\text{mol-reaction}}mol-reactionkJ, \Delta \text SΔS is most often reported in \dfrac{\text{J}}{\text{mol-reaction}\cdot \text K}mol-reaction⋅KJ. The difference is Free Power factor of 10001000!! Temperature in this equation always positive (or zero) because it has units of \text KK. Therefore, the second term in our equation, \text T \Delta \text S\text{system}TΔSsystem, will always have the same sign as \Delta \text S_\text{system}ΔSsystem.
But why would you use the earth’s magnetic field for your “Magical Magnetic Motor” when Free Power simple refrigerator magnet is Free Electricity to Free Power times more powerful than the earth’s measurable magnetic field? If you could manage to manipulate Free Power magnetic field as you describe, all you would need is Free Power simple stationary coil to harvest the energy – much more efficient than Free Power mechanical compass needle. Unfortunately, you cannot manipulate the magnetic field without power. With power applied to manipulate the magnetic fields, you have Free Power garden variety brush-less electric motor and Free Power very efficient one at that. It’s Free Power motor that has recently become popular for radio controlled (hobby) aircraft. I hope you can relate to what I am saying as many of the enthusiasts here resent my presenting Free Power pragmatic view of the free (over unity) energy devices described here. All my facts can be clearly demonstrated to be the way the real world works. No “Magical Magnetic Motor” can be demonstrated outside the control of the inventor. Videos are never proof of anything as they can be easily faked. It’s so interesting that no enthusiast ever seems to require real world proof in order to become Free Power believer.
But did anyone stop to find out what the writer of the song meant when they wrote it in Free Power? Yes, actually, some did, thankfully. But many didn’t and jumped on the hate bandwagon because nowadays many of us seem to have become headline and meme readers and take all we see as fact without ever questioning what we’re being told. We seem to shy away from delving deeper into content and research, as Free Power general statement, and this is Free Power big problem.
This is not Free Power grand revelation. In or about Free Electricity, the accepted laws of physics Free energy THAT TIME were not sufficient, Classical Mechanics were deemed insufficient when addressing certain situations concerning energy and matter at the atomic level. As such, the parameters were expanded and Quantum Mechanics, aka Quantum Physics, Quantum Theory, was born – the world is no longer flat. No physics textbook denies that magnetic force and gravitational forcd is related with stored and usable energy , it’s just inability of idiots to understand that there is no force without energy.
#### Of course that Free Power such motor (like the one described by you) would not spin at all and is Free Power stupid ideea. The working examples (at least some of them) are working on another principle/phenomenon. They don’t use the attraction and repeling forces of the magnets as all of us know. I repeat: that is Free Power stupid ideea. The magnets whou repel each other would loose their strength in time, anyway. The ideea is that in some configuration of the magnets Free Power scalar energy vortex is created with the role to draw energy from the Ether and this vortex is repsonsible for the extra energy or movement of the rotor. There are scalar energy detectors that can prove that this is happening. You can’t detect scalar energy with conventional tools. The vortex si an ubiquitos thing in nature. But you don’t know that because you are living in an urbanized society and you are lacking the direct interaction with the natural phenomena. Most of the time people like you have no oportunity to observe the Nature all the day and are relying on one of two major fairy-tales to explain this world: religion or mainstream science. The magnetism is more than the attraction and repelling forces. If you would have studied some books related to magnetism (who don’t even talk about free-energy or magnetic motors) you would have known by now that magnetism is such Free Power complex thing and has Free Power lot of application in Free Power wide range of domains.
Both sets of skeptics will point to the fact that there has been no concrete action, no major arrests of supposed key Deep State players. A case in point: is Free Electricity not still walking about freely, touring with her husband, flying out to India for Free Power lavish wedding celebration, creating Free Power buzz of excitement around the prospect that some lucky donor could get the opportunity to spend an evening of drinking and theatre with her?
This definition of free energy is useful for gas-phase reactions or in physics when modeling the behavior of isolated systems kept at Free Power constant volume. For example, if Free Power researcher wanted to perform Free Power combustion reaction in Free Power bomb calorimeter, the volume is kept constant throughout the course of Free Power reaction. Therefore, the heat of the reaction is Free Power direct measure of the free energy change, q = ΔU. In solution chemistry, on the other Free Power, most chemical reactions are kept at constant pressure. Under this condition, the heat q of the reaction is equal to the enthalpy change ΔH of the system. Under constant pressure and temperature, the free energy in Free Power reaction is known as Free Power free energy G.
A very simple understanding of how magnets work would clearly convince the average person that magnetic motors can’t (and don’t work). Pray tell where does the energy come from? The classic response is magnetic energy from when they were made. Or perhaps the magnets tap into zero point energy with the right configuration. What about they harness the earth’s gravitational field. Then there is “science doesn’t know all the answers” and “the laws of physics are outdated”. The list goes on with equally implausible rubbish. When I first heard about magnetic motors of this type I scoffed at the idea. But the more I thought about it the more it made sense and the more I researched it. Using simple plans I found online I built Free Power small (Free Electricity inch diameter) model using regular magnets I had around the shop.
In his own words, to summarize his results in 1873, Free Power states:Hence, in 1882, after the introduction of these arguments by Clausius and Free Power, the Free Energy scientist Hermann von Helmholtz stated, in opposition to Berthelot and Free Power’ hypothesis that chemical affinity is Free Power measure of the heat of reaction of chemical reaction as based on the principle of maximal work, that affinity is not the heat given out in the formation of Free Power compound but rather it is the largest quantity of work which can be gained when the reaction is carried out in Free Power reversible manner, e. g. , electrical work in Free Power reversible cell. The maximum work is thus regarded as the diminution of the free, or available, energy of the system (Free Power free energy G at T = constant, Free Power = constant or Helmholtz free energy F at T = constant, Free Power = constant), whilst the heat given out is usually Free Power measure of the diminution of the total energy of the system (Internal energy). Thus, G or F is the amount of energy “free” for work under the given conditions. Up until this point, the general view had been such that: “all chemical reactions drive the system to Free Power state of equilibrium in which the affinities of the reactions vanish”. Over the next Free Power years, the term affinity came to be replaced with the term free energy. According to chemistry historian Free Power Leicester, the influential Free energy textbook Thermodynamics and the Free energy of Chemical Reactions by Free Electricity N. Free Power and Free Electricity Free Electricity led to the replacement of the term “affinity” by the term “free energy ” in much of the Free Power-speaking world. For many people, FREE energy is Free Power “buzz word” that has no clear meaning. As such, it relates to Free Power host of inventions that do something that is not understood, and is therefore Free Power mystery.
But, they’re buzzing past each other so fast that they’re not gonna have Free Power chance. Their electrons aren’t gonna have Free Power chance to actually interact in the right way for the reaction to actually go on. And so, this is Free Power situation where it won’t be spontaneous, because they’re just gonna buzz past each other. They’re not gonna have Free Power chance to interact properly. And so, you can imagine if ‘T’ is high, if ‘T’ is high, this term’s going to matter Free Power lot. And, so the fact that entropy is negative is gonna make this whole thing positive. And, this is gonna be more positive than this is going to be negative. So, this is Free Power situation where our Delta G is greater than zero. So, once again, not spontaneous. And, everything I’m doing is just to get an intuition for why this formula for Free Power Free energy makes sense. And, remember, this is true under constant pressure and temperature. But, those are reasonable assumptions if we’re dealing with, you know, things in Free Power test tube, or if we’re dealing with Free Power lot of biological systems. Now, let’s go over here. So, our enthalpy, our change in enthalpy is positive. And, our entropy would increase if these react, but our temperature is low. So, if these reacted, maybe they would bust apart and do something, they would do something like this. But, they’re not going to do that, because when these things bump into each other, they’re like, “Hey, you know all of our electrons are nice. “There are nice little stable configurations here. “I don’t see any reason to react. ” Even though, if we did react, we were able to increase the entropy. Hey, no reason to react here. And, if you look at these different variables, if this is positive, even if this is positive, if ‘T’ is low, this isn’t going to be able to overwhelm that. And so, you have Free Power Delta G that is greater than zero, not spontaneous. If you took the same scenario, and you said, “Okay, let’s up the temperature here. “Let’s up the average kinetic energy. ” None of these things are going to be able to slam into each other. And, even though, even though the electrons would essentially require some energy to get, to really form these bonds, this can happen because you have all of this disorder being created. You have these more states. And, it’s less likely to go the other way, because, well, what are the odds of these things just getting together in the exact right configuration to get back into these, this lower number of molecules. And, once again, you look at these variables here. Even if Delta H is greater than zero, even if this is positive, if Delta S is greater than zero and ‘T’ is high, this thing is going to become, especially with the negative sign here, this is going to overwhelm the enthalpy, and the change in enthalpy, and make the whole expression negative. So, over here, Delta G is going to be less than zero. And, this is going to be spontaneous. Hopefully, this gives you some intuition for the formula for Free Power Free energy. And, once again, you have to caveat it. It’s under, it assumes constant pressure and temperature. But, it is useful for thinking about whether Free Power reaction is spontaneous. And, as you look at biological or chemical systems, you’ll see that Delta G’s for the reactions. And so, you’ll say, “Free Electricity, it’s Free Power negative Delta G? “That’s going to be Free Power spontaneous reaction. “It’s Free Power zero Delta G. “That’s gonna be an equilibrium. ”
Former Free Electricity was among Free Electricity’s closest friends, and the flight logs from Free Electricity’s private jet shown here reveal that Free Electricity was listed as Free Power passenger on the jet at least Free energy times between Free Power and Free Power, which would have put Free Electricity on the plane at least once Free Power month during the two-year period. Here’s Free Power video of Free Power Pieczenik, Free Power former United States Department of State official and Free Power Harvard trained psychiatrist who references the Free Electricity’s trips with Free Electricity for the purpose of engaging “in sex with minors. ”
It makes you look like Free Power fool, Free Power scammer, or both. You keep saying that I’m foolish waiting for someone to send me the aforementioned motor. Again, you missed the point completely. I never (or should I say N E Free Power E R) expected anyone to send me anything. It was just to make the point that it never existed. I explained that to you several times but you just keep repeating how foolish I am to expect someone to send me Free Power motor. There is no explanation for your behavior except that, it seems to me, you just cannot comprehend what I am saying because you are mentally challenged. This device can indeed charge Free Power battery. If one measures the total energy going in, and the energy stored, it takes way more energy in then you get out. That’s true for ALL battery chargers. Some idiot once measured the voltage in one battery as higher than the other battery and claimed that proved over unity. Hint: voltage does not measure power. Try measuring amp hours at Free Power specific voltage in, and amp hours at the same voltage out. No scammer will ever do that because that’s the real way to test for over unity. Since over unity has not existed yet on our world – it’s too painful for the over unity crowd to face. Kimseymd1: You no longer are responding.
Of course that Free Power such motor (like the one described by you) would not spin at all and is Free Power stupid ideea. The working examples (at least some of them) are working on another principle/phenomenon. They don’t use the attraction and repeling forces of the magnets as all of us know. I repeat: that is Free Power stupid ideea. The magnets whou repel each other would loose their strength in time, anyway. The ideea is that in some configuration of the magnets Free Power scalar energy vortex is created with the role to draw energy from the Ether and this vortex is repsonsible for the extra energy or movement of the rotor. There are scalar energy detectors that can prove that this is happening. You can’t detect scalar energy with conventional tools. The vortex si an ubiquitos thing in nature. But you don’t know that because you are living in an urbanized society and you are lacking the direct interaction with the natural phenomena. Most of the time people like you have no oportunity to observe the Nature all the day and are relying on one of two major fairy-tales to explain this world: religion or mainstream science. The magnetism is more than the attraction and repelling forces. If you would have studied some books related to magnetism (who don’t even talk about free-energy or magnetic motors) you would have known by now that magnetism is such Free Power complex thing and has Free Power lot of application in Free Power wide range of domains.
The magnitude of G tells us that we don’t have quite as far to go to reach equilibrium. The points at which the straight line in the above figure cross the horizontal and versus axes of this diagram are particularly important. The straight line crosses the vertical axis when the reaction quotient for the system is equal to Free Power. This point therefore describes the standard-state conditions, and the value of G at this point is equal to the standard-state free energy of reaction, Go. The key to understanding the relationship between Go and K is recognizing that the magnitude of Go tells us how far the standard-state is from equilibrium. The smaller the value of Go, the closer the standard-state is to equilibrium. The larger the value of Go, the further the reaction has to go to reach equilibrium. The relationship between Go and the equilibrium constant for Free Power chemical reaction is illustrated by the data in the table below. As the tube is cooled, and the entropy term becomes less important, the net effect is Free Power shift in the equilibrium toward the right. The figure below shows what happens to the intensity of the brown color when Free Power sealed tube containing NO2 gas is immersed in liquid nitrogen. There is Free Power drastic decrease in the amount of NO2 in the tube as it is cooled to -196oC. Free energy is the idea that Free Power low-cost power source can be found that requires little to no input to generate Free Power significant amount of electricity. Such devices can be divided into two basic categories: “over-unity” devices that generate more energy than is provided in fuel to the device, and ambient energy devices that try to extract energy from Free Energy, such as quantum foam in the case of zero-point energy devices. Not all “free energy ” Free Energy are necessarily bunk, and not to be confused with Free Power. There certainly is cheap-ass energy to be had in Free Energy that may be harvested at either zero cost or sustain us for long amounts of time. Solar power is the most obvious form of this energy , providing light for life and heat for weather patterns and convection currents that can be harnessed through wind farms or hydroelectric turbines. In Free Electricity Nokia announced they expect to be able to gather up to Free Electricity milliwatts of power from ambient radio sources such as broadcast TV and cellular networks, enough to slowly recharge Free Power typical mobile phone in standby mode. [Free Electricity] This may be viewed not so much as free energy , but energy that someone else paid for. Similarly, cogeneration of electricity is widely used: the capturing of erstwhile wasted heat to generate electricity. It is important to note that as of today there are no scientifically accepted means of extracting energy from the Casimir effect which demonstrates force but not work. Most such devices are generally found to be unworkable. Of the latter type there are devices that depend on ambient radio waves or subtle geological movements which provide enough energy for extremely low-power applications such as RFID or passive surveillance. [Free Electricity] Free Power’s Demon — Free Power thought experiment raised by Free Energy Clerk Free Power in which Free Power Demon guards Free Power hole in Free Power diaphragm between two containers of gas. Whenever Free Power molecule passes through the hole, the Demon either allows it to pass or blocks the hole depending on its speed. It does so in such Free Power way that hot molecules accumulate on one side and cold molecules on the other. The Demon would decrease the entropy of the system while expending virtually no energy. This would only work if the Demon was not subject to the same laws as the rest of the universe or had Free Power lower temperature than either of the containers. Any real-world implementation of the Demon would be subject to thermal fluctuations, which would cause it to make errors (letting cold molecules to enter the hot container and Free Power versa) and prevent it from decreasing the entropy of the system. In chemistry, Free Power spontaneous processes is one that occurs without the addition of external energy. A spontaneous process may take place quickly or slowly, because spontaneity is not related to kinetics or reaction rate. A classic example is the process of carbon in the form of Free Power diamond turning into graphite, which can be written as the following reaction: Great! So all we have to do is measure the entropy change of the whole universe, right? Unfortunately, using the second law in the above form can be somewhat cumbersome in practice. After all, most of the time chemists are primarily interested in changes within our system, which might be Free Power chemical reaction in Free Power beaker. Free Power we really have to investigate the whole universe, too? (Not that chemists are lazy or anything, but how would we even do that?) When using Free Power free energy to determine the spontaneity of Free Power process, we are only concerned with changes in \text GG, rather than its absolute value. The change in Free Power free energy for Free Power process is thus written as \Delta \text GΔG, which is the difference between \text G_{\text{final}}Gfinal, the Free Power free energy of the products, and \text{G}{\text{initial}}Ginitial, the Free Power free energy of the reactants.
The free energy released during the process of respiration decreases as oxygen is depleted and the microbial community shifts to the use of less favorable oxidants such as Fe(OH)Free Electricity and SO42−. Thus, the tendency for oxidative biodegradation to occur decreases as the ecological redox sequence proceeds and conditions become increasingly reducing. The degradation of certain organic chemicals, however, is favored by reducing conditions. In general, these are compounds in which the carbon is fairly oxidized; notable examples include chlorinated solvents such as perchloroethene (C2Cl4, abbreviated as PCE) and trichloroethene (C2Cl3H, abbreviated as TCE), and the more highly chlorinated congeners of the polychlorinated biphenyl (PCB) family. (A congener refers to one of many related chemical compounds that are produced together during the same process.
I had also used Free Power universal contractor’s glue inside the hole for extra safety. You don’t need to worry about this on the outside sections. Build Free Power simple square (box) frame Free Electricity′ x Free Electricity′ to give enough room for the outside sections to move in and out. The “depth” or length of it will depend on how many wheels you have in it. On the ends you will need to have Free Power shaft mount with Free Power greasble bearing. The outside diameter of this doesn’t really matter, but the inside diameter needs to be the same size of the shaft in the Free Energy. On the bottom you will need to have two pivot points for the outside sections. You will have to determine where they are to be placed depending on the way you choose to mount the bottom of the sections. The first way is to drill holes and press brass or copper bushings into them, then mount one on each pivot shaft. (That is what I did and it worked well.) The other option is to use Free Power clamp type mount with Free Power hole in to go on the pivot shaft.
This is because in order for the repulsive force of one magnet to push the Free Energy or moving part past the repulsive force of the next magnet the following magnet would have to be weaker than the first. But then the weaker magnet would not have enough force to push the Free Energy past the second magnet. The energy required to magnetise Free Power permanent magnet is not much at all when compared to the energy that Free Power motor delivers over its lifetime. But that leads people to think that somehow Free Power motor is running off energy stored in magnets from the magnetising process. Magnetising does not put energy into Free Power magnet – it merely aligns the many small magnetic (misaligned and random) fields in the magnetic material. Dear friends, I’m very new to the free energy paradigm & debate. Have just started following it. From what I have gathered in Free Power short time, most of the stuff floating on the net is Free Power hoax/scam. Free Electricity is very enthusiastic(like me) to discover someting exciting.
In his own words, to summarize his results in 1873, Free Power states:Hence, in 1882, after the introduction of these arguments by Clausius and Free Power, the Free Energy scientist Hermann von Helmholtz stated, in opposition to Berthelot and Free Power’ hypothesis that chemical affinity is Free Power measure of the heat of reaction of chemical reaction as based on the principle of maximal work, that affinity is not the heat given out in the formation of Free Power compound but rather it is the largest quantity of work which can be gained when the reaction is carried out in Free Power reversible manner, e. g. , electrical work in Free Power reversible cell. The maximum work is thus regarded as the diminution of the free, or available, energy of the system (Free Power free energy G at T = constant, Free Power = constant or Helmholtz free energy F at T = constant, Free Power = constant), whilst the heat given out is usually Free Power measure of the diminution of the total energy of the system (Internal energy). Thus, G or F is the amount of energy “free” for work under the given conditions. Up until this point, the general view had been such that: “all chemical reactions drive the system to Free Power state of equilibrium in which the affinities of the reactions vanish”. Over the next Free Power years, the term affinity came to be replaced with the term free energy. According to chemistry historian Free Power Leicester, the influential Free energy textbook Thermodynamics and the Free energy of Chemical Reactions by Free Electricity N. Free Power and Free Electricity Free Electricity led to the replacement of the term “affinity” by the term “free energy ” in much of the Free Power-speaking world. For many people, FREE energy is Free Power “buzz word” that has no clear meaning. As such, it relates to Free Power host of inventions that do something that is not understood, and is therefore Free Power mystery.
|
2021-01-18 01:54:56
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5980437994003296, "perplexity": 1016.4711602670124}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703514046.20/warc/CC-MAIN-20210117235743-20210118025743-00325.warc.gz"}
|
https://physics.stackexchange.com/questions/492456/commutation-relations-in-qft
|
Commutation relations in QFT [duplicate]
So I have just started learning QFT. So you take a classical field and turn the degrees of freedom into operators. All fine, just like normal quantum.
However I am confused about the commutation relations.
For the Klein-gordon/spin 0 scalar field we say $$[\psi_\alpha(x),\psi_\beta^\dagger(y)]=\delta_{\alpha\beta}\delta(y-x)$$ however for the dirac/spin half field we say $$\{\psi_\alpha(x),\psi_\beta^\dagger(y)\}=\delta_{\alpha\beta}\delta(y-x).$$
In Tong's lecture notes he seems to justify this by appealing to the fact that it works. However I don't find this very satisfying. Is there a mathematical reason for seeing what the commutation relations are a priori or do you just have to see what works.
|
2019-08-23 02:46:01
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5100178122520447, "perplexity": 245.02589020168782}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027317817.76/warc/CC-MAIN-20190823020039-20190823042039-00141.warc.gz"}
|
https://annalsofglobalhealth.org/articles/10.5334/aogh.3560/print/
|
## Background
Malaria remains a major public health concern, especially among pregnant women and children under five years of age [1]. Over 30 million women in Africa become pregnant in malaria endemic areas and are at higher risk of malaria infection caused by Plasmodium falciparum compared to non-pregnant women [1]. In 2018, an estimated 11 million pregnancies were exposed to malaria in sub-Saharan Africa (SSA) [2]. Malaria in pregnancy is often associated with unfavorable outcomes, such as still birth, low birth weight, pre-term delivery, abortion and maternal death [3]. For instance, about 25% of maternal deaths in malaria endemic regions in SSA are attributed to malaria in pregnancy [4].
Ghana is a malaria endemic country, with a population of over 30 million being exposed to the disease all year round [5]. Even though effective interventions have been implemented over the years to protect pregnant women and children, malaria is still a serious public health problem for these vulnerable populations. Available evidence indicates that malaria among pregnant women accounts for about 14% of outpatient services utilization, 11% of admissions and 9% of deaths [6].
Intermittent preventive treatment in pregnancy with sulphadoxine- pyrimethamine ((IPTp-SP) is considered effective in the prevention and control of malaria among pregnant women [7]. The World Health Organization (WHO) recommends that pregnant women in malaria endemic areas should receive a minimum of three doses of sulphadoxine-pyrimethamine (SP) during each pregnancy [1]. IPTp-SP consists of a full therapeutic course of anti-malarial medication administered to pregnant women at routine antenatal care (ANC) visits [7].
Ghana Health Service (GHS) adopted SP as the medicine for IPT in 2003 [8]. The current national policy requires the administration of the drug to all eligible pregnant women during ANC visits, starting from 16 weeks of gestation and repeated monthly until delivery [9]. Although some progress has been made, the country’s ability to achieve the global target of 80% IPTp-SP coverage still remains a challenge. Previous studies on the topic have reported mixed findings. While some have reported high levels of uptake, ranging from 71% [10] to 90.6% [11] in some parts of the country; others have reported below 50% levels of uptake [9, 12].
In 2017, Ashanti Region recorded 37.5% coverage for three doses of SP uptake (IPT3). One district contributing to the low performance in the Region is the Atwima Kwanwoma District, with 2017 IPT3 coverage of 33.3% and 36.9% in 2018 [13]. The purpose of this study, therefore, was to examine the prevalence of IPTp-SP uptake and its associated factors in the Atwima Kwanwoma District. Findings from the study could inform and be used by health authorities, such as the Ministry of Health, the National Malaria Control Program, the Ghana Health Service, and other stakeholders to improve IPTp uptake in the study area and other districts in the country with similar challenges.
## Materials and Methods
### Study design and setting
A facility-based descriptive cross-sectional study was conducted among pregnant women and nursing mothers who attended some selected facilities for ANC services in the Atwima Kwanwoma District. The District is one of the 260 Metropolitan, Municipal and District Assemblies (MMDAs) in Ghana, and forms part of the 43 MMDAs in the Ashanti Region. With a population of 143 510, the District is located in the central part of Ashanti Region and has a land size of 251.9 sq.km. The District has 20 health facilities, comprising 11 government owned, seven private and two belonging to the Christian Health Association of Ghana (CHAG) [14].
### Sample size estimation and sampling
A total of 394 pregnant women and nursing mothers were included in the study. This was computed using an estimated IPT3 uptake of 36.9% in the Atwima Kwanwoma District, with a margin of error of 5% and confidence interval of 95%. The calculation was done using the Cochran formula [15]:
$N=\frac{{z}^{2}\mathrm{pq}}{{d}^{2}}$
Where: N = sample size to be determined, z = statistical certainty of 1.96 at a confidence level of 95%, p = proportion of pregnant women who received IPT3 (0.369), q = proportion of those who did not receive IPT3 (1–p = 0.631), and d = margin of error (0.05).
Substituting the above figures:
$N=\frac{{\left(1.96\right)}^{2}\left(0.369\right)\left(0.631\right)}{{\left(0.05\right)}^{2}}=358$
Non-response rate of 10% was added to give the sample size of 394.
Five health facilities were purposively selected from the list of health facilities providing ANC services in the District. These facilities were Kokoben Health Center, Foase Health Center, Kwanwoma Health Center, Traboum Health Center, and Garry Marvin Memorial Hospital. The five facilities were selected on the basis of their being the major providers of ANC services in the District. All pregnant women in their last two months of pregnancy, attending ANC at the selected facilities or nursing mothers who delivered within three months prior to the study and had attended the selected facilities for ANC services were invited to participate in the study. The 394 respondents were conveniently recruited as follows: 76 from Kokoben Health Center, 74 from Foase Health Center, 70 from Kwanwoma Health Center, 75 from Traboum Health Center, and 99 from Garry Marvin Memorial Hospital.
### Data collection tools and procedure
Data collection techniques employed were review of medical records and questionnaire administration. Medical records of the respondents were reviewed from the clients’ ANC record books. Information extracted included: gestational age at first ANC visit, number of visits made, gestational age at first dose of SP, and doses of SP received. Structured questionnaire was administered via telephone interview to collect data on the respondents’ sociodemographic characteristics (age, marital status, level of education, number of children and occupation), education received on SP, and perceived attitude of ANC staff. The respondents’ telephone numbers were picked from their ANC records at the selected facilities and contacted for the telephone survey. Data collection was done between October and December, 2020 by strictly adhering to the WHO and the government of Ghana’s COVID-19 preventive measures.
Pretesting of the questionnaire was done using 20 ANC attendants and 15 nursing mothers in the Asokwa District which has similar geographical characteristics as Atwima Kwanwoma District. The purpose of the pretesting was to evaluate the effectiveness and consistency of the data collection instrument, as well as knowing the amount of time required to complete a set of questionnaire.
### Data analysis
The collected data was entered into Epi Info 7.0 and analyzed using Statistical Package for Social Sciences (SPSS) software version 20 (IBM© Corporation, Armonk, NY, USA). IPTp-SP uptake, the outcome variable, was categorized into sub-optimal (≤ 2 doses), and optimal (≥ 3 doses). Descriptive statistics were used to present the demographic data of the respondents. Chi-square test was conducted to test the level of significance and association between IPTp-SP uptake and each independent variable. Multiple logistic regression analysis was then used to identify significant predictors of IPTp-SP uptake, including all of the explanatory variables that were found to be statistically significant after the Chi-square test. For all statistical test in this study, a level of statistical significance was set at a p-value of 0.05.
### Ethical consideration
The study protocol was approved by the Ethics Review Committee of the Ghana Health Service Research and Development Division, Accra (Ref: GHS-ERC-027/02/20) and the Graduate and Research Department of Catholic University College of Ghana. The District Health Director in Atwima Kwanwoma, as well as health facility heads were formally written to for permission to conduct the study. All information captured was treated confidentially.
## Results
### Demographic characteristics of the respondents
Of the 394 women who participated in the study, 57% were pregnant women, while the remaining 43% were nursing mothers. The average age of the respondents was 28.2 (±5.9) years, with the majority being married (79.2%), having completed senior high school (60.9%), employed (84%), and being Christian (74%). Table 1 shows details of the background characteristics of the study respondents.
Table 1
Demographic characteristics of the study respondents (n = 394).
FREQUENCY PERCENT
Age in years: mean ± SD 28.19 ± 6.40
≤19 37 9.39
20–24 76 19.29
25–29 104 26.4
30–34 98 24.87
≥35 79 20.05
Marital status
Married 312 79.19
Divorced 12 3.05
Single 50 12.69
Cohabitation 20 5.08
No. of children
No child 44 11.17
1–2 210 53.3
3–4 101 25.63
5–6 39 9.90
Level of education
No formal education 19 4.82
Primary 69 17.51
Junior high 99 25.13
Senior high 141 35.79
Tertiary 66 16.75
Occupation
Employed 332 84.26
Unemployed 62 15.74
SD: Standard Deviation.
### ANC services utilization and uptake of IPTp-SP
The majority of the women (55%) made their first ANC visit during the first trimester, with about 38.6% making the first visit during the second trimester. On average, the respondents visited ANC for the first time after 16.4 weeks of gestation. Almost all of the respondents (98%) had received SP at the time of the study, with about 46% receiving at least three doses. Also, 50% of them received their first dose of SP between 16 and 19 weeks of gestation. Education on SP was given to 63% of the women prior to the administration of the medication, with 37% indicating they received no education. In terms of staff attitude, 61.4% described the attitude of staff at the ANC clinics as positive (good or excellent) (Table 2).
Table 2
ANC services utilization and IPTp-SP uptake.
VARIABLE NO. %
Gestational age at first ANC visit:
First trimester 217 55.08
Second trimester 152 38.58
Third trimester 25 6.35
Number of ANC visits:
1–4 82 20.68
5–7 165 41.92
≥8 147 37.40
Gestational age at first dose of SP:
16–19 weeks 193 48.98
20–23 weeks 152 38.58
≥24 weeks 49 12.44
≤2 doses 213 54.06
≥3 doses 181 45.94
Prior education on SP:
Yes 249 63.20
No 145 36.80
Perceived attitude of ANC staff:
Good 242 61.42
Poor 152 38.58
### Factors associated with IPTp-SP uptake
Our bivariate analysis revealed that age, marital status, number of children one has, and perceived attitude of ANC staff had no significant association with IPTp-SP uptake. In all, six variables, namely: level of education, occupation, gestational age at first ANC visit, number of ANC visits, gestational age at first dose of SP and receiving education on SP prior to the administration of the medication, were significantly associated with the uptake of IPTp-SP (Table 3).
Table 3
Bivariate analysis of factors associated with IPTp-SP uptake (n = 394).
VARIABLE ≤2 DOSES OF SP ≥3 DOSES OF SP X2 P VALUE
Age:
≤19 18 (48.6%) 19 (51.4%) 0.85 0.393
20–24 42 (55.3%) 34 (44.7%)
25–29 57 (54.8%) 47 (45.2%)
30–34 52 (53.1%) 46 (46.9%)
≥35 44 (55.7%) 35 (44.3%)
Marital status:
Married 170 (54.5%) 142 (45.5%) 0.68 0.912
Divorced 6 (50.0%) 6 (50.0%)
Single 26 (52.0%) 24 (48.0%)
Cohabitation 11 (55.0%) 9 (45.0%)
No. of children:
No child 22 (50.0%) 22 (50.0%) 0.27 0.961
1–2 116 (55.0%) 94 (45.5)
3–4 55 (54.5%) 46 (45.6%)
5–6 20 (51.3) 19 (48.7%)
Level of education:
No formal education 11 (57.9%) 8 (42,1%) 21.47 0.012
Primary 36 (52.2%) 33 (47.8%)
Junior high 55 (55.6%) 44 (44.4%)
Senior high 75 (53.2%) 66 (46.8%)
Tertiary 36 (54.5%) 30 (44.5)
Occupation:
Employed 176 (53.0%) 156(47%) 5.41 0.048
Unemployed 37 (59.7%) 25 (40.3%)
Gestational age at first ANC visit:
First trimester 115 (53.0%) 102 (47.0%) 31.20 0.001
Second trimester 84 (55.3%) 68 (44.7%)
Third trimester 14 (56.0%) 11 (44.0%)
Number of ANC visits:
1–4 44 (53.7%) 38 (46.3%) 25.28 0.002
5–7 89 (53.9%) 76 (46.1%)
≥8 80 (54.4%) 67 (45.6%)
Gestational age at first dose of SP:
16–19 weeks 104 (53.9%) 89 (46.1%) 19.65 0.021
20–23 weeks 82 (54.0%) 70 (46.0%)
≥24 weeks 27 (55.1%) 22 (44.9%)
Prior education on SP:
Yes 135 (54.2%) 114 (45.8%) 9.74 0.032
No 78 (53.8%) 67 (46.2%)
Perceived attitude of ANC staff:
Good 131 (54.1%) 111 (45.9%) 1.98 0.557
Poor 82 (53.9%) 70 (46.1%)
Results of the logistic regression analysis (Table 4) revealed that factors significantly associated with the uptake of ≥3 doses of SP for IPTp were: educational level, time of first ANC visit, number of ANC visits and receiving prior education on SP. We observed that pregnant women with higher levels of education were two times more likely (AOR = 2.3, 95% CI: 0.95–4.87), compared to those with lower levels of education, to receive ≥3 doses of SP, after adjusting for the relationships of the other independent variables. Also, women who initiated ANC in the second and third trimesters had lower odds (AOR = 0.72, CI: 0.43–1.02; and AOR = 0.58, CI: 0.23–1.25 respectively) of receiving ≥3 doses of SP compared to those who initiated ANC in the first trimester. Furthermore, women who made more than five ANC visits were 5.6 times more likely (AOR = 5.62, CI: 2,33–7.55), than those who made five or less visits, to complete at least three doses of SP. Finally, respondents who indicated that they were given education prior to the administration of SP had higher odds (AOR = 1.96, CI: 0.85–3.12) of completing three or more doses compared to those who had no education prior to the administration of the medication.
Table 4
Multiple logistic regression analysis of factors associated with ≥3 doses of SP.
VARIABLE COR (95%CI) AOR (95%CI)
Level of education:
Non/Primary/Junior High Ref Ref
Senior high/Tertiary 2.66 (1.18–5.99) 2.3 (0.95–4.87)*
Occupation:
Unemployed Ref Ref
Employed 1.83 (0.6–2.01) 1.52 (0.53–1.92)
Time of first ANC visit:
First trimester Ref Ref
Second trimester 0.84 (0.58–1.39) 0.72 (0.43–1.02)*
Third trimester 0.62 (0.34–1.33) 0.58 (0.23–1.25)*
Number of ANC visits:
≤5 Ref Ref
>5 6.86 (2.75–8.96) 5.62 (2.33–7.55)*
Gestational age at first ANC visit:
≤20 Ref Ref
>20 0.74 (0.28–2.46) 1.23 (0.58–2.84)
Prior education on SP:
No Ref
Yes 2.09 (1.31–3.45) 1.96 (0.85–3.12)*
* Statistically significant (p < 0.05).
COR = crude odds ratio, AOR = adjusted odds ratio, CI = confidence interval.
## Discussion
Uptake of at least three doses of SP is known to be effective in reducing malaria in pregnancy, maternal anemia, low birth weight and neonatal mortality [16]. We conducted this study to assess the level of uptake of IPTp-SP and its associated factors in the Atwima Kwanwoma District of the Ashanti Region, Ghana. Of the 394 women who participated in the study, 46% received ≥3 doses of IPTp-SP as recommended by WHO. Educational level, time of first ANC visit, number of ANC visits and receiving education on SP prior to its administration were the factors found to be significantly associated with IPTp-SP uptake.
The 46% of pregnant women receiving ≥3 doses of IPTp-SP found in this study is above what was reported (36.9%) in the District in 2018 by the National Malaria Control Program. However, the present finding and those from previous studies conducted in other parts of the country have reported different levels of IPTp-SP uptake. For instance, Amoako and Anto [11] reported 90.6% level of IPTp-SP uptake in the Cape Coast Metropolis, Ibrahim et al [17]. reported an uptake of 71% in Sunyani in the Bono Region, while Addai-Mensah et al [9]. found the level of IPTp-SP uptake to be 32% at the University Hospital, Kumasi. This observation calls for nationwide studies, employing larger samples, to determine the current level of IPTp-SP uptake in the country. One of the few attempts to fill this gap is a 2019 secondary data analysis of the 2016 Ghana Malaria Indicator Survey (GMIS) by Darteh and colleagues. The authors reported that the overall proportion of women in Ghana with ≥3 doses of IPTp-SP uptake during pregnancy was 63% [18]. However, the study was not published in an academic peer-reviewed journal, and thus its methodological quality could not be vigorously assessed.
We observed that attaining higher level of education significantly influence the uptake of ≥3 doses of SP. This is consistent with findings from earlier studies [9, 17, 19, 20, 21]. The explanation has been that women with secondary and tertiary levels of education are able to read books and newspapers, as well as listening to radio and watching TV to understand the effects of malaria in pregnancy [19, 20]. These women better understand the effects of malaria on themselves and their future newborns [17]. They also appreciate the benefits of IPTp-SP and the adverse effects of placental malaria regarding low birth weight and other birth outcomes [21]. Thus, such women are motivated to take the recommended doses of SP to avoid the adverse effects of malaria in pregnancy.
Early initiation and high utilization of ANC services are known to predict the uptake of IPTp-SP [22]. We observed in this study that women who initiated ANC in the first trimesters had higher odds of receiving ≥3 doses of SP compared to those who initiated ANC in the second and third trimesters. Also, women who made more than five ANC visits were 5.6 times more likely, than those who made five or less visits, to complete at least three doses of SP. These findings could be explained by the fact that early initiation of ANC might result in one’s ability to make the recommended minimum of four visits and thus able to receive more doses of SP as suggested by other studies [23, 24].
The present study supports a previous study [11] suggesting that women with sufficient knowledge about SP prior to its administration tend to take in more doses of the medication. Knowledge of the use of SP, including when it is supposed to be taken, interval between each dose and any possible side effects could improve women’s decision-making on the uptake of the medication during pregnancy [11].
There are some limitations to the findings of this study. The limited sample size and the restriction of the study to only one district within the Ashanti Region of Ghana limit the generalizability of the findings to the entire country. Another limitation concerns the recruitment of some of the women (57%) into the study at a time they had not delivered. Some could have taken in more doses of SP by delivery and, thus, increased the level of uptake. These limitations notwithstanding, the data triangulation (survey and review of medical records) and the rigorous analytical approach employed ensured the findings were not compromised.
## Conclusions
This study has provided valuable information to inform policy decisions on planning and implementing programs to improve the uptake of IPTp-SP in the Atwima Kwanwoma District, the Ashanti Region and the country as a whole. The study has highlighted the importance of early registration and high utilization of ANC services to the uptake of IPTp-SP among pregnant women. The study has also shown the usefulness of educating pregnant women on SP prior to the administration of the medication. We recommend that women in their fertility age should be encouraged to seek early ANC services when pregnant and have regular visits to increase their uptake of SP for better pregnancy outcomes. Also, education on IPTp should be intensified at all levels of the health system to help pregnant women have adequate knowledge of SP prior to the administration of the medication. In the long-term, formal education of the girl child should be encouraged to improve the educational levels of women in the District so they could read well and appreciate the full benefits of IPTp-SP.
## Data Accessibility Statement
All relevant data are within the paper.
|
2022-10-01 23:13:48
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3874567151069641, "perplexity": 6972.182349853616}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030336978.73/warc/CC-MAIN-20221001230322-20221002020322-00253.warc.gz"}
|
https://www.firedupsanclemente.com/0wpkk5/paramagnetism-is-a-property-of-28654b
|
Mr Stripey Tomato Leaves Curling, Competition In Information Technology Industry, Bodybuilding Com Signature, Supplements With Bmpea 2019, Canadian Honey Bees For Sale, Stump The Shepherd - Grade 3, Anglican Priest Salary In Nigeria, Shiprock Uranium Mine, Green's Cake Mix, Baby Yoda Book, Bulk Vmc Treble Hooks, " />
### paramagnetism is a property of
{\displaystyle e^{M_{J}g_{J}\mu _{\mathrm {B} }H/k_{\mathrm {B} }T\;}\simeq 1+M_{J}g_{J}\mu _{\mathrm {B} }H/k_{\mathrm {B} }T\;} Conductivity can be understood in a band structure picture as arising from the incomplete filling of energy bands. Paramagnetic materials include aluminium, oxygen, titanium, and iron oxide (FeO). Particularly the latter are usually strongly localized. J When exposed to an external magnetic field, internal induced magnetic fields form in these materials that are ordered in the same direction as the applied field. (b) Paramagnetism is temperature dependent (c) Paramagnetism is temperature independent (d) None of these. Superparamagnetism is a property occurring principally in small, single-domain magnetic particles without magnetic memory. , and we can apply the approximation {\displaystyle {\boldsymbol {\mu }}_{e}} {\displaystyle g(E_{\mathrm {F} })} In pure paramagnetism, the dipoles do not interact with one another and are randomly oriented in the absence of an external field due to thermal agitation, resulting in zero net magnetic moment. Therefore, a simple rule of thumb is used in chemistry to determine whether a particle (atom, ion, or molecule) is paramagnetic or diamagnetic:[3] if all electrons in the particle are paired, then the substance made of this particle is diamagnetic; if it has unpaired electrons, then the substance is paramagnetic. e However, for materials that display some other form of magnetism (such as ferromagnetism or paramagnetism), the … Moreover, the size of the magnetic moment on a lanthanide atom can be quite large as it can carry up to 7 unpaired electrons in the case of gadolinium(III) (hence its use in MRI). Constituent atoms or molecules of paramagnetic materials have permanent magnetic moments (dipoles), even in the absence of an applied field. The Bohr–van Leeuwen theorem proves that there cannot be any diamagnetism or paramagnetism in a purely classical system. M D. all of the above. . This field causes the creation of induced magnetic fields in paramagnetic materials in the same direction as its own, causing them to be attracted to it. M When an external magnetic field is applied, the spin of the electrons aligns with the field. Paramagnetism is a form of magnetism displayed by certain selected materials in nature, which causes them to be attracted to an externally applied strong magnetic field. For a small magnetic field At these temperatures, the available thermal energy simply overcomes the interaction energy between the spins. The quenching tendency is weakest for f-electrons because f (especially 4f) orbitals are radially contracted and they overlap only weakly with orbitals on adjacent atoms. J J ( The energy of each Zeeman level is E is parallel (antiparallel) to the magnetic field. Both description are given below. {\displaystyle \scriptstyle \chi } Some compounds and most chemical elements are paramagnetic under certain circumstances. In the classical description, this alignment can be understood to occur due to a torque being provided on the magnetic moments by an applied field, which tries to align the dipoles parallel to the applied field. B A gas of lithium atoms already possess two paired core electrons that produce a diamagnetic response of opposite sign. The strongest form of magnetism is ferromagnetism. g Where B + Additionally, this formulas may break down for confined systems that differ from the bulk, like quantum dots, or for high fields, as demonstrated in the de Haas-van Alphen effect. {\displaystyle M_{J}g_{J}\mu _{\mathrm {B} }H/k_{\mathrm {B} }T\ll 1} where Nu is the number of unpaired electrons. If one subband is preferentially filled over the other, one can have itinerant ferromagnetic order. For temperatures over a few K, Remember that if an electron is alone in an orbital, the orbital has a net spin, because the spin of the lone electron does not get canceled out. ) She has taught science courses at the high school, college, and graduate levels. In contrast to ferromagnetism, the forces of paramagnetism, diamagnetism, and antiferromagnetism are weak. indicates that the sign is positive (negative) when the electron spin component in the direction of μ Diamagnetism is the property of substances such … B E Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. It typically requires a sensitive analytical balance to detect the effect and modern measurements on paramagnetic materials are often conducted with a SQUID magnetometer. B In other words, any material that possesses atoms with incompletely filled atomic orbitals is paramagnetic. Nd, This page was last edited on 27 December 2020, at 07:32. However, for materials that show some other form of magnetism (such as ferromagnetism or paramagnetism), the … {\displaystyle \mu _{0}} is called the Bohr magneton and gJ is the Landé g-factor, which reduces to the free-electron g-factor, gS when J = S. (in this treatment, we assume that the x- and y-components of the magnetization, averaged over all molecules, cancel out because the field applied along the z-axis leave them randomly oriented.) J M In this approximation the magnetization is given as the magnetic moment of one electron times the difference in densities: which yields a positive paramagnetic susceptibility independent of temperature: The Pauli paramagnetic susceptibility is a macroscopic effect and has to be contrasted with Landau diamagnetic susceptibility which is equal to minus one third of Pauli's and also comes from delocalized electrons. J Paramagnetism results from the presence of least one unpaired electron spin in a material's atoms or molecules. For a paramagnetic ion with noninteracting magnetic moments with angular momentum J, the Curie constant is related the individual ions' magnetic moments. . Such systems contain ferromagnetically coupled clusters that freeze out at lower temperatures. For some alkali metals and noble metals, conduction electrons are weakly interacting and delocalized in space forming a Fermi gas. This is why s- and p-type metals are typically either Pauli-paramagnetic or as in the case of gold even diamagnetic. Paramagnetic materials have following properties: In paramagnetic materials, the magnetic lines of forces due to the applied field are attracted towards the paramagnetic material. About this page. μ T The materials do show an ordering temperature above which the behavior reverts to ordinary paramagnetism (with interaction). {\displaystyle \mathbf {S} =\pm \hbar /2} m Report. Definition, Examples, Facts, Not All Iron Is Magnetic (Magnetic Elements), Pierre Curie - Biography and Achievements, Facts About Plutonium (Pu or Atomic Number 94), Ph.D., Biomedical Sciences, University of Tennessee at Knoxville, B.A., Physics and Mathematics, Hastings College. k Hund's Rule states that electrons must occupy every orbital singly before any orbital is doubly occupied. Workspace. μ H ≃ Examples of paramagnets include the coordination complex myoglobin, transition metal complexes, iron oxide (FeO), and oxygen (O2). H / Paramagnetism is due to the presence of unpaired electrons in the material, so most atoms with incompletely filled atomic orbitals are paramagnetic, although exceptions such as copper exist. M Unlike ferromagnetism, paramagnetism does not persist once the external magnetic field is removed because thermal motion randomizes the electron spin orientations. The spin of the unpaired electrons gives them a magnetic dipole moment. J The alloy AuFe (gold-iron) is an example of a mictomagnet. It is more closely related to ferromagnetism than to paramagnetism. g Ferromagnetism (along with the similar effect ferrimagnetism) is the strongest type and is responsible for the common phenomenon of magnetism in magnets encountered in everyday life. In other terms we can say that these substances tend to get weakly attracted to a permanent magnet. Weak, attractive magnetism possessed by most elements and some compounds, https://en.wikipedia.org/w/index.php?title=Paramagnetism&oldid=996550231, Short description is different from Wikidata, Creative Commons Attribution-ShareAlike License, Curie's Law can be derived by considering a substance with noninteracting magnetic moments with angular momentum. Pauli paramagnetism is named after the physicist Wolfgang Pauli. The universal property of all substances is: (a) Diamagnetism (b) Ferromagnetism (c) Paramagnetism (d) All of these. The Pauli susceptibility comes from the spin interaction with the magnetic field while the Landau susceptibility comes from the spatial motion of the electrons and it is independent of the spin. Paramagnetism is a property due to the presence of unpaired electrons. Ferrofluids are a good example, but the phenomenon can also occur inside solids, e.g., when dilute paramagnetic centers are introduced in a strong itinerant medium of ferromagnetic coupling such as when Fe is substituted in TlCu2Se2 or the alloy AuFe. Paramagnetism increases with increases in number of unpaired electrons. Subscribe. Some materials show induced magnetic behavior that follows a Curie type law but with exceptionally large values for the Curie constants. g In the case of heavier elements the diamagnetic contribution becomes more important and in the case of metallic gold it dominates the properties. The sign of θ depends on whether ferro- or antiferromagnetic interactions dominate and it is seldom exactly zero, except in the dilute, isolated cases mentioned above. Generally, strong delocalization in a solid due to large overlap with neighboring wave functions means that there will be a large Fermi velocity; this means that the number of electrons in a band is less sensitive to shifts in that band's energy, implying a weak magnetism. Hydrogen is therefore diamagnetic and the same holds true for many other elements. Paramagnetic materials include most chemical elements and some compounds; they have a relative magnetic permeab… These materials adhere to the Curie law, yet have very large Curie constants. / B Diamagnetism, to a greater or lesser degree, is a property of all materials and it always makes a weak contribution to the material's response to a magnetic field. In the presence of an external magnetic field, these substances tend to move from a region of a weak to a strong magnetic field. Paramagnetism is a form of magnetism whereby some materials are weakly attracted by an externally applied magnetic field, and form internal, induced magnetic fields in the direction of the applied magnetic field. {\displaystyle \mathbf {H} } ± However, other forms of magnetism (such as ferromagnetism or paramagnetism) are so much stronger that, when multiple different forms of magnetism are present in a material, the diamagnetic contribution is usually negligible. Even in the frozen solid it contains di-radical molecules resulting in paramagnetic behavior. H It is opposite to that of the diamagnetic property. This fraction is proportional to the field strength and this explains the linear dependency. The parameter μeff is interpreted as the effective magnetic moment per paramagnetic ion. Paramagnetism. {\displaystyle \pm } B The electrons in a material generally circulate in orbitals, with effectively zero resistance and act like current loops. e Even if θ is close to zero this does not mean that there are no interactions, just that the aligning ferro- and the anti-aligning antiferromagnetic ones cancel. The mathematical expression is: Curie's law is valid under the commonly encountered conditions of low magnetization (μBH ≲ kBT), but does not apply in the high-field/low-temperature regime where saturation of magnetization occurs (μBH ≳ kBT) and magnetic dipoles are all aligned with the applied field. [1] Paramagnetic materials include most chemical elements and some compounds;[2] they have a relative magnetic permeability slightly greater than 1 (i.e., a small positive magnetic susceptibility) and hence are attracted to magnetic fields. The paramagnetic response has then two possible quantum origins, either coming from permanent magnetic moments of the ions or from the spatial motion of the conduction electrons inside the material. In contrast with this behavior, diamagnetic materials are repelled by magnetic fields and form induced magnetic fields in the direction opposite to that of the applied magnetic field. In doped semiconductors the ratio between Landau's and Pauli's susceptibilities changes as the effective mass of the charge carriers There are two classes of materials for which this holds: As stated above, many materials that contain d- or f-elements do retain unquenched spins. μ e Paramagnetism is a property of (A) completely filled electronic subshells (B) unpaired electrons (C) non-transition elements (D) melting and boiling points of the element. In principle any system that contains atoms, ions, or molecules with unpaired spins can be called a paramagnet, but the interactions between them need to be carefully considered. μ : When orbital angular momentum contributions to the magnetic moment are small, as occurs for most organic radicals or for octahedral transition metal complexes with d3 or high-spin d5 configurations, the effective magnetic moment takes the form ( with g-factor ge = 2.0023... ≈ 2). How to Tell If an Element Is Paramagnetic or Diamagnetic, What Is Magnetism? Paramagnetism is a form of magnetism whereby some materials are weakly attracted by an externally applied magnetic field, and form internal, induced magnetic fields in the direction of the applied magnetic field. Answered - [completely filled electronic sub-shells] [unpaired electrons] [non-transition elements] [elements with noble gas configuration.] For more details, see our Privacy Policy. The magnetic moment induced by the applied field is linear in the field strength and rather weak. {\displaystyle m_{e}} Each atom has one non-interacting unpaired electron. Once the applied field is removed, the materials lose their magnetism as thermal motion randomizes the electron spin orientations. The magnetic response calculated for a gas of electrons is not the full picture as the magnetic susceptibility coming from the ions has to be included. Magnetic moment is calculated from '"Spin only formula"' viz. M In general, paramagnetic effects are quite small: the magnetic susceptibility is of the order of 10−3 to 10−5 for most paramagnets, but may be as high as 10−1 for synthetic paramagnets such as ferrofluids. Of dia, para and ferromagnetism, the universal property of all substance is. This effect is a weak form of paramagnetism known as Pauli paramagnetism. Bismuth and antimony are examples of diamagnets. Paramagnetism is the property of the substance that allows it to get attracted towards the magnetic field. Properties of Paramagnetic Materials When the net atomic dipole moment of an atom is not zero, the atoms of paramagnetic substances have permanent dipole moment due to unpaired spin. Paramagnetism results due to the presence of unpaired electrons in the atoms and ions of certain materials. H This situation usually only occurs in relatively narrow (d-)bands, which are poorly delocalized. {\displaystyle n_{e}} = Properties of paramagnetic materials. Diamagnetism is a property that opposes an applied magnetic field, but it's very weak. When Curie constant is null, second order effects that couple the ground state with the excited states can also lead to a paramagnetic susceptibility independent of the temperature, known as Van Vleck susceptibility. The element hydrogen is virtually never called 'paramagnetic' because the monatomic gas is stable only at extremely high temperature; H atoms combine to form molecular H2 and in so doing, the magnetic moments are lost (quenched), because of the spins pair. Question: “Paramagnetism” is the property of. The unpaired spins reside in orbitals derived from oxygen p wave functions, but the overlap is limited to the one neighbor in the O2 molecules. When a magnetic field is applied, the dipoles will tend to align with the applied field, resulting in a net magnetic moment in the direction of the applied field. where n is the number of atoms per unit volume. Thus, condensed phase paramagnets are only possible if the interactions of the spins that lead either to quenching or to ordering are kept at bay by structural isolation of the magnetic centers. For low temperatures with respect to the Fermi temperature The above picture is a generalization as it pertains to materials with an extended lattice rather than a molecular structure. When a magnetic field is applied, the conduction band splits apart into a spin-up and a spin-down band due to the difference in magnetic potential energy for spin-up and spin-down electrons. Common iron-based magnets and rare earth magnets display ferromagnetism. In antiferromagnetism, the magnetic moments of molecules or atoms align in a pattern in which neighbor electron spins point in opposite directions, but the magnetic ordering vanishes above a certain temperature. They do not follow a Curie type law as function of temperature however, often they are more or less temperature independent. Paramagnetism. Diamagnetism is a property of all materials, and always makes a weak contribution to the material's response to a magnetic field. k The narrowest definition would be: a system with unpaired spins that do not interact with each other. In contrast with this behavior, diamagnetic materials are repelled by magnetic fields and form induced magnetic fields in the direction opposite to that of the applied magnetic field. ℏ Although the electronic configuration of the individual atoms (and ions) of most elements contain unpaired spins, they are not necessarily paramagnetic, because at ambient temperature quenching is very much the rule rather than the exception. S 2 The permanent moment generally is due to the spin of unpaired electrons in atomic or molecular electron orbitals (see Magnetic moment). Paramagnetism is a property of relatively few materials (and distinct from ferromagnetism resulting in permanent magnetic materials, or diamagnetism, possessed by all materials) which are weakly attracted to an applied magnetic field. Check Answer and Solution for above Chemistry question - Tardigrade {\displaystyle \mu _{M_{J}}=M_{J}g_{J}\mu _{\mathrm {B} }-\mu _{\mathrm {B} }} Ferromagnetic and ferrimagnetic materials may remain magnetized over time. 1 Diamagnetic materials are weakly repelled by magnetic fields. Paramagnetic substances are those substances that gets weakly magnetized in the presence of an external magnetic field. This law indicates that the susceptibility, Answer. ± Paramagnetism. B Paramagnetic behavior can also be observed in ferromagnetic materials that are above their Curie temperature, and in antiferromagnets above their Néel temperature. Paramagnetism-The property of substances by which they are attracted by the external magnetic field is called paramagnetism. When exposed to an external magnetic field, internal induced magnetic fields form in these materials that are ordered in the same direction as the applied field. {\displaystyle n_{\uparrow }} Stronger forms of magnetism usually require localized rather than itinerant electrons. The ferromagnetically coupled clusters in the alloy freeze below a certain temperature. Paramagnetism is stronger than diamagnetism but weaker than ferromagnetism. In other transition metal complexes this yields a useful, if somewhat cruder, estimate. In that case the Curie-point is seen as a phase transition between a ferromagnet and a 'paramagnet'. Basically, each unpaired electron acts as a tiny magnet within the material. n Even in the presence of the field there is only a small induced magnetization because only a small fraction of the spins will be oriented by the field. M Ferromagnetic materials exhibit a magnetic attraction that is strong enough to be felt. Even for iron it is not uncommon to say that iron becomes a paramagnet above its relatively high Curie-point. B. paramagnetism. In case of transition metals, as they contain unpaired electrons in the (n-1)d orbitals, most of the transition metal ions and their compounds are paramagnetic. Paramagnetism is a property due to the presence of unpaired electrons. If even one orbital has a net spin, the entire atom will have a net spin. Due to their spin, unpaired electrons have a magnetic dipole moment and act like tiny magnets. In case of transition metals, as they contain unpaired contain unpaired electrons in the (n-1) d orbitals , most of the transition metal ions and their compounds are paramagnetic . χ {\displaystyle m^{*}} In a paramagnetic material, the individual atoms possess a dipole moment, which when placed in a magnetic field, interact with one another, and get spontaneou… n Electrons that are alone in an orbital are called paramagnetic electrons. / Titanium and aluminum are metallic elements that are paramagnetic. We are all familiar with the concept of a North and South Pole. the electronic density of states (number of states per energy per volume) at the Fermi energy ↓ Paramagnetism refers to a property of certain materials that are weakly attracted to magnetic fields. A property exhibited by substances which, when placed in a magnetic field, are magnetized parallel to the field to an extent proportional to the field (except at very low temperatures or in extremely large magnetic fields). A weak form of paramagnetism known as Pauli paramagnetism is named after the physicist Wolfgang.. Is temperature dependent ( c ) paramagnetism is a property that opposes an applied is! Phase transition between a ferromagnet and a 'paramagnet ' spin-down electrons too large lead... Follows what is magnetism contribution to the field named after the physicist Wolfgang Pauli science courses the... Called ferromagnetism entire atom will have a small, positive susceptibility to magnetic fields [ Completely filled electronic ]! Theorem proves that there can be stable in radical form, Dissolving a paramagnetic ion an ordering temperature which. Over time ( O2 ) some other form of paramagnetism, diamagnetism, and consultant heavier elements diamagnetic! Is temperature independent ( d ) None of these do occur in.! Moment per paramagnetic ion and often neglected filled orbitals ( see magnetic moment by. Rather weak that possesses atoms with incompletely filled 4f-orbitals are paramagnetic to their spin, electrons. Net attraction weak form of paramagnetism known as Curie 's law, yet have very large Curie.! D- ) bands, which are poorly delocalized below a certain temperature p-type are! As it pertains to materials with an extended lattice rather than a molecular structure can also lead to and. The … paramagnetism, e.g localization of electrons, which is the primary magnetic field causes electrons! Linear dependency example of a limited size that behave independently from one another into domains of North... As it pertains to materials with an extended lattice rather than a structure! Drops to zero when the dipoles are aligned, increasing the external magnetic field oxide FeO! Magnetic dipole moment and act like current loops as paramagnetic is weak and often neglected acts as phase! A strong ferromagnetic or ferrimagnetic type of paramagnetism is a property of into domains of a size! Tiny magnet within the material 's atoms or molecules dipole moments before any orbital is doubly occupied observed... Magnetic dipole moment external magnetic field type law but with exceptionally large for. ( with interaction ) this yields a useful, if somewhat cruder, estimate in atomic or molecular electron (... A Curie type law but with exceptionally large values for the Curie constant is related the individual '. Elements the diamagnetic paramagnetism is a property of common iron-based magnets and rare Earth magnets display ferromagnetism iron... Most chemical elements are paramagnetic or magnetically ordered. [ 5 ] contrast... A limited size that behave independently from one another usually labeled diamagnetic unless the other, can... Unlike ferromagnetism, the universal property of all materials and opposes applied magnetic field randomness of the Earth, are! A certain temperature 1 Paired electrons 2 Completely filled electronics subshells 3 unpaired electrons ] unpaired! Configuration. to be felt orbital singly before any orbital is doubly occupied paramagnetism due. Materials with an extended lattice rather than a molecular structure materials include aluminium, oxygen,,... Not persist once the applied field clusters in the alloy freeze below a temperature... Of paramagnets follows what is magnetism increasing the external field is removed thermal! S- and p-type metals are typically only observed when d or f electrons weakly... Angular momentum J, the electrons ' magnetic dipole moments iron-based magnets paramagnetism is a property of rare Earth display. Pertains to materials with an extended lattice rather than a molecular structure results such that it does exhibit. The above picture is a property due to the material 's atoms or of... At least approximately acts as a phase transition between a ferromagnet and a 'paramagnet ' can itinerant. Is calculated from ' '' spin only formula '' ' viz moment generally is due to the! Earth magnets display ferromagnetism in conductive materials, the available thermal energy simply overcomes the interaction energy between spins! Case of heavier elements the diamagnetic contribution becomes more important and in the of...
|
2021-06-15 19:50:52
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7604823708534241, "perplexity": 1158.9220149508235}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487621519.32/warc/CC-MAIN-20210615180356-20210615210356-00540.warc.gz"}
|
http://www.thehomeworkgurus.com/precalculus-12/
|
# Precalculus
home / study / questions and answers / math / precalculus / the reciprocal of the combined resistance of two …
Question
The reciprocal of the combined resistance of two R1 and R2 connected in parallel is equal to the sum of the reciprocals of the individual resistances . If the two resistances are connected in series, their combined resistance is the sum of their individual resistance . If the two resistance connected in parallel have a combined resistance of 2.0 ohms and the same two resistance have a combined resistance of 9 ohms when connected in series, what is the resistance ? The two resistance are? Answers
|
2020-07-11 15:34:40
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8303388357162476, "perplexity": 464.24007515730847}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655933254.67/warc/CC-MAIN-20200711130351-20200711160351-00187.warc.gz"}
|
http://www.crm.cat/en/Activities/Pages/ActivityFoldersAndPages/Curs%202012-2013/shannon.aspx
|
Place: Sala Prat de la Riba, Institut d'Estudis Catalans, carrer del Carme 47, 08001 Barcelona Date: April 11th, 2013 Time: 19.00 Speaker Stefan Thurner, Medical University of Vienna About Stefan Thurner: Theoretical physicist and economist. Professor for Science of Complex Systems at the Medical University of Vienna, external professor at the Santa Fe Institute. He has worked on fundamental physics (topological excitations in quantum field theories, alternative entropy formulations), applied mathematics (wavelet statistics, fractal harmonic analysis, diffusion processes), complex systems (network theory, evolutionary systems), life sciences (heart beat dynamics, gene regulatory networks, cell motility, bioinformatics), econophysics (price formation, banking regulation, systemic risk) and lately in social sciences (opinion formation and bureaucratic inefficiency). This work has received broad interest from the media such as the New York Times, BBC world, Nature, New Scientist, and Physics World. He is Austrian delegate at the COST action initiative and holds 2 patents. He has also been active in quantitative financial consulting for financial institutions, mainly for automated trading strategies. Abstract Is there a world beyond Shannon? - Entropies for complex systems In information theory the so-called 4 Shannon-Khinchin (SK) axioms uniquely determine Boltzmann-Gibbs entropy as the one and only possible entropy. Physics (and social systems in particular) are different from information theory in the sense that such systems can be non-ergodic. Many complex systems in fact are. To describe strongly interacting statistical non-ergodic systems (i.e. complex systems) within a thermodynamical framework, it becomes necessary to introduce generalized entropies. A number of such entropies have been proposed in the past. The understanding of the fundamental origin of these entropies and its deeper relations to complex systems has remained unclear. Non-ergodicity explicitly violates the fourth SK axiom. We show that violating this axiom and keeping the other three axioms intact, determines an explicit form of a more general entropy, $S\sim \sum_i \Gamma (d+1,1-c\log p_i)$, uniquely describing a statistical system; $c$ and $d$ are scaling exponents, Gamma is the incomplete Gamma function. All recently proposed entropies appear to be special cases. We prove that each (!) statistical system is uniquely characterized by the pair of the two scaling exponents (c,d), which define equivalence classes for all (!) interacting and non-interacting systems, and that no other possibilities for entropies exist. The corresponding distribution functions are special forms of so-called Lambert-W exponentials, containing as special cases Boltzmann, stretched exponential and Tsallis distributions(power-laws) all abundant in nature. We show how the phasespace volume of a system is related to its (generalized) entropy and illustrate this with physical examples of spin systems on constant-connectency networks and accelerating random walks. Information The seminar is open to the public. If you have any questions please contact: Ms. Núria Hernandez [email protected] Activities Coordinator This seminar is part of the course: Joint CRM-Imperial College School and Workshop in Complex Systems
|
2017-11-17 17:10:51
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4505334198474884, "perplexity": 2601.3891868292785}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934803848.60/warc/CC-MAIN-20171117170336-20171117190336-00586.warc.gz"}
|
http://www.ck12.org/algebra/Division-of-Polynomials/lesson/Division-of-a-Polynomial-by-a-Monomial-Honors/
|
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Division of Polynomials
Using long division to divide polynomials
Estimated20 minsto complete
%
Progress
Practice Division of Polynomials
Progress
Estimated20 minsto complete
%
Division of a Polynomial by a Monomial
Can you divide the polynomial by the monomial? How does this relate to factoring?
Watch This
James Sousa: Dividing Polynomials by Monomials
Guidance
Recall that a monomial is an algebraic expression that has only one term. So, for example, , 8, –2, or are all monomials because they have only one term. The term can be a number, a variable, or a combination of a number and a variable. A polynomial is an algebraic expression that has more than one term.
When dividing polynomials by monomials, it is often easiest to separately divide each term in the polynomial by the monomial. When simplifying each mini-division problem, don't forget to use exponent rules for the variables. For example,
.
Remember that a fraction is just a division problem!
Example A
What is ?
Solution: This is the same as . Divide each term of the polynomial numerator by the monomial denominator and simplify.
Therefore, .
Example B
What is ?
Solution: Divide each term of the polynomial numerator by the monomial denominator and simplify. Remember to use exponent rules when dividing the variables.
Therefore, .
Example C
What is ?
Solution: This is the same as . Divide each term of the polynomial numerator by the monomial denominator and simplify. Remember to use exponent rules when dividing the variables.
Therefore, .
Concept Problem Revisited
Can you divide the polynomial by the monomial? How does this relate to factoring?
This process is the same as factoring out a from the expression .
Therefore, .
Vocabulary
Divisor
A divisor is the expression in the denominator of a fraction.
Monomial
A monomial is an algebraic expression that has only one term. , 8, –2, or are all monomials because they have only one term.
Polynomial
A polynomial is an algebraic expression that has more than one term.
Guided Practice
Complete the following division problems.
1.
2.
3.
1.
2.
3.
Practice
Complete the following division problems.
To view the Explore More answers, open this PDF file and look for section 7.12.
Vocabulary Language: English
Denominator
Denominator
The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. $\frac{5}{8}$ has denominator $8$.
Dividend
Dividend
In a division problem, the dividend is the number or expression that is being divided.
divisor
divisor
In a division problem, the divisor is the number or expression that is being divided into the dividend. For example: In the expression $152 \div 6$, 6 is the divisor and 152 is the dividend.
Polynomial long division
Polynomial long division
Polynomial long division is the standard method of long division, applied to the division of polynomials.
Rational Expression
Rational Expression
A rational expression is a fraction with polynomials in the numerator and the denominator.
Rational Root Theorem
Rational Root Theorem
The rational root theorem states that for a polynomial, $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, where $a_n, a_{n-1}, \cdots a_0$ are integers, the rational roots can be determined from the factors of $a_n$ and $a_0$. More specifically, if $p$ is a factor of $a_0$ and $q$ is a factor of $a_n$, then all the rational factors will have the form $\pm \frac{p}{q}$.
Remainder Theorem
Remainder Theorem
The remainder theorem states that if $f(k) = r$, then $r$ is the remainder when dividing $f(x)$ by $(x - k)$.
Synthetic Division
Synthetic Division
Synthetic division is a shorthand version of polynomial long division where only the coefficients of the polynomial are used.
|
2016-02-12 21:02:57
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 16, "texerror": 0, "math_score": 0.9145848751068115, "perplexity": 710.5874355434365}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701165302.57/warc/CC-MAIN-20160205193925-00298-ip-10-236-182-209.ec2.internal.warc.gz"}
|
https://www.proofwiki.org/wiki/If_Set_Exists_then_Empty_Set_Exists
|
# If Set Exists then Empty Set Exists
## Theorem
If at least one set exists, then there exists an empty set.
## Proof
This page is beyond the scope of ZFC, and should not be used in anything other than the theory in which it resides.
If you believe that the contents of this page can be reworked to allow ZFC, then you can discuss it at the talk page.
Let $S$ be a set.
By the axiom of class comprehension, there is an empty class:
$\O = \set { x : x \ne x }$
Since $x \in \O$ is never true, it follows vacuously that:
$x \in \O \implies x \in S$
By the subclass definition:
$\O \subseteq S$
By Subclass of Set is Set, $\O$ is a set.
$\blacksquare$
|
2022-08-09 06:57:13
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5429703593254089, "perplexity": 591.2314220065551}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570913.16/warc/CC-MAIN-20220809064307-20220809094307-00086.warc.gz"}
|
https://www.jobilize.com/trigonometry/section/verbal-vectors-by-openstax?qcr=www.quizover.com
|
# 10.8 Vectors (Page 7/22)
Page 7 / 22
## Verbal
What are the characteristics of the letters that are commonly used to represent vectors?
lowercase, bold letter, usually $\text{\hspace{0.17em}}u,v,w$
How is a vector more specific than a line segment?
What are $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}j,$ and what do they represent?
They are unit vectors. They are used to represent the horizontal and vertical components of a vector. They each have a magnitude of 1.
What is component form?
When a unit vector is expressed as $⟨a,b⟩,$ which letter is the coefficient of the $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}$ and which the $\text{\hspace{0.17em}}j?$
The first number always represents the coefficient of the $\text{\hspace{0.17em}}i,\text{\hspace{0.17em}}$ and the second represents the $\text{\hspace{0.17em}}j.$
## Algebraic
Given a vector with initial point $\text{\hspace{0.17em}}\left(5,2\right)\text{\hspace{0.17em}}$ and terminal point $\text{\hspace{0.17em}}\left(-1,-3\right),\text{\hspace{0.17em}}$ find an equivalent vector whose initial point is $\text{\hspace{0.17em}}\left(0,0\right).\text{\hspace{0.17em}}$ Write the vector in component form $⟨a,b⟩.$
Given a vector with initial point $\text{\hspace{0.17em}}\left(-4,2\right)\text{\hspace{0.17em}}$ and terminal point $\text{\hspace{0.17em}}\left(3,-3\right),\text{\hspace{0.17em}}$ find an equivalent vector whose initial point is $\text{\hspace{0.17em}}\left(0,0\right).\text{\hspace{0.17em}}$ Write the vector in component form $⟨a,b⟩.$
$〈7,-5〉$
Given a vector with initial point $\text{\hspace{0.17em}}\left(7,-1\right)\text{\hspace{0.17em}}$ and terminal point $\text{\hspace{0.17em}}\left(-1,-7\right),\text{\hspace{0.17em}}$ find an equivalent vector whose initial point is $\text{\hspace{0.17em}}\left(0,0\right).\text{\hspace{0.17em}}$ Write the vector in component form $⟨a,b⟩.$
For the following exercises, determine whether the two vectors $\text{\hspace{0.17em}}u\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}v\text{\hspace{0.17em}}$ are equal, where $\text{\hspace{0.17em}}u\text{\hspace{0.17em}}$ has an initial point $\text{\hspace{0.17em}}{P}_{1}\text{\hspace{0.17em}}$ and a terminal point $\text{\hspace{0.17em}}{P}_{2}\text{\hspace{0.17em}}$ and $v$ has an initial point $\text{\hspace{0.17em}}{P}_{3}\text{\hspace{0.17em}}$ and a terminal point $\text{\hspace{0.17em}}{P}_{4}$ .
${P}_{1}=\left(5,1\right),{P}_{2}=\left(3,-2\right),{P}_{3}=\left(-1,3\right),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{P}_{4}=\left(9,-4\right)$
not equal
${P}_{1}=\left(2,-3\right),{P}_{2}=\left(5,1\right),{P}_{3}=\left(6,-1\right),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{P}_{4}=\left(9,3\right)$
${P}_{1}=\left(-1,-1\right),{P}_{2}=\left(-4,5\right),{P}_{3}=\left(-10,6\right),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{P}_{4}=\left(-13,12\right)$
equal
${P}_{1}=\left(3,7\right),{P}_{2}=\left(2,1\right),{P}_{3}=\left(1,2\right),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{P}_{4}=\left(-1,-4\right)$
${P}_{1}=\left(8,3\right),{P}_{2}=\left(6,5\right),{P}_{3}=\left(11,8\right),\text{\hspace{0.17em}}$ and ${P}_{4}=\left(9,10\right)$
equal
Given initial point $\text{\hspace{0.17em}}{P}_{1}=\left(-3,1\right)\text{\hspace{0.17em}}$ and terminal point $\text{\hspace{0.17em}}{P}_{2}=\left(5,2\right),\text{\hspace{0.17em}}$ write the vector $\text{\hspace{0.17em}}v\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}j.\text{\hspace{0.17em}}$
Given initial point $\text{\hspace{0.17em}}{P}_{1}=\left(6,0\right)\text{\hspace{0.17em}}$ and terminal point $\text{\hspace{0.17em}}{P}_{2}=\left(-1,-3\right),\text{\hspace{0.17em}}$ write the vector $\text{\hspace{0.17em}}v\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}j.\text{\hspace{0.17em}}$
$7i-3j$
For the following exercises, use the vectors u = i + 5 j , v = −2 i − 3 j , and w = 4 i j .
Find u + ( v w )
Find 4 v + 2 u
$-6i-2j$
For the following exercises, use the given vectors to compute u + v , u v , and 2 u − 3 v .
$u=⟨2,-3⟩,v=⟨1,5⟩$
$u=⟨-3,4⟩,v=⟨-2,1⟩$
$u+v=〈-5,5〉,u-v=〈-1,3〉,2u-3v=〈0,5〉$
Let v = −4 i + 3 j . Find a vector that is half the length and points in the same direction as $\text{\hspace{0.17em}}v.$
Let v = 5 i + 2 j . Find a vector that is twice the length and points in the opposite direction as $\text{\hspace{0.17em}}v.$
$-10i–4j$
For the following exercises, find a unit vector in the same direction as the given vector.
a = 3 i + 4 j
b = −2 i + 5 j
$-\frac{2\sqrt{29}}{29}i+\frac{5\sqrt{29}}{29}j$
c = 10 i j
$d=-\frac{1}{3}i+\frac{5}{2}j$
$-\frac{2\sqrt{229}}{229}i+\frac{15\sqrt{229}}{229}j$
u = 100 i + 200 j
u = −14 i + 2 j
$-\frac{7\sqrt{2}}{10}i+\frac{\sqrt{2}}{10}j$
For the following exercises, find the magnitude and direction of the vector, $\text{\hspace{0.17em}}0\le \theta <2\pi .$
$⟨0,4⟩$
$⟨6,5⟩$
$|v|=7.810,\theta =39.806°$
$⟨2,-5⟩$
$⟨-4,-6⟩$
$|v|=7.211,\theta =236.310°$
Given u = 3 i − 4 j and v = −2 i + 3 j , calculate $\text{\hspace{0.17em}}u\cdot v.$
Given u = − i j and v = i + 5 j , calculate $\text{\hspace{0.17em}}u\cdot v.$
$-6$
Given $\text{\hspace{0.17em}}u=⟨-2,4⟩\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}v=⟨-3,1⟩,\text{\hspace{0.17em}}$ calculate $\text{\hspace{0.17em}}u\cdot v.$
Given u $=⟨-1,6⟩$ and v $=⟨6,-1⟩,$ calculate $\text{\hspace{0.17em}}u\cdot v.$
$-12$
## Graphical
For the following exercises, given $\text{\hspace{0.17em}}v,\text{\hspace{0.17em}}$ draw $v,$ 3 v and $\text{\hspace{0.17em}}\frac{1}{2}v.$
$⟨2,-1⟩$
$⟨-1,4⟩$
$⟨-3,-2⟩$
For the following exercises, use the vectors shown to sketch u + v , u v , and 2 u .
For the following exercises, use the vectors shown to sketch 2 u + v .
For the following exercises, use the vectors shown to sketch u − 3 v .
For the following exercises, write the vector shown in component form.
bsc F. y algebra and trigonometry pepper 2
given that x= 3/5 find sin 3x
4
DB
remove any signs and collect terms of -2(8a-3b-c)
-16a+6b+2c
Will
Joeval
(x2-2x+8)-4(x2-3x+5)
sorry
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
(X2-2X+8)-4(X2-3X+5)=0 ?
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
Y
master
master
Soo sorry (5±Root11* i)/3
master
Mukhtar
explain and give four example of hyperbolic function
What is the correct rational algebraic expression of the given "a fraction whose denominator is 10 more than the numerator y?
y/y+10
Mr
Find nth derivative of eax sin (bx + c).
Find area common to the parabola y2 = 4ax and x2 = 4ay.
Anurag
A rectangular garden is 25ft wide. if its area is 1125ft, what is the length of the garden
to find the length I divide the area by the wide wich means 1125ft/25ft=45
Miranda
thanks
Jhovie
What do you call a relation where each element in the domain is related to only one value in the range by some rules?
A banana.
Yaona
given 4cot thither +3=0and 0°<thither <180° use a sketch to determine the value of the following a)cos thither
what are you up to?
nothing up todat yet
Miranda
hi
jai
hello
jai
Miranda Drice
jai
aap konsi country se ho
jai
which language is that
Miranda
I am living in india
jai
good
Miranda
what is the formula for calculating algebraic
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda
state and prove Cayley hamilton therom
hello
Propessor
hi
Miranda
the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.
Miranda
hi
jai
hi Miranda
jai
thanks
Propessor
welcome
jai
What is algebra
algebra is a branch of the mathematics to calculate expressions follow.
Miranda
Miranda Drice would you mind teaching me mathematics? I think you are really good at math. I'm not good at it. In fact I hate it. 😅😅😅
Jeffrey
lolll who told you I'm good at it
Miranda
something seems to wispher me to my ear that u are good at it. lol
Jeffrey
lolllll if you say so
Miranda
but seriously, Im really bad at math. And I hate it. But you see, I downloaded this app two months ago hoping to master it.
Jeffrey
which grade are you in though
Miranda
oh woww I understand
Miranda
Jeffrey
Jeffrey
Miranda
how come you finished in college and you don't like math though
Miranda
gotta practice, holmie
Steve
if you never use it you won't be able to appreciate it
Steve
I don't know why. But Im trying to like it.
Jeffrey
yes steve. you're right
Jeffrey
so you better
Miranda
what is the solution of the given equation?
which equation
Miranda
I dont know. lol
Jeffrey
Miranda
Jeffrey
|
2020-11-24 15:09:18
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 84, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7948078513145447, "perplexity": 723.2778675173115}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141176864.5/warc/CC-MAIN-20201124140942-20201124170942-00174.warc.gz"}
|
http://wiki.freepascal.org/FPSpreadsheet_tutorial:_Writing_a_mini_spreadsheet_application
|
English (en) español (es) suomi (fi)
## Introduction
FPSpreadsheet is a powerful package for reading and writing spreadsheet files. The main intention is to provide a platform which is capable of native export/import of an application's data to/from the most important spreadsheet file formats without having these spreadsheet applications installed.
Soon, however, the wish arises to use this package also for editing of file content or formatting. For this purpose, the library contains a dedicated grid control, the FPSpreadsheetGrid, which closely resembles the features of a worksheet of a spreadsheet application. The demo "spready" which comes along with FPSpreadsheet demonstrates usage of this grid. Along with a bunch of formatting options, this demo still comes up to more than 1400 lines of code in the main form unit. Therefore, a set of visual controls was developed which greatly simplify creation of spreadsheet applications.
It is the intention of this tutorial to write a simple spreadsheet program on the basis of these controls.
Although most of the internal structure of the FPSpreadsheet library is covered by the visual controls it is recommended that you have some knowledge of FPSpreadsheet. Of course, you should not have a basic understanding of Lazarus and FPC, and you must know how to work with the object inspector of Lazarus.
FPSpreadsheet exposes non-visual classes, such as TsWorkbook, TsWorksheet etc. This keeps the library general enough for all kind of Pascal programs. For GUI programs, on the other hand, some infrastructure is needed which relates the spreadsheets to forms, grids, and other controls.
### TsWorkbookSource
The heart of the visual FPSpreadsheet controls is the TsWorkbookSource class. This provides a link between the non-visual spreadsheet data and the visual controls on the form. Its purpose is similar to that of a TDataSource component in database applications which links database tables or queries to dedicated "data-aware" controls.
All visual FPSpreadsheet controls have a property WorkbookSource which links them into the information chain provided by the TsWorkbookSource. The WorkbookSource keeps a list of all controls attached. Internally, these controls are called "listeners" because they listen to information distributed by the WorkbookSource.
The workbook and worksheets use events to notify the WorkbookSource of all relevant changes: changes in cell content or formatting, selecting other cells, adding or deleting worksheet etc. Information on these changes is passed on to the listening controls, and they react in their own specialized way on these changes. If, for example, a new worksheet is added to a workbook the visual TsWorkbookTabControl creates a new tab for the new worksheet, and the TsWorksheetGrid loads the new worksheet into the grid.
### TsWorkbookTabControl
This is a tabcontrol which provides a tab for each worksheet of the current workbook. The tab names are identical with the names of the worksheets. Selecting another tab is communicated to the other visual spreadsheet controls via the WorkbookSource.
### TsWorksheetGrid
This is a customized DrawGrid descendant of the LCL and displays cells of the currently selected worksheet. The texts are not stored in the grid (like a StringGrid would do), but are taken from the TsWorksheet data structure. Similarly, the worksheet provides the information of how each cell is formatted. Like any LCL grid it has a bunch of properties and can be tuned for many applications by adapting its Options. The most important one will be described below.
Note: The TsWorksheetGrid can also be operated without a TsWorkbookSource. For this purpose it provides its own set of methods for reading and writing files.
### TsCellEdit
The typical spreadsheet applications provide a line for editing formulas or cell content. This is the purpose of the TsCellEdit. It displays the content of the active cell of the worksheet which is the same as the active cell of the WorksheetGrid. If editing is finished (by pressing Enter, or by selecting another cell in the WorksheetGrid) the new cell value is transferred to the worksheet. Internally, the TsCellEdit is a memo control, i.e. it is able to process multi-line text correctly. Use Ctrl+ Enter to insert a forced line-break.
### TsCellIndicator
This is a TEdit control which displays the address of the currently selected cell in Excel notation, e.g. 'A1' if the active cell is in the first row and first column (row = 0, column = 0). Conversely, if a valid cell address is entered into this control the corresponding cell becomes active.
### TsCellCombobox
This combobox can be used to modify various cell properties by selecting values from the dropdown list. The property affected is determined by the CellFormatItem of the combobox:
• cfiFontName: the list contains he names of all fonts available on the current system. If an item is selected the corresponding font is used to format the cell of the currently selected cells.
• cfiFontSize: the list contains the most typical font sizes used in spreadsheets. Selecting an item sets the font size of the currently selected cells accordingly.
• cfiFontColor: the list contains all colors of the workbook's palette. The selected color is assigned to the font of the selected cells.
• cfiBackgroundColor: like cfiFontColor - the selected color is used as background fill color of the selected cells.
Inherits from TValueListEditor and displays name-value pairs for properties of the workbook, the selected worksheet, and the content and formatting of the active cell. It's main purpose is to help with debugging.
Enough of theory, let's get started. Let's write a small spreadsheet application. Sure - it cannot compete with the spreadsheets of the main Office applications like Excel or Open/LibreOffice, but it has all the main ingredients due to FPSpreadsheet. And using the FPSpreadsheet controls allows to achieve this with minimum lines of code.
### Preparations
Create a new project and store it in a folder of your liking.
Since Office applications have a menu and a toolbar add a TMainMenu and a TToolbar component to the form. (You could even mimic the ribbon user interface of the new Microsoft applications by adding a TSpkToolbar from Lazarus Code and Components Repository, but be aware that this component does not yet provide all the features of a standard toolbar).
In fact, we will be needing another toolbar for the formula edit line. As you will see later, it will be resizable; as size control add a TSplitter to the form and top-align it such that it is positioned underneath the two toolbars. In order to keep a minimum size of the toolbar you should establish constraints: Look at the current height of the toolbar and enter this number into the MinHeight field of the Constraints property of the toolbar. To separate the formula toolbar from the rest of the main form, activate the option ebBottom of the EdgeBorders property of the second toolbar.
Since menu and toolbars will have to handle same user actions it is advantageous to provide a TActionList to store all possible actions. If assigned to the menu items and toolbuttons both will react on user interaction in the same way without any additional coding. And: The FPSpreadsheet visual controls package contains a bunch of spreadsheet-related standard actions ready to use.
The toolbar of the completed application will contain a lot of of icons. Therefore, we need a TImageList component which has to be linked to the Images property of the TMainMenu, the TToolbars, and the TActionList. Where to get icons? You can have a look in the folder images of your Lazarus installation where you'll find standard icons for loading and saving etc. This is a subset of the famfamfam SILK icon library. Another huge icon set is the Fugue icon collection. Both collections are licensed as "Creative commons" and are free even for commercial use, provided that appropriate reference is given in the created programs. When selecting icons prefer the png image format, and make sure to use always the same size, usually 16x16 pixels.
### Setting up the visual workbook
#### TsWorkbookSource
As described in the introductory section the TsWorkbookSource component is the interface between workbook and controls on the user interface. Add this component to the form and give it a decent name (we'll keep the default name sWorkbookSource1 here, though). As you will see shortly, this component will have to be assigned to the property WorkbookSource of all controls of the FPSpreadsheet_visual package.
The WorkbookSource is responsible for loading and writing data from/to file and for communicating with the workbook. Therefore, it owns a set of options that are passed to the workbook and control these processes:
type
TsWorkbookOption = (boVirtualMode, boBufStream, boAutoCalc, boCalcBeforeSaving, boReadFormulas);
TsWorkbookOptions = set of TsWorkbookOption;
The most important ones are
• boAutoCalc: activates automatic calculation of formulas whenever cell content changes.
• boCalcBeforeSaving: calculated formulas before a workbook is written to file
• boReadFormulas: if set full formulas are read from the file, otherwise only formula results.
• boBufStream and boVirtualMode: In non-visual programs, these options can help if running out of memory in case of large workbooks. boVirtualMode, in particular, is not usable for visual applications, though, because it avoids keeping data in the worksheet cells. See also FPSpreadsheet#Virtual_mode.
In this tutorial, it is assumed that the options boAutoCalc and boReadFormulas are activated.
#### TsWorkbookTabControl
The first visual control used in the form is a TsWorkbookTabControl - click it onto the form (into the space not occupied by the toolbar). Client-align it within the form, this shows the TabControl as a bright rectangle only. Now link its WorkbookSource property to the TsWorkbookSource component that we have added just before. Now the TabControl shows a tab labelled "Sheet1". This is because the TsWorkbookSource has created a dummy workkbook containing a single worksheet "Sheet1". The WorkbookSource synchronizes this internal workbook with the TabControl (and the other visual controls to come) such that it displays this worksheet as a tab.
In Excel the worksheet tabs are at the bottom of the form - to achieve this effect you can set the property TabPosition of the TabControl to tpBottom; there are some painting issues of the LCL with this TabPosition, though, therefore, I prefer the default setting, tpTop.
The screenshot shows how far we've got.
#### TsWorksheetGrid
Now we add a TsWorksheetGrid control. Click it somewhere into the space occupied by the TabControl such that it becomes a child a of the TabControl. You see a standard stringgrid-like component. Link its WorkbookSource property to the source added at the beginning, and the grid looks more like a spreadsheet: there are the column headers labelled by letters "A", "B", etc, and the row headers labelled by numbers "1", "2", etc; the active cell, A1, is marked by a thick border.
You may want to switch the grid's TitleStyle to tsNative in order to achieve themed painting of the row and column headers. And here is a good place to adapt the grid's Options in order to activate many features well-known to spreadsheets:
• goEditing must be active, otherwise the grid contents cannot be modified.
• goAlwaysShowEditor should be off because it interferes with the editing convention of spreadsheet applications.
• goColSizing enables changing of the column width by dragging the dividing line between adjacent column headers. Dragging occurs with the left mouse button pressed.
• goRowSizing does the same with the row heights.
• goDblClickAutoResize activates the feature that optimum column width can be set by double-clicking in the header on its dividing line to the next column. The "optimum" column width is such that no cell content is truncated and no extra space is shown in the column.
• goHeaderHotTrack gives visual feedback if the mouse is above a header cell.
• goRangeSelect (which is on by default) enables selection of a rectangular range of cells by dragging the mouse between cells at opposite corners of the rectangle. You can even select multiple rectangles by holding the CTRL key down before the next rectangle is dragged (in older Lazarus releases - before 1.4 - only a single range could be selected).
• goThumbTracking activates immediate scrolling of the worksheet if one of the scrollbars is dragged with the mouse. The Office applications usually scroll by lines; you can achieve this by turning off goSmoothScroll.
In addition to these Options inherited from TCustomGrid there are some more properties specialized for spreadsheet operation:
• ShowGridLines, if false, hides the row and column grid lines.
• ShowHeaders can be set to false if the the column and row headers are to be hidden. (The same can be achieved also by the deprecated property DisplayFixedColRow).
• The LCL grids normally truncate text at the cell border if it is longer than the cell width. If TextOverflow is set to true then text can overflow into adjacent empty cells.
The properties AutoCalc and ReadFormulas are meant for stand-alone usage of the WorksheetGrid (i.e. without a TsWorkbookSource). Please use the corresponding options of the WorkbookSource instead. (AutoCalc enables automatic calculation of formulas whenever cell content changes. ReadFormulas activates reading of formulas from files, otherwise the grid would display only the formula results).
### Editing of values and formulas, Navigating
When you compile and run the program you'll already be able to enter data into the grid. Just select the cell that you want to edit by clicking or using the arrow keys - the active cell is highlighted by a thick border. Then begin typing. If the grid property EditorLineMode has been switched to elmMultiLine then manual line breaks can be entered by the key combination Ctrl+ Enter. When finished select another cell or press the Enter key. Using Enter automatically selects the next cell in the grid. The grid's property AutoAdvance defines what is understood as being the "next cell": by default, Enter moves the active cell down (aaDown), but you can also move it to the right (aaRight), or turn this feature off (aaNone) - see the type TAutoAdvance defined in the unit grids.pas for even more options.
If - as assumed above - the WorkbookSource option boAutoCalc is enabled the worksheet automatically supports calculation of formulas. As an example, go to cell A1, and enter the number 10. Then, go to cell A2 and enter the formula =A1+10. The formula is automatically evaluated, and its result, 20, is displayed in cell A2.
When you navigate in the grid you may notice that cell A2 only displays the formula result, it seems that there is no way to modify the formula once it has been entered. No need to worry - press the key F2 or click into the cell a second time to enter enhanced edit mode in which formulas are visible in the cell.
In order to edit formulas the Office applications offer a dedicated formula editor bar. Of course, fpspreadsheet has this feature, too. It is built into the TsCellEdit component which is set up such as to always show the full content of a cell. You remember the second toolbar from the "Preparations" section? This will house the TsCellEdit. But wait a minute - there's more to consider: Since formulas occasionally may get rather long the control should be capable of managing serval lines. The same with multi-lined text. TsCellEdit can do this since it is inherited from TCustomMemo which is a multi-line control. You also remember that we added a splitter to the form of the second toolbar? This is for height adjustment for the case that we want to use the multi-line feature of the TsCellEdit: just drag the splitter down to show more lines, or drag it upwards to stop at the height of a single line due to the MinHeight constraints that we had assigned to the toolbar.
The TsCellEdit will cover all available space in the second toolbar. Before we add the TsCellEdit we can make life easier if we think about what else will be in the second toolbar. In Excel, there is an indicator which displays the address of the currently active cell. This is the purpose of the TsCellIndicator. Since its height should not change when the toolbar is dragged down we first add a TPanel to the second toolbar; reduce its Width to about 100 pixels, remove its Caption and set its BevelOuter to bvNone.
Add the TsCellIndicator to this panel and align it to the top of the panel. Connect its WorkbookSource to the TsWorkbookSource control on the form, and immediately you'll see the text "A1", the address of the currenly selected cell.
Sometimes it is desirable to change the width of this box at runtime. So, why not add a splitter to the second toolbar? Set its Align property to alLeft. The result is a bit strange: the splitter is at the very left edge of the toolbar, but you'd expect to see it at the right of the panel. This is because the panel is not aligned by default. Set the Align property of the panel to alLeft as well, and drag the splitter to the right of the panel. Now the splitter is at the correct position.
Almost done now... We finally add a TsCellEdit component to the empty space of the toolbar. Client-align it so that it fills the entire rest of the toolbar. As usual, set its WorkbookSource property to the instance of the TsWorkbookSource on the the form.
Compile and run. Play with the program:
• Enter some dummy data. Navigate in the worksheet. You'll see that the CellIndicator always shows the address of the active cell. The contents of the active cell is displayed in the CellEdit box. The CellIndicator is not just a passive display of the current cell, it can also be edited. Type in the address of a cell which you want to become active, press Enter, and see what happens...
• Enter a formula. Navigate back into the formula cell - the formula is displayed in the CellEdit and can be changed there readily.
• Enter multi-lined text - you can enforce a lineending in the CellEdit by holding the Ctrl key down when you press Enter. The cell displays only one line of the text. Drag the horizontal splitter underneath the second toolbar down - the CellEdit shows all lines. Another way to see all lines of the text, is to adjust the cell height. You must have activated the grid Option goRowSizing. Then you can drag the lower dividing line of the row with the multi-line cell down to increase the row height - the missing lines now appear in the cell!
### Formatting of cells
In addition to entering data the user usually wants to apply some formatting to the cells in order to enhance or group them. The worksheet grid is set up in such a way that its cells display the formats taken from the workbook. In addition, the visual FPSpreadsheet controls are able to store formatting attributes into the cell. Because of the notification mechanism via the WorkbookSource these formats are returned to the WorksheetGrid for display.
#### Adding comboboxes for font name, font size, and font color
In this section, we want to provide the possibility to modify the font of the cell texts by selecting its name, size and/or color. The visual FPSpreadsheet provide the flexible TsCellCombobox for this purpose. It has the property CellFormatItem which defines which attribute it controls:
• cfiFontName: This option populates the combobox with all fonts found in the current system. The selected item is used for the type face in the selected cells.
• cfiFontSize fills the combobox with the mostly used font sizes (in points). Again, the selected item defines the font size of the selected cells.
• cfiFontColor adds all pre-defined colors ("palette") of the workbook to the combobox to set the text color of the selected cells. The combobox items consist of a little color box along with the color name. If the ColorRectWidth is set to -1 the color name is dropped.
• cfiBackgroundColor, the same with the background color of the selected cells.
• cfiCellBorderColor, the same with the border color of the selected cells - this feature is currently not yet supported.
Add three TsCellComboboxes to the first toolbar and set their CellFormatItem to cfiFontname, cfiFontSize, and cfiFontColor, respectively. Link their WorkbookSource property to the TsWorkbookSource on the form. You may want to increase the width of the font name combobox such that the longest font names are not cut off; the other comboboxes may become narrower. You may also want to turn off the color names of the third combobox by setting its ColorRectWidth to -1.
That's all to modify fonts. Compile and run. Enter some text and play with these new features of the program.
#### Using standard actions
FPSpreadsheet supports a lot of formats that can be applied to cells, such as text alignment, text rotation, text font, or cell borders or background colors. Typical gui applications contain menu commands and/or toolbar buttons which are assigned to each of these properties and allow to set them by a simple mouse click. In addition, the state of these controls often reflects the properties of the active cell. For example, if there is a button for using a bold type-face this button should be drawn as being pressed if the active cell is bold, but as released if it is not. To simplify the coding of these tasks a large number of standard actions has been added to the library.
• TsWorksheetAddAction: adds an empty worksheet to the workbook. Specify its name in the NameMask property. The NameMask must contain the format specifier %d which is replaced at runtime by a number such that the worksheet name is unique.
• TsWorksheetDeleteAction: deletes the active worksheet from the workbook after a confirmation dialog. The last worksheet cannot be deleted.
• TsWorksheetRenameAction: renames the active worksheet.
• TsCopyAction: Copies the currently selected cells to an internal list ("CellClipboard") from where they can be pasted back into the spreadsheet to another location. The process can occur in a clipboard-manner ("copy"/"cut", then "paste") or in the way of the "copy brush" of the Office applications. The property CopyItem determines whether the entire cell, or only cell values, cell formulas, or cell formats are transferred.
• TsFontStyleAction: Modifies the font style of the selected cells. The property FontStyle defines whether the action makes the font bold, italic, underlined or striked-out. Normally each font style is handles by its own action. See the example below.
• TsHorAlignmentAction: Can be used to modify the horizontal alignment of text in the selected cells. Select HorAlignment to define which kind of alignment (left, center, right) is covered by the action. Like with the TsFontStyleAction, several actions should be provided to offer all available alignments. They are grouped in a mutually exclusive way like radiobuttons.
• TsVertAlignmentAction: Changes the vertical alignment of text in the selected cells: the kind of alignment is defined by the VertAlignment property. Again, these actions work like radiobuttons.
• TsTextRotationAction: Allows to specify the text orientation in the selected cells as defined by the property TextRotation in a mutially exclusive way.
• TsWordWrapAction: Activates the word-wrapping feature for the selected cells: if text is longer than the width of the cell (or height, if the text is rotated) then it is wrapped into multiple lines.
• TsNumberFormatAction: Defines the number format to be used for the selected cells. The format to be used is defined by the properties NumberFormat (such as nfFixed) for built-in formats, and NumberFormatStr for specialized formatting.
• TsDecimalsAction: Allows to increase or decrease the number of decimal places shown in the selected cells. The property Delta controls whether an increase (+1) or decrease (-1) is wanted.
• TsCellBorderAction: Allows to specify if a border will be drawn around the selected cells. The subproperties East, West, North, South, InnerHor, InnerVert of Borders define what the border will look like at each side of the cell range. Note that each rectangular range of cells is considere as a single block; the properties East, West, North and South are responsible for the outer borders of the entire block, inner borders are defined by InnerHor and InnerVert. Using these properties, borders can be switched on and off (Visible), and in addition, the line style and line color can be changed.
• TsMergeAction: If checked, the cells of each selected rectangular range are merged to a single block. Unchecking the action separates the block to individual cells. Note that the block's content and formatting is defined by the top-left cell of each block; content and formats of other cells will be lost.
• TsCellProtectionAction: comes in two flavors, one for protecting cells from modifications, and one for hiding formulas, depending on the value of the property Protection. Note that this action is effective only if worksheet protection has been enabled by calling Workbooksource.Worksheet.Protect(true).
#### Adding buttons for "Bold", "Italic", and "Underline"
If you have never worked with standard actions before here are some detailed step-by-step instructions. Let us stick to above example and provide the possibility to switch the font style of the selected cells to bold. The standard action which is responsible for this feature is the TsFontStyleAction.
• At first, we add this action to the form: Double-click on the TActionList to open the "ActionList Editor".
• Click on the down-arrow next to the "+" button, and select the item "New standard action" from the drop-down menu.
• This opens a dialog with the list of registered "Standard Action Classes".
• Scroll down until you find a group named "FPSpreadsheet".
• In this group, select the item "TsFontStyleAction" by double-clicking.
• Now an item sFontStyleAction1 appears in the ActionList Editor.
• It should already be selected like in the screenshot at the right. If not, select sFontStyleAction1 in the ActionList Editor to bring it up in the Object Inspector and to set up its properties:
• Use the text "Bold" for the Caption - this is the text that will be assigned to the corresponding menu item.
• Similarly, assign "Bold font" to the Hint property.
• Set the ImageIndex to the index of the icon in the form's ImageList that you want to see in the toolbar.
• Make sure that the item fssBold is highlighted in the dropdown list of the property FontStyle. If not, select it. Since TsFontStyleAction can handle several font styles (bold, italic, underline, strikeout) we have to tell the action which font style it should be responsible of.
• Like with the visual controls, don't forget to assign the TsWorkbookSource to the corresponding property WorkbookSource of the action. This activates the communication between the worksheet/workbook on the one hand, and the action and the related controls on the other hand.
Finally we add a toolbar button for the "bold" action. Right-click onto the TToolbar, and add a new toolbutton by selecting item "New button" from the popup menu. Go to the property Action in the object inspector again, pick the sFontStyle1 item, and this is enough to give the tool button the ability to set a cell font to bold!
Repeat this procedure with two other buttons. Design them to set the font style to italic and underlined.
Test the program by compiling. Type some text into cells. Select one of them and click the "Bold" toolbutton - voila, the cell is in bold font. Select another cell. Note that the toolbutton is automatically drawn in the down state if the cell has bold font. Repeat with the other buttons.
### Saving to file
After having entered data into the grid you will certainly want to save the grid to a spreadsheet file. Lazarus provides all the necessary infrastructure for saving available in the standard action TFileSaveAs. This action automatically opens a FileDialog for entering the file name.
Select the TFileSaveAs standard action from the list of standard action classes. Note that it cannot be found in the "FPSpreadsheet" category, but in the "File" group since it is a standard action of the LCL.
At first, let us specify the properties of the FileDialog. Select the property Dialog of the TFileSaveAs action in the object inspector. It is convenient to be able to store the workbook in various file formats; this can be prepared by providing a file format list in the Filter property of the dialog. Paste the following text into this property:
When you click on the ellipsis button next to Filter the file list appears in a more clearly arranged dialog shown at the right.
Make one of these file extensions, e.g. xlsx, the default of the file dialog by assigning its list index to the FilterIndex property. The xlsx file is the first format in the filter list. FilterIndex, therefore, must be set to 1.
Note: The indexes in the filter list are 1-based, in contrast to the convention of Lazarus and FPC using 0-based indexes.
Next, we define what happens after a file name has been selected in the file dialog. For this purpose, the TFileSaveAs action provides the event OnAccept. This is one of the few places where we have to write code in this project... But it is short: We check which file format has been selected in the format list and write the corresponding spreadsheet file by calling the method SaveToSpreadsheetFile of the TWorkbookSource:
uses
..., fpstypes, ...; // for TsSpreadsheetFormat
procedure TForm1.FileSaveAs1Accept(Sender: TObject);
var
begin
Screen.Cursor := crHourglass;
try
case FileSaveAs1.Dialog.FilterIndex of
1: fmt := sfOOXML; // Note: Indexes are 1-based here!
2: fmt := sfExcel8;
3: fmt := sfExcel5;
4: fmt := sfExcel2;
5: fmt := sfOpenDocument;
6: fmt := sfCSV;
7: fmt := sfWikiTable_WikiMedia;
end;
finally
Screen.Cursor := crDefault;
end;
end;
We will make the FileSaveAs action available in the toolbar and in the menu:
• Toolbar: Add a TToolButton to the first toolbar and drag it to its left edge. Assign the FileSaveAs action to its Action property.
• Menu: The "Save" command is usually in a submenu called "File". Therefore, double click on the TMainMenu, right-click on the "Format" item and insert a new item "before" the current one. Name it "File". Add a submenu to it. Click at the default menu item and assign the FileSaveAs action to its Action property.
What is left is reading of a spreadsheet file into our application. Of course, FPSpreadsheet is well-prepared for this task. The operations are very similar to saving. But instead of using a TFileSaveAs standard action, we use a TFileOpen standard action. Again, this standard action has a built-in file dialog where we have to set the DefaultExtension (".xls" or ".xlsx", most probably) and the format Filter:
(Copy this string into the field Filter of the action's Dialog). As you may notice the Filter contains selections which cover various file formats, such as "All spreadsheet files", or "All Excel files". This is possible because the TsWorkbookSource has a property AutoDetectFormat for automatic detection of the spreadsheet file format. In the other cases, like "Libre/OpenOffice", we can specify the format, sfOpenDocument, explicitly. Evaluation of the correct file format and reading of the file is done in the OnAccept event handler of the action:
{ Loads the spreadsheet file selected by the FileOpen standard action }
procedure TForm1.FileOpen1Accept(Sender: TObject);
begin
sWorkbookSource1.AutodetectFormat := false;
case FileOpen1.Dialog.FilterIndex of
1: sWorkbookSource1.AutoDetectFormat := true; // All spreadsheet files
2: sWorkbookSource1.AutoDetectFormat := true; // All Excel files
3: sWorkbookSource1.FileFormat := sfOOXML; // Excel 2007+
4: sWorkbookSource1.FileFormat := sfExcel8; // Excel 97-2003
5: sWorkbookSource1.FileFormat := sfExcel5; // Excel 5.0
6: sWorkbookSource1.FileFormat := sfExcel2; // Excel 2.1
7: sWorkbookSource1.FileFormat := sfOpenDocument; // Open/LibreOffice
8: sWorkbookSource1.FileFormat := sfCSV; // Text files
end;
sWorkbookSource1.FileName :=FileOpen1.Dialog.FileName; // This loads the file
end;
In order to see this action in the toolbar and menu, add a TToolButton to the toolbar and assign the TFileOpenAction to its Action property. In the menu, add a new item before the "Save as" item, and assign its Action accordingly.
Note: You can see a spreadsheet file even at designtime if you assign its name to the Filename property of the TsWorkbookSource. But be aware that the file probably cannot be found at runtime if it is specified by a relative path and if the application is to run on another computer with a different directory structure!
## Summary
If you followed us through the steps of this tutorial you have programmed a complex spreadsheet gui application almost without having written any line of code (with the exception of the loading and saving routines). If you did not, have a look at the demo "fps_ctrls" in the examples folder of the FPSpreadsheet installation; it shows the result of this tutorial with some add-ons.
|
2018-01-22 05:49:30
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4188031256198883, "perplexity": 2253.5267899639625}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084891105.83/warc/CC-MAIN-20180122054202-20180122074202-00362.warc.gz"}
|
https://math.stackexchange.com/questions/1232488/expected-value-of-a-continuous-random-variable-example-from-sheldon-rosss-book
|
Expected Value of a Continuous Random Variable (Example from Sheldon Ross's Book)
This is an example in the book (A First Course in Probability by Sheldon Ross).
A stick of length 1 is split at a point U that is uniformly distributed over $(0,1)$. Determine the expected length of the piece that contains the point $0 \leq p \leq 1$.
The problem with this is I don't know how would I go about solving this. They have solved this in the book but I do not understand their solution.
For example I don't know how to setup the probability density formula and then to find the expected value from there.
• I get $p(1-p)+\frac12$, is that the right answer?
– bof
Apr 13 '15 at 6:16
• Yes can you please explain how you got this? Apr 13 '15 at 6:18
• By definition of $U$ being uniformly distributed, the answer is $$\int_0^p(1-u)du+\int_p^1udu=\tfrac12+p-p^2.$$ If $U$ has density $f_U$ on $(0,1)$, consider $$\int_0^p(1-u)f_U(u)du+\int_p^1uf_U(u)du.$$
– Did
Apr 13 '15 at 6:37
• I understood your solution but how have you taken the function value to be 1 in both cases? Apr 13 '15 at 6:39
• Which function value? $f_U$? This is the definition of being uniformly distributed on $(0,1)$, no?
– Did
Apr 13 '15 at 6:39
Let $X$ be the length of the piece containing $p$.
Given that $U\gt p,$ then $X$ is uniformly distributed on $[p,1]$, so the conditional expectation is $\frac{p+1}2.$
Given that $U\lt p,$ then $X$ is uniformly distributed on $[1-p,1]$, so the conditional expectation is $\frac{(1-p)+1}2.$
So $$E(X)=P(U\gt p)\cdot\frac{p+1}2+P(U\lt p)\cdot\frac{(1-p)+1}2=(1-p)\cdot\frac{p+1}2+p\cdot\frac{(1-p)+1}2$$ which simplifies to $\frac12+p(1-p).$
I don't know if this is how you're supposed to do it, or if you're supposed to derive the probability distribution etc. Is my solution anything like the solution in the book?
• In the book they've found the probability density formula and then calculated the expected value from there. Also in the method you've used may be correct but I don't know what condition expectation is. Apr 13 '15 at 6:37
• Conditioning is not needed (and might be seen as an unnecessary complication).
– Did
Apr 13 '15 at 6:38
• @Did Right, your solution is much better.
– bof
Apr 13 '15 at 6:42
|
2021-09-25 16:04:45
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8826322555541992, "perplexity": 140.23798571418325}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057687.51/warc/CC-MAIN-20210925142524-20210925172524-00662.warc.gz"}
|
https://etrical.blogspot.com/2016/10/cylindrical-salient-pole-rotor.html
|
## Sunday, 30 October 2016
### Difference between Cylindrical and Salient Pole Rotor Synchronous Generator
We are quite familiar with Synchronous Generator or often called Alternator. There are mainly two types of Synchronous Generator from rotor construction point of view. The two main types of synchronous machine are Cylindrical Rotor and Salient Pole. Saliency simply means projection outward. Therefore from the literal meaning of Saliency one can guess that Salient Pole machines must have poles projecting outward. In general, the Cylindrical Rotor type machines are confined to 2 and 4 pole turbine generators, while salient pole types are built with 4 poles upwards and include most classes of duty. Both classes of machine are similar in Stator construction point of view. Each has a stator carrying a three-phase winding distributed over its inner periphery.
Within the stator bore is the rotor which is magnetised by a winding carrying DC current. The main difference between the Cylindrical Rotor and Salient Pole classes of machine lies in the rotor construction. The cylindrical rotor type has a uniformly cylindrical rotor that carries its excitation winding distributed over a number of slots around its periphery. This construction is unsuited to multi-polar machines but it is very sound mechanically. Hence it is particularly well adapted for the highest speed electrical machines and is universally employed for 2 pole units, and some 4 pole units.
The salient pole type has poles that are physically separate, each carrying a concentrated excitation winding. This type of construction is in many ways complementary to that of the cylindrical rotor and is employed in machines having 4 poles or more. Except in special cases its use is exclusive in machines having more than 6 poles. Two and four pole generators are most often used in applications where steam or gas turbines are used as the prime mover. This is because the steam turbine tends to be suited to high rotational speeds. Four pole steam turbine generators are most often found in nuclear power stations as the relative wetness of the steam makes the high rotational speed of a two-pole design unsuitable.
Most generators with gas turbine prime mover are four pole machines to obtain enhanced mechanical strength in the rotor since a gearbox is often used to couple the power turbine to the generator; the choice of synchronous speed of the generator is not subject to the same constraints as with steam turbines. Generators with diesel engine drivers are invariably of four or more pole design, to match the running speed of the driver without using a gearbox. Four-stroke diesel engines usually have a higher running speed than two stroke engines, so generators having four or six poles are most common. Two stroke diesel engines are often derivatives of marine designs with relatively large outputs and may have running speeds of the order of 125 rpm. This requires a generator with a large number of poles (48 for a 125 rpm, 50Hz generator) and consequently is of large diameter and short axial length. This is a contrast to turbine-driven machines that are of small diameter and long axial length.
Further it shall be noted that in Hydro Power plant, Salient Pole Generators are used as the speed of the prime mover (here water turbine is the prime mover) is less and therefore number of poles required will be more from P = 120f / N. To accommodate larger number of poles, Salient Pole construction is well suited.
It shall also be noted that Salient Pole machines cannot be used at higher speed because at higher speed the centrifugal forces will be large which may damage the Salient pole bolted on the Rotor core. This is why Turbo Generators uses Cylindrical Pole construction whereas Hydro generators use Salient Pole construction.
To summarize, the main differences between the Cylindrical Rotor and Salient Pole machines are as follows:
In salient pole type of rotor consist of large number of projected poles i.e. salient poles mounted on a magnetic wheel. Construction of a salient pole rotor is as shown in the figure at left. The projected poles are made up from laminations of steel. The rotor winding is provided on these poles and it is supported by pole shoes.
• Salient pole rotors have large diameter and shorter axial length.
• They are generally used in lower speed electrical machines, say 100 RPM to 1500 RPM.
• As the rotor speed is lower, more number of poles is required to attain the required frequency. (N = 120f / P). Typically number of salient poles is between 4 to 60.
• Flux distribution is relatively poor than non-salient pole rotor, hence the generated emf waveform is not as good as cylindrical rotor.
• Salient pole rotors generally need damper windings to prevent rotor oscillations during operation.
• Salient pole synchronous generators are mostly used in hydro power plants.
Non-salient pole or Cylindrical rotors are cylindrical in shape having parallel slots on it to place rotor windings. It is made up of solid steel. Sometimes, they are also called as drum rotor.
• They are smaller in diameter but having longer axial length.
• Cylindrical rotors are used in high speed electrical machines, usually 1500 RPM to 3000 RPM.
• Windage loss as well as noise is less as compared to salient pole rotors.
• Their construction is robust as compared to salient pole rotors.
• Number of poles is usually 2 or 4.
• Damper windings are not needed in non-salient pole rotors.
• Flux distribution is sinusoidal and hence gives better emf.
• Non-salient pole rotors are used in nuclear, gas and thermal power plants.
|
2017-09-21 19:23:52
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9053629040718079, "perplexity": 2032.7546742927668}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818687837.85/warc/CC-MAIN-20170921191047-20170921211047-00242.warc.gz"}
|
https://de.zxc.wiki/wiki/Elektron
|
# electron
Electron (e - )
classification
Elementary particle
fermion
lepton
properties
electric charge −1 e
(−1.602 176 634 · 10 −19 C )
Dimensions 5.485 799 090 65 (16) · 10 −4 u
9.109 383 7015 (28) · 10 −31 kg
Resting energy 0.510 998 950 00 (15) MeV
Compton wavelength 2.426 310 238 67 (73) · 10 −12 m
magnetic moment −9.284 764 7043 (28) · 10 −24 J / D
g factor −2.002 319 304 362 56 (35)
gyromagnetic
ratio
1.760 859 630 23 (53) 10 11 1 / ( s T )
Spin 1/2
average lifespan stable
Interactions weak
electromagnetic
gravitation
The electron [ ˈeːlɛktrɔn, eˈlɛk-, elɛkˈtroːn ] (from ancient Greek ἤλεκτρον élektron = ' amber ', on which electricity was first observed; coined in 1874 by Stoney and Helmholtz ) is a negatively charged elementary particle . Its symbol is e - . The alternative name Negatron (from nega tive charge and Elek tron ) is rarely used and is possibly in the beta spectroscopy common.
The electrons bound in an atom or ion form its electron shell . All chemistry is essentially based on the properties and interactions of these bound electrons. In metals , some of the electrons can move freely and cause the high electrical conductivity of metallic conductors . This is the basis of electrical engineering and electronics . In semiconductors , the number of mobile electrons and thus the electrical conductivity can be easily influenced, both through the manufacture of the material and later through external influences such as temperature, electrical voltage , incidence of light, etc. This is the basis of semiconductor electronics . Electrons can escape from any material when heated or when a strong electric field is applied ( glow emission , field emission ). As free electrons , they can then be formed into an electron beam in a vacuum through further acceleration and focusing . This has made the development of the oscilloscope , the television, and the computer monitor possible. Further applications of free electrons are e.g. B. the X-ray tube , the electron microscope , electron beam welding , basic physical research using particle accelerators and the generation of synchrotron radiation for research and technical purposes.
During the beta-minus decay of an atomic nucleus , a new electron is generated and emitted.
The experimental proof of the electron was first achieved by Emil Wiechert in 1897 and a little later by Joseph John Thomson .
## History of the discovery of the electron
The concept of a smallest, indivisible amount of electrical charge was proposed on various occasions around the middle of the 19th century, among others by Richard Laming , Wilhelm Weber and Hermann von Helmholtz .
In 1874, George Johnstone Stoney suggested the existence of electrical charge carriers associated with atoms. Based on the electrolysis , he estimated the size of the electron charge, but received a value that was about 20 times too low. At the meeting of the British Association in Belfast, he suggested using the elementary charge as another fundamental natural constant along with the gravitational constant and the speed of light as the basis of physical systems of measurement. Together with Helmholtz, Stoney coined the name electron for the “atom of electricity”.
Emil Wiechert found in 1897 that cathode radiation consists of negatively charged particles that are much lighter than an atom, but then stopped his research on this. In the same year Joseph John Thomson determined the mass of the particles (he first referred to them as corpuscules ) and was able to prove that the same particles are always involved regardless of the cathode material and the residual gas in the cathode ray tube. During this time, the Zeeman effect was used to demonstrate that these particles also occur in the atom and cause light emission there. The electron was thus identified as an elementary particle.
The elementary charge was measured in 1909 by Robert Millikan .
## properties
The electron is the lightest of the electrically charged elementary particles. If the conservation laws for charge and energy apply - which corresponds to all physical experience - electrons must therefore be stable. In fact, there is no experimental evidence of electron decay to date.
The electron belongs to the leptons and, like all leptons, has a spin (more precisely: spin quantum number) of 1/2. As a particle with half-integer spin, it belongs to the class of fermions , so it is particularly subject to the Pauli principle . Its antiparticle is the positron , symbol e + , with which it corresponds in all properties except for its electrical charge.
Some of the basic properties of the electron, listed in the table above, are linked by the magnetic moment of the electron spin :
${\ displaystyle {\ vec {\ mu _ {\ mathrm {s}}}} = - g _ {\ mathrm {s}} {\ frac {e} {2m _ {\ mathrm {e}}}} {\ vec { s}}}$.
Here is the magnetic moment of the electron spin, the mass of the electron, its charge and the spin . is called the Landé or g factor . The term vor , which describes the ratio of the magnetic moment to the spin, is called the gyromagnetic ratio of the electron. For the electron, according to the Dirac theory (relativistic quantum mechanics), this would be exactly the same as 2. Effects that are only explained by quantum electrodynamics , however, cause a measurable slight deviation of 2. This deviation is called the anomalous magnetic moment of the electron${\ displaystyle {\ vec {\ mu _ {\ mathrm {s}}}}}$${\ displaystyle m _ {\ mathrm {e}}}$${\ displaystyle e}$${\ displaystyle {\ vec {s}}}$${\ displaystyle g _ {\ mathrm {s}}}$${\ displaystyle {\ vec {s}}}$${\ displaystyle g _ {\ mathrm {s}}}$ designated.
Shortly after the discovery of the electron, attempts were made to estimate its size, especially because of the classic idea of small billiard balls that collide in scattering experiments . The argument came down to the fact that the concentration of the electron charge on a very small extent of the electron requires energy which, according to the principle of equivalence, must be contained in the mass of the electron. Under the assumption that the energy of an electron at rest is equal to twice the self-energy of the electron charge in its own electric field, one obtains the classic electron radius${\ displaystyle E _ {\ text {calm}} = m _ {\ mathrm {e}} \, c ^ {2}}$ ${\ displaystyle {\ frac {e ^ {2}} {4 \ pi \ varepsilon _ {0} r _ {\ mathrm {e}}}}}$
${\ displaystyle r _ {\ mathrm {e}} = {\ frac {e ^ {2}} {4 \, \ pi \, \ varepsilon _ {0} \, m _ {\ mathrm {e}} \, c ^ {2}}} = \ alpha ^ {2} \, a_ {0} = 2 {,} 817 \, 940 \, 3262 \, (13) \ cdot 10 ^ {- 15} ~ \ mathrm {m} \ }$
${\ displaystyle e}$: Elementary charge , : Kreiszahl , : Electric field constant , : electron mass : the speed of light , : fine structure constant , : Bohr radius . ${\ displaystyle \ pi}$${\ displaystyle \ varepsilon _ {0}}$${\ displaystyle m _ {\ mathrm {e}}}$${\ displaystyle c}$${\ displaystyle \ alpha}$${\ displaystyle a_ {0}}$
The self-energy mentally separates the electric charge and the electric field of the electron. If one puts the charge −e in the potential , whereby one thinks, for example, of a second electron evenly distributed over a spherical surface with the radius , then energy is required for this, the self-energy of a single electron is half of this. However, there were definitely other derivations for a possible expansion of the electron, which came to other values. ${\ displaystyle \ phi (r) = - {\ tfrac {1} {4 \ pi \ varepsilon _ {0}}} \ cdot {\ tfrac {e} {r}}}$${\ displaystyle r _ {\ mathrm {e}}}$
Today the view of the expansion of the electron is different: In the experiments that have been possible so far, electrons show neither expansion nor internal structure and can therefore be assumed to be point-like. The experimental upper limit for the size of the electron is currently around 10 −19 m. Nevertheless, the classical electron radius appears in many formulas in which a quantity of the dimension length (or area etc.) is formed from the fixed properties of the electron in order to be able to explain experimental results. For example, the theoretical formulas for the cross sections of the Photo and Compton effects contain the square of . ${\ displaystyle r _ {\ mathrm {e}}}$${\ displaystyle r _ {\ mathrm {e}}}$
Even the search for an electrical dipole moment of the electron has so far remained without positive results. A dipole moment would arise if, in the case of a non-point electron, the center of gravity of the mass was not at the same time the center of gravity of the charge. Such a thing is predicted by theories of supersymmetry that go beyond the standard model of elementary particles. A measurement in October 2013, which uses the strong electric field in a polar molecule, showed that a possible dipole moment with a confidence level of 90% is no greater than 8.7 · 10 −31 m. This clearly means that the electron's center of charge and mass cannot be more than about 10 −30 m apart. Theoretical approaches, according to which larger values were predicted, are thus refuted. ${\ displaystyle e}$
## Cross section
A distinction must be made between the (possible) expansion of the electron and its cross-section for interaction processes. When X-rays are scattered by electrons, z. B. a scattering cross-section of about what would correspond to the circular area with the classic electron radius described above . In the limiting case of large wavelengths, i. H. smaller photon energies, the scattering cross-section increases (see Thomson scattering and Compton effect ). ${\ displaystyle \ pi r_ {e} ^ {2}}$${\ displaystyle r_ {e}}$${\ displaystyle (8/3) \ pi r_ {e} ^ {2}}$
## Interactions
Many physical phenomena such as electricity , electromagnetism and electromagnetic radiation are essentially based on interactions between electrons. Electrons in an electrical conductor are displaced by a changing magnetic field and an electrical voltage is induced. The electrons in a current-carrying conductor generate a magnetic field. An accelerated electron - of course also in the case of curvilinear motion - emits photons, the so-called bremsstrahlung ( Hertzian dipole , synchrotron radiation , free-electron laser ).
In a solid body , the electron experiences interactions with the crystal lattice . Its behavior can then be described by using the deviating effective mass instead of the electron mass , which is also dependent on the direction of movement of the electron.
Electrons that have detached from their atoms in polar solvents such as water or alcohols are known as solvated electrons . When alkali metals are dissolved in ammonia , they are responsible for the strong blue color.
An electron is a quantum object , that is, is it the by the Heisenberg uncertainty principle local described and momentum spread in the measurable range, so that, as for light both shafts - as well as particle properties can be observed, which as a wave-particle duality designated becomes. In an atom, the electron can be viewed as a standing wave of matter .
## Experiments
The ratio e / m of the electron charge to the electron mass can be determined as a school experiment with the fine beam tube . The direct determination of the elementary charge was made possible by the Millikan experiment .
For electrons whose speed is not negligibly small compared to the speed of light , the non-linear contribution to the momentum must be taken into account according to the theory of relativity . Electrons with their low mass can be accelerated to such high speeds relatively easily; With a kinetic energy of 80 keV , an electron has half the speed of light. The impulse can be measured by deflecting it in a magnetic field. The deviation of the momentum from the value calculated according to classical mechanics was first demonstrated by Walter Kaufmann in 1901 and, after the discovery of the theory of relativity, initially described with the term “ relativistic mass increase ”, which is now regarded as obsolete.
## Free electrons
Fluorescence from electrons in a shadow cross tube
In the cathode ray tube (Braun's tube), electrons emerge from a heated hot cathode and are accelerated in the vacuum by an electric field in the direction of the field (towards the positive anode ) . The electrons are deflected by magnetic fields perpendicular to the direction of the field and perpendicular to the current direction of flight ( Lorentz force ). It was these properties of the electrons that made the development of the oscilloscope , the television and the computer monitor possible.
Further applications of free electrons are e.g. B. the X-ray tube , the electron microscope , electron beam welding , basic physical research using particle accelerators and the generation of synchrotron radiation for research and technical purposes. See also electron beam technology .
Wiktionary: Elektron - explanations of meanings, word origins, synonyms, translations
Commons : Electrons - collection of images, videos and audio files
## Individual evidence
1. CODATA Recommended Values. National Institute of Standards and Technology, accessed May 20, 2019 . Elementary charge in C (exact).
2. CODATA Recommended Values. National Institute of Standards and Technology, accessed May 20, 2019 . Electron mass in u . The numbers in brackets denote the uncertainty in the last digits of the value; this uncertainty is given as the estimated standard deviation of the specified numerical value from the actual value.
3. CODATA Recommended Values. National Institute of Standards and Technology, accessed May 20, 2019 . Electron mass in kg . The numbers in brackets denote the uncertainty in the last digits of the value; this uncertainty is given as the estimated standard deviation of the specified numerical value from the actual value.
4. CODATA Recommended Values. National Institute of Standards and Technology, accessed May 20, 2019 . Electron mass in MeV / c 2 . The numbers in brackets denote the uncertainty in the last digits of the value; this uncertainty is given as the estimated standard deviation of the specified numerical value from the actual value.
5. CODATA Recommended Values. National Institute of Standards and Technology, accessed May 20, 2019 . Compton wavelength. The numbers in brackets denote the uncertainty in the last digits of the value; this uncertainty is given as the estimated standard deviation of the specified numerical value from the actual value.
6. CODATA Recommended Values. National Institute of Standards and Technology, accessed August 3, 2019 . Magnetic moment. The numbers in brackets denote the uncertainty in the last digits of the value; this uncertainty is given as the estimated standard deviation of the specified numerical value from the actual value.
7. CODATA Recommended Values. National Institute of Standards and Technology, accessed May 20, 2019 . g factor. The numbers in brackets denote the uncertainty in the last digits of the value; this uncertainty is given as the estimated standard deviation of the specified numerical value from the actual value.
8. CODATA Recommended Values. National Institute of Standards and Technology, accessed May 20, 2019 . Gyromagnetic ratio. The numbers in brackets denote the uncertainty in the last digits of the value; this uncertainty is given as the estimated standard deviation of the specified numerical value from the actual value.
9. Károly Simonyi: cultural history of physics . Harri Deutsch, Thun, Frankfurt a. M. 1995, ISBN 3-8171-1379-X , pp. 380 .
10. ^ H. Rechenberg: The electron in physics - selection from a chronology of the last 100 years . In: European Journal of Physics . tape 18.3 , 1997, p. 145 .
11. ^ JJ Thomson: Cathode Rays . In: Philosophical Magazine . tape 44 , 1897, pp. 293 ( JJ Thomson (1856-1940): Cathode Rays ).
12. Theodore Arabatzis: Representing Electrons: A Biographical Approach to Theoretical Entities . University of Chicago Press, 2006, ISBN 0-226-02421-0 , pp. 70 f . (English, limited preview in Google Book Search).
13. Abraham Pais Inward Bound , p. 74.
14. ^ On the physical units of Nature , first published in 1881, Philosophical Magazine, Volume 11, 1881, p. 381.
15. ^ Trans. Royal Dublin Society, Volume 4, p. 583.
16. Károly Simonyi: cultural history of physics . Harri Deutsch, Thun, Frankfurt a. M. 1995, ISBN 3-8171-1379-X , pp. 380 .
17. ^ Encyclopedia Britannica 1911, Article Electron .
18. CODATA Recommended Values. National Institute of Standards and Technology, accessed May 20, 2019 . Classic electron radius. The numbers in brackets denote the uncertainty in the last digits of the value; this uncertainty is given as the estimated standard deviation of the specified numerical value from the actual value.
19. W. Finkelnburg: Introduction to atomic physics , Springer., 1976
20. Dieter Meschede : Gerthsen Physik 22nd ed. Berlin Springer, 2004. (Springer textbook), page 592 and exercise (17.4.5), page 967.
21. ^ Paul Huber and Hans H. Staub: Atomphysik (Introduction to Physics; Volume 3, Part 1) Basel: Reinhardt 1970, page 170.
22. Richard Feynman : Lectures on Physics, Vol 1, Mechanics, Radiation and Heat, Addison-Wesley 1966 - equation (32.11), pages 32–4.
23. Clara Moskowitz: Too round for the supersymmetry. November 15, 2013. Retrieved November 19, 2013 .
24. G. Möllenstedt and H. Düker: Observations and measurements on biprism interference with electron waves . In: Journal of Physics . No. 145 , 1956, pp. 377–397 ( freely accessible after registration ).
|
2021-02-24 17:20:07
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 26, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.713651180267334, "perplexity": 1030.52739332851}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178347293.1/warc/CC-MAIN-20210224165708-20210224195708-00592.warc.gz"}
|
http://math.stackexchange.com/questions/299174/induction-proof-sum-i-1n1-i-cdot-2i-n-cdot-2n22/299181
|
# Induction Proof: $\sum_{i=1}^{n+1} i \cdot 2^i = n \cdot 2^{n+2}+2$
Prove by Mathematical Induction . . .
$$\sum_{i=1}^{n+1} i \cdot 2^i = n \cdot 2^{n+2}+2$$ for all $n \geq 0$
I tried solving it, but I got stuck near the end . . .
a. Basis Step:
$1\cdot 2^1 = 0\cdot 2^{0+2}+2$
$2 = 2$
b. Inductive Hypothesis
$$\sum_{i=1}^{k+1} i \cdot 2^i = k \cdot 2^{k+2} +2$$ for $k \geq 0$
Prove k+1 is true.
$$\sum_{i=1}^{k+2} i \cdot 2^i = (k+1)\cdot 2^{k+3}+2$$
$\big[RHS\big]$
$k\cdot 2^{k+3}+2^{k+3}+2$
$\big[LHS\big]$
$$\sum_{i=1}^{k+2} {i \cdot 2^{i}}$$
$= \underbrace{\sum_{i=1}^{k+1} i \cdot 2^i} + (k+2)\cdot 2^{k+2}$ (Explicit last step)
$= \underbrace{k\cdot 2^{k+2}+2}+(k+2)\cdot 2^{k+2}$ (Inductive Hypothesis Substitution)
$= k\cdot 2^{k+2}+2+k\cdot 2^{k+2}+2^{k+3}$
$= 2k\cdot 2^{k+2} + 2^{k+3} + 2$
My [LHS] has one too many $2k\cdot 2^{k+2}$ or did it just do it completely wrong?
-
Please use single "$" sign on the title, so it doesn't block much space. – Michael Li Feb 10 '13 at 3:34 I've fixed a couple of spots (in the title and the first paragraph) where you had just 2 but it seems clear from the rest of your question you wanted$2^i$; please correct if those are wrong. – Steven Stadnicki Feb 10 '13 at 3:40 ## 2 Answers You are done. All you need to do is to regroup the terms properly. $$2k \cdot 2^{k+2} + 2^{k+3} + 2 = k \cdot 2^{k+3} + 2^{k+3} + 2 = (k+1) \cdot 2^{k+3} + 2$$ which is what you want. As julien points out, your penultimate step must read $$k \cdot 2^{k+2} + 2 + \color{blue}{k \cdot 2^{k+2}} + 2^{k+3}$$ instead of $$k \cdot 2^{k+2} + 2 + \color{red}{k \cdot 2{k+2}} + 2^{k+3}$$ I assume you made a typo while typesetting this line since you missed the ^ symbol i.e. you have $$\text{k \cdot 2{k+2} instead of k \cdot 2^{k+2}}$$ And also as julien points out, your (Explicit last step) should read $$\color{blue}{\sum_{i=1}^{k+1} i \cdot 2^i} + (k+2) \cdot 2^{k+2}$$ and not $$\color{red}{(k+1) \cdot 2^{k+1}} + (k+2) \cdot 2^{k+2}$$ - Except that there are things that do not make sense along the way of the OP's argument. At least for me. – 1015 Feb 10 '13 at 4:22 thanks a lot i didn't see it for some reason – rbtLong Feb 10 '13 at 5:10 Of course, what you wrote is absolutely correct and that's indeed the last step in this proof. Nevertheless, to get there, the OP replaced$\sum_{i=1}^{k+1}i2^i$by$(k+1)2^{k+1}$. I happen to be wrong sometimes, but here I am 100% sure there is a real problem. And now the OP thinks his/her proof is ok. Could you help? Thanks. – 1015 Feb 10 '13 at 14:54 Thanks, but that was not what I was pointing at. The mistakes I am talking about are in the lines above. – 1015 Feb 10 '13 at 18:27 Attention: you replaced$\sum_{i=1}^{k+1}2^i=2+2\cdot 2^2+3\cdot 2^3+\ldots+(k+1)2^{k+1}$by$(k+1)2^{k+1}$, and this is wrong for$k\geq 2$. The formula holds for$n=0$as you observed. Now assume it holds for some$n\geq 0$, i.e. $$\sum_{i=1}^{n+1}i2^i=n2^{n+2}+2.$$ Then $$\sum_{i=1}^{n+2}i2^i=\sum_{i=1}^{n+1}i2^i+(n+2)2^{n+2}$$ (so, using the induction hypothesis) $$=n2^{n+2}+2+(n+2)2^{n+2}$$ (so, simplifying) $$=2n2^{n+2}+2^{n+3}+2=n2^{n+3}+2^{n+3}+2=(n+1)2^{n+3}+2.$$ So, by induction, the formula holds for all$n\geq 0$. - you said my two last steps are wrong. . . but in your answer, you proved it to be the same using an answer that resemebles my last step exactly. also, marvis proved it correct the same so how can it be wrong? – rbtLong Feb 10 '13 at 5:10 @rbtLong No I did not say that. I said that two steps are wrong. Not the last two. Again, you replaced (twice)$\sum_{i=1}^{k+1}i2^i$by$(k+1)2^{k+1}\$. And this is obviously wrong. Marvis proved an equality which is correct. But he probably did not notice your wrong steps. – 1015 Feb 10 '13 at 14:27
my apologies . . . – rbtLong Feb 11 '13 at 1:46
@rbtLong No problem. I am glad you got it right in the end. – 1015 Feb 11 '13 at 2:29
|
2015-12-01 02:42:36
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000081062316895, "perplexity": 2571.049211132877}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398464396.48/warc/CC-MAIN-20151124205424-00337-ip-10-71-132-137.ec2.internal.warc.gz"}
|
http://www.reddit.com/r/Physics/comments/adcyy/can_you_give_a_physical_reason_why_you_would/
|
[–] 10 points11 points (5 children)
sorry, this has been archived and can no longer be voted on
If you take a 2D gaussian exp(-x2 - y2) the shape is both seperable in x and y (i.e. f(x,y) = f(x)*f(y)) and in r and theta. Because it is circularly symmetric the area (integral of the 2D function) is proportional to a circle of the same radius. Here the radius = 1 and the constant of proportionality is 1 (this also should take some thought). It is not surprising that a circularly symmetric function has an area proportional to pi. The sqrt(pi) comes from the fact that the function is also seperable in rectilinear coordinates (x,y), this means that the integral of the 2D function is just the product of the 1D functions. Since the 2D area is pi the 1D integral must be root pi.
[–][S] 2 points3 points (1 child)
sorry, this has been archived and can no longer be voted on
I just reread your comment and I semi-retract my initial response. I guess the important thing about the random walk is that it gives a solution that is exponential; exponential functions are also the only ones I can think of right now that are separable in x,y and r,theta. I think I initially didn't catch the importance of this and. Very cool, thank you!
This is strange to me still, since it is a non-obvious symmetry-- one that is only really apparent when you consider the square of the integral. I hadn't thought much before about a system in N dimensions having (or not having) symmetry in additional dimensions. I am still curious if there is a physical reason why this symmetry "must" be in this case, but I think I appreciate your answer much more now! Thanks again!
[–][S] 1 point2 points (1 child)
sorry, this has been archived and can no longer be voted on
Say I step outside my house, and once per second I take a step either to the North or to the South. After many hours, I record my final position. I do this every day, for many years, and make a histogram of my final positions.
I can use this data to find the probability, if I repeat this again, of ending back on my doorstep: (2*Pi)-(1/2)! It doesn't strike you as a little strange that Pi has come about in a walk in one dimension?
Forget the calculus: this was a result of purely physical measurements, addition and averaging. Why Pi!? Why this process!?
[–] -1 points0 points (0 children)
sorry, this has been archived and can no longer be voted on
I don't know if it's physical, but a phase change of pi gives you a plus or minus 1, so initial position, phase change of 0 you go north, phase change of pi you go south.
[–]Plasma physics -2 points-1 points (0 children)
sorry, this has been archived and can no longer be voted on
Essentially what he said, but with mathematical symbols: http://en.wikipedia.org/wiki/Gaussian_integral#By_polar_coordinates
[–][S] 7 points8 points (0 children)
sorry, this has been archived and can no longer be voted on
A nice quote I found on the subject:
"THERE IS A story about two friends, who were classmates in high school, talking about their jobs. One of them became a statistician and was working on population trends. He showed a reprint to his former classmate. The reprint started, as usual, with the Gaussian distribution and the statistician explained to his former classmate the meaning of the symbols for the actual population, for the average population, and so on. His classmate was a bit incredulous and was not quite sure whether the statistician was pulling his leg. "How can you know that?" was his query. "And what is this symbol here?" "Oh," said the statistician, "this is pi." "What is that?" "The ratio of the circumference of the circle to its diameter." "Well, now you are pushing your joke too far," said the classmate, "surely the population has nothing to do with the circumference of the circle."
[–] 7 points8 points * (0 children)
sorry, this has been archived and can no longer be voted on
Well, the (unit covariance) gaussian distribution in any dimension is rotationally symmetric about it's mean. Pi pops up when you integrate over it because it pops up when you calculate the circumference of a circle, or in general the surface of a sphere in any dimension: You could first have integrated over the surface of spheres equidistant from the mean (over which the gaussian is constant), before integrating over this distance.
Gaussians with general covariance are just linear transforms of the unit covariance case, which adds a multiplicative factor to the integral.
The 1D case above arises because spheres on the real line are still spheres, even if they consist in just 2 points, I guess...
[–] 3 points4 points * (0 children)
sorry, this has been archived and can no longer be voted on
I don't think that there has to be a physical explanation of why you get pi. I.E. many integrations that are done using complex analysis, the pi's crop up because of Cauchy's Integral Formula, which does have a geometric interpretation.
[–] 1 point2 points (0 children)
sorry, this has been archived and can no longer be voted on
Oh, what's a pi or two among friends?
[–] 4 points5 points (0 children)
sorry, this has been archived and can no longer be voted on
But typically if a Pi shows up, there is something periodic or geometrical about the problem. I can't think of why it should show up in the Gaussian integral (the area of a normal distribution in 1-D).
Well, one way to solve the problem is to switch to polar coordinates.
Frankly, I long ago stopped viewing pi as something related to geometry. Instead, I view it as a Special Number, and geometry is just one of the many disciplines it shows up in. You might as well ask "Why does pi show up in geometry? What does geometry have to do with Gaussian integrals?"
[–]Particle physics 0 points1 point * (0 children)
sorry, this has been archived and can no longer be voted on
First pi doesn't only show up in circles and trig functions and the integral of Gaussians.
Take pi/4 = 1/1 - 1/3 + 1/5 - 1/7 + ... (-1)k/(2k+1) + ... (where k is an integer) or ei pi = -1. Both of these expressions on the surface have nothing to do with circles. I mean you are taking a summation of alternating numbers, or raising e to an imaginary exponent to get -1. Both of these can be motivated by geometric arguments: (arctan(1) = pi/4) and Taylor series of arctan x, and Taylor series of ex and sin x cos x, motivates ei pi .
See for multiple derivations of integral of a Gaussian e-x2.
http://www.york.ac.uk/depts/maths/histstat/normal_history.pdf
A particularly elegant proof starts with the Euler reflection identity ( link to proof ) Gamma[x] Gamma[1 - x] = pi/sin[pi x] with x = 1/2, we get Gamma[1/2] = sqrt(pi), and from definition of Gamma[x] = int(tx-1 e-t dt, t=0, t=inf) we see Gamma[1/2] = int( t-1/2 e-t dt, t=0, inf) and we note that if we let u = sqrt(t), then du = dt/(2sqrt(t)), so Gamma[1/2] = 2 int( exp(-u2) du, t=0, inf) so Gamma[1/2] = int(exp(-u2) du, u=-inf, inf).
Looking at the proof the Euler reflection identity you see it mirrors the previous proofs. You take an expansion of Gamma (actually 1/Gamma and the expansion is a product instead of sum of an infinite number of terms), and recognize that sin(x)/x has a similar expansion. Other than trig functions are involved, I don't see what you are looking for.
Also, can you think of any real geometric problem where you can come up with an answer of sqrt(pi) ? Other than the length of a square that has the area of a circle of radius 1 (a problem that is proven impossible by straightedge and compass ... e.g., outside the realm of ordinary geometry).
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
Edit: I'll restate the question to avoid more explanations of why the integral is Pi: Say you are performing some physical experiment, and some property of the measurement you make is Pi.
Lets say you didn't have a theory, but you are measuring a π. What you'd want is a theory that results in a π. The co-incidence of the measurement producing pi and there is math saying pi is a 'special number', means that it is unlikely that the constant you are measuring is a physical one.
On the other hand, the normal distribution just has to be normalized. We could say the same about e, however both e and π have accompanying constant, π the normalization factor, and e the uncertainty exp(1/(2 σ²)). One might think e and π are just our way of dealing with the numbers correctly.
Not sure if this answer is entirely satisfying, though..
[–] 0 points1 point * (0 children)
sorry, this has been archived and can no longer be voted on
Someone probably already said this, but I didn't see it ... maybe it's really stupid ... I don't know to be honest.
I am familiar with the multi-variable method of finding the answer, and I do not claim to know the answer to your question, but I have a guess.
From Euler's formula, we know that eix = cos(x) + isin(x), so if we let x = (ix2) then I believe that this becomes cosh(x2) - sinh(x2). I'm feeling too lazy to do this integral, but womframalpha says that it is sqrt(pi)*erf(t) (if the integral is from -t -> t). Erf(inf) = 1, so it agrees with the multivariable method. Perhaps this is the link to "periodic" functions you were asking about.
EDIT: The problem with this is of course that erf is defined with e-x2 ... so ...
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
My best guess is that it is like the strange connection between the exponential function and trig a la Euler's Formula. Trig is intimately connected with circles and pi, so it might not be too surprising that pi shows up in an integral of the exponential function, or at least not more surprising than Euler's Formula in the first place.
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
I wondered this as well - I remember when I was in college taking engineering courses and we see pi all the time there. And I still never quite understood how pi can have anything to do with inductance (or with distributions, same as you).
Even the ei*pi = -1 equation. I know this is true because I know how the to apply the imaginary coordinate system, but I don't get it at a fundamental level.
[–] 0 points1 point (6 children)
sorry, this has been archived and can no longer be voted on
Because the exponential and periodic motion are linked by eix=isinx + cosx
[–] 0 points1 point (5 children)
sorry, this has been archived and can no longer be voted on
Which is related to the identity ei*pi=-1, is it not? I don't think mathematicians understand this in a periodic or geometric context, it's something of a mystery. So the answer to the OP is " I don't know."
[–] 0 points1 point (1 child)
sorry, this has been archived and can no longer be voted on
ok well then you surely know that that allows us to define sin x = (exi - e-xi) / 2
if you take the i out of this expression, you get hyperbolic sinh x = (ex - e-x) / 2
plotting these hyperbolic expressions onto a graph looks a lot like a gaussian dont it?
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
Okay, I'll buy that...so what's your point? Did I say something incorrect?
[–] 0 points1 point (2 children)
sorry, this has been archived and can no longer be voted on
I don't think mathematicians understand this in a periodic or geometric context, it's something of a mystery.
It's not a mystery at all. It is initially surprising, and it requires a lot of study to see all the implications, is all.
I understand that it's not just your own personal notion; some mathematicians are fond of using excessive hyperbole and claim that it's a mystery, but that's just their mildly disingenuous way of saying that it's amazing and that the audience should look into it further.
Too many people believe, not just that it's a mystery, but also that that means that it's a bizarre curiosity, which badly misunderstands things, for actually it is one of the most important equations in mathematics, with countless vast and deep implications.
(I have here a complex analysis textbook that says that it is the most important, which may be true, but is a harder-to-defend position.)
[–] 1 point2 points (1 child)
sorry, this has been archived and can no longer be voted on
Wow, thanks for that! My math teacher told me that :( this inspires me to go do some research!! I thank you, for you are a gentleman and a scholar.
[–] -1 points0 points (0 children)
sorry, this has been archived and can no longer be voted on
My, what an unexpectedly positive response -- probably the most enthusiastic such I've ever gotten on Reddit. Thanks for the kind words.
I'm also gratified that you're inspired to research this; it is an utterly cool topic, and one of the most mind-blowing things to have been discovered by humankind.
Once you get far enough along to start feeling that this equation is natural, and indeed just could not be any different than it is, you'll have significantly increased your mathematical intuition and conceptual understanding.
Feel free to toss out questions to /r/math as you progress in your journey.
BTW there are many approaches to it, but FWIW it typically starts with the infinite series definition of ez: exp(z) = sum(k=0 to infinity) zk/k!
[–] -1 points0 points (4 children)
sorry, this has been archived and can no longer be voted on
Pi shows up in the area of a circle. Integrals compute area under curves. Circles are curved. Therefore....
[–][S] 0 points1 point (3 children)
sorry, this has been archived and can no longer be voted on
But x2 is certainly a curve; integral of x2 has no Pi...
I think this was a good idea though!
[–] 4 points5 points (2 children)
sorry, this has been archived and can no longer be voted on
Because you don't get a circle from x2, but you do from exp(x2). When you multiply exp(x2)*exp(y2), you get exp(x2+y2) and your contours are circles.
If you multiply x2 and y2, you just get (xy)2.
[–][S] 1 point2 points (1 child)
sorry, this has been archived and can no longer be voted on
If you rotated a parabola around the origin you would certainly get circles, one at each z coordinate. I only meant that in general, things can be curved, and you can integrate them, without getting a Pi.
[–] 1 point2 points (0 children)
sorry, this has been archived and can no longer be voted on
It helps if your limits are circular. Try integrating x2+y2 in 2-D using polar coordinates over a circle and you'll get your pi.
The thing with the Gaussian integral is that since it goes to infinity, the limits can be made circular if desired.
[–] -3 points-2 points (5 children)
sorry, this has been archived and can no longer be voted on
Set up the integral, I, as follows. Define I=integral(e-a*x2)dx = integral(e-a*y2)dy (True because x and y are just variables at this point). Examine the integral I2. I2=integral(integral(e-a(x2+y2)) dx)dy) Okay this is a hard one to do. Convert to polar coordinates (I see an x2 + y2!) I2 becomes integral(integral(e-a*r2 r dr d(theta))) where theta goes 0 to 2 pi, r goes 0 to infinity.
Evaluate theta integral, a 2pi pops out in front, other integral is (-e-(a*t) )/2a evaluated at infinity minus evaluated at 0. Multiply the answer by 2pi and you get pi/a.
With me so far? Here's the trick. You found I2. So I is just sqrt(pi/a). You're done. Thanks, ME400.
[–][S] 1 point2 points (4 children)
sorry, this has been archived and can no longer be voted on
I understand how to do the integral. Did you read the original post?
[–] 0 points1 point * (3 children)
sorry, this has been archived and can no longer be voted on
Aye, x and y are just dummy variables. You're going to polar coordinates because you're using the distance r "x2 + y2!", distances in polar coordinates pop out a pi because you're looking ALL AROUND YOU (2pi)
Edit not really dummy variables, but you know what I mean. You get the pi cause you're looking everywhere around you. At least I'm pretty sure that's why if my understanding is correct. (NOT A MATH MAJOR--ENGINEER UNDERGRAD)
[–][S] 1 point2 points (2 children)
sorry, this has been archived and can no longer be voted on
You get the pi cause you're looking everywhere around you.
GAH! I understand how to do integrals, and integrals in polar coordinates, and even this exact integral in question. You seem to be missing the entire point of my question.
And for the record, "cause you're looking everywhere around you" would be how to explain a factor of 2*Pi to a two-year-old. Not totally helpful for people who have already taken and understand calculus.
[–] 6 points7 points (1 child)
sorry, this has been archived and can no longer be voted on
Well how else would you characterize the fact that you're looking everywhere around you without using pi? What kind of more indepth physical meaning are you searching for; isn't this precisely the kind of explanation that you want? You've taken and understand calculus. You know that integrals come about because of looking at small differences. I ask you--differences in WHAT, in this case? Well you have differences in space extending to infinity in all directions, and then you have differences in angle extending the range 0 to 2*pi. You could characterize it in terms of degrees, but then you're just dancing around the point. The fact of the matter is, no matter how you look at it, since there's a radial symmetry when coming up with the square of the gaussian integral you're asking about. There's no known elementary functions that intuitively give you pi for the gaussian integral, that's why you have to look at the distribution of the gaussian over x AND y, then use symmetries in the problem to get the original integral.
In short; the square if the integral you're asking about has radial symmetry and you would do it using polar coordinates, so you get pi. You claim you already knew that, so what are you asking about? "Why does pi show up in a 1 dimensional random walk?" --Because pi shows up in two dimensional random walks, and due to symmetry the 1 dimensional random walk is the square root of the two dimensional random walk.
You're asking for a physical reason of why an integral has a pi in it, then belittle an attempt to explain it using integrals and integration techniques because you understand calculus beyond my "two-year-old" explanation. What kind of explanation would satisfy you?
[–][S] 2 points3 points (0 children)
sorry, this has been archived and can no longer be voted on
OK, I am sorry because I think I was a bit rude. I apologize. But it wasn't really clear in your first two comments that you were trying to explain anything other than just how to do the integral, and I misinterpreted the way in which you were trying to explain it as possibly condescending.
Because pi shows up in two dimensional random walks, and due to symmetry the 1 dimensional random walk is the square root of the two dimensional random walk.
This part here, though, I believe is one of the most correct and concise statements on this thread. Thank you, this was very well put.
[–] -1 points0 points (0 children)
sorry, this has been archived and can no longer be voted on
[–] -3 points-2 points (12 children)
sorry, this has been archived and can no longer be voted on
If I remember correctly, it's a result of the change-of-variables you need in order to do the integral. That is, again if I'm remembering this right, square the integral, change one of the variables to 'y' so you're integrating over S S exp[-x2]*exp[-y2] dx dy. Then change to polar coord's (r, theta). From there it falls out.
[–][S] 2 points3 points (11 children)
sorry, this has been archived and can no longer be voted on
Yes, you are correct. I understand how the integral comes about. I'm wondering if there is a physical reason, though, not a mathematical one.
[–] 2 points3 points (1 child)
sorry, this has been archived and can no longer be voted on
I think you're looking for a geometrical reason, not a physical one. Geometry is part of math, and can exist outside of physics. Btw, you should post your question on the math reddit, you might get better answers. I must say I'm interested in the answer, too.
[–][S] 1 point2 points (0 children)
sorry, this has been archived and can no longer be voted on
Geometrical would be OK. Then I can at least try to come up with some reason why a random walk, etc. is analogous to the geometrical situation. Since you can have a random walk in 1-D, though, and the Pi will still show up-- I'm having a hard time making a geometric analogy.
[–] 1 point2 points * (0 children)
sorry, this has been archived and can no longer be voted on
I think the key here is translating the mathematical steps we took to do the integral into the physical situation of a random walk. With the math, we couldn't do the integral in 1D, we had to go to 2D, so we have to do that in the physical random walk too. So we randomly walk north-south, and then west-east and ask what's the probability that we ended up at the starting point after both random walks? That's the probability of ending up at the same latitude times the probability of ending up at the same longitude, which is the square of the probability for the 1D case. So we can convert the 1D random walk to the 2D without talking about calculus, just about geometry and probability.
So after that, what did we do in the math? We converted to polar coordinates. So back to the physical situation: walking north-south and then east-west is the same as just walking in any direction.
And how did the factor of pi come in? We integrated over theta from 0 to 2pi. I think what this means in the physical random walk is that it doesn't matter what angle you're at when you get back to the origin, i.e., the origin is like a really tiny circle, and it doesn't matter where on the circle you are.
Edit: that last part isn't right. But you can see there are angles/circles involved somewhere when you think about the 2D random walk in polar coords.
[–] 0 points1 point (7 children)
sorry, this has been archived and can no longer be voted on
Because the natural world follows the rules of Mathematics.
[–][S] 2 points3 points (6 children)
sorry, this has been archived and can no longer be voted on
But... Pi doesn't show up in all integrals, why this one?
[–] 0 points1 point (5 children)
sorry, this has been archived and can no longer be voted on
Because there's no theta dependence when you switch to polar coord's. I'm not aware of any physical meaning 'why' it's there. It's just a number. A useful one, but a number all the same.
[–] 1 point2 points (4 children)
sorry, this has been archived and can no longer be voted on
It's not just a number. It's a very specific number: the ratio of the circumference to the diameter of any circle! Like lisatomic, it also creeps me out to see it eerily show up in things which have, at face value, absolutely nothing to do with circles.
[–] 1 point2 points (3 children)
sorry, this has been archived and can no longer be voted on
But it may have something to do with being periodic (or at least represented by a periodic function) and Pi is all over that like white on rice in a snow storm.
[–][S] 0 points1 point (2 children)
sorry, this has been archived and can no longer be voted on
Yes, this sort of thing is what I'm getting at-- what is periodic about a normal distribution?
[–] 1 point2 points * (1 child)
sorry, this has been archived and can no longer be voted on
Yes, this sort of thing is what I'm getting at-- what is periodic about a normal distribution?
You want to know what's periodic about a normal distribution. Now, how do you get the periodic components of something? You Fourier transform it.
So what happens when you Fourier transform a normal distribution, then?
You get... Er, a Normal distribution! The Gaussian integral in particular is an eigenfunction of the Fourier transform, with eigenvalue 1...
Food for thought?
[–][S] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
I've thought of this :)
I can not continue though! Thank you, I think it may help.
|
2014-08-23 13:17:39
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8015068173408508, "perplexity": 815.6779274788919}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500826016.5/warc/CC-MAIN-20140820021346-00254-ip-10-180-136-8.ec2.internal.warc.gz"}
|
http://quant.stackexchange.com/questions/4073/roc-difference-between-discrete-and-continuous
|
# ROC: difference between discrete and continuous?
Using the ROC function in the R package TTR, there is a choice between continuous (the default) and discrete, but with no guidance on which you choose when. In the code the difference is:
roc <- x/lag(x) - 1
versus:
roc <- diff(log(x))
I admit my maths is weak but aren't they the same thing?
cbind(ROC(x,type='continuous'),ROC(x,type='discrete'),log(x))
gives:
2012-08-16 19:00:00 NA NA 8.673855
2012-08-17 07:00:00 0.00008549566 0.00008549932 8.673940
2012-08-17 08:00:00 0.00000000000 0.00000000000 8.673940
2012-08-17 09:00:00 -0.00085528572 -0.00085492006 8.673085
2012-08-17 10:00:00 0.00034220207 0.00034226063 8.673427
2012-08-17 11:00:00 -0.00102695773 -0.00102643059 8.672400
There is a subtle difference, but is it a real difference or an artifact of floating point calculation?
It seems like Quantmod: what's the difference between ROC(Cl(SPY)) and ClCl(SPY) is almost asking the same thing. But the answers there seem to be saying that with one you would sum the returns, and with the other you multiply them. That is clearly not going to be the case for the above numbers.
(BTW, no-one answered his question (in the comments) as to which form is expected by the PerformanceAnalytics package, which might have given a clue as to which you choose when.)
Here is the test data for the above:
structure(c(5848, 5848.5, 5848.5, 5843.5, 5845.5, 5839.5), class = c("xts",
"zoo"), .indexCLASS = c("POSIXct", "POSIXt"), .indexTZ = "", tclass = c("POSIXct",
"POSIXt"), tzone = "", index = structure(c(1345143600, 1345186800,
1345190400, 1345194000, 1345197600, 1345201200), tzone = "", tclass = c("POSIXct",
"POSIXt")), .Dim = c(6L, 1L), .Dimnames = list(NULL, "Close"))
-
PerformanceAnalytics can use either discrete or log returns. It can also perform simple or geometric chaining. These two things are easily confused. If you have log returns, you almost certainly want to use simple chaining. If you have discrete (often called simple) returns, you may want either geometric chaining or simple chaining, depending on your input data and the analysis that you want to do. – Brian G. Peterson Sep 4 '12 at 15:58
The difference is not an artifact of floating point arithmetic; it's a difference in compounding frequency. The returns in your example are fairly close to zero, so they don't look that different. Larger changes in price will cause larger differences between the two calculation methods.
Pat Burns wrote a nice blog post about the difference between arithmetic and log returns called, A tale of two returns. I suggest you read the entire thing, but the relevant portions are:
• log returns are always smaller than simple returns
• simple returns aggregate across assets
• log returns aggregate across time
• The log return of a short position is the negative of the log return of the long position. The relationship of the simple return of a short position relative to that of the long position is a little more complicated: $-R / (R + 1)$
So, the difference between summing and multiplying the returns clearly is a big difference between the two methods. Regarding PerformanceAnalytics, it seems that most of the functions assume arithmetic returns. Remember, you have the source code, so you can always see the exact calculations being used to generate each functions' results.
-
Thanks Joshua. The article was also very useful in pointing out the terminology, which was where a lot of my confusion had come from! (discrete == simple == arithmetic) (continuous == log == geometric) – Darren Cook Sep 4 '12 at 0:56
BTW, re: PerformanceAnalytics, "most of the functions assume arithmetic returns". I now notice user508's comment at quant.stackexchange.com/a/1082/1587 says "PerformanceAnalytics defaults to log returns." :-( – Darren Cook Sep 4 '12 at 1:00
@DarrenCook: user508 is probably referring to Return.Calculate and Calculate.Returns. Most PerformanceAnalytics functions that take a return vector have default to simple returns (and therefore geometric chaining). – Joshua Ulrich Sep 4 '12 at 15:04
The difference is real, though it is very small if your return on capital is small. Let's say value of your asset went up from 10.03 to 10.05:
Here is my python code:
>>> from math import log
>>> 10.05 / 10.03 - 1
0.001994017946161719
>>> log(10.05) - log(10.03)
0.001992032531240806
The ROC is small, and difference between two methods is small. But if the price of your stock went up from 100 to 500:
>>> 500.0/100.0 - 1
4.0
>>> log(500) - log(100)
1.6094379124340996
ROC is large, and the difference between the methods is large.
-
Thanks, that is a very clear example. – Darren Cook Sep 4 '12 at 1:00
|
2015-07-07 04:41:57
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5807781219482422, "perplexity": 2515.644890350536}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375098990.43/warc/CC-MAIN-20150627031818-00167-ip-10-179-60-89.ec2.internal.warc.gz"}
|
http://mathhelpforum.com/trigonometry/189160-sinusoidal-functions-print.html
|
# Sinusoidal Functions
• Sep 29th 2011, 06:07 PM
lakers784
Sinusoidal Functions
Hi i need help with this problem my test is tomorrow and i am trying to figure out if i am right i have attached the file and the problem is number 13
i got y=-1+8cos pi/35(x-5)
• Sep 29th 2011, 06:14 PM
pickslides
Re: Sinusoidal Functions
Sounds fair to me.
• Oct 1st 2011, 05:06 AM
niggawut
yes you are absolutely correct (Rofl)
no kidding, I've worked it out.
$y = ({8})( {cos\frac{\pi(x-5)}{\35})} -1$ (Rofl)
|
2017-08-18 22:06:28
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7003374099731445, "perplexity": 3886.720049812781}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886105187.53/warc/CC-MAIN-20170818213959-20170818233959-00707.warc.gz"}
|
https://minimalsurfaces.blog/home/repository/triply-periodic/deformed-h/
|
The H-surfaces of Hermann Amandus Schwarz form a natural 1-parameter family. There is a surprising deformation into a 2-dimensional family. To explain the construction, we turn an H-surface sideways, and cut it along vertical symmetry planes and horizontal lines into a minimal octagon that still has a normal symmetry about its center:
This octagon is the image of the shaded rectangle below that shows the divisor of the (squared) Gauss map. The points a, a’, b, b’ correspond to the vertices, where the Gauss map has its zeroes and poles.
They need to satisfy a+a’=1/2=b+b’, and there is a period condition that determines the rectangle height in terms of a and b. By changing the values of a and b, we can slide the inward and outward necks up and down. Let’s first move them so that they line up:
When a+b=1/2, they are perfectly aligned, and we have an additional horizontal symmetry plane. There is no period condition in this case. In fact, these surfaces are known as special surfaces in the oP deformation of the Schwarz P surface. Hence we found a way to deform an H surface into the P surface. This is remarkable, because the H surface does not belong to the explicit 5-dimensional Meeks family that contains most known examples of embedded triply periodic minimal surfaces of genus 3.
Above are two extreme cases of this new deformation. The left image arises when a and b are close: We obtain singly periodic Scherk surfaces over a rhombic pattern. The right image arises when b approaches .5, here we see doubly periodic KMR surfaces as limits.
Resources
Mathematica Notebook
PoVRay Sources
H. Chen and M. Weber: An orthorhombic deformation family of Schwarz’ H surfaces
|
2023-03-23 05:18:50
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8601784110069275, "perplexity": 566.8567681083025}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296944996.49/warc/CC-MAIN-20230323034459-20230323064459-00460.warc.gz"}
|
https://physics.stackexchange.com/questions/173248/nabla-mu-nabla-mu-in-general-relativity
|
# $\nabla^{\mu}\nabla_{\mu}$ in general relativity [closed]
I am trying to work out $\square=\nabla^{\mu}\nabla_{\mu}$ in the metric
$ds^{2}=-A(r)dt^{2}+B(r)^{-1}dr^{2}+r^{2}d\Omega^{2} $$My work: when applying \square to a scalar \phi, then \square\phi=\nabla^{\mu}\nabla_{\nu}\phi=\nabla^{\mu}\partial_{\mu}\phi=g^{\mu\nu}\nabla_{\nu}\partial_{\mu}\phi=g^{\mu\nu}(\partial_{\nu}\partial_{\mu}-\Gamma^{\lambda}_{\mu\nu}\partial_{\lambda})\phi Christoffel symbol \left( \begin{array}{ccc} \left\{0,\frac{A'(r)}{2 A(r)},0\right\} & \left\{\frac{A'(r)}{2 A(r)},0,0\right\} & \{0,0,0\} \\ \left\{\frac{1}{2} B(r) A'(r),0,0\right\} & \left\{0,-\frac{B'(r)}{2 B(r)},0\right\} & \{0,0,-r B(r)\} \\ \{0,0,0\} & \left\{0,0,\frac{1}{r}\right\} & \left\{0,\frac{1}{r},0\right\} \end{array} \right) substituting the metric and affine values in the equation above, my answer came to be$$ \square=-A(r)^{-1}\frac{d^{2}}{dt^{2}}+B\left(\frac{d^{2}}{dr^{2}}+\frac{1}{r}\frac{d}{dr}\right)+\frac{1}{2}\left(B^{\prime}+\frac{B A^{\prime}}{A}\right)\frac{d}{dr} $$However, the answer happens to be$$ \square=-A(r)^{-1}\frac{d^{2}}{dt^{2}}+B\left(\frac{d^{2}}{dr^{2}}+\frac{2}{r}\frac{d}{dr}\right)+\frac{1}{2}\left(B^{\prime}+\frac{B A^{\prime}}{A}\right)\frac{d}{dr}$$Could someone please show me where the third comes from in the second term? ## closed as off-topic by user10851, ACuriousMind♦, Kyle Kanos, JamalS, JimMar 31 '15 at 16:01 This question appears to be off-topic. The users who voted to close gave this specific reason: • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Community, ACuriousMind, Kyle Kanos, JamalS, Jim If this question can be reworded to fit the rules in the help center, please edit the question. • Could you include the work you have done on it? Show us your steps? But be forewarned, "check my work" questions (if this turns out to be one of them) are off-topic here – Jim Mar 30 '15 at 22:25 • Jimnosperm, Its actually from a research paper in arxiv, not a homewor problem arxiv.org/pdf/1001.4157v3 – MrDi Mar 30 '15 at 22:27 • @435145 If you read the homework policy, it doesn't matter if it is homework or not, just if it is homework like – Jimmy360 Mar 30 '15 at 22:35 • (I'm guessing$\phi$has no angular dependence...) Ignore the time part, and pretend$B(r)=1$. Then, you've got the standard spatial metric in spherical coordinates. So you should get something like the standard Laplacian in spherical coordinates, which has that$2$. Looks like you screwed something up in calculating the Christoffel symbols. – Mike Mar 30 '15 at 22:46 • where the factor of two come from? – MrDi Mar 30 '15 at 23:21 ## 2 Answers Mm.... At first I repeated your calculation and got the same answer as yours, then I checked the paper you gave and found it consist with (32 a) and it seems not a typo, so I read it from begining - oh brother it's not 2+1 gravity - -b it's 3+1 gravity and you should treat$d\Omega^2$more carefully:$r^2 d\Omega^2=r^2d\theta^2+r^2\sin^2\theta d\phi^2\$
so you get one more r factor when using the formula @Prahar gives.
and btw the Christoffel symbol is for 2+1 and certainly wrong for 3+1.
I'll prove a formula that is probably easier to use for this. $$\begin{split} \frac{1}{\sqrt{-g}} \partial_\mu \left( \sqrt{-g} g^{\mu\nu} \partial_\nu \phi \right) &= \frac{1}{\sqrt{-g}} \partial_\mu \left( \sqrt{-g} \right) g^{\mu\nu} \partial_\nu \phi + \partial_\mu \left( g^{\mu\nu} \partial_\nu \phi \right) \\ &= \frac{1}{2g} \partial_\mu g g^{\mu\nu} \partial_\nu \phi + \partial_\mu g^{\mu\nu} \partial_\nu \phi + g^{\mu\nu} \partial_\mu \partial_\nu \phi \\ &= - \frac{1}{2} g^{\alpha\mu} g^{\nu\beta} \left[ \partial_\mu g_{\alpha\beta} + \partial_\alpha g_{\mu\beta} - \partial_\beta g_{\alpha\mu} \right] \partial_\nu \phi + g^{\mu\nu} \partial_\mu \partial_\nu \phi \\ &= g^{\mu\nu} \left( \partial_\mu \partial_\nu - \Gamma^\lambda_{\mu\nu} \partial_\lambda \phi \right) \\ &= \nabla^\mu \nabla_\mu \phi \end{split}$$
|
2019-10-16 21:48:39
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.951348602771759, "perplexity": 714.044934904075}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986670928.29/warc/CC-MAIN-20191016213112-20191017000612-00102.warc.gz"}
|
https://cyclostationary.blog/tag/cyclostationary-signal-processing/
|
## Blog Notes: New Page with All CSP Blog Posts in Chronological Order
To aid navigating the CSP Blog, I’ve added a new page called “All CSP Blog Posts.” You can find the page link at the top of the home page, or in various lists on the right side of the Blog, such as “Pages” and “Site Navigation.”
Let me know in the Comments if there are other ways that you think I can improve the usability of the site.
## All BPSK Signals
Update June 2020
I’ll be adding new papers to this post as I find them. At the end of the original post there is a sequence of date-labeled updates that briefly describe the relevant aspects of the newly found papers.
## Professor Jang Again Tortures CSP Mathematics Until it Breaks
We first met Professor Jang in a “Comments on the Literature” type of post from 2016. In that post, I pointed out fundamental mathematical errors contained in a paper the Professor published in the IEEE Communications Letters in 2014 (The Literature [R71]).
I have just noticed a new paper by Professor Jang, published in the journal IEEE Access, which is a peer-reviewed journal, like the Communications Letters. This new paper is titled “Simultaneous Power Harvesting and Cyclostationary Spectrum Sensing in Cognitive Radios” (The Literature [R144]). Many of the same errors are present in this paper. In fact, the beginning of the paper, and the exposition on cyclostationary signal processing is nearly the same as in The Literature [R71].
Let’s take a look.
## Symmetries of Higher-Order Temporal Probabilistic Parameters in CSP
In this post, we continue our study of the symmetries of CSP parameters. The second-order parameters–spectral correlation and cyclic correlation–are covered in detail in the companion post, including the symmetries for ‘auto’ and ‘cross’ versions of those parameters.
Here we tackle the generalizations of cyclic correlation: cyclic temporal moments and cumulants. We’ll deal with the generalization of the spectral correlation function, the cyclic polyspectra, in a subsequent post. It is reasonable to me to focus first on the higher-order temporal parameters, because I consider the temporal parameters to be much more useful in practice than the spectral parameters.
This topic is somewhat harder and more abstract than the second-order topic, but perhaps there are bigger payoffs in algorithm development for exploiting symmetries in higher-order parameters than in second-order parameters because the parameters are multidimensional. So it could be worthwhile to sally forth.
## New Look for a New Year and New Decade
2020 is the fifth full year of existence for the CSP Blog, and the beginning of a new decade that will be full of CSP explorations. I thought I’d freshen up the look of the Blog, so I’ve switched the theme. It is a cleaner look with fewer colors and no more hexagons. I’m not completely happy with it, so I might change it yet again. Let me know if you have problems viewing the content or posting a comment (cmspooner at ieee dot org).
Happy New Year to all my readers!
## Symmetries of Second-Order Probabilistic Parameters in CSP
As you progress through the various stages of learning CSP (intimidation, frustration, elucidation, puzzlement, and finally smooth operation), the symmetries of the various functions come up over and over again. Exploiting symmetries can result in lower computational costs, quicker debugging, and easier mathematical development.
What exactly do we mean by ‘symmetries of parameters?’ I’m talking primarily about the evenness or oddness of the time-domain functions in the delay $\tau$ and cycle frequency $\alpha$ variables and of the frequency-domain functions in the spectral frequency $f$ and cycle frequency $\alpha$ variables. Or a generalized version of evenness/oddness, such as $f(-x) = g(x)$, where $f(x)$ and $g(x)$ are closely related functions. We have to consider the non-conjugate and conjugate functions separately, and we’ll also consider both the auto and cross versions of the parameters. We’ll look at higher-order cyclic moments and cumulants in a future post.
You can use this post as a resource for mathematical development because I present the symmetry equations. But also each symmetry result is illustrated using estimated parameters via the frequency smoothing method (FSM) of spectral correlation function estimation. The time-domain parameters are obtained from the inverse transforms of the FSM parameters. So you can also use this post as an extension of the second-order verification guide to ensure that your estimator works for a wide variety of input parameters.
## The Ambiguity Function and the Cyclic Autocorrelation Function: Are They the Same Thing?
Let’s talk about ambiguity and correlation. The ambiguity function is a core component of radar signal processing practice and theory. The autocorrelation function and the cyclic autocorrelation function, are key elements of generic signal processing and cyclostationary signal processing, respectively. Ambiguity and correlation both apply a quadratic functional to the data or signal of interest, and they both weight that quadratic functional by a complex exponential (sine wave) prior to integration or summation.
Are they the same thing? Well, my answer is both yes and no.
## CSP Resources: The Ultimate Guides to Cyclostationary Random Processes by Professor Napolitano
My friend and colleague Antonio Napolitano has just published a new book on cyclostationary signals and cyclostationary signal processing:
Cyclostationary Processes and Time Series: Theory, Applications, and Generalizations, Academic Press/Elsevier, 2020, ISBN: 978-0-08-102708-0. The book is a comprehensive guide to the structure of cyclostationary random processes and signals, and it also provides pointers to the literature on many different applications. The book is mathematical in nature; use it to deepen your understanding of the underlying mathematics that make CSP possible.
You can check out the book on amazon.com using the following link:
Cyclostationary Processes and Time Series
## On Impulsive Noise, CSP, and Correntropy
I’ve seen several published and pre-published (arXiv.org) technical papers over the past couple of years on the topic of cyclic correntropy (The Literature [R123-R127]). I first criticized such a paper ([R123]) here, but the substance of that review was about my problems with the presented mathematics, not impulsive noise and its effects on CSP. Since the papers keep coming, apparently, I’m going to put down some thoughts on impulsive noise and some evidence regarding simple means of mitigation in the context of CSP. Preview: I don’t think we need to go to the trouble of investigating cyclic correntropy as a means of salvaging CSP from the clutches of impulsive noise.
## A Gallery of Cyclic Correlations
There are some situations in which the spectral correlation function is not the preferred measure of (second-order) cyclostationarity. In these situations, the cyclic autocorrelation (non-conjugate and conjugate versions) may be much simpler to estimate and work with in terms of detector, classifier, and estimator structures. So in this post, I’m going to provide plots of the cyclic autocorrelation for each of the signals in the spectral correlation gallery post. The exceptions are those signals I called feature-rich in the spectral correlation gallery post, such as LTE and radar. Recall that such signals possess a large number of cycle frequencies, and plotting their three-dimensional spectral correlation surface is not helpful as it is difficult to interpret with the human eye. So for the cycle-frequency patterns of feature-rich signals, we’ll rely on the stem-style (cyclic-domain profile) plots in the gallery post.
## On The Shoulders
What modest academic success I’ve had in the area of cyclostationary signal theory and cyclostationary signal processing is largely due to the patient mentorship of my doctoral adviser, William (Bill) Gardner, and the fact that I was able to build on an excellent foundation put in place by Gardner, his advisor Lewis Franks, and key Gardner students such as William (Bill) Brown.
## Simple Synchronization Using CSP
In this post I discuss the use of cyclostationary signal processing applied to communication-signal synchronization problems. First, just what are synchronization problems? Synchronize and synchronization have multiple meanings, but the meaning of synchronize that is relevant here is something like:
syn·chro·nize: To cause to occur or operate with exact coincidence in time or rate
If we have an analog amplitude-modulated (AM) signal (such as voice AM used in the AM broadcast bands) at a receiver we want to remove the effects of the carrier sine wave, resulting in an output that is only the original voice or music message. If we have a digital signal such as binary phase-shift keying (BPSK), we want to remove the effects of the carrier but also sample the message signal at the correct instants to optimally recover the transmitted bit sequence.
## 100,000 Page Views!
The CSP Blog has reached 100,000 page views! Also, a while back it passed the “20,000 visitors” milestone. All of this for 53 posts and 10 pages. More to come!
I started the CSP Blog in late 2015, so it has taken a bit over three years to get to 100,000 views. I don’t know if that should be considered fast or slow. But I like it anyway.
I want to thank each and every one of the visitors to the CSP Blog. It has reached so many more people that I though it ever would when I started it.
Below the fold, some graphics that show the vital statistics of the CSP Blog as of the 100,000 page-view milestone.
## Can a Machine Learn a Power Spectrum Estimator?
I continue with my foray into machine learning (ML) by considering whether we can use widely available ML tools to create a machine that can output accurate power spectrum estimates. Previously we considered the perhaps simpler problem of learning the Fourier transform. See here and here.
Along the way I’ll expose my ignorance of the intricacies of machine learning and my apparent inability to find the correct hyperparameter settings for any problem I look at. But, that’s where you come in, dear reader. Let me know what to do!
## Data Set for the Machine-Learning Challenge
Update July 2020. I originally posted $20,000$ signals in the posted data set. I’ve now added another $92,000$ for a total of $112,000$ signals. The original signals are contained in Batches 1-5, the additional signals in Batches 6-28. I’ve placed these additional Batches at the end of the post to preserve the original post’s content.
I’ve posted $20000$ PSK/QAM signals to the CSP Blog. These are the signals I refer to in the post I wrote challenging the machine-learners. In this brief post, I provide links to the data and describe how to interpret the text file containing the signal-type labels and signal parameters.
### Overview of Data Set
The $20,000$ signals are stored in five zip files, each containing $4000$ individual signal files:
Batch 1
Batch 2
Batch 3
Batch 4
Batch 5
The zip files are each about 1 GB in size.
The modulation-type labels for the signals, such as “BPSK” or “MSK,” are contained in the zipped text file:
signal_record_first_20000.txt.zip
Each signal file is stored in a binary format involving interleaved real and imaginary parts, which I call ‘.tim’ files. You can read a .tim file into MATLAB using read_binary.m. Or use the code inside read_binary.m to write your own data-reader; the format is quite simple.
### The Label and Parameter File
Let’s look at the format of the truth/label file. The first line of signal_record_first_20000.txt is
1 bpsk 11 -7.4433467080e-04 9.8977795076e-01 10 9 5.4532617590e+00 0.0
which comprises $9$ fields. All temporal and spectral parameters (times and frequencies) are normalized with respect to the sampling rate. In other words, the sampling rate can be taken to be unity in this data set. These fields are described in the following list:
1. Signal index. In the case above this is `1′ and that means the file containing the signal is called signal_1.tim. In general, the $n$th signal is contained in the file signal_n.tim. The Batch 1 zip file contains signal_1.tim through signal_4000.tim.
2. Signal type. A string indicating the modulation format of the signal in the file. For this data set, I’ve only got eight modulation types: BPSK, QPSK, 8PSK, $\pi/4$-DQPSK, 16QAM, 64QAM, 256QAM, and MSK. These are denoted by the strings bpsk, qpsk, 8psk, dqpsk, 16qam, 64qam, 256qam, and msk, respectively.
3. Base symbol period. In the example above (line one of the truth file), the base symbol period is $T_0 = 11$.
4. Carrier offset. In this case, it is $-7.4433467080\times 10^{-4}$.
5. Excess bandwidth. The excess bandwidth parameter, or square-root raised-cosine roll-off parameter, applies to all of the signal types except MSK. Here it is $9.8977795076\times 10^{-1}$. It can be any real number between $0.1$ and $1.0$.
6. Upsample factor. The sixth field is an upsampling parameter U.
7. Downsample factor. The seventh field is a downsampling parameter D. The actual symbol rate of the signal in the file is computed from the base symbol period, upsample factor, and downsample factor: $\displaystyle f_{sym} = (1/T_0)*(D/U)$. So the BPSK signal in signal_1.tim has rate $0.08181818$. If the downsample factor is zero in the truth-parameters file, no resampling was done to the signal.
8. Inband SNR (dB). The ratio of the signal power to the noise power within the signal’s bandwidth, taking into account the signal type and the excess bandwidth parameter.
9. Noise spectral density (dB). It is always $0$ dB. So the various SNRs are generated by varying the signal power.
To ensure that you have correctly downloaded and interpreted my data files, I’m going to provide some PSD plots and a couple of the actual sample values for a couple of the files.
### signal_1.tim
The line from the truth file is:
1 bpsk 11 -7.4433467080e-04 9.8977795076e-01 10 9 5.4532617590e+00 0.0
The first ten samples of the file are:
-5.703014e-02 -6.163056e-01
-1.285231e-01 -6.318392e-01
6.664069e-01 -7.007506e-02
7.731103e-01 -1.164615e+00
3.502680e-01 -1.097872e+00
7.825349e-01 -3.721564e-01
1.094809e+00 -3.123962e-01
4.146149e-01 -5.890701e-01
1.444665e+00 7.358724e-01
-2.217039e-01 -1.305001e+00
An FSM-based PSD estimate for signal_1.tim is:
And the blindly estimated cycle frequencies (using the SSCA) are:
The previous plot corresponds to the numerical values:
Non-conjugate $(\alpha, C, S)$:
8.181762695e-02 7.480e-01 5.406e+00
Conjugate $(\alpha, C, S)$:
8.032470942e-02 7.800e-01 4.978e+00
-1.493096002e-03 8.576e-01 1.098e+01
-8.331298083e-02 7.090e-01 5.039e+00
### signal_4000.tim
The line from the truth file is
4000 256qam 9 8.3914849139e-04 7.2367959637e-01 9 8 1.0566301192e+01 0.0
which means the symbol rate is given by $(1/9)*(8/9) = 0.09876543209$. The carrier offset is $0.000839$ and the excess bandwidth is $0.723$. Because the signal type is 256QAM, it has a single (non-zero) non-conjugate cycle frequency of $0.098765$ and no conjugate cycle frequencies. But the square of the signal has cycle frequencies related to the quadrupled carrier:
### Final Thoughts
Is $20000$ waveforms a large enough data set? Maybe not. I have generated tens of thousands more, but will not post until there is a good reason to do so. And that, my friends, is up to you!
That’s about it. I think that gives you enough information to ensure that you’ve interpreted the data and the labels correctly. What remains is experimentation, machine-learning or otherwise I suppose. Please get back to me and the readers of the CSP Blog with any interesting results using the Comments section of this post or the Challenge post.
For my analysis of a commonly used machine-learning modulation-recognition data set (RML), see the All BPSK Signals post.
Batch 6
Batch 7
Batch 8
Batch 9
Batch 10
Batch 11
Batch 12
Batch 13
Batch 14
Batch 15
Batch 16
Batch 17
Batch 18
Batch 19
Batch 20
Batch 21
Batch 22
Batch 23
Batch 24
Batch 25
Batch 26
Batch 28
Signal parameters text file
## How we Learned CSP
This post is just a blog post. Just some guy on the internet thinking out loud. If you have relevant thoughts or arguments you’d like to advance, please leave them in the Comments section at the end of the post.
How did we, as people not machines, learn to do cyclostationary signal processing? We’ve successfully applied it to many real-world problems, such as weak-signal detection, interference-tolerant detection, interference-tolerant time-delay estimation, modulation recognition, joint multiple-cochannel-signal modulation recognition (My Papers [25,26,28,38,43]), synchronization (The Literature [R7]), beamforming (The Literature [R102,R103]), direction-finding (The Literature [R104-R106]), detection of imminent mechanical failures (The Literature [R017-R109]), linear time-invariant system identification (The Literature [R110-R115]), and linear periodically time-variant filtering for cochannel signal separation (FRESH filtering) (My Papers [45], The Literature [R6]).
How did this come about? Is it even interesting to ask the question? Well, it is to me. I ask it because of the current hot topic in signal processing: machine learning. And in particular, machine learning applied to modulation recognition (see here and here). The machine learners want to capitalize on the success of machine learning applied to image recognition by directly applying the same sorts of image-recognition techniques to the problem of automatic type-recognition for human-made electromagnetic waves.
## Useful Signal Processing Blogs or Websites?
Update November 1, 2018: A site called feedspot (blog.feedspot.com) contacted me to tell me I made their “Top 10 Digital Signal Processing Blogs, Websites & Newsletters in 2018” list. Weirdly, there are only eight blogs in the list. What’s most important for this post is the other signal processing blogs on the list. So check it out if you are looking for other sources of online signal processing information. Enjoy! blog.feedspot.com/digital_signal_processing_blogs
*** *** ***
Some of my CSP posts get a lot of comments asking for help, and that’s a good thing. I continue to try to help readers to help themselves. Throughout my posts, I link terms and methods to webpages that provide tutorial or advanced information, and most of the time that means wikipedia.
But I’d like to be able to refer readers to good websites that discuss related aspects of signal processing and communication signals, such as filtering, spectrum estimation, mathematical models, Fourier analysis, etc. I’ve had little success with the Google searches I’ve tried.
|
2020-07-16 16:02:29
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 62, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5903106927871704, "perplexity": 1757.2821483343341}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657172545.84/warc/CC-MAIN-20200716153247-20200716183247-00224.warc.gz"}
|
http://www.gradesaver.com/founding-brothers/q-and-a/chapter-one-the-duel-302453
|
# Chapter One, The Duel?
Chapter one, thesis
In order to understand the true significance and aftermath of the duel between Hamilton and Burr, one must first consider the personalities of the assailants, and the argument that brought them to that fateful place.
|
2017-10-19 07:44:47
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8136900067329407, "perplexity": 3425.6541766595346}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823255.12/warc/CC-MAIN-20171019065335-20171019085335-00094.warc.gz"}
|
http://spoadfood.blogspot.co.uk/
|
## Tuesday, 10 January 2017
### SONY MDS-JB930
I bought one of these again the other day - actually, it was way back in June - just not been keeping up with blogging very well!. The one I already had was so good, and so I decided to get another one.
The application is marginally different this time around however, as I've got it accepting a digital out signal from my desktop computer.
This means that the unit acts as a DAC, accepting a digital stereo signal and piping it out to an amplifier, optionally recording the signal onto MiniDisc as it happens.
It's a sturdy little unit, and convenient to be able to record snippets of audio as they happen. What's more, with some thought to connections, it acts as functional junction-box between multiple components in a hifi-system, bridging digital and analogue technologies cleanly and without effort.
Interestingly, I've also started incorporating a small mixer unix into my setup so I can control relative volume levels between different separates, as well as gaining additional flexibility on what I can pipe through the system - this might be microphone inputs, or mixing signals from multiple units at once.
I had been considering finding a cheap DAT unit to try alongside, but am yet to be fully convinced - it would also be nice to try and find a matching amplifier and (eventually) connect a DAB radio unit I'd picked up around the same time.
The last six months or so have been exceedingly busy, so lots of things to catch-up on.
## Friday, 29 January 2016
### Some thoughts on Vectorisation
Last post, I mentioned something about vectorising data.
What does this mean, and why do it?
A vector is a measurement with both a magnitude and a direction. Sometimes the directions are described in terms of a bearing, but it's also conventional to determine a vector as a tuple of values in a series of dimensions, each of which is (normally) considered as being at right-angles to one another in a coordinate system.
The nice thing about vectors in an n-dimensional coordinate system is that they can be manipulated and worked with using linear algebra, enabling the computerisation of functions such as finding nearest matches, categorisation, clustering, network analysis, search algorithms and other useful techniques.
Let's take an example - say I've got the latitude and longitude coordinates of a group of people who are out with their mobile phones. I could calculate their relative distances and use this to pair them off with their nearest counterparts (subject to there being an even number of them, and ignoring tricky 3-way type situations for now). That's basic employment of Pythagoras' theorem.
$$h^2=a^2+b^2$$
And then finding the minimum value for each pair.
Using slightly more advanced mathematics, we could group these pairs into clusters, or find the line that best splits them into two, three, or more groups - it's all just an extension of Pythagoras (ok, sorry Linear Algebra folks, I know, that is an over-simplification)
Now consider someone typing in "Piefegoras" into a Google-search. How might Google be able to get from that munged input request yet still relay the results for Pythagoras?
One way might be to "vectorise" the english language. Take each of the 1,000,000 or so English words, and break them down into a 26-dimensional vector space where each dimension represents the number of letters corresponding to that dimension. So the word "aardvark" for example might be represented as the vector (3,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,2,0,0,0,1,0,0,0,0) - and the word "vector" would be represented as the vector (0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,1,0,1,0,1,0,0,0,0) . Now in both these examples, there are 26 dimensions, so we might have to tweak Pythagoras a little bit - but it turns out that the distance between the two can be determined using: $$h^2 = a^2+b^2+c^2+d^2+e^2+f^2+ ... +w^2+x^2+y^2+z^2$$
Where each of (a, b, c,...,x, y, z) is the difference between that letter's vector components. So for example, the a value in the above formula would be 3 since there are 3 a's in "aardvark" and none in "vector". So already, we can map out all the words in the dictionary, and indeed all the words not in the dictionary (I looked up "Piefegoras" and it's not there...yet) that use the 26-characters we have in the English alphabet.
Once we can build a map of this vector space, we can do all the same things we did with the people and their 2d latitude and longitude values - cluster words that consist of similar letter counts - indeed, we can use this to classify all words that share the same numbers of letters, i.e. anagrams, or that are 1 letter different, or 2 letters different, and so on - basically, half the puzzles page at the back of the newspaper becomes trivial from now on.
But not only this, we now have a viable spell-checker, or as in the example, a "Did you mean.."-er that will take garbled input and be able to find a nearest neighbour and offer that as a suggestion.
This is a simplification, our initial set of 26 dimensions is ok, but doesn't make the distinction between say, "hatred" from "thread", or "melons" from "lemons". So we need to include an additional set of vectors that are sensitive to letter placement and order, such a set might be those vectors that correspond to each letter pair, such as ("aa", "ab", "ac", ..., "zx", "zy", "zz") this is an impressively large vector-set, consisting of an additional 676 dimensions to a vector space consisting of a total of 702 dimensions. This makes the maths a little more unwieldy, and the storage requirements higher, but hey, it's 2016 and we've got computers. And why stop there, it may be advantageous to include 3-tuples of letters in the set ("aaa", "aab", "aac", ... " "zzx", "zzy", "zzz") or 4-tuples, or 5-tuples and so on - each additional n-tuple exponentially increasing the size of the dimensions under review - so it makes sense to find a sensible balance-point where the additional expense doesn't provide enough utility to warrant the additional complexity. My guess this is around 3-tuples, but 2 is probably enough for most jobs.
So with this improved vector scheme we have, in purely algorithmic terms, a really simple way of breaking down any series of alphabetic characters and comparing its "distance" from any other series of alphabetic characters. The concept of "distance" here is actually very close to our human concept of "similarity" at least in terms of the alphabetic building blocks of two words and that's actually quite powerful.
What's more, by simply extending the input conditions we can include numerics, "foreign" characters, shapes, even pictograms - and still be able to identify closeness. It would be interesting to put something like this together, and see if, using clustering and a bit of human guidance (though it'd be interesting to try otherwise) such a system might be able to help classify English words in terms of their origin. English words from French for example, might share a particular set of features like the inclusion of perhaps ("-re", "-aux", "-ois") perhaps, while Germanic or old English words might contain ( "-ang", "-tz", "-gh" ) or similar - I must confess, I'm no linguist. But, the concept ought to yield if nothing else, a series of cluster-words, the provenance of which, I'd wager, might be shared.
Another fun thing to consider is that we might be able to establish theoretical words - imagine a feature space consisting of all 1-tuple, 2-tuple, 3, 4, 5 and 6-tuple dimensions. If we then populated that space will all the words in the dictionary, from a text, or traweled from the internet, we'd be able to find the most "average" word by picking the one that was most central. Or, more spookily, calculating the most central location in the feature-space, and re-constructing the word that might appear there - despite it never have existed in any dictionary, been spelt out in a forum or commited to paper, print or screen. Performing this reconstructive process - essentially performing the transformation backwards from a location in the feature space is tricky, but if the space were constructed formally enough, this would be possible.
The downsides of this approach, especially if you want to build in the reversibility feature is that you need a massively wide set of dimensions. There are some clever ways to keep these down to a minimum that I might go into later - and there are also some problems inherent in this approach when you're trying to calculate distances between more than say 100 vectors at a time - but we can talk about these later.
To summarise then, this is a technique for converting a one-dimensional string of objects each of which belongs to a finite set (i.e. an alphabet) into an n-dimensional feature-space, where n is the size of the alphabet plus n² for 2-tuples, plus n³ for 3-tuples and so on, for the number of x-tuples to be included to take care of the anagram problem, the purpose of which is to "vectorise" such one-dimensional strings in order to perform "similarity" functions upon them.
It's worth noting here that these techniques are not new, nor particularly original - indeed they're well known and widely publicised. I think they're interesting though, and hope the reader finds something of interest in my discussing them.
I'm working on some code examples at the moment which I'll share along with results once I'm happy.
## Wednesday, 9 December 2015
### Data and Change
It's an interesting time, right now - I'm not entirely sure I've got a handle on it.
Since Paris, I've done a little freelance work, but mostly been looking for the next long-term job, and in doing so, making something of a minor course-correction.
You see, for a while now, I've been working very much in a niche function of a niche (if global) industry. Even more nichely, the focus has been around a particular software vendor's product and its support. It's served me well, and given me a depth and breadth of experience I might otherwise have found difficult to find elsewhere. I've also benefited from a fairly close-knit network of colleagues, each of whom have a similar set of experiences built up over time. But as time goes by, I'm finding myself interested in areas that don't neatly overlap with recs.
Interestingly, that's quite a statement - there's not much that doesn't somehow share some degree of contextual intersection with software modelled reconciliations - but that's another story.
So in this most recent job-search, I've been trying to take a step away from recs, broaden my horizons, and do something different - ideally, something linked to data-science, analytics or similar. To assist, countless recruitment consultants have called, sent over specs and discussed skills, salaries and start-dates. Sometimes, I've enjoyed the process, and at others, I've found it deeply frustrating. I don't envy the recruitment consultant's role in life. Some of them are good, decent and honest people. Others, and I'm sure these were just having a bad day, didn't fill me full of confidence. Which is fine under normal circumstances - but when you're responsible for a family, you don't want to place your future in the hands of someone who lies at the drop of the hat - least-so, one who lies so unconvincingly.
As I write, there is an offer in the works, for a job whose title hasn't yet been decided, the working arrangements of which are not yet known, but for which I'm very excited. It will mean working with data, and helping others work effectively with data too - and, to an extent not yet entirely known, it may also mean working on new technology and methodologies within an environment of like-minded people.
I'm not sure if it's fair, but this could be my Solsbury Hill moment, which for me suggests a kind of homecoming/becoming thing. I'm sure I'm over-romanticising here.
Upcoming things to think about -
• What is 'data'
• What common problems exist when handling/managing data
• How can we break those problems down into their most abstract form?
• What tools can we build around those abstract ideas that will generate practical value?
• How can we leverage the usage differences between mutable and immutable data records?
• What standards are out there for taxonomies, classifications, models and forms?
• What are the practical pros and cons of those standards?
• How can different data be 'vectorised' i.e. turned into a vector?
I'm sure there will be more, but these are what's interesting at the moment.
## Thursday, 23 July 2015
### Auf Wiedersehen Paris
After an interesting few months, it looks as though my time in Paris is coming to an end.
It's been an interesting assignment, and I've been able to work on developing lots of interesting ideas that hopefully I can take forward into my next role - wherever that will be.
I don't look forward to restarting the process of finding paying work, but it's important to approach these things flexibly, and positively - otherwise, things can quickly spiral into negativity - which is no fun, and is rarely warranted.
I'd like to focus more on data analytics in the future, drawing information, narrative and decision from raw data - and helping others get to grips with the kinds of techniques and applications that can make this happen.
I'd also like to continue working with python - I'm still impressed with the speed with which thorny problems can be tackled - and would like to apply some of these techniques in a more commercial setting. So far, much of my work has been to provide tools, the users of which don't necessarily see or know the internal workings - only the outputs.
So, that's what I'd like - but at some point may have to take something slightly off-track, got to pay those bills.
Yesterday, I started applying to nearly every job pinged over to my email by the JobServe search engine, and so far, not received a sausage in terms of a human reply. Either there's some glaring gap in my CV, or I should relax a little (for now) and wait to see what comes in later - although, characteristically, I notice that I'm faced with a data-analysis problem here
I don't want to apply for any job more than once - but the job-boards are populated with entries from different agencies - more, there are multiple jobs-boards. This means that any particular position is likely to be posted to more than one board, and by more than one agent. This potentially results in duplication - so the question is - how to systematically keep track of what jobs have been applied for, in what areas, and via which agencies?
I think I may have found my next working project...
If last time was anything to go by, finding the next thing could take something like 2 months - but fingers crossed it's quicker than that.
Oh and Paris? Yes, it's been nice knowing you.
|
2017-05-25 21:47:09
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5239251852035522, "perplexity": 959.0311800811677}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463608617.6/warc/CC-MAIN-20170525214603-20170525234603-00418.warc.gz"}
|
https://tex.stackexchange.com/questions/299660/how-do-i-break-a-long-equation
|
# How do I break a long equation?
I have a long equation that it was hard for me to break so that it appears professional. How can I do it?
$$\begin{split} &\mathbb{P}\left( \left. Z_{1}\in dx_{1},Z_{2}\in dx_{2},\cdots ,Z_{n}\in dx_{n}\right\vert Z_{n+1}=x_{n+1}\right) \\ \qquad &=f_{Z_{n+1}}\left(x_{n+1}\right) ^{-1}\left[ \mathbb{P}\left( \left\{ Z_{1}\in dx_{1},Z_{2}\in dx_{2},\cdots ,Z_{n}\in dx_{n}\right\} \cap \left\{ Z_{n+1}=x_{n+1}\right\} \right) \right] \\ =f_{Z_{n+1}}\left( x_{n+1}\right) ^{-1}\times \mathbb{\pi } \\ =e^{-x_{n+1}}dx_{1}\cdots dx_{n+1}/f_{Z_{n+1}}\left( x_{n+1}\right) \\ =n!x^{-n}dx_{1}\cdots dx_{n} \end{split}$$
You're not far. Remove all \left and \right tokens that only do bad, in this context; there's just a pair of brackets that asks for being a bit larger (with \bigl and \bigr).
Also some help is needed for the dots in the final two lines; use \mid instead of \vert, for better spacing and don't forget \, after the factorial.
Last thing: I added empty superscripts to f_{Z_{n+1}}, so the subscript is slightly moved down.
\documentclass{article}
\usepackage{mathtools}
\usepackage{amsfonts} % for \mathbb
\DeclareMathOperator{\PR}{\mathbb{P}}
\begin{document}
$$\begin{split} &\PR(Z_{1}\in dx_{1},Z_{2}\in dx_{2},\dots ,Z_{n}\in dx_{n}\mid Z_{n+1}=x_{n+1}) \\ &\quad =f^{}_{Z_{n+1}}(x_{n+1}) ^{-1}\bigl[ \PR( \{ Z_{1}\in dx_{1},Z_{2}\in dx_{2},\dots,Z_{n}\in dx_{n}\} \\ &\qquad\qquad \cap\{Z_{n+1}=x_{n+1}\})\bigr] \\ &\quad =f^{}_{Z_{n+1}}( x_{n+1}) ^{-1}\times \pi \\ &\quad =e^{-x_{n+1}}dx_{1}\dotsm dx_{n+1}/f^{}_{Z_{n+1}}(x_{n+1}) \\ &\quad =n!\,x^{-n}dx_{1}\dotsm dx_{n} \end{split}$$
\end{document}
• why not an outer multline with all lines but the first either split or aligned? Mar 18 '16 at 14:33
• @barbarabeeton I prefer the equals signs to be near the left side, like if the first line were X&=\PR(...) Mar 18 '16 at 14:35
Some suggestions:
• Keep using a split environment, but provide alignment points on all lines
• Use an aligned environment to the right of the = symbol in row 2 to break up that line into two separate lines
• Don't overuse \left and \right; in fact, for the equations at hand, they're not needed at all.
• Don't use \cdots for the equations at hand. Instead, use \dots in the first two rows and \dotsm ("multiplicate dots") in the final two rows.
• There is no "Blackboard Bold" version of \pi; hence, \mathbb{\pi } is the same as \pi. If you must produce a bold version of the symbol, write \boldsymbol{\pi}; alternatively, load the bm package and write \bm{\pi}.
\documentclass{article}
\usepackage{amsmath,amsfonts,bm}
\begin{document}
\begin{split} \mathbb{P}( &Z_{1}\in dx_{1},Z_{2}\in dx_{2},\dots, Z_{n}\in dx_{n} \mid Z_{n+1}=x_{n+1}) \\ &=\!\begin{aligned}[t] f_{Z_{n+1}}^{} (x_{n+1})^{-1} \bigl[ \mathbb{P}(\{ &Z_{1}\in dx_{1}, Z_{2}\in dx_{2}, \dots , Z_{n}\in dx_{n}\} \\ &\cap \{ Z_{n+1}=x_{n+1}\} ) \bigr] \end{aligned}\\ &=f_{Z_{n+1}}( x_{n+1})^{-1} \times \bm{\pi} \\ &=e^{-x_{n+1}}\,dx_{1}\dotsm dx_{n+1} / f_{Z_{n+1}}( x_{n+1}) \\ &=n!\,x^{-n}\,dx_{1}\dotsm dx_{n} \end{split}
\end{document}
I propose a solution with multline and aligned. The geometry package yields more sensible margins, if you don't use margin notes. Finally, I simplified your code, using the \DeclarePairedDelimiter commad form mathtools: I defined a \Prob+\given command for conditional probabilities, and a \set command, which have variable-sized delimiters if used in their starred version, or with an option (\big, \Big, &c.):
\documentclass{article}
\usepackage{mathtools,amsfonts} % for 'dcases' environment
\usepackage{xparse} \usepackage[showframe]{geometry}
\DeclarePairedDelimiterXPP\Prob[1]{\mathbb P}(){}{
\newcommand\given{\nonscript\:\delimsize\vert\nonscript\:}
#1}
\DeclarePairedDelimiterX{\set}[1]\{\}{\setargs{#1}}
\NewDocumentCommand{\setargs}{>{\SplitArgument{1}{;}}m} {\setargsaux#1}
\NewDocumentCommand{\setargsaux}{mm}
{\IfNoValueTF{#2}{\nonscript\,#1\nonscript\,}{\nonscript\,#1\nonscript\;{;}
\nonscript\:\allowbreak #2\nonscript\,}}
\begin{document}
\begin{multline}
\Prob{ Z_{1} \in dx_{1},Z_{2} \in dx_{2}, \dots ,Z_{n}\in
dx_{n}\given Z_{n+1} =x_{n+1}} \\
\begin{aligned}
& =f_{Z_{n+1}}\left(x_{n+1}\right) ^{-1}\left[ \Prob*{\set*{ Z_{1} \in dx_{1},Z_{2} ∈
dx_{2}, \dots ,Z_{n} \in dx_{n}} \cap \set*{ Z_{n+1}=x_{n+1}}} \right] \\
& =f_{Z_{n+1}}\left( x_{n+1}\right) ^{-1} \times \mathbb{\pi } \\
& =e^{-x_{n+1}}dx_{1}\dotsm dx_{n+1}/f_{Z_{n+1}}\left( x_{n+1}\right) \\
& =n!\,x^{-n}dx_{1} \dotsm dx_{n}
\end{aligned}
\end{multline}
\end{document}
• I don't think it's typographically appropriate to use \cdots in comma-delimited lists -- see rows 1 and 2. Consider using either \dots or \dotsc ("dots between commas"). Also, \mathbb{\pi} is suspect.
– Mico
Mar 18 '16 at 15:12
• @Mico: I just took the code from the O.P and didn't notice. But you're righht, I'll update my answer. Thanks for pointing it Mar 18 '16 at 15:21
|
2022-01-19 09:22:42
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 3, "x-ck12": 0, "texerror": 0, "math_score": 0.9997311234474182, "perplexity": 7480.513977933747}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320301264.36/warc/CC-MAIN-20220119064554-20220119094554-00561.warc.gz"}
|
https://latexref.xyz/_005cmakebox-_0028picture_0029.html
|
Next: , Previous: , Up: `picture` [Contents][Index]
#### 8.19.13 `\makebox` (picture)
Synopsis:
```\makebox(rec-width,rec-height){text}
\makebox(rec-width,rec-height)[position]{text}
```
Make a box to hold text. This command fits with the `picture` environment, although you can use it outside of there, because rec-width and rec-height are numbers specifying distances in terms of the `\unitlength` (see `picture`). This command is similar to the normal `\makebox` command (see `\mbox` & `\makebox`) except here that you must specify the width and height. This command is fragile (see `\protect`).
This makes a box of length 3.5 times `\unitlength` and height 4 times `\unitlength`.
```\put(1,2){\makebox(3.5,4){...}}
```
The optional argument `position` specifies where in the box the text appears. The default is to center it, both horizontally and vertically. To place it somewhere else, use a string with one or two of these letters.
`t`
Puts text the top of the box.
`b`
Put text at the bottom.
`l`
Put text on the left.
`r`
Put text on the right.
|
2022-01-29 13:03:32
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.853683590888977, "perplexity": 2155.0596697785754}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320306181.43/warc/CC-MAIN-20220129122405-20220129152405-00438.warc.gz"}
|
http://zbmath.org/?q=an:0768.11009
|
zbMATH — the first resource for mathematics
Examples
Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.
Operators
a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
Fields
any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)
Irregular primes to one million. (English) Zbl 0768.11009
The authors have calculated the irregular pairs $\left(p,k\right)$ for all primes $p<{10}^{6}$. Recall that a pair $\left(p,k\right)$ is called irregular if $k$ is an even integer with $2\le k\le p-3$ such that $p$ divides the numerator of the Bernoulli number ${B}_{k}$. Previous computations of the irregular pairs, covering the range $p<150\phantom{\rule{4pt}{0ex}}000$ [see J. W. Tanner and S. S. Wagstaff, ibid. 48, 341–350 (1987; Zbl 0613.10012)], used algorithms requiring $O\left({p}^{2}\right)$ arithmetic operations for each prime $p$. The present authors were able to reduce this number to $O\left(plogp\right)$. They computed Bernoulli numbers modulo $p$ basically from the formula
$\sum _{k=0}^{\infty }{B}_{k}{x}^{k}/k!={\left(1+x/2!+{x}^{2}/3!+\cdots \right)}^{-1},$
performing the power series inversion by algorithms based on the fast Fourier transform and multisectioning of power series. The maximum number of irregular pairs $\left(p,k\right)$ found in this range was 6, occurring for $p=527377$.
The authors also used their results to verify that Fermat’s “Last Theorem” and Vandiver’s conjecture are true for the primes $p<{10}^{6}$.
{A subsequent work by the first two authors together with R. Ernvall and the reviewer [ibid., July 1993 issue] extends the above calculations to all primes below four million and moreover gives the ordinary cyclotomic invariants for these primes}.
MSC:
11B68 Bernoulli and Euler numbers and polynomials 11D41 Higher degree diophantine equations 65Y20 Complexity and performance of numerical algorithms 11R18 Cyclotomic extensions 11Y55 Calculation of integer sequences 68Q25 Analysis of algorithms and problem complexity
|
2013-12-05 04:08:25
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 16, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9076002836227417, "perplexity": 4154.838213450383}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386163039753/warc/CC-MAIN-20131204131719-00018-ip-10-33-133-15.ec2.internal.warc.gz"}
|
https://mathematica.stackexchange.com/revisions/161524/2
|
2 of 2 edited tags
# log plot problems
Why does the following code give me two different curves?
a = 100;
b = -0.1;
s = a n^b;
LogLogPlot[{s, a n^b}, {n, 1000, 1000000}]
|
2020-07-14 07:04:15
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6088894605636597, "perplexity": 12924.605918710904}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657149205.56/warc/CC-MAIN-20200714051924-20200714081924-00280.warc.gz"}
|
https://forum.bebac.at/forum_entry.php?id=19096&page=1&category=1&descasc=DESC&view=thread
|
## Lower bound for power of two combined TOST? [Power / Sample Size]
Dear All!
Occasionally we have discussed here the impact on overall power if we decide BE based on two (or more) PK metrics, combined via 'and'. For instance here.
PowerTOST has a function power.2TOST() to deal with that problem.
But crucial is here the correlation argument, which is difficult to estimate. See this lengthy thread.
Quite recently I discovered something in that direction in the book
Patterson, Jones
Bioequivalence and Statistics in Clinical Pharmacology,
Second Edition, CRC Press, Boca Raton 2017
Chapter 5.7 "Determining Trial Size", page 134
Quote:
It is assumed for the purposes of this discussion that within-subject variability estimates are available, for both AUC and Cmax, to determine the trial size. For this purpose the larger of the two pooled estimates is of primary interest in calculations, for obvious reasons (i.e., power will be greater, or alternatively the probability of a Type 2 error will be lower, for the endpoint with smaller variation). However, the degree of this increase should be estimated using appropriate code (just switching the estimate of variability) to ensure adequate overall power for the study, as it is known [918] that
Power >= PAUC + PCmax - (2 - 1)
where PAUC is the estimate of power for AUC and PCmax is the estimate of power for Cmax. In the event that the overall power falls below the desired level, sample size may be increased to compensate, resulting in the desired level of power. For example, if power for Cmax is 0.90, and for AUC 0.95, the resulting overall study power is at least
0.9 + 0.95 - 1 = 0.85.
They use that formula or an analogous one also in other context for combining powers. Search for the reference [918] in the Patterson/Jones book.
My question(s): Does anybody knows where that formula came from?
Does anybody own the reference and can enlighten me?
[918] Nauta, J. (2010) Statistics in Clinical Vaccine Trials. Springer, London.
Regards,
Detlew
### Complete thread:
Admin contact
21,082 posts in 4,397 threads, 1,468 registered users;
online 6 (0 registered, 6 guests [including 3 identified bots]).
Forum time: Monday 13:56 CEST (Europe/Vienna)
A central lesson of science is that to understand complex issues
(or even simple ones), we must try to free our minds of dogma and
to guarantee the freedom to publish, to contradict, and to experiment.
Arguments from authority are unacceptable. Carl Sagan
The Bioequivalence and Bioavailability Forum is hosted by
Ing. Helmut Schütz
|
2020-09-28 11:56:37
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6345115900039673, "perplexity": 3297.8138166704525}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600401600771.78/warc/CC-MAIN-20200928104328-20200928134328-00299.warc.gz"}
|
https://sivertlindblom.se/y4sv8pf/2e4d3a-sufficient-statistic-for-bernoulli-distribution
|
) y θ ) , X i ∼ Binomial(n,θ) Prove that T (X ) is sufficient for X by deriving the distribution of X | T (X ) = t. Example 2. {\displaystyle \theta } , Note that this distribution does not depend on . Once the sample mean is known, no further information about μ can be obtained from the sample itself. and that {\displaystyle P_{\theta }} Stephen Stigler noted in 1973 that the concept of sufficiency had fallen out of favor in descriptive statistics because of the strong dependence on an assumption of the distributional form (see Pitman–Koopman–Darmois theorem below), but remained very important in theoretical work.[3]. n 2 X This property is mathematically expressed as one of the results of the theory of statistical decision making which says … ≤ If {\displaystyle \theta } {\displaystyle y_{1},\dots ,y_{n}} ) , , the above likelihood can be rewritten as. J ) . X The idea roughly is to trap the CDF of X n by the CDF of Xwith an interval whose length converges to 0. y 1 {\displaystyle X_{1}^{n}=(X_{1},\dots ,X_{n})} θ The link function is given by. [1] In particular, a statistic is sufficient for a family of probability distributions if the sample from which it is calculated gives no additional information than the statistic, as to which of those probability distributions is the sampling distribution. 1 ) In this case $$\bs X$$ is a random sample from the common distribution. n are unknown parameters), then Y n β ( ( ) … n T ( β ( ) x X Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. The Bernoulli model admits a complete statistic. Complete statistics. ) X [8] However, under mild conditions, a minimal sufficient statistic does always exist. where 1{...} is the indicator function. of independent identically distributed data conditioned on an unknown parameter through the function , If ) Thanks for contributing an answer to Cross Validated! ) − , x , θ θ {\displaystyle T(\mathbf {X} )} 1 {\displaystyle T} 0 1. sufficient for θ. Without prior information, ... (which are the sufficient statistics for the Bernoulli distribution). ] 1 Let Y1 = u1(X1, X2, ..., Xn) be a statistic whose pdf is g1(y1; θ). In other words, S(X) is minimal sufficient if and only if[7]. , X With the first equality by the definition of pdf for multiple variables, the second by the remark above, the third by hypothesis, and the fourth because the summation is not over 1 {\displaystyle T(x_{1}^{n})=\left(\prod _{i=1}^{n}x_{i},\sum _{i=1}^{n}x_{i}\right),}, the Fisher–Neyman factorization theorem implies where is the natural parameter, and is the sufficient statistic. denote the conditional probability density of How were drawbridges and portcullises used tactically? More formally, define ν to be counting measure on {0,1}, and define the following density function with respect to ν: p(x|π) = πx(1−π)1−x (8.5) = exp ˆ log π 1−π x+log(1−π) ˙. θ X n u , x θ X {\displaystyle X_{1}^{n}=(X_{1},\ldots ,X_{n})} s 1 − n 2 . x , , ( Examples [edit | edit source] Bernoulli distribution … X 1 Actually, we can understand sufficient statistic from two views: (1). max 2 and 1 n X α , Y Info; Current issue; All issues; Search ← Previous article; TOC; Next article → Bernoulli; Volume 6, Number 6 (2000), 1121-1134. are independent and normally distributed with expected value = X ) Sufficient statistics are most easily recognized through the following fundamental result: A statistic T = t(X) is sufficient for θ if and only if the family of densities can be factorized as f(x;θ) = h(x)k{t(x);θ}, x ∈ X,θ ∈ Θ, (1) i.e. n 1.Under weak conditions (which are almost always true, a complete su cient statistic is also minimal. X ) 1 where $\sigma(T)$ denotes the sigma generated by T and i , 1 $\sigma(S)$ denotes the sigma generated by S. Since $\sigma(S)\subset \sigma(T)$ (the information in $T$ is more than $S$) ,$S$ is a minimal sufficient statistic and $S$ is a function of $T$ ,hence $T$ is a sufficient statistic(But not a minimal one). ) , ∑ The sufficient statistic of a set of independent identically distributed data observations is simply the sum of individual sufficient statistics, and encapsulates all the information needed to describe the posterior distribution of the parameters, given the data (and hence to … . n \right. I calculated and found out $X_1+X_2$ as a sufficient statistic for $p$. ( 1 From this factorization, it can easily be seen that the maximum likelihood estimate of For example(*1). Since In short, we claim to have a over the probability , which represents our prior belief. . y ^ Active 9 months ago. y f X {\displaystyle X_{1},\dots ,X_{n}} 2 [ over , with the natural parameter , sufficient statistic , log partition function and . i Due to the factorization theorem (see below), for a sufficient statistic (), the joint distribution can be written as () = (, ()). x Do I need my own attorney during mortgage refinancing? min 1 = 1 = ( … E Let T = X 1 + 2 X 2 , S = X 1 + X 2. If X1, ...., Xn are independent and uniformly distributed on the interval [0,θ], then T(X) = max(X1, ..., Xn) is sufficient for θ — the sample maximum is a sufficient statistic for the population maximum. {\displaystyle \beta } n | ) In fact, the minimum-variance unbiased estimator (MVUE) for θ is. ) {\displaystyle Y_{1}} {\displaystyle \Theta } Specifically, if the distribution of X is a k-parameter exponential family with the natural sufficient statistic U=h(X) then U is complete for θ (as well as minimally sufficient for θ). and in all cases it does not depend of the parameter. , As this is the same in both cases, the dependence on θ will be the same as well, leading to identical inferences. is a sufficient statistic for , [12], A concept called "linear sufficiency" can be formulated in a Bayesian context,[13] and more generally. are unknown parameters of a Gamma distribution, then , ( while the response function is given by the logistic function. α , ≤ The concept is due to Sir Ronald Fisher in 1920. , is a two-dimensional sufficient statistic for Since $T \equiv X_1+X_2$ is a sufficient statistic, the question boils down to whether or not you can recover the value of this sufficient statistic from the alternative statistic $T_* \equiv X_1 + 2 X_2$. ) 1 ( Since [11] A range of theoretical results for sufficiency in a Bayesian context is available. , {\displaystyle T(X_{1}^{n})=\left(\prod _{i=1}^{n}X_{i},\sum _{i=1}^{n}X_{i}\right)} On the other hand, for an arbitrary distribution the median is not sufficient for the mean: even if the median of the sample is known, knowing the sample itself would provide further information about the population mean. *3 & t=2 \\ δ(X ) may be inefficient ignoring important information in X that is relevant to θ. δ(X ) may be needlessly complex using information from X that is irrelevant to θ. . over , with the natural parameter , sufficient statistic , log partition function and . ( 1 The Bernoulli distribution A Bernoulli random variable X assigns probability measure π to the point x = 1 and probability measure 1 − πto x= 0. {\displaystyle T(X_{1}^{n})=\left(\prod _{i=1}^{n}{X_{i}},\sum _{i=1}^{n}X_{i}\right)} ( , x X x n n Note: One should not be surprised that the joint pdf belongs to the exponen-tial family of distribution. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. {\displaystyle \mathbf {X} } n 1 P(X_1=x_1,X_2=x_2|T=0)= ( ( n Thus the density takes form required by the Fisher–Neyman factorization theorem, where h(x) = 1{min{xi}≥0}, and the rest of the expression is a function of only θ and T(x) = max{xi}. 1 \right. ) \end{eqnarray} {\displaystyle \theta } x ( X ) : X →A Issue. 0 & O.W. . 1 ; that is, it is the conditional pdf n … Roughly, given a set of independent identically distributed data conditioned on an unknown parameter , a sufficient statistic is a function () whose value contains all the information needed to compute any estimate of the parameter (e.g. y . It follows a Gamma distribution. = 1 = *1 & t=0 \\ = X = x ≤ {\displaystyle \alpha } While it is hard to find cases in which a minimal sufficient statistic does not exist, it is not so hard to find cases in which there is no complete statistic. ( ≤ {\displaystyle f_{\theta }(x,t)=f_{\theta }(x)} is a function of θ De nition I Typically, it is important to handle the case where the alternative hypothesis may be a composite one I It is desirable to have the best critical region for testing H 0 against each simple hypothesis in H 1 I The critical region C is uniformly most powerful (UMP) of size against H 1 if it is so against each simple hypothesis in H 1 I A test de ned by such a regions is a uniformly most n {\displaystyle t=T(x)} Use MathJax to format equations. , The other answer by Masoud gives you the information you need to construct such a mapping, so use this to have a go constructing a function of this kind. ( h X As an example, the sample mean is sufficient for the mean (μ) of a normal distribution with known variance. ) {\displaystyle T(X_{1}^{n})=\left(\min _{1\leq i\leq n}X_{i},\max _{1\leq i\leq n}X_{i}\right),}. Distributions discussed above a fair coin from a biased coin effect of the parameter λ interacts with data. Gamma distribution with both parameter unknown, where the natural parameter, and the sufficient statistic with MIPS to... Special cases of a generalized linear model ) $given$ T=X_1+2X_2 $on... Function deals with individual finite data ; the related notion there is no minimal sufficient statistic is sufficient! T ( X ). } functions, called a jointly sufficient from! N by the definition of sufficient statistics are to show that T ( X ) the... X_1$ and $X_2$ be iid random variables from a biased coin to: Suppose that (,. N iid U [ 0 ] for distinguishing a fair coin from a $(... The logistic function a sequence of independent Bernoulli random variables from a$ Bernoulli ( p $... What would be the most efficient and cost effective way to show that distributions. Ones, conditional probability and Expectation 2 the bias, and the sufficient statistics are, \beta.! Theorem stated above they occur to 0 and test and the sufficient statistic xwhere... Efficiently captures all possible information about μ can be compared by the probabilities function which does involve! N ; ) distribution ;::: ; Xn ). } this to. With both parameter unknown, where the natural parameter, sufficient statistic concrete application, this a! S = X 1 + X 2 which of the sufficient statistic, log partition function and the latter is.$ ( X_1, X_2 ) $distribution the data only through its sum T ( X ) continuous. Μ at all they occur Suppose that ( X_1, X_2 )$ distribution, e.g for set. ( X ) in the discrete case could envision keeping only T and throwing away all the in... 1987 that caused a lot of travel complaints bias, and beta distributions discussed above of! Statements based on opinion ; back them up with references or personal experience Bernoulli.! The possible values of the data, e.g over, with the only! Personal experience did DEC develop Alpha instead of continuing with MIPS of the followings can be obtained the! Statistic and the sufficient statistics for the Bernoulli ( theta ) distribution X. Statistic is a sufficient statistic if is discrete or has a density function if the statistic $X_1+2X_2,. Of theoretical results for sufficiency in a Bayesian context is available 1987 caused! You agree to our terms of service, privacy sufficient statistic for bernoulli distribution and cookie policy X... Will be the number of trials up to the flrst success called a jointly sufficient statistic obvious once you the... Following the de nition First page ; references ; Abstract an MSS is also sufficient a CSS is also CSS. Our prior belief following question: is there a better way to stop star... Unbiased estimator ( MVUE ) for θ$ X_1+2X_2 $is sufficient for$ p.. Success occurs with probability µ interest can be written as a sufficient statistic most efficiently captures all information! $X_1+2X_2$ is sufficient = p n i=1 X i is a sample from Bernoulli. Θ is only in conjunction with T ( X1,..., Xn ) = ( n. i. X. i ) is sufficient statistic for bernoulli distribution algorithmic sufficient statistic may be a set of functions, called jointly...: different individuals may assign different probabilities to the Fisher-Neyman factorisation to show that T ( X =! Surprised that the distributions corresponding to different values of $p$ or not may assign probabilities. Normal and Bernoulli models ( and many others ) are special cases of a statistic taking values in set. Likelihood ratios is a minimal sufficient statistics Bernardo and Smith for fuller treatment of foun-dational issues for... Depends on X through T ( X ). } a range of results... 2020 Stack Exchange Inc ; user contributions licensed under cc sufficient statistic for bernoulli distribution something happen in 1987 that caused lot.:: ; Xn be independent Bernoulli random variables with same parameter µ simpler... Develop Alpha instead of continuing with MIPS is known, no further information the. Sample maximum T ( X ) in the theorem is called the natural parameter, statistic... R\ ). } the distributions corresponding to different values of the past trials will wash out should! A 1-1 function of X n by the definition of sufficient statistics for Bernoulli, Poisson, and Exponential a... May be a set of functions, called a jointly sufficient statistic is minimal sufficient statistic, log partition and. ( MVUE ) for θ is only in the sample mean is sufficient for the mean ( μ ) a... Logo © 2020 Stack Exchange Inc ; user contributions licensed under cc by-sa ; ) distribution with. Are almost always true, a success occurs with probability µ let X be the same event, even they. ^ = X 1 ) for θ is is unknown statistic by nonzero... Is called the natural parameter, and is the maximum likelihood estimator θ! Is unknown corresponding to different values of the past trials will wash.!: ( 1 ). } set condition '' ( OSC ). } do i need my attorney. There are parameters the theorem is called the natural parameters are, and the test in ( c ) minimal. And cookie policy enough to rule out the possibility of $T$ and $X_2 be! Corresponding to different values of the followings can be compared, there are parameters in which there no! Distribution Consider a sequence of independent Bernoulli trials ) of a statistic taking values in a set of,! You can then appeal directly to the Fisher-Neyman factorisation to show that (... This RSS feed, copy and paste this URL into Your RSS reader ima '' in... Are distinct as there are parameters, leading to identical inferences the factorization criterion the... Of em '' correct for the bias, and Exponential ] is unknown function and could keeping... Words, S = X 1 + 2 X 2 cost effective way to show that θ ^ = 1! Contains an open set in Rk i a contains a k-dimensional ball the following question: is any! Obvious once you note the parameter be a statistic, log partition and... The probability, which represents our prior belief statistic does always exist, \beta ). } away the... Have a over the probability, which represents our prior belief Smith fuller... An MSS is also minimal 1 ; X n iid U [ 0, 1 ] unknown. Happen in 1987 that caused a lot of travel complaints ( which are almost always true, a sufficient..., copy and paste this URL into Your RSS reader and Expectation 2 trials to... Pdf belongs to the flrst success, although it applies only in conjunction with T ( Xn =... The indicator function later remarks ). } not involve µ at all that all! To random samples from the sample about µ ( b ) is a su statistic... Depends on$ p $in relation to a model for a set of functions, a! Data, e.g for Gamma distribution with both parameter unknown, where the natural parameter is and MVUE! ) 1.The statistic T ( Xn ). } followings can be written as product. Develop Alpha instead of continuing with MIPS statements based on opinion ; back them up with or... Xi without losing any information both parameter unknown, where the natural su cient for! Results for sufficiency in a Bayesian context is available statistic from two views: ( 1 ). } of... Where 1 {... } is the indicator function statistic if is discrete or has a density.... Other sufficient statistic, i.e the parameter λ interacts with the natural parameter, sufficient statistic log... Roughly is to trap the CDF of Xwith an interval whose length converges to 0 multiply a sufficient may... Ships on remote ocean planet instead of continuing with MIPS distribution, with h ( ). It ensures that the joint probability density function of a normal distribution with parameter! Eqnarray } and find * 1,..., X. n. be iid n ( θ σ!, conditional probability and Expectation 2 get another sufficient statistic may be set! Is parameterized by the factorization criterion, with h ( X ) \ be! Distributions corresponding to different values of the followings can be represented as a from. May assign different probabilities to the same in both cases, the sufficient statistic 8.6 ) 1.The statistic T X! And thus T { \displaystyle T } is the sufficient statistic which follows a Binomial.$ or not interacts with the data have to respect checklist order ( c ) a... Only if [ 7 ] calculated and found out $X_1+X_2$ as a sufficient statistic may a! Right-Tailed test. beta distributions discussed above and is the sufficient statistic i i... Later remarks ). } in conjunction with T ( X ). } a 's... Copy and paste this URL into Your RSS reader could envision keeping only T and throwing away the... In particular we can understand sufficient statistic, log partition function and distributions! } and find * 1,... ( which are the sufficient statistic most efficiently captures all possible about... Weak conditions ( which are almost always true, a minimal sufficient statistic way... } depend only upon X 1 \alpha \,,\, \beta ). },. The logistic function back them up with references or personal experience choose an as ( H+T goes.
|
2023-03-27 07:40:21
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9476061463356018, "perplexity": 792.677675962552}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948609.41/warc/CC-MAIN-20230327060940-20230327090940-00129.warc.gz"}
|
http://wiki.stat.ucla.edu/socr/index.php?title=SOCR_EduMaterials_AnalysesCommandLineTwoIndependentKruskalWallis
|
# SOCR EduMaterials AnalysesCommandLineTwoIndependentKruskalWallis
This page includes the information on how to access the Two Independent Kruskal Wallis Test library via shell-based command-line interface on local machines. More information about other SOCR Analyses command-line interfaces is available here.
## Introduction
In addition to the graphical user interfaces, via a web-browser, all SOCR Analyses allow command-line shell execution on local systems.
## General Usage
• Get the latest SOCR JAR files from the SOCR page (http://socr.ucla.edu/htmls/jars/).
• The command-line interface to SOCR Analyses generally uses EXAMPLE 1 from the list of example data files for the corresponding analysis.
• All Input files are ASCII (see examples within each of the specific analyses).
• a -h flag at the end of the command-line indicates that the first row in all ASCII input data files is a HEADER row (so it's not interpreted as data)
• Number of variables have to be indicated at the end (after -h flag). If no number of variables is specified, 4 is set as default.
## Two Independent Kruskal Wallis Test Usage
• Generic Setting:
java -cp [SOCRjar_location]/SOCR_core.jar:[SOCRjar_location]/SOCR_plugin.jar edu.ucla.stat.SOCR.analyses.command.TwoIndependentKruskalWallisCSV [data_location]/kw.txt -h [number_of_variables]
• Example: Edit a new file (TwoIndependentKruskalWallisCSV.csh) using any editor and paste this inside (make sure the file has executable permissions). Some operating systems/platforms may require variants of this (C-shell) script.
#!/bin/csh
date
java -cp /ifs/ccb/CCB_SW_Tools/others/Statistics/SOCR_Statistics/bin/SOCR_core.jar:/ifs/ccb/CCB_SW_Tools/others/Statistics/SOCR_Statistics/bin/SOCR_plugin.jar edu.ucla.stat.SOCR.analyses.command.TwoIndependentKruskalWallisCSV /ifs/ccb/CCB_SW_Tools/others/Statistics/SOCR_Statistics/SOCR_CSV_test_Scripts_Data/kw.txt -h 4
date
exit
## Example Input data files
One test datafile is included with the SOCR analyses command-line distribution (kw.txt). The ASCII content of each of these is included below. Note that the first lines in these files are column headers. This requires the "-h" flag at the end of the command line execution so that these first lines are interpreted as column headers.
kw.txt
A B C D
59.8 59.8 60.7 61.0
60.0 60.2 60.7 60.8
60.8 60.4 60.5 60.6
60.8 59.9 60.9 60.5
59.8 60.0 60.3 60.5
|
2022-05-28 19:32:47
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4714887738227844, "perplexity": 7122.71307830556}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663019783.90/warc/CC-MAIN-20220528185151-20220528215151-00738.warc.gz"}
|
https://www.nature.com/articles/s41598-020-63206-1?error=cookies_not_supported&code=e497e73e-e8f7-4f3d-ba9b-885317274245
|
Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript.
# A Galactic centre gravitational-wave Messenger
## Abstract
Our existence in the Universe resulted from a rare combination of circumstances. The same must hold for any highly developed extraterrestrial civilisation, and if they have ever existed in the Milky Way, they would likely be scattered over large distances in space and time. However, all technologically advanced species must be aware of the unique property of the galactic centre: it hosts Sagittarius A* (Sgr A*), the closest supermassive black hole to anyone in the Galaxy. A civilisation with sufficient technical know-how may have placed material in orbit around Sgr A* for research, energy extraction, and communication purposes. In either case, its orbital motion will necessarily be a source of gravitational waves. We show that a Jupiter-mass probe on the retrograde innermost stable circular orbit around Sgr A* emits, depending on the black hole spin, at a frequency of fGW = 0.63–1.07 mHz and with a power of PGW = 2.7 × 1036–2.0 × 1037 erg/s. We discuss that the energy output of a single star is sufficient to stabilise the location of an orbiting probe for a billion years against gravitational wave induced orbital decay. Placing and sustaining a device near Sgr A* is therefore astrophysically possible. Such a probe will emit an unambiguously artificial continuous gravitational wave signal that is observable with LISA-type detectors.
## Introduction
It is conceivable that a civilisation advanced enough to see itself in a cosmological context would think of a way to make itself known to the Galaxy. We humans, for instance, sent in 1977 with Voyager 1 and 2 two postcards from Earth out into the vast depths of outer space. Destination: “addressee unknown”. The Voyagers carry not only instruments designed to study the planets in the outer Solar System, but also two golden records with messages intended for future humans or extraterrestrial lifeforms. Chances are that other civilisations would also try to announce their presence to the Galaxy. Using probes with instruments that emit radio waves aboard is one of the most obvious strategies to us. Indeed SETI, our most elaborated search programme for extraterrestrial intelligence, is based primarily on analysing radio waves1. The SETI Breakthrough Listen Initiative regularly releases data from surveys of the radio spectrum between 1 and 12 gigahertz (GHz) to the public. And we are not only searching – on 17 November 1974 the largest radio antenna on Earth sent the famous “Arecibo message” to the globular cluster M13.
Of course, there is no particular reason to limit searching and announcing activities to radio waves. One may also think of signals encoded in the X-ray or $$\gamma$$-ray waves, or even in neutrinos. In addition, gravitational wave phenomena are omnipresent in the Universe. As opposed to electromagnetic waves, gravitational waves travel through space virtually undamped by matter along their way. In the case of the Advanced LIGO and Advanced Virgo detectors, waves emitted during the last stages of the inspiral and merger of binary systems of compact objects – stellar mass black holes and neutron stars – are firmly detectable from distances up to a few gigaparsec (Gpc, $$1{\rm{p}}{\rm{c}}=3.086\times {10}^{18}\,{\rm{c}}{\rm{m}}$$)2,3,4,5,6,7.
A particular problem with searching for extraterrestrial civilisations is often summarised in Enrico Fermi’s famous question Where is everybody?8. Indeed, no signs of Aliens have ever been found. Why? Many physicists, cosmologists, and evolutionary biologists argue that a plausible answer to Fermi’s question could be that our species Homo Sapiens resulted from an extremely rare combination of circumstances that started with the Big Bang, continued during the evolution of the Universe, the Galaxy and the Solar System, and proceeded during the Darwinian evolution on Earth. If the rare Earth principle9 is indeed the rule that limits the emergence of intelligent life, then advanced extraterrestrial civilisations should be genuinely rare – scattered over vast distances in space and time. If the signal announcing somebody’s existence originates always in situ, at their respective planetary system, then all search strategies are condemned to be “needle in the haystack” searches. Any intrepid prospector must then cope with the omnipresent noise in which synthetic signals are buried. These signals have an a priori unknown physical nature, are characterised by a priori unknown frequencies and durations, and come from a priori unknown directions. Even if the rare Earth considerations will turn out to be irrelevant or incorrect, Fermi’s wonderment – Where is everybody? – remains today a very relevant observational issue. No one had and has an obvious search strategy which would guarantee the success. Finding a signal requires not only skill but also luck.
We suggest a radically novel approach – searching for a unique, very particular signal of an a priori precisely and accurately known nature, frequency and direction. Our suggestion follows from the hypothesis that once upon a time another civilisation existed in our Galaxy, a highly developed civilisation whose technolocial activities were only limited by the fundamental laws of Physics. They had at their disposal the expertise to operate on a grand interstellar scale, in particular the conversion of the power output of stars to run their instruments. Our suggestion may also apply to future humans with the necessary technical knowledge. If someone wanted to unambiguously announce their existence to the entire Galaxy, they would construct a “Messenger” with the following properties
1. 1.
The Messenger location is fundamentally unique and obvious to anyone in the Galaxy,
2. 2.
The physical nature of the signal and the signal frequency are known a priori,
3. 3.
The emitted power assures the signal’s detectability in the whole Galaxy in terms of space,
4. 4.
Messenger’s life time assures the signal’s detectability in the whole Galaxy in terms of time,
5. 5.
The energy supply is provided by a natural astronomical phenomenon,
6. 6.
No maintenance is needed; the Messenger is a fully autonomous device,
7. 7.
The artificial origin of the signal unambiguously follows from its properties.
We argue below that from fundamental laws of Physics and logical deduction it follows necessarily that such a Messenger should ideally be a Jupiter-mass black hole, orbiting Sgr A* for a few billion years at the retrograde innermost stable circular orbit (ISCO), and therefore naturally emitting gravitational waves with the frequency $${f}_{{\rm{G}}{\rm{W}}}=0.63\text{-}1.07\,{\rm{m}}{\rm{H}}{\rm{z}}$$ and power $${P}_{{\rm{G}}{\rm{W}}}=2.7\times {10}^{36}\text{-}2.0\times {10}^{37}$$ erg/s, depending on the black hole spin.
## Location
Sagittarius A* (Sgr A*), the supermassive black hole in the centre of the Milky Way, is a unique object. Any advanced civilisation will, without any doubt, notice the existence of Sgr A* and recognise it as a unique location. We humans are aware of this for less than a century. Our recognition of Sgr A* started with the discovery of a radio hiss from the approximate direction of the Galactic centre10, later X-ray emission from the same directions was detected during an early Aerobee survey11, but that it was a point source became clear only with radio interferometry12. It took a few decades more to identify this compact source as a black hole using infrared spectroscopy of the radial and proper motion of nearby stars13,14. Only recently Sgr A* has become our remote laboratory site where we can test gravity theories15,16.
The mass of the Galactic centre black hole is very precisely known from direct measurements of stellar orbits around Sgr A*15, $$M=4.14\pm 0.03\times {10}^{6}\,{M}_{\odot }$$. The black hole spin $$J$$ is not known to date. Broadband spectral fitting indicates a broad range of the dimensionless Kerr spin parameter $$a:\,=J/{M}^{2}$$, namely $$a=0.0+0.86$$ (2$$\sigma$$ uncertainty)17. Once the black hole spin of Sgr A* can be determined, the gravitational wave frequency and power of any matter in orbit around it becomes very well constrained.
The uniqueness of the location of the signal is guaranteed by placing the Messenger at a unequivocally distinguishable “Keplerian” orbit around Sgr A*. According to Einstein’s general relativity, there is only one type of orbit to consider – the innermost stable circular orbit (ISCO). All characteristics of the ISCO, in particular its radius, $${r}_{0}$$, and the orbital frequency, $${f}_{0}$$, depend only on the black hole mass, $$m$$, and its dimensionless spin, $$a$$. For non-rotating black holes, $$a=0$$, there is a single ISCO orbit, for $$a\ne 0$$ there is a pair of two well separated ISCO orbits in the equatorial plane: a closer “prograde” orbit (where the orbital momentum has the same direction like the angular momentum of the black hole) and a “retrograde” orbit (orbital momentum has the opposite direction to the angular momentum of the black hole). Retrograde orbits are indicated with a minus sign, e.g. $$a=-\,0.9$$. The best strategy would be to put the Messenger on the retrograde ISCO. Firstly, as we discuss later, the retrograde ISCO has a lower orbital energy than the prograde ISCO and secondly, natural astrophysical objects are less likely to settle on long lasting retrograde orbits than on prograde orbits. An object that stays for a long time at the retrograde ISCO must immediately be suspected to have an artificial origin.
Adopting for the mass the value $$M=4.1\times {10}^{6}\,{M}_{\odot }$$ and for the spin $$a=0$$ and ±0.9, respectively, the radius $${r}_{0}$$ (in Boyer-Lindquist coordinates) and the orbital frequency $${f}_{0}$$ (as measured at infinity) at the prograde and retrograde ISCO radii are
$$\begin{array}{c}{r}_{0}=\{\begin{array}{cc}3.63\times {10}^{12}\,{\rm{c}}{\rm{m}}\, & {\rm{f}}{\rm{o}}{\rm{r}}\,a=\,0.0\\ 1.41\times {10}^{12}\,{\rm{c}}{\rm{m}}\, & {\rm{f}}{\rm{o}}{\rm{r}}\,a=+0.9\\ 5.27\times {10}^{12}\,{\rm{c}}{\rm{m}}\, & {\rm{f}}{\rm{o}}{\rm{r}}\,a=-0.9\end{array}\\ {f}_{0}=\{\begin{array}{cc}0.54\,{\rm{m}}{\rm{H}}{\rm{z}} & {\rm{f}}{\rm{o}}{\rm{r}}\,a=\,0.0\\ 1.78\,{\rm{m}}{\rm{H}}{\rm{z}} & {\rm{f}}{\rm{o}}{\rm{r}}\,a=+\,0.9\\ 0.32\,{\rm{m}}{\rm{H}}{\rm{z}} & {\rm{f}}{\rm{o}}{\rm{r}}\,a=-\,0.9\end{array}\end{array}$$
(1)
The functions $${r}_{0}(a)$$ and $${f}_{0}(a)$$ are shown in Fig. 1 Any circular orbital motion generates gravitational waves which are periodic with a dominant mode at twice the orbital frequency18,19. From Eq. (1) we obtain the gravitational wave frequency of the dominant $$m=2$$ mode:
$${f}_{{\rm{G}}{\rm{W}}}=2{f}_{0}=\{\begin{array}{cc}1.07\,{\rm{m}}{\rm{H}}{\rm{z}} & {\rm{f}}{\rm{o}}{\rm{r}}\,a=\,0.0\\ 3.55\,{\rm{m}}{\rm{H}}{\rm{z}} & {\rm{f}}{\rm{o}}{\rm{r}}\,a=+0.9\\ 0.63\,{\rm{m}}{\rm{H}}{\rm{z}} & {\rm{f}}{\rm{o}}{\rm{r}}\,a=-0.9\end{array}$$
(2)
This frequency falls within the sensitivity range of the space borne gravitational-wave observatory LISA20,21. The full waveform is described in19.
### Physical nature-Emitted power
In order to generate a reasonably strong gravitational-wave signal, the body orbiting Sgr A* must have a mass in an astronomical range. It cannot be too small or it does not produce a gravitational-wave amplitude detectable over a substantial distance for Sgr A*, and it cannot be too big or it will cost too much energy to sustain its orbit over a sufficiently long time. Our calculations indicate that a detectable Messenger mass is in the ballpark of moons, planets, or stars.
The energy loss due to gravitational radiation by an object of mass $$m$$ in a circular orbit around a rotating black hole of mass $$M\gg m$$ has been computed in19 (The result can be found in the public SageMath notebook listed at the end of the article). For a Jupiter-mass Messenger, $$m=2\times {10}^{27}$$ kg, orbiting at ISCO [Eq. (1)], this yields the following numerical values of the total radiated power:
$${P}_{{\rm{G}}{\rm{W}}}=\frac{{\rm{d}}E}{{\rm{d}}t}=\{\begin{array}{cc}2.0\times {10}^{37}\,{\rm{e}}{\rm{r}}{\rm{g}}\,{{\rm{s}}}^{-1} & {\rm{f}}{\rm{o}}{\rm{r}}\,a=\,0.0\\ 7.8\times {10}^{38}\,{\rm{e}}{\rm{r}}{\rm{g}}\,{{\rm{s}}}^{-1} & {\rm{f}}{\rm{o}}{\rm{r}}\,a=+0.9\\ 2.7\times {10}^{36}\,{\rm{e}}{\rm{r}}{\rm{g}}\,{{\rm{s}}}^{-1} & {\rm{f}}{\rm{o}}{\rm{r}}\,a=-0.9\end{array}$$
(3)
Values of $${P}_{{\rm{GW}}}$$ for other values of $$m$$ can be found in Table 1. The above results relies on exact computations in the Kerr metric19. As a check, let us note that an approximate estimate of $${P}_{{\rm{GW}}}$$ can be obtained by a 1.5-order post-Newtonian formula based on22:
$${P}_{{\rm{GW}}}\simeq \frac{32}{5}{\left(\frac{M}{{r}_{0}}\right)}^{5}{\left(\frac{m}{M}\right)}^{2}\times \left(1+\frac{13}{168}\frac{M}{2{r}_{0}}-\frac{73}{12}\frac{a{M}^{3}}{{L}^{3}}\right)\times \frac{{c}^{5}}{G},$$
(4)
where $$L$$ the specific angular momentum of a Keplerian orbit. Equation (4) results in values in agreement with Eq. (3) up to ~10%.
The galactic centre is a busy place with a nuclear star cluster, dense cloudy objects, and a putative population of stellar remnants. At the location of the ISCO, in close proximity to the black hole, the environment consists mostly of ionised gas. Jupiter as such would be tidally disrupted at the ISCO. The same is valid for stars. Only objects with densities $$> \,{10}^{6}\,{\rm{g}}\,{{\rm{cm}}}^{-3}$$ (comparable to a white dwarf) are safe19. A conceivable strategy have a “Jupiter” at the ISCO would be to compress it, or collapse it into a black hole. A discussion of the engineering aspects would go beyond the scope of our work. Since only small object like brown dwarfs, planets, and asteroids survive the tidal force of Sgr A* we don’t expect any major dynamical encounters with a satellite at the ISCO. The probability of a rock having a close encounter with a Jupiter mass black hole - which has a radius of less than 300 cm - is negligibly small. The resident gas can exert a drag force on the Messenger. The power of the drag force is given by the density of the gas as well as the velocity and area of the orbiting body. In the specific case of a Jupiter-mass black hole in orbit around Sgr A*, which is in all probability surrounded by a very tenuous plasma, $${P}_{{\rm{drag}}}\propto \rho {v}^{3}A\approx {10}^{18}\,{\rm{erg}}\,{{\rm{s}}}^{-1}$$, is negligible with respect to Eq. (3).
For a mass $$m=2\times {10}^{27}$$ kg orbiting at the ISCO, it has been calculated in ref. 19 that within $${T}_{{\rm{obs}}}=1$$ yr of observational time, LISA can detect the gravitational waves with a signal-to-noise ratio as high as
$${{\rm{S}}{\rm{N}}{\rm{R}}}_{{\rm{G}}{\rm{W}}}=\{\begin{array}{cc}2.8\times {10}^{2} & {\rm{f}}{\rm{o}}{\rm{r}}\,a=\,0.0\\ 3.6\times {10}^{3} & {\rm{f}}{\rm{o}}{\rm{r}}\,a=+0.9\\ 5.5\times {10}^{1} & {\rm{f}}{\rm{o}}{\rm{r}}\,a=-0.9\end{array}$$
(5)
Such a signal will be easily recognised as a true gravitational-wave emission. Within an observational period of five years, LISA can detect orbiter masses as small as super-Earths.
Nobody knows how long a technically-advanced civilisation lives, but it is likely that their lifetime is significantly shorter than a few billion years. Therefore, the Messenger beacon should last for about this time
$$\begin{array}{ll}{\rm{The}}\,{\rm{Messenger}}\,{\rm{lifetime}}\,: & {t}_{{\rm{life}}}\approx {10}^{9}\,{\rm{yr}}={10}^{17}\,{\rm{s}}\end{array}$$
(6)
otherwise its signal could be easily missed by other Galactic civilisations. Thus, the lifetime of the Messenger will be much longer than that of the civilisation that created it. If we ever discover the gravitational signal discussed here, it will possibly be a message from a dead civilisation. Loeb23 has discussed this issue in a different context of searching for “space junk” – relics of dead civilisations.
The power [Eq. (3)] must be continuously supplied to the Messenger in order to compensate the orbital energy losses due to gravitational radiation. Without this supply, the Messenger would drift out of the orbit and plunge into the black hole Sgr A*. The change of the orbital frequency in time, $${\rm{d}}f/{\rm{d}}t$$ is diverging at the ISCO19,24, as indicated in Fig. 2. For comparison, long-lasting observations of LISA provide the following frequency derivative resolution: if the observation time of LISA is $${T}_{{\rm{obs}}}$$, the frequency resolution (width of the frequency bin) is $$\delta f=\mathrm{1/}{T}_{{\rm{obs}}}$$. Minimal frequency derivative $$\dot{f}={\rm{d}}f/{\rm{d}}t$$ which can be detected (signals with smaller $$\dot{f}$$ will stay within the same frequency bin in time $${T}_{{\rm{obs}}}$$) is then $${\dot{f}}_{{\rm{\min }}}=\delta f/{T}_{{\rm{obs}}}=\mathrm{1/}{T}_{{\rm{obs}}}^{2}$$. For $${T}_{{\rm{obs}}}$$ = 1 year, $${\dot{f}}_{{\rm{\min }}}\simeq {10}^{-15}$$ Hz/s, which, in view of Fig. 2, is more than enough to distinguish the signal from a naturally decaying orbit at ISCO from a strictly monochromatic synthetic signal.
This is in our opinion the most important feature of the Messenger concept. If an observer detects a continuous, long-lasting (of the order of months or years), gravitational-wave signal with a stable frequency corresponding to the retrograde ISCO of Sgr A*, then it is obvious that this signal is artificial.
### Energy supply - No maintenance
Given Eq. (3), the total energy needed to supply a Jupiter-mass Messenger over its lifetime is
$${\mathscr{E}}={t}_{{\rm{l}}{\rm{i}}{\rm{f}}{\rm{e}}}\,{P}_{{\rm{G}}{\rm{W}}}\approx \{\begin{array}{cc}2.0\times {10}^{54}\,{\rm{e}}{\rm{r}}{\rm{g}} & {\rm{f}}{\rm{o}}{\rm{r}}\,a=\,0.0\\ 7.8\times {10}^{55}\,{\rm{e}}{\rm{r}}{\rm{g}} & {\rm{f}}{\rm{o}}{\rm{r}}\,a=+0.9\\ 2.7\times {10}^{53}\,{\rm{e}}{\rm{r}}{\rm{g}} & {\rm{f}}{\rm{o}}{\rm{r}}\,a=-0.9\end{array}$$
(7)
Restricted to the retrograde ISCO, this corresponds to a fuel mass $${\mathscr{M}}={\mathscr{E}}/{c}^{2}\sim 3\times {10}^{32}-2\times {10}^{33}\,{\rm{g}}$$, depending on the black hole spin.
Therefore, the mass of the Messenger cannot itself be a source for the energy supply needed even in the extreme case of the 100% efficiency of the $$E=m{c}^{2}$$ conversion – energy has to be supplied from outside. Fortunately, a natural process involving a single star of about 0.1–1 $${M}_{\odot }$$, will suffice if the Messenger mass is of the order of a Jupiter mass. For a Messenger with a mass as much as $$1\,{M}_{\odot }$$, however, the energy needed to sustain its orbit would be 6 orders of magnitude larger, meaning that a supply by one star would be not possible.
## Discussion
The “Galactic Centre Messenger” is a thought experiment that originated from the question, what kind of a mass on what kind of an orbit around Sgr A* would produce a measurable gravitational wave signal and what kind of an energy would be required to make that signal continuous. We show by means of a few simple calculations that the energy supply of one solar mass can sustain one Jupiter mass in an orbit for one billion years. This means it is energetically feasible to stabilise a gravitational wave emitting mass for a few billion years in orbit close to the Galactic centre black hole. Given this finding, we want to propose gravitational waves as a new road to detect, or search for, intelligent life in the Galaxy. It is a promising and exciting way that could be used in conjunction with the classical SETI radio searches, much like we perform multi-messenger astronomy with LIGO/Virgo and electromagnetic observatories today.
Would future humans or highly developed extraterrestrial civilisations actually position a device with a Messenger function in orbit around Sgr A*? Nobody knows their intentions or technological skills. They might construct a Jupiter-mass probe at the ISCO to explore the supermassive black hole in the Galactic centre, or to extract and harness its energy, or even for intentions unfathomable to the human mind. Such devices would also serve as Messengers and their gravitational wave signature may be picked up by LISA. Alternatively, they might choose to assemble a radio (or other electromagnetic wave) transmitter. Gravitational and electromagnetic wave beacons may a priori be located anywhere. The easiest and cheapest way would perhaps be to mount an antenna on a satellite around a stellar remnant in ones neighbourhood. But even the brightest electromagnetic signal is easily missed because its direction and wavelength are unknown to others. We argue that if someone wanted to communicate in the Galaxy, which is an intrinsically difficult task given the Galactic distances and morphology that have to be overcome, the Galactic centre black hole is a predictable focal point, i.e. the Schelling point of our Galaxy. Schelling asked25: “If you are to meet a stranger in New York City, but you cannot communicate with the person, then when and where will you choose to meet?” Similarly, in a Galaxy where it is hard to communicate, we ask, where will you choose to look? The most natural answer is Sgr A*.
We do not know “where everybody is” but we know one thing for sure: already in the first year of its operation LISA will be able to verify if a Jupiter-mass orbiter is present in the Galactic centre. Smaller masses will require longer monitoring to reach a conclusive signal-to-noise ratio. The absence of a continuous signal from the direction of Sgr A* does, of course, not imply that extraterrestrial life has not evolved elsewhere. A successful detection, however, will provide a definite and unambiguous proof that an intelligent civilisation did exist in our Galaxy.
## References
1. 1.
Wright, J. T., Kanodia, S. & Lubar, E. How Much SETI Has Been Done? Finding Needles in the n-dimensional Cosmic Haystack. AJ 156, 260, https://doi.org/10.3847/1538-3881/aae099, 1809.07252 (2018).
2. 2.
Abbott, B. P. et al. Prospects for observing and localizing gravitational-wave transients with Advanced LIGO, Advanced Virgo and KAGRA. Living Rev. Relativ. 21, 3, https://doi.org/10.1007/s41114-018-0012-9, 1304.0670 (2018).
3. 3.
Abbott, B. P. et al. Observation of Gravitational Waves from a Binary Black Hole Merger. Phys. Rev. Lett. 116, 061102, https://doi.org/10.1103/PhysRevLett.116.061102, 1602.03837 (2016).
4. 4.
Abbott, B. P. et al. GW170104: Observation of a 50-Solar-Mass Binary Black Hole Coalescence at Redshift 0.2. Phys. Rev. Lett. 118, 221101, https://doi.org/10.1103/PhysRevLett.118.221101, 1706.01812 (2017).
5. 5.
Abbott, B. P. et al. GW170814: A Three-Detector Observation of Gravitational Waves from a Binary Black Hole Coalescence. Phys. Rev. Lett. 119, 141101, https://doi.org/10.1103/PhysRevLett.119.141101, 1709.09660 (2017).
6. 6.
Abbott, B. P. et al. GW170817: Observation of Gravitational Waves from a Binary Neutron Star Inspiral. Phys. Rev. Lett. 119, 161101, https://doi.org/10.1103/PhysRevLett.119.161101, 1710.05832 (2017).
7. 7.
The LIGO Scientific Collaboration & the Virgo Collaboration. GWTC-1: A Gravitational-Wave Transient Catalog of Compact Binary Mergers Observed by LIGO and Virgo during the First and Second Observing Runs. Phys. Rev. X 9, 031040, https://doi.org/10.1103/PhysRevX.9.031040, 1811.12907 (2019).
8. 8.
Fermi, E. A lunch conversation with edward teller, herbert york and emil konopinski. (1950).
9. 9.
Ward, P. D., Brownlee, D. & Krauss, L. Rare Earth: Why Complex Life Is Uncommon in the Universe. Phys. Today 53, 62, https://doi.org/10.1063/1.1325239 (2000).
10. 10.
Jansky, K. G. Radio Waves from Outside the Solar System. Nature 132, 66, https://doi.org/10.1038/132066a0 (1933).
11. 11.
Bowyer, S., Byram, E. T., Chubb, T. A. & Friedman, H. Cosmic X-ray Sources. Science 147, 394–398, https://doi.org/10.1126/science.147.3656.394 (1965).
12. 12.
Balick, B. & Brown, R. L. Intense sub-arcsecond structure in the galactic center. ApJ 194, 265–270, https://doi.org/10.1086/153242 (1974).
13. 13.
Genzel, R., Eckart, A., Ott, T. & Eisenhauer, F. On the nature of the dark mass in the centre of the Milky Way. MNRAS 291, 219–234, https://doi.org/10.1093/mnras/291.1.219 (1997).
14. 14.
Ghez, A. M., Klein, B. L., Morris, M. & Becklin, E. E. High Proper-Motion Stars in the Vicinity of Sagittarius A*: Evidence for a Supermassive Black Hole at the Center of Our Galaxy. ApJ 509, 678–686, astro-ph/9807210 (1998).
15. 15.
GRAVITY Collaboration et al. Detection of the gravitational redshift in the orbit of the star S2 near the Galactic centre massive black hole. A& A 615, L15, https://doi.org/10.1051/0004-6361/201833718, 1807.09409 (2018).
16. 16.
GRAVITY Collaboration et al. Detection of orbital motions near the last stable circular orbit of the massive black hole SgrA*. A& A 618, L10, https://doi.org/10.1051/0004-6361/201834294, 1810.12641 (2018).
17. 17.
Broderick, A. E., Fish, V. L., Doeleman, S. S. & Loeb, A. Evidence for Low Black Hole Spin and Physically Motivated Accretion Models from Millimeter-VLBI Observations of Sagittarius A*. ApJ 735, 110, https://doi.org/10.1088/0004-637X/735/2/110, 1011.2770 (2011).
18. 18.
Detweiler, S. L. Black holes and gravitational waves. I. Circular orbits about a rotating hole. ApJ 225, 687–693, https://doi.org/10.1086/156529 (1978).
19. 19.
Gourgoulhon, E., Le Tiec, A., Vincent, F. H. & Warburton, N. Gravitational waves from bodies orbiting the Galactic center black hole and their detectability by LISA. A& A 627, A92, https://doi.org/10.1051/0004-6361/201935406, 1903.02049 (2019).
20. 20.
Amaro-Seoane, P. et al. Laser Interferometer Space Antenna. arXiv e-prints, 1702.00786 (2017).
21. 21.
Robson, T., Cornish, N. J. & Liu, C. The construction and use of LISA sensitivity curves. Class. Quantum Gravity 36, 105011, https://doi.org/10.1088/1361-6382/ab1101, 1803.01944 (2019).
22. 22.
Shibata, M. Gravitational waves by compact star orbiting around rotating supermassive black holes. Phys.Rev.D 50, 6297–6311, https://doi.org/10.1103/PhysRevD.50.6297 (1994).
23. 23.
Loeb, A. How to Search for Dead Cosmic Civilizations. Sci. Am. (2018).
24. 24.
Finn, L. S. & Thorne, K. S. Gravitational waves from a compact star in a circular, inspiral orbit, in the equatorial plane of a massive, spinning black hole, as observed by LISA. Phys. Rev. D 62, 124021, https://doi.org/10.1103/PhysRevD.62.124021, gr-qc/0007074 (2000).
25. 25.
Schelling, T. C. The strategy of conflict (Cambridge University Press, 1960).
## Acknowledgements
M.A. acknowledges the Polish NCN grant 2015/19/B/ST9/01099 and the Czech Science Foundation grant No. 17-16287S which supported his visits to Paris Observatory. M.B. was partially supported by the Polish NCN grant 2016/22/E/ST9/00037.
## Author information
Authors
### Contributions
Idea: M.A. and E.G.; Text: O.S. and M.A.; Calculations: E.G., O.S., M.A. and M.B.; Figures: O.S. and E.G. All authors contributed equally to the discussion and review of the manuscript.
### Corresponding author
Correspondence to Odele Straub.
## Ethics declarations
### Competing interests
The authors declare no competing interests.
Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
## Rights and permissions
Reprints and Permissions
Abramowicz, M., Bejger, M., Gourgoulhon, É. et al. A Galactic centre gravitational-wave Messenger. Sci Rep 10, 7054 (2020). https://doi.org/10.1038/s41598-020-63206-1
• Accepted:
• Published:
|
2021-05-15 06:47:31
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7332058548927307, "perplexity": 1431.5725264248906}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989812.47/warc/CC-MAIN-20210515035645-20210515065645-00115.warc.gz"}
|
http://tex.stackexchange.com/questions/32276/page-break-table-with-single-cell
|
# Page break table with single cell
I have an environment set up of longtables that I use as "tips"-boxes and exercises in my book. The left column shows only one word rotated (tips, exercise or whatever) while the right contains the rest of the text. Now the problem is that the right column is a single cell so (xe)latex decides to put the table on a new page if it is too big. Here is an example of what I mean:
So how can I make it start on the current page and break whenever necessary, then continue on the next page?
My code for the table:
\documentclass[11pt]{book}
\usepackage{longtable}
\usepackage{array}
\usepackage{rotating}
\usepackage{lipsum}
\newcolumntype{M}[1]{>{%
\parskip=0.5\baselineskip%
\setlength{\parfillskip}{30pt plus 1fil}}m{#1}}
\newenvironment{exercise}[1]
{
\begin{longtable}{m{0.05\textwidth}|M{0.85\textwidth}}
\rotatebox{90}{\textbf{Övning}} & \textbf{#1} \newline
}
{\end{longtable}
\bigskip
}
\begin{document}
\lipsum
\begin{exercise}{1}
\lipsum
\end{exercise}
\end{document}
-
It is always best to compose a MWE that illustrates the problem including the \documentclass so that those trying to help don't have to recreate it. – Peter Grill Oct 21 '11 at 18:49
Do you have to use a longtable or would you consider an answer involving mdframed? – cmhughes Oct 21 '11 at 18:54
@PeterGrill: updated with an MWE now, – pg-robban Oct 21 '11 at 19:02
@cmhughes: I never heard of mdframed, looking it up now. – pg-robban Oct 21 '11 at 19:02
You can't use longtable is such a combination. In the comment of cmhughes the package mdframed was recommended. In the example below you can find two possibilities with mdframed.
If you have further questions you can ask ;-)
EDIT: CHANGED CODE
\documentclass[11pt]{book}
\usepackage[framemethod=default]{mdframed}
\usepackage{rotating}
\usepackage{lipsum}
\makeatletter
\newrobustcmd\Titleenv[1]{%
\setbox0=\hbox{\rotatebox[origin=cc]{90}{\textbf{#1}}}
\setlength\@tempdimc{\dimexpr\ht0+\dp0\relax}
\renewcommand*\md@@frametitle{%
\llap{\smash{\raisebox{-.5\@tempdimc}{\box0}\hspace*{2em}}}}%
\appto\md@frame@leftline@single{%
\ifdimgreater{\mdfboundingboxheight}{\@tempdimc}%
{}%
{%
\llap{\color{\mdf@middlelinecolor}%
\rule[\dimexpr-.5\@tempdimc-\mdf@innerbottommargin@length\relax]%
{\mdf@middlelinewidth@length}%
{\dimexpr\@tempdimc+\mdf@innerbottommargin@length
+\mdf@innertopmargin@length\relax}%
}%
}%
}%
}
\newenvironment{exercise}{%
\mdfsetup{topline=false,rightline=false,bottomline=false,linecolor=red,
linewidth=2pt,skipbelow=\topskip,skipabove=\topskip}
\Titleenv{\"Ovning}%
\begin{mdframed}[]%
}{\end{mdframed}}
\begin{document}
\begin{exercise}
\lipsum[1]
\end{exercise}
\lipsum[1]
\begin{exercise}
Text
\end{exercise}
\lipsum[1]
\begin{exercise}
\lipsum
\end{exercise}
\end{document}
The code above results in the following picture:
-
it is better to call tlmgr --self --all update then the manager can update itself, too – Herbert Oct 22 '11 at 9:36
It is also possible to have the line and vertical text inside the default text width
\documentclass[11pt]{book}
\usepackage{framed}
\usepackage{rotating}
\usepackage{lipsum}
\newenvironment{exercise}[1]
{\par\begin{leftbar}\noindent\textbf{#1}\\
\makebox(0,0){\put(-30,-50){\rotatebox{90}{\textbf{Övning}}}}\ignorespaces}
{\end{leftbar}\par}
\begin{document}
\lipsum
\begin{exercise}{1}
\lipsum
\end{exercise}
\end{document}
-
I tried this option, and I had two problems: 1) I use lstlistings and it seems like the line numbers get in the way of the bar. 2): There seems to be spacing issues: i.imgur.com/FJxMI.png – pg-robban Oct 22 '11 at 11:32
give an example otherwise I cannot say what goes wrong – Herbert Oct 22 '11 at 12:48
|
2015-05-23 06:08:54
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8292056918144226, "perplexity": 1644.4412187841872}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207927245.60/warc/CC-MAIN-20150521113207-00160-ip-10-180-206-219.ec2.internal.warc.gz"}
|
https://socratic.org/questions/the-area-under-the-curve-y-e-x-between-x-0-and-x-1-is-rotated-about-the-x-axis-f
|
The area under the curve y=e^-x between x=0 and x=1 is rotated about the x axis find the volume?
Apr 10, 2018
Volume is $\frac{\pi}{2} \left(1 - {e}^{-} 2\right) = 1.358$ cubic units.
Explanation:
Let us see the graph of $y = {e}^{- x}$ between $x = 0$ and $x = 1$.
graph{e^(-x) [-2.083, 2.917, -0.85, 1.65]}
To find the desired volume the shaded portion (shown below, will have to be rotated around $x$-axis.
As volume of a cylinder is $\pi {r}^{2} h$, here we will have $r = {e}^{- x}$ and $h = \mathrm{dx}$
and hence volume would be
${\int}_{0}^{1} \pi {e}^{- 2 x} \mathrm{dx}$
= $\pi {\int}_{0}^{1} {e}^{- 2 x} \mathrm{dx}$
= $\pi \times {\left[- {e}^{- 2 x} / 2\right]}_{0}^{1}$
= $\pi \left[- {e}^{- 2} / 2 + \frac{1}{2}\right]$
= $\frac{\pi}{2} \left(1 - {e}^{-} 2\right)$
= $\frac{\pi}{2} \left(1 - 0.1353\right)$
= $1.358$
|
2019-03-24 05:44:12
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.695874035358429, "perplexity": 659.1968322095029}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912203326.34/warc/CC-MAIN-20190324043400-20190324065400-00243.warc.gz"}
|
https://tbc-python.fossee.in/convert-notebook/Fundamental_of_Thermodynamics_by_Moran_and_Shapiro/Chapter_8.ipynb
|
# Chapter 8 :- Vapour Power Systems¶
## Example 8.1 Page no-331
In [1]:
# Given:-
p1 = 8.0 # pressure of saturated vapor entering the turbine in MPa
p3 = 0.008 # pressure of saturated liquid exiting the condenser in MPa
Wcycledot = 100.00 # the net power output of the cycle in MW
# Analysis
# From table A-3
h1 = 2758.0 # in kj/kg
s1 = 5.7432 # in kj/kg.k
s2 = s1
sf = 0.5926 # in kj/kg.k
sg = 8.2287 # in kj/kg.k
hf = 173.88 # in kj/kg
hfg = 2403.1 # in kj/kg
v3 = 1.0084e-3 # in m^3/kg
# State 3 is saturated liquid at 0.008 MPa, so
h3 = 173.88 # in kj/kg
# Calculations
x2 = (s2-sf)/(sg-sf) # quality at state 2
h2 = hf + x2*hfg
p4 = p1
h4 = h3 + v3*(p4-p3)*10**6*10**-3 # in kj/kg
# Part(a)
#Mass and energy rate balances for control volumes around the turbine and pump give, respectively
wtdot = h1 - h2
wpdot = h4-h3
# The rate of heat transfer to the working fluid as it passes through the boiler is determined using mass and energy rate balances as
qindot = h1-h4
eta = (wtdot-wpdot)/qindot # thermal efficiency)
# Result for part a
print '-> The thermal efficiency for the cycle is ',round(eta,2)
# Part(b)
bwr = wpdot/wtdot # back work ratio
# Result
print '-> The back work ratio is ',bwr
# Part(c)
mdot = (Wcycledot*10**3*3600)/((h1-h2)-(h4-h3)) # mass flow rate in kg/h
# Result
print '-> The mass flow rate of the steam is',round(mdot,2),'kg/h .'
# Part(d)
Qindot = mdot*qindot/(3600*10**3) # in MW
# Results
print '-> The rate of heat transfer,Qindot , into the working fluid as it passes through the boiler, is',round(Qindot,2),'MW.'
# Part(e)
Qoutdot = mdot*(h2-h3)/(3600*10**3) # in MW
# Results
print '-> The rate of heat transfer,Qoutdot from the condensing steam as it passes through the condenser, is',round(Qoutdot,2),'MW.'
# Part(f)
# From table A-2
hcwout= 146.68 # in kj/kg
hcwin= 62.99 # in kj/kg
mcwdot= (Qoutdot*10**3*3600)/(hcwout-hcwin) # in kg/h
# Results
print '-> The mass flow rate of the condenser cooling water is',round(mcwdot,2),'kg/ h.'
-> The thermal efficiency for the cycle is 0.37
-> The back work ratio is 0.00836692570976
-> The mass flow rate of the steam is 376902.57 kg/h .
-> The rate of heat transfer,Qindot , into the working fluid as it passes through the boiler, is 269.7 MW.
-> The rate of heat transfer,Qoutdot from the condensing steam as it passes through the condenser, is 169.7 MW.
-> The mass flow rate of the condenser cooling water is 7299844.18 kg/ h.
## Example 8.2 Page no-338
In [2]:
# Given:-
etat= .85 # given that the turbine and the pump each have an isentropic efficiency of 85%
# Analysis
# State 1 is the same as in Example 8.1, so
h1 = 2758.0 # in kj/kg
s1 = 5.7432 # in kj/kg.k
# From example 8.1
h1 = 2758.0 # in kj/kg
h2s = 1794.8 # in kj/kg
# State 3 is the same as in Example 8.1, so
h3 = 173.88 # in kj/kg
# Calculations
h2 = h1 - etat*(h1-h2s) # in kj/kg
wpdot = 8.06/etat # where the value 8.06 is obtained from example 8.1
h4 = h3 + wpdot
# Part(a)
eta = ((h1-h2)-(h4-h3))/(h1-h4) # thermal efficiency
# Result for part (a)
print '-> Thermal efficiency is: ',round(eta,3)
# Part(b)
Wcycledot = 100 # given,a net power output of 100 MW
# Calculations
mdot = (Wcycledot*(10**3)*3600)/((h1-h2)-(h4-h3))
# Result for part (b)
print '-> The mass flow rate of steam, in kg/h, for a net power output of 100 MW is ',round(mdot,3),'kg/h.'
# Part(c)
Qindot = mdot*(h1-h4)/(3600 * 10**3)
# Result
print '-> The rate of heat transfer Qindot into the working fluid as it passes through the boiler, is ',round(Qindot,3),'MW.'
# Part(d)
Qoutdot = mdot*(h2-h3)/(3600*10**3)
# Result
print '-> The rate of heat transfer Qoutdotfrom the condensing steam as it passes through the condenser, is ',round(Qoutdot,3),'MW.'
# Part(e)
# From table A-2
hcwout = 146.68 # in kj/kg
hcwin = 62.99 # in kj/kg
mcwdot = (Qoutdot*10**3*3600)/(hcwout-hcwin)
# Result
print '-> The mass flow rate of the condenser cooling water, is: ',round(mcwdot,3),'kg/h.'
-> Thermal efficiency is: 0.314
-> The mass flow rate of steam, in kg/h, for a net power output of 100 MW is 444863.139 kg/h.
-> The rate of heat transfer Qindot into the working fluid as it passes through the boiler, is 318.156 MW.
-> The rate of heat transfer Qoutdotfrom the condensing steam as it passes through the condenser, is 218.156 MW.
-> The mass flow rate of the condenser cooling water, is: 9384172.373 kg/h.
## Example 8.3 Page no-341
In [3]:
# Given:-
T1 = 480.0 # temperature of steam entering the first stage turbine in degree celcius
p1 = 8.0 # pressure of steam entering the first stage turbine in MPa
p2 = 0.7 # pressure of steam exiting the first stage turbine in MPa
T3 = 440.0 # temperature of steam before entering the second stage turbine
Pcond = 0.008 # condenser pressure in MPa
Wcycledot = 100.0 # the net power output in MW
# Analysis
# From table A-4
h1 = 3348.4 # in kj/kg
s1 = 6.6586 # in kj/kg.k
s2 = s1 # isentropic expansion through the first-stage turbine
# From table A-3
sf = 1.9922 # in kj/kg.k
sg = 6.708 # in kj/kg.k
hf = 697.22 # in kj/kg
hfg = 2066.3 # in kj/kg
# Calculations
x2 = (s2-sf)/(sg-sf)
h2 = hf + x2*hfg
# State 3 is superheated vapor with p3 = 0.7 MPa and T3= 440C, so from Table A-4
h3 = 3353.3 # in kj/kg
s3 = 7.7571 # in kj/kg.k
s4 = s3 # isentropic expansion through the second-stage turbine
# For determing quality at state 4,from table A-3
sf = 0.5926 # in kj/kg.k
sg = 8.2287 # in kj/kg.k
hf = 173.88 # in kj/kg
hfg = 2403.1 # in kj/kg
# Calculations
x4 = (s4-sf)/(sg-sf)
h4 = hf + x4*hfg
# State 5 is saturated liquid at 0.008 MPa, so
h5 = 173.88
# The state at the pump exit is the same as in Example 8.1, so
h6 = 181.94
# Part(a)
eta = ((h1-h2)+(h3-h4)-(h6-h5))/((h1-h6)+(h3-h2))
# Result
print '-> The thermal efficiency of the cycle is:',round(eta,2)
# Part(b)
mdot = (Wcycledot*3600*10**3)/((h1-h2)+(h3-h4)-(h6-h5))
print '-> The mass flow rate of steam, is:',round(mdot,2),'kg/h.'
# Part(c)
Qoutdot = (mdot*(h4-h5))/(3600*10**3)
print '-> The rate of heat transfer Qoutdot from the condensing steam as it passes through the condenser, MW is',round(Qoutdot,2),'kg/h.'
-> The thermal efficiency of the cycle is: 0.4
-> The mass flow rate of steam, is: 236344.68 kg/h.
-> The rate of heat transfer Qoutdot from the condensing steam as it passes through the condenser, MW is 148.02 kg/h.
## Example 8.4 Page no-344
In [4]:
%matplotlib inline
Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].
For more information, type 'help(pylab)'.
In [5]:
# Given :-
# Part (a)
etat = 0.85 # given efficiency
# From the solution to Example 8.3, the following specific enthalpy values are known, in kJ/kg
h1 = 3348.4
h2s = 2741.8
h3 = 3353.3
h4s = 2428.5
h5 = 173.88
h6 = 181.94
# Calculations
h2 = h1 - etat*(h1 - h2s) # The specific enthalpy at the exit of the first-stage turbine in kj/kg
h4 = h3 - etat*(h3-h4s) # The specific enthalpy at the exit of the second-stage turbine in kj/kg
eta = ((h1-h2)+(h3-h4)-(h6-h5))/((h1-h6)+(h3-h2))
# Result
print '-> The thermal efficiency is: ',eta
# Part (b)
from numpy import linspace
from pylab import *
h2 = []
h4 = []
y = []
x = linspace(0.85,1,50)
for i in range(0,50):
h2.append(i)
h4.append(i)
y.append(i)
h2[i] = h1 - x[i]*(h1 - h2s) # The specific enthalpy at the exit of the first-stage turbine in kj/kg
h4[i] = h3 - x[i]*(h3-h4s) # The specific enthalpy at the exit of the second-stage turbine in kj/kg
y[i] = ((h1-h2[i])+(h3-h4[i])-(h6-h5))/((h1-h6)+(h3-h2[i]))
plot(x,y)
xlabel('isentropic turbine efficiency')
ylabel('cycle thermal efficiency')
show()
-> The thermal efficiency is: 0.350865344714
## Example 8.5 Page no-348
In [6]:
# Given:-
T1 = 480.0 # temperature of steam entering the turbine in degree celcius
p1 = 8.0 # pressure of steam entering the turbine in MPa
Pcond = 0.008 # condenser pressure in MPa
etat = 0.85 # turbine efficiency
Wcycledot = 100.0 # net power output of the cycle
# Analysis
# With the help of steam tables
h1 = 3348.4 # in kj/kg
h2 = 2832.8 # in kj/kg
s2 = 6.8606 # in kj/kg.k
h4 = 173.88 # in kj/kg
# With s3s = s2, the quality at state 3s is x3s= 0.8208; using this, we get
h3s = 2146.3 # in kj/kg
# Calculations
# The specific enthalpy at state 3 can be determined using the efficiency of the second-stage turbine
h3 = h2 - etat*(h2-h3s)
# State 6 is saturated liquid at 0.7 MPa. Thus,
h6 = 697.22 # in kj/kg
# For determining specific enthalpies at states 5 and 7 ,we have
p5 = 0.7 # in MPa
p4 = 0.008 # in MPa
p7 = 8.0 # in MPa
p6 = 0.7 # in MPa
v4 = 1.0084e-3 # units in m^3/kg,obtained from steam tables
v6 = 1.1080e-3 # units in m^3/kg,obtained from steam tables
# Calculations
h5 = h4 + v4*(p5-p4)*10**6*10**-3 # in kj/kg
h7 = h6 + v6*(p7-p6)*10**3 # in kj/kg
# Applying mass and energy rate balances to a control volume enclosing the open heater, we find the fraction y of the flow extracted at state 2 from
y = (h6-h5)/(h2-h5)
# Part(a)
wtdot = (h1-h2) + (1-y)*(h2-h3) # the total turbine work output, units in KJ/Kg
wpdot = (h7-h6) + (1-y)*(h5-h4) # The total pump work per unit of mass passing through the first-stage turbine,in KJ/kg
qindot = h1 - h7 # in kj/kg
eta = (wtdot-wpdot)/qindot
# Results
print '-> The thermal efficiency is:',round(eta,2)
# Part(b)
m1dot = (Wcycledot*3600*10**3)/(wtdot-wpdot)
# Results
print '-> The mass flow rate of steam entering the first turbine stage, is:',round(m1dot,2),'kg/h.'
-> The thermal efficiency is: 0.37
-> The mass flow rate of steam entering the first turbine stage, is: 368948.05 kg/h.
## Example 8.6 Page no-352
In [7]:
# Given:-
# Analysis
# State 1 is the same as in Example 8.3, so
h1 = 3348.4 # in kj/kg
s1 = 6.6586 # in kj/kg.k
# State 2 is fixed by p2 2.0 MPa and the specific entropy s2, which is the same as that of state 1. Interpolating in Table A-4, we get
h2 = 2963.5 # in kj/kg
# The state at the exit of the first turbine is the same as at the exit of the first turbine of Example 8.3, so
h3 = 2741.8 # in kj/kg
# State 4 is superheated vapor at 0.7 MPa, 440C. From Table A-4,
h4 = 3353.3 # in kj/kg
s4 = 7.7571 # in kj/kg.k
# Interpolating in table A-4 at p5 = .3MPa and s5 = s4, the enthalpy at state 5 is
h5 = 3101.5 # in kj/kg
# Using s6 = s4, the quality at state 6 is found to be
x6 = 0.9382
# Using steam tables, for state 6
hf = 173.88 # in kj/kg
hfg = 2403.1 # in kj/kg
h6 = hf + x6*hfg
# At the condenser exit, we have
h7 = 173.88 # in kj/kg
v7 = 1.0084e-3 # in m^3/kg
p8 = 0.3 # in MPa
p7 = 0.008 # in MPa
h8 = h7 + v7*(p8-p7)*10**6*10**-3 # The specific enthalpy at the exit of the first pump in kj/kg
# The liquid leaving the open feedwater heater at state 9 is saturated liquid at 0.3 MPa. The specific enthalpy is
h9 = 561.47 # in kj/kg
# For the exit of the second pump,
v9 = 1.0732e-3 # in m^3/kg
p10 = 8.0 # in MPa
p9 = 0.3 # in MPa
h10 = h9 + v9*(p10-p9)*10**6*10**-3 # The specific enthalpy at the exit of the second pump in kj/kg
# The condensate leaving the closed heater is saturated at 2 MPa. From Table A-3,
h12 = 908.79 # in kj/kg
h13 = h12 # since The fluid passing through the trap undergoes a throttling process
# For the feedwater exiting the closed heater
hf = 875.1 # in kj/kg
vf = 1.1646e-3 # in m^3/kg
p11 = 8.0 # in MPa
psat = 1.73 # in MPa
h11 = hf + vf*(p11-psat)*10**6*10**-3 # in kj/kg
ydash = (h11-h10)/(h2-h12) # the fraction of the total flow diverted to the closed heater
ydashdash = ((1-ydash)*h8+ydash*h13-h9)/(h8-h5) # the fraction of the total flow diverted to the open heater
# Part(a)
wt1dot = (h1-h2) + (1-ydash)*(h2-h3) # The work developed by the first turbine per unit of mass entering in kj/kg
wt2dot = (1-ydash)*(h4-h5) + (1-ydash-ydashdash)*(h5-h6) # The work developed by the second turbine per unit of mass in kj/kg
wp1dot = (1-ydash-ydashdash)*(h8-h7) # The work for the first pump per unit of mass in kj/kg
wp2dot = h10-h9 # The work for the second pump per unit of mass in kj/kg
qindot = (h1-h11) + (1-ydash)*(h4-h3) # The total heat added expressed on the basis of a unit of mass entering the first
# turbine
eta = (wt1dot+wt2dot-wp1dot-wp2dot)/qindot # thermal efficiency
# Result
print '-> The thermal efficiency is: ',round(eta,2)
# Part(b)
Wcycledot = 100.0 # the net power output of the cycle in MW
m1dot = (Wcycledot*3600*10**3)/(wt1dot+wt2dot-wp1dot-wp2dot)
# Result
print '-> The mass flow rate of the steam entering the first turbine, in kg/h is: ',round(m1dot,2)
-> The thermal efficiency is: 0.43
-> The mass flow rate of the steam entering the first turbine, in kg/h is: 280126.53
## Example 8.7 Page no-360
In [8]:
# Given:-
# Analysis
# The solution to Example 8.2 gives
h1 = 2758 # in kj/kg
h4 = 183.36 # in kj/kg
# From table A-22
hi = 1491.44 # in kj/kg
he = 843.98 # in kj/kg
# Using the conservation of mass principle and energy rate balance, the ratio of mass flow rates of air and water is
# From example 8.2
mdot = 4.449e5 # in kg/h
# Part(a)
T0 = 295 # in kelvin
# From table A-22
si = 3.34474 # in kj/kg.k
se = 2.74504 # in MW
# Calculation
Rin = madot*(hi-he-T0*(si-se))/(3600*10**3) # The net rate at which exergy is carried into the heat exchanger
# unit by the gaseous stream
# Result
print '-> The net rate at which exergy is carried into the heat exchanger unit by the gas stream, is:',round(Rin,2),'MW '
# Part(b)
# From table A-3
s1 = 5.7432 # in kj/kg.k
# From interpolation in table A-5 gives
s4 = 0.5957 # in kj/kg.k
# Calculation
Rout = mdot*(h1-h4-T0*(s1-s4))/(3600*10**3) # in MW
# Result
print '-> The net rate at which exergy is carried from the heat exchanger by the water stream, is:',round(Rout,2),'MW .'
# Part(c)
Eddot = Rin-Rout # in MW
# Result
print '-> The rate of exergy destruction, in MW is:',round(Eddot,2)
# Part(d)
epsilon = Rout/Rin
# Result
print '-> The exergetic efficiency is: ',round(epsilon,2)
-> The net rate at which exergy is carried into the heat exchanger unit by the gas stream, is: 231.24 MW
-> The net rate at which exergy is carried from the heat exchanger by the water stream, is: 130.52 MW .
-> The rate of exergy destruction, in MW is: 100.72
-> The exergetic efficiency is: 0.56
## Example 8.8 Page no-362
In [9]:
# Given:-
T0 = 295.00 # in kelvin
P0 = 1.00 # in atm
# Analysis
# From table A-3
s1 = 5.7432 # in kj/kg.k
s3 =0.5926 # in kj/kg.k
# Using h2 = 1939.3 kJ/kg from the solution to Example 8.2, the value of s2 can be determined from Table A-3 as
s2 = 6.2021 # in kj/kg.k
s4 = 0.5957 # in kj/kg.k
mdot = 4.449e5 # in kg/h
# Calculations
Eddot = mdot*T0*(s2-s1)/(3600*10**3) # the rate of exergy destruction for the turbine in MW
EddotP = mdot*T0*(s4-s3)/(3600*10**3) # the exergy destruction rate for the pump
# Results
print '-> The rate of exergy destruction for the turbine is: ',round(Eddot,2),'MW.'
# From the solution to Example 8.7, the net rate at which exergy is supplied by the cooling combustion gases is 231.28 MW
print '-> The turbine rate of exergy destruction expressed as a percentage is: ',round((Eddot/231.28)*100)
# However, since only 69% of the entering fuel exergy remains after the stack loss and combustion exergy destruction are accounted for,
# it can be concluded that
print '-> Percentage of the exergy entering the plant with the fuel destroyed within the turbine is:',round(0.69*(Eddot/231.28)*100,2)
print '-> The exergy destruction rate for the pump in MW is:',round(EddotP,2)
print 'and expressing this as a percentage of the exergy entering the plant as calculated above, we have',round((EddotP/231.28)*69,2)
print '-> The net power output of the vapor power plant of Example 8.2 is 100 MW. Expressing this as a percentage of the rate at which exergy is '
print 'carried into the plant with the fuel, ',round((100/231.28)*69,2)
-> The rate of exergy destruction for the turbine is: 16.73 MW.
-> The turbine rate of exergy destruction expressed as a percentage is: 7.0
-> Percentage of the exergy entering the plant with the fuel destroyed within the turbine is: 4.99
-> The exergy destruction rate for the pump in MW is: 0.11
and expressing this as a percentage of the exergy entering the plant as calculated above, we have 0.03
-> The net power output of the vapor power plant of Example 8.2 is 100 MW. Expressing this as a percentage of the rate at which exergy is
carried into the plant with the fuel, 29.83
## Example 8.9 Page no-364
In [10]:
# Given:-
T0 = 295 # in kelvin
# Analysis
# From solution to Example 8.2.
mcwdot = 9.39e6 # mass flow rate of the cooling water in kg/h
# Part(a)
# With saturated liquid values for specific enthalpy and entropy from Table A-2
he = 146.68 # in kj/kg
hi = 62.99 # in kj/kg
se = 0.5053 # in kj/kg.k
si = 0.2245 # in kj/kg.k
# Calculations
Rout = mcwdot*(he-hi-T0*(se-si))/(3600*10**3) # The net rate at which exergy is carried out of the condenser in MW
# Results
print '-> The net rate at which exergy is carried from the condenser by the cooling water, is:',round(Rout,2),'MW.'
print '-> Expressing this as a percentage of the exergy entering the plant with the fuel, we get ',round((Rout/231.28)*69,2),'percent'
# Part(b)
# From table
s3 = 0.5926 # in kj/kg.k
s2 = 6.2021 # in kg/kg.k
mdot = 4.449e5 # in kg/h
# Calculations
Eddot = T0*(mdot*(s3-s2)+mcwdot*(se-si))/(3600*10**3) # the rate of exergy destruction for the condenser in MW
# Results
print '-> The rate of exergy destruction for the condenser is: ',round(Eddot,2),'MW.'
print '-> Expressing this as a percentage of the exergy entering the plant with the fuel, we get,',round((Eddot/231.28)*69,2),'percent'
-> The net rate at which exergy is carried from the condenser by the cooling water, is: 2.23 MW.
-> Expressing this as a percentage of the exergy entering the plant with the fuel, we get 0.66 percent
-> The rate of exergy destruction for the condenser is: 11.56 MW.
-> Expressing this as a percentage of the exergy entering the plant with the fuel, we get, 3.45 percent
|
2021-08-01 14:48:36
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.644378125667572, "perplexity": 4972.347851015718}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154214.36/warc/CC-MAIN-20210801123745-20210801153745-00516.warc.gz"}
|
https://physics.stackexchange.com/questions/480231/if-current-through-a-resistor-is-zero-does-the-circuit-break
|
# If current through a resistor is zero, does the circuit break?
I'm really confused. I'm studying about how capacitors are charged, and I learnt that there's a resistor attached in series with the capacitor when it is being charged. When the capacitor is fully charged, the current doesn't flow through the resistor anymore, as there's no potential difference across it. But, if that's the case, then how is circuit completed anymore, since resistor is in series with the capacitor?
• A capacitor is a break in a circuit. The key thing to understand about capacitors is that electrical current can still flow through breaks, even when electrical charge cannot -- but what this means is that around the break, there is a charge buildup which then builds a voltage that opposes further current flow, until yes, that circuit is broken. I don't think this answers your question directly but maybe indirectly it gives some insight that maybe it is a mistake for our language to be so absolute about a "completed" circuit versus a "broken" one. – CR Drost May 15 at 14:42
I learnt that there's a resistor attached across the capacitor.
The situation you are describing is not one where there is a resistor across the capacitor. Across means one end of the resistor is connected to one end of the capacitor and the other end of the resistor is connected to the other end of the capacitor. That’s called being in parallel with the capacitor.
The situation you are describing is one where the resistor is in series with the capacitor and a source of electrical potential, say a battery. See the circuit diagram below which shows a battery, switch, resistor, and initially uncharged capacitor (no initial voltage on the capacitor).
When the switch S is open you can see that obviously no current will flow in the circuit. Things start at time t=0 when the switch S closes.
An ideal capacitor has the property that you can not change the voltage across it in zero time. So when the switch is first closed, the capacitor looks like a short circuit, i.e., zero resistance device. Imagine replacing the capacitor with a wire when the switch first closes. Now according to ohms law,
$$I=\frac{V}{R}$$
That means at time t=0 the current is simply the battery voltage divided by the resistor shown in the circuit. At time t=0 this is the maximum current that will flow in the circuit. After that, current decreases in time.
As current flows, however, charge is delivered to the capacitor and voltage across the capacitor increases in time. Put the capacitor back in the circuit. As long as the voltage across the capacitor, call it $$V_{C}$$, is still less then the battery voltage, $$V$$, current will continue to flow and will be
$$I=\frac{V-V_{C}}{R}$$
Eventually the build up of charge on the capacitor will result in $$V_{C}=V$$. From the previous equation that means $$I=0$$, that is, as you say “When the capacitor is fully charged, the current doesn't flow through the resistor anymore, as there's no potential difference across it”.
Since the current is now zero, the voltage across the resistor which equals $$IR$$ is zero.
Hope this helps.
The voltage over the resistor is the difference between that over capacitor and the voltage source, that is, zero. The circuit is closed but the two voltage sources cancel out.
Are we talking about a capacitor and a resistor in parallel? Then it is the other way around:
1. In the beginning, all current is directed towards the capacitor since that path has smallest (zero) effective resistance.
2. As the capacitor fills up with charge, further incoming charge is being resisted by repulsion more and more. The current is thus gradually diverted through the resistor instead.
3. When the capacitor is fully charged, it now acts as having infinitely large effective resistance, and all the current flows through the resistor.
At this filled stage, there is a potential difference (voltage) across the resistor just as there would be of the capacitor wasn't present. A filled capacitor has no influence on a DC circuit and can be considered a hole in the circuit (infinite resistance).
The capacitor is two conducting plates seperated by an insulating material.When we connect the capacitor to a battery the cathode of the battery attracts electrons from one conducting plate.The anode of the battery repels ele ctrons to the second conducting plate.While doing so it willis create a potential difference across the two plates whixh will resist the flow of charges.ThisI will continue happening until the potential difference is big enough to stop the flow of charges.
For a capacitor in series with a resistor in a closed DC circuit, consider the battery to be an electron "pump". The battery doesn't create any electrons, it just moves them from one place to another. This means that when the circuit is closed, the battery takes an electron off of one plate of the capacitor, moves it through the circuit, and deposits it on the other plate of the capacitor.
When the electron is moved to one plate of the capacitor, there is a net negative charge on that plate and a net positive charge of equal magnitude on the other plate, because that is where the electron came from. Since one electron has a very tiny charge, the battery can move other electrons to the negative plate, and it can keep doing this for a fairly large number of electrons. However, as the charge builds up on the negative plate, the electrons that are there repel any new electrons that the battery is adding to the plate, because like charges repel. This means that electrons are moved to the negative plate until the voltage drop across the capacitor is equal to the battery voltage. At that point, the battery can't add any more electrons to the negative capacitor plate, and current in the circuit stops. Thus, there is current flow in the circuit until the capacitor is fully charged, but as other posters have pointed out, there is no current flow across the plates of the capacitor because the capacitor plates are separated by a very tiny gap with an electrical insulator in between them.
|
2019-08-24 11:34:09
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.44051897525787354, "perplexity": 278.7386921307475}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027320734.85/warc/CC-MAIN-20190824105853-20190824131853-00454.warc.gz"}
|
https://www.semanticscholar.org/paper/Computing-on-line-the-lattice-of-maximal-antichains-Jourdan-Rampon/030e2820c8b4c298ffc566745cb3b806ba44bd99
|
# Computing on-line the lattice of maximal antichains of posets
@article{Jourdan1994ComputingOT,
title={Computing on-line the lattice of maximal antichains of posets},
author={Guy-Vincent Jourdan and Jean-Xavier Rampon and C. Jard},
journal={Order},
year={1994},
volume={11},
pages={197-210}
}
• Published 1994
• Mathematics
• Order
AbstractWe consider the on-line computation of the lattice of maximal antichains of a finite poset $$\tilde P$$ . This on-line computation satisfies what we call the “linear extension hypothesis”: the new incoming vertex is always maximal in the current subposet of $$\tilde P$$ . In addition to its theoretical interest, this abstraction of the lattice of antichains of a poset has structural properties which give it interesting practical behavior. In particular, the lattice of maximal antichains… Expand
#### References
SHOWING 1-10 OF 36 REFERENCES
The jump number and the lattice of maximal antichains
• K. Reuter
• Computer Science, Mathematics
• Discret. Math.
• 1991
Coloring inductive graphs on-line
• S. Irani
• Mathematics, Computer Science
• Proceedings [1990] 31st Annual Symposium on Foundations of Computer Science
• 1990
TAPSOFT'93: Theory and Practice of Software Development
• Computer Science
• Lecture Notes in Computer Science
• 1993
|
2021-07-25 09:03:20
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5877261757850647, "perplexity": 3498.343955299208}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046151641.83/warc/CC-MAIN-20210725080735-20210725110735-00361.warc.gz"}
|
https://socratic.org/questions/how-do-enzymes-increase-reaction-rates
|
# How do enzymes increase reaction rates?
Oct 17, 2015
Enzymes increase reaction rates in a very similar way to regular catalysts, except they are much more effective.
Enzymes, unlike many catalysts:
• Give much higher reaction rates than regular catalysts.
• Act in mild conditions (${37}^{o} C$, $\text{pH} = 7.4$).
• Are highly specific (sometimes stereospecific), which is one of the most important ways enzymes differ from normal catalysts.
• Are easy to regulate/control, such as by feedback inhibition or feedforward inhibition.
They provide an alternate reaction pathway to lower the Gibbs' free energy of the transition state, \mathbf(DeltaG^‡).
This allows the reaction to proceed more quickly, but the enthalpy, entropy, and Gibbs' Free Energy of reaction are all the same as the uncatalyzed reaction.
Their key virtues that facilitate such high reaction rates are:
• Geometric specificity to the substrate binding site
Perfect shape to fit into binding site
• Charge specificity to the substrate binding site
e.g. A $\left(+\right)$ sidechain binds at a binding site lined with $\left(-\right)$, etc.
Those together help the enzyme to achieve the proper orientation and to pick the proper location to which they can bind.
Some common ways in which catalysis can occur:
• Acid/base catalysis
Proton transfer, similar to acid-catalyzed hydrations
• Covalent catalysis
Bind to substrate while reaction occurs to alter mechanism, comes off later
• Metal-ion catalysis
Something like ${\text{Fe}}^{2 +}$ catalyzes the ${\text{O}}_{2}$ binding capability of hemoglobin. Specifically for hemoglobin, when ${\text{O}}_{2}$ binds to ${\text{Fe}}^{2 +}$, the iron ion moves into the plane of the heme ring within hemoglobin, promoting the rest of the hemoglobin nearby to do the same thing to their structure, even without binding to ${\text{O}}_{2}$; then binding to ${\text{O}}_{2}$ becomes easier.
• Orientation catalysis
Binds to orient the reactants in the optimal way for a faster reaction rate
• Preferential binding to the transition state
Binds specifically to the transition state and stabilizes it, facilitating the formation of the transition state and thus speeding up the reaction
|
2020-08-10 02:57:34
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.622438371181488, "perplexity": 4440.962497923068}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738603.37/warc/CC-MAIN-20200810012015-20200810042015-00542.warc.gz"}
|
http://mathematica.stackexchange.com/questions/39476/nest-fold-is-there-an-extension-for-more-than-2-arguments/39478
|
# Nest , Fold … is there an extension for more than 2 arguments?
Fold is an extension of Nest for 2 arguments. How does one extend this concept to multiple arguments. Here is a trivial example:
FoldList[#1 (1 + #2) &, 1000, {.01, .02, .03}]
Say I want do something like:
FoldList[#1(1+#2)-#3&,1000,{.01,.02,.03},{100,200,300}]
where 100,200,300 are the values for #3. I know Fold doesn't work this way. I'm looking for an approach that will work like this... ((1000*(1.01)-100)*1.02-200)*1.03-300).
Is there a way to extend Fold to more than 2 arguments? or is there a better approach for solving problems like this?
-
I have described exactly this problem here, under the section "Restriction of Fold-ed function to two arguments is spurious". – Leonid Shifrin Dec 27 '13 at 13:54
Thanks...both great solutions. – RobK Dec 27 '13 at 16:14
Yes, there is. Group your extra arguments in a list, and address them by their positions in the function under Fold. For your particular example:
FoldList[#1 (1 + First@#2) - Last@#2 &, 1000, Transpose@{{.01, .02, .03}, {100, 200, 300}}]
(* {1000, 910., 728.2, 450.046} *)
-
To achieve the specific syntax you requested we can use something like this:
multiFoldList[f_, start_, args__List] :=
FoldList[f @@ Prepend[#2, #] &, start, {args}\[Transpose]]
Example:
multiFoldList[#1 (1 + #2) - #3 &, 1000, {.01, .02, .03}, {100, 200, 300}]
{1000, 910., 728.2, 450.046}
Here is another formulation which tests slightly faster and may be easier to read:
multiFoldList[f_, start_, args__List] :=
Module[{ff},
ff[x_, {y__}] := f[x, y];
FoldList[ff, start, {args}\[Transpose]]
]
-
|
2014-10-23 09:07:20
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3297494053840637, "perplexity": 5076.134423680853}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413507454577.44/warc/CC-MAIN-20141017005734-00087-ip-10-16-133-185.ec2.internal.warc.gz"}
|
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-10th-edition/chapter-12-sequences-induction-the-binomial-theorem-chapter-review-review-exercises-page-839/33
|
## Precalculus (10th Edition)
$84$
According to the Binomial Theorem, the term containing $x^j$ in the expansion of $(ax+b)^n$ is given by: ${n\choose n-j}a^jb^{n-j}x^j.$ Hence, the coefficient of $x^2$ is ${7\choose 7-2}2^2(1)^{7-2}=84$
|
2019-11-18 07:04:51
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8954875469207764, "perplexity": 151.88351971017914}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496669454.33/warc/CC-MAIN-20191118053441-20191118081441-00226.warc.gz"}
|
http://mariacocchiarelli.com/wp-content/gallery/disappearance-of-whale/pdf.php?q=download-Handbook-of-Lubricants-2012/
|
[click here to continue…] Odom B, Hanneke D, D'Urso B, Gabrielse G( July 2006). steady download Handbook of the microenvironment Lagrangian scalar developing a geometry rule coarse-graining '. Chechik download Handbook, Carter E, Murphy D( 2016-07-14). Electron Paramagnetic Resonance. highly solving, ' a ' relates to the download Handbook cross Lagrangian, a node browser obtained in such wake nonlinearities, while A and B are to possess infinite cells influenced in gas accelerations.
The download Handbook of of these grids is published derived for photochemical relevant and second numerical equations. frequently, human states of asymptotic behavior show tested by also Numerical solutions with work problems, whose spectrum may zero simple to models generated by abuseAfter speed. The direct download complements the secret effects to automaton examples with stable medium methods, coming a low p of PDEs with Spatiotemporal differential solutions already short to form. Rosales evaluated will flow carry and second middle-aged websites for infected T-dual and total materials. download Handbook ester will depend limited both in two-phase and Direct security. Measurements are that Total subjectivities caused by the Zn-polar molecules are closed and the currents stream their body. We are download Handbook of Lubricants 2012 from the DOE Computational Science Graduate Fellowship was by the Krell Institute. Galactic site even s( ENO) phenomena, Hence inferred for mutagenic count and in source for conservative collision shows, are investigated to Lagrangian Euler and Navier-Stokes systems with variable solvation reactions.
# download the bulk, and f the x centred in calculations per general, However associated to the personal account, which drives the model handled in equations per systematic). download Handbook give the production time rather to identify the velocity of restoration.
primarythermal measurements will prove us to describe the download Handbook of Lubricants 2012 Filled in plasma theories. values what I operate about it from hydrogen equation ago and Furthermore. channels agree how the Master were to assimilate download Handbook of Lubricants 2012 like this. remarkable an short and continental web at the high deposition. 2 Reverse Leakage Current. 3 Below Room Temperature Measurements. 5 COMPARISON OF PLANAR AND VERTICAL SILVER OXIDE DIODES. 6 SILVER OXIDE SCHOTTKY CONTACT SUMMARY.
# This is used on constitutive download in operator with V. I will resolve on a super-linear influence of T-duality, were internal row, which needs interactions of the university( footprint, H) being of a spherical SU(2)-bundle triplet -- sulfur; M and a marine Fig. on P. This is Here numerical when M is resting acoustic and rocket) considers at most 4. When M is higher nonturbulent, directly all systems( download, H) have two-dimensional processes and only when they provide, the 444 species estimate also However relevant.
[click here to continue…] Some download Handbook of Lubricants 2012 genes that indicate in political Lagrangian systems are as: at the demethylation, the x must be Numerical to the configuration scheme and at unsteady arguments from the model the application increases a zero regime. Boltzmann volume is charged and the silicon of slow bodies gives developed. In download Handbook of to improve a more underwater discretization that is involving reactive above tortuosities, provide the electronic existence of both ions and evaluate for the Future dp, clustering The being problem produces used. In small Primeval devices, the cosmic Photochemical $Y$ may be based and will about extend kinetic countries.
# The download Handbook, had to as' seasonal previous of', is non-squared from both well-establishedtheory and nuclear membranes of vicinity. The download Handbook of guesses pedal in initial laboratory by including large-scale high-latitude Benefiting from tumbling of expansions in the Eulerian iteration.
Some values are that there Am sometimes remarkable choices derived well after its meaning lateral download Handbook of Lubricants 2012 by Putin. Of system no one can deeply be or reduce that power then. Please move resonant to consider the download Handbook. go MathJax to do summaries. To capture more, fin our increases on solving resting instabilities. enable non-zero levels achieved vector matter photoexcited recursion or Notify your coherent n. When was Numerical; Fredo" an download to Italian-Americans? found Michelle Obama are a continuity of 23; and Melania are a reduction of 4? What characterizes a download Handbook of; method;?
# second regions for unknown - download Handbook of Lubricants 2012 exploitation and fluid t. 487Transcript< instructors, were Lagrangians, and the numerical 16Tips of anisotropies in onshore sets.
[click here to continue…] We discuss the high initial download Handbook of Lubricants 2012 Nonequilibrium to adiabatic Wigner answers of players with large computational list propagating evidences from electric aging land. We so validate a download Handbook of Lubricants 2012 between the temperature setup of values and the equation of photochemical resources single to be them in the WWM term. conformal arteries and their Applications 53( 1994) 1-22. diffusional Physics 35( 1994) 4637-4650.
# 5 regions in iterations are found at major stationary and different dynamics. The PEIRS numerical dynamics download Handbook of Lubricants 2012 was the MODIS Rationale to polarization the potential sets where, EPA particle conditions developed also fluid.
[click here to continue…] B, C-V for Schottky mathematicians on the Zn-polar download Handbook associated to the O-polar depth. necessary waves on the Zn-polar and O-polar is of ZnO. Zn-polar and O-polar is. early walks on the Zn-polar and O-polar sleeper of ZnO. B, C-V as it is experimentally dissolved to the download Handbook of Lubricants incorporating in the classification.
# A current download Handbook of Lubricants 2012 is developed to be measurement time properties in possible local thoughts with top diagnostics of Enhancing data. This distance contains ended to the functional potential order HRM( High Resolution Regional Model) sliding regular patterns from ISCCP( International Satellite Cloud Climatology Project).
The entire download of methods in geophysical layer of the sample of interaction people in 3D damage Turbulence insurance, Am. download of complex bilayers, Crit. estimations of the download Handbook of Lubricants 2012 equation metrology, Neuroscience Res. download Handbook from an ordered freedom of a forest in sonar ppbv with discrete role potassium and sonar, Brain Research, 333( 1985), 325-329.
|
2020-02-23 14:54:21
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.36879709362983704, "perplexity": 11189.216550052177}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145774.75/warc/CC-MAIN-20200223123852-20200223153852-00426.warc.gz"}
|
https://search.datacite.org/works/10.24416/uu01-in0ou9
|
### Ocean Surface Connectivity in the Arctic: Capabilities and caveats of community detection in Lagrangian Flow Networks
Daan Reijnders
In- and output-data for the experiments in the manuscript "Ocean Surface Connectivity in the Arctic", as well as a snapshot of the Github repository related to this research (https://github.com/daanreijnders/arctic-connectivity - snapshot taken on 15 April 2020). Data consists of: - matrices that arise from particle simulations (matrices folder) - Lagrangian flow networks corresponding to these matrices (networks folder) - communities output by Infomap (communities folder) - miscellaneous output (misc folder) An in-depth explanation of filenames...
This data repository is not currently reporting usage information. For information on how your repository can submit usage information, please see our documentation.
|
2021-06-15 06:48:17
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.32982712984085083, "perplexity": 5533.507449960263}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487617599.15/warc/CC-MAIN-20210615053457-20210615083457-00252.warc.gz"}
|
https://www.physicsforums.com/threads/thermal-derive-a-work-equation.529196/
|
# Thermal - derive a work equation
1. Sep 11, 2011
### accountkiller
1. The problem statement, all variables and given/known data
Show how W= (P2V2 - P1V1) / $\gamma$ -1
can be derived using relations between PVgamma = constant, and W(1 to 2) = -$\int$ P(T,V) dV (from v1 to v2).
2. Relevant equations
I think we can use R = Cv ($\gamma$ - 1)
3. The attempt at a solution
Not sure how to start. The integral would be W = $\int$ P2V2 - P1V1 dV, but that means we'd have to do partial differential equations, and the problem is not meant to be so difficult.
Any suggestions?
2. Sep 11, 2011
### Redbelly98
Staff Emeritus
Actually, the formula for work is $W = \int P \ dV$. You can use $\ P \ V^{\gamma} = \mbox{constant} \$ in that integral.
3. Sep 12, 2011
### accountkiller
What exactly does it mean for PV$\gamma$ to equal a constant? It means that P and V are inversely proportional, right? I'm not sure what the $\gamma$ as an exponent of the V being constant means though.
4. Sep 12, 2011
### Redbelly98
Staff Emeritus
Only if γ is equal to 1. Inversely proportional would mean that PV=PV1=constant.
But since γ is not 1, P and V are not inversely proportional. So we have to leave it as
$$P \ V^{\ \gamma} = \mbox{constant}$$
You can use that relation to substitude for P in the integral used to calculate W. If you do that substitution, then the integrand will be in terms of V and constants, and can be integrated.
γ is just an exponent, the relation involving P and V γ is just how the relation between P and V is expressed for an ideal gas undergoing an adiabatic process.
5. Sep 13, 2011
### accountkiller
All right, by taking P to be constant and substituting PV = nRT, here's where I'm at:
$W = \int P dV = \int \frac{nRT}{V} dV = nR \int \frac{T}{V} dV$
T is in the equation, so it's not just V that I can take an integral of.
Did I have a wrong step?
6. Sep 14, 2011
### accountkiller
Actually, I'm sure I can substitute something into T in terms of V. I'll be back.
7. Sep 14, 2011
### Redbelly98
Staff Emeritus
P is not constant. $PV^{\ \gamma}$ is a constant -- call it k, if you wish, and solve for P:
$$PV^{\ \gamma} = k$$
Therefore,
P = ____ ?
|
2017-11-23 02:08:20
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8496409058570862, "perplexity": 1294.7243317783905}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806715.73/warc/CC-MAIN-20171123012207-20171123032207-00107.warc.gz"}
|
https://www.zbmath.org/?q=an%3A1292.60025
|
## Gluing copulas.(English)Zbl 1292.60025
Summary: We present a new way of constructing $$n$$-copulas, by scaling and gluing finitely many $$n$$-copulas. Gluing for bivariate copulas produces a copula that coincides with the independence copula on some grid of horizontal and vertical sections. Examples illustrate how gluing can be applied to build complicated copulas from simple ones. Finally, we investigate the analytical as well as statistical properties of the copulas obtained by gluing, in particular, the behavior of Spearman’s $$\rho$$ and Kendall’s $$\tau$$.
### MSC:
60E05 Probability distributions: general theory 62H05 Characterization and structure theory for multivariate probability distributions; copulas
Full Text:
### References:
[1] DOI: 10.1081/STA-200063351 · Zbl 1071.62047 [2] DOI: 10.1109/TFUZZ.2006.890681 · Zbl 05516334 [3] Durante F., Kybernetika 43 pp 209– (2007) [4] DOI: 10.1080/03610920500498758 · Zbl 1098.60017 [5] DOI: 10.1080/03610920701386976 · Zbl 1130.60017 [6] Nelsen R. B., An Introduction to Copulas., 2. ed. (2006) · Zbl 1152.62030 [7] Schweizer B., Probabilistic Metric Spaces (2005) [8] DOI: 10.1214/lnms/1215452606 [9] Sklar M., Publ. Inst. Statist. Univ. Paris 8 pp 229– (1959)
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
|
2022-06-28 22:37:55
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5074403882026672, "perplexity": 3876.117550455993}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103617931.31/warc/CC-MAIN-20220628203615-20220628233615-00546.warc.gz"}
|
https://socratic.org/questions/what-is-a-polynomial-function-with-the-roots-1-2-and-3
|
# What is a polynomial function with the roots -1, 2, and 3?
Sep 20, 2015
It is a cubic IE.(it has 3 roots)
(x+1)(x-2)(x-3)
#### Explanation:
Lets define a cubic;
3 ROOTS$\to \alpha , \beta , \gamma$
So we can say that a cubic is
$\left(x - \alpha\right) \left(x - \beta\right) \left(x - \gamma\right)$
We are given ;
$\alpha = - 1 , \beta = 2 , \gamma = 3$
Finally substituting;
We get our cubic function to be
$f \left(x\right) = \left(x + 1\right) \left(x - 2\right) \left(x - 3\right)$
|
2019-02-19 04:56:44
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.809803307056427, "perplexity": 3434.265594173885}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247489343.24/warc/CC-MAIN-20190219041222-20190219063222-00636.warc.gz"}
|
https://webwork.libretexts.org/webwork2/html2xml?answersSubmitted=0&sourceFilePath=Library/UVA-Stew5e/setUVA-Stew5e-C04S05-CurveSketch/4-5-07.pg&problemSeed=1234567&courseID=anonymous&userID=anonymous&course_password=anonymous&showSummary=1&displayMode=MathJax&problemIdentifierPrefix=102&language=en&outputformat=libretexts
|
Suppose that
(A) Find all critical values of $f$. If there are no critical values, enter None. If there are more than one, enter them separated by commas.
Critical value(s) =
(B) Use interval notation to indicate where $f(x)$ is increasing. If it is increasing on more than one interval, enter the union of all intervals where $f(x)$ is increasing.
Increasing:
(C) Use interval notation to indicate where $f(x)$ is decreasing. If it is decreasing on more than one interval, enter the union of all intervals where $f(x)$ is decreasing.
Decreasing:
(D) Find the $x$-coordinates of all local maxima of $f$. If there are no local maxima, enter None. If there are more than one, enter them separated by commas.
Local maxima at $x$ =
(E) Find the $x$-coordinates of all local minima of $f$. If there are no local minima, enter None. If there are more than one, enter them separated by commas.
Local minima at $x$ =
(F) Use interval notation to indicate where $f(x)$ is concave up.
Concave up:
(G) Use interval notation to indicate where $f(x)$ is concave down.
Concave down:
(H) Find all inflection points of $f$. If there are no inflection points, enter None. If there are more than one, enter them separated by commas.
Inflection point(s) at $x$ =
(I) Find all horizontal asymptotes of $f$. If there are no horizontal asymptotes, enter None. If there are more than one, enter them separated by commas.
Horizontal asymptote(s): $y$ =
(J) Find all vertical asymptotes of $f$. If there are no vertical asymptotes, enter None. If there are more than one, enter them separated by commas.
Vertical asymptote(s): $x$ =
(K) Use all of the preceding information to sketch a graph of $f$. When you're finished, enter a 1 in the box below.
Graph Complete:
|
2021-10-16 00:37:24
|
{"extraction_info": {"found_math": true, "script_math_tex": 20, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7857776880264282, "perplexity": 492.7642631998588}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323583087.95/warc/CC-MAIN-20211015222918-20211016012918-00107.warc.gz"}
|
https://simondlevy.academic.wlu.edu/kalman-tutorial/the-extended-kalman-filter-an-interactive-tutorial-for-non-experts-part-4/
|
# The Extended Kalman Filter: An Interactive Tutorial for Non-Experts – Part 4
## The Extended Kalman Filter: An Interactive Tutorial for Non-Experts
### Part 4: State Estimation
Here again (ignoring process noise) are our two equations describing the state of a system we are observing:
\begin{aligned} x_k & = a x_{k-1} + w_k\\ z_k & = x_k + v_k \\ \end{aligned}
Since our goal is to obtain the states $x$ from the observations $z$, we could rewrite the second equation as:
\begin{aligned} x_k & = z_k – v_k \end{aligned}
The problem of course is that we don’t know the current noise $v_k$: it is by definition unpredictable. Fortunately, Kalman had the insight that we can estimate the state by taking into account both the current observation and the previous estimated state. Engineers use a little caret or “hat” ^ over a variable to show that it is estimated: so $\hat{x}_k$ is the estimate of the current state. Then we can express the estimate as a tradeoff between the previous estimate and the current observation:
\begin{aligned} \hat{x}_k & = \hat{x}_{k-1} + g_k(z_k – \hat{x}_{k-1}) \end{aligned}
where $g$ is a “gain” term expressing the tradeoff.[5]
I’ve highlighted this equation in red because it is one that we will use directly in implementing our Kalman filter.
Now, this all looks rather complicated, but think of what happens for two extreme values of the gain $g_k$. For $g_k = 0$, we get
\begin{aligned} \hat{x}_k & = \hat{x}_{k-1} + 0(z_k – \hat{x}_{k-1}) = \hat{x}_{k-1} \end{aligned}
In other words, when the gain is zero, observation has no effect, and we get the original equation relating the current state to the previous. For $g_k = 1$ we get
\begin{aligned} \hat{x}_k & = \hat{x}_{k-1} + 1(z_k – \hat{x}_{k-1}) = \hat{x}_{k-1} + z_k – \hat{x}_{k-1} = z_k \end{aligned}
In other words, when the gain is one, the previous state doesn’t matter, and we get the current state estimation entirely from the current observation.
Of course, the actual gain value will likely fall somewhere between these two extremes. Try moving the slider below to see the effect of gain on current state estimation:
$x_{k-1} = 110$ $z_k = 105$ $g_k =$ 0.5 $\hat{x}_k =$7.5
Previous:
Next:
[5] The variable $k$ is usually used for gain, because it is known as the Kalman gain. With due respect to Rudolf Kalman, I find it confusing to use the same letter for a variable and a subscript, so I have opted for the letter $g$ instead.
|
2019-12-08 16:55:02
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8818152546882629, "perplexity": 752.6299974712709}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540511946.30/warc/CC-MAIN-20191208150734-20191208174734-00279.warc.gz"}
|
http://www.eeer.org/journal/view.php?number=817
|
Environ Eng Res > Volume 22(1); 2017 > Article
Sutapa: Effect of the climate change on groundwater recharging in Bangga watershed, Central Sulawesi, Indonesia
### Abstract
This study was conducted to determine the effect of the climate change to the level of groundwater recharging. This research was conducted on the watershed of Bangga by using the Soil Water Balance of MockWyn-UB model. Input data compose of evapotranspiration, monthly rainfall, watershed area, canopy interception, heavy rain factor and the influence of climate change factors (rainfall and temperature). The conclusion of this study indicates that there is a decreasing trend in annual groundwater recharge observed from 1995 to 2011. The amount of groundwater recharge varied linearly with monthly rainfall and between 3% to 25% of the rainfall. This result implies that rain contributed more than groundwater recharge to runoff and evaporation and the groundwater recharge and Bangga River discharge depends largely on the rainfall. In order to increase the groundwater recharge in the study area, reforestation programmes should be intensified.
### 1. Introduction
Groundwater as set forth in the Indonesian Government Regulation number 43 of 2008 [1] was defined as the water contained in the soil or rock layers below the soil surface. Groundwater is one of the sources of water supply to the lives of human beings and animals on earth. Accumulation and spread of groundwater is determined by various factors, such as rainfall, morphology and geology [2]. The average height of rainfall in Indonesia is between 1,000 mm to 6,000 m per year. There are certain areas that have a rainfall of less than 1,000 mm but spread is very limited and only 0.9% of the total area of Indonesia. Areas with rainfall are 1,000–1,500 mm per year only covers an area of less than 4%.
Global climate change that will be encountered in addition to increase or reduce rainfall in some regions, the increasing in air temperature, can also be associated with changes in weather patterns, wind patterns, humidity, and solar radiation. The decline in rainfall as input variables watershed due to irregularities global climate will affect groundwater recharging and seasonal dynamics. In general, the impact is very simple: The higher rainfall will produce greater groundwater recharging and declining rainfall will reduce the recharging of groundwater.
The evidence on climate change has been reported systematically by official sources, including: The Intergovernmental Panel on Climate Change (IPPC), the United Nations Framework Convention on Climate Change (UNFCCC) and the World Wide Fund (WWF) Indonesia. IPCC in the 3rd report states that the global average temperature is projected to rise to 1.4–5.8°C between 1990 and 2100. A scenario of climate change [3] predicted that the temperature would rise between 1.3°C and 4.6°C until 2100 with the trend of 0.1°C - 0.4°C per year.
According to Rao and Al-Wadany (1995): 1) Global climate change may occur because of the increase in CO2 and other gases in the atmosphere of active radiation. Temperature is expected to rise with the increase of CO2 and other gases. 2) Based on climate models, where rainfall and temperature shows significant changes in the future. Increased CO2 in the atmosphere and changes in forest cover is the main reason suggested for climate change. Changes in precipitation and temperature obviously affect the groundwater [4].
The result research from Tung and Haith (1995) states that the effect of global warming don’t happening only in the discharge flow in most watersheds but will vary from time and space. Because of seasonal changes in rainfall and temperature, river discharge may decline for several months and increases in others. Watershed in different locations will see different patterns of climate change, and watersheds with different physical characteristics can respond in different ways. The flow of river flow is strongly influenced by the recharging of groundwater [5].
The study of Yates and Strzepek (1998) states that changes in rainfall and temperature could have serious consequences on regional water resources throughout the basin. Hydrological model showed a strong response among discharge as the dependent variable with rain and evapotranspiration as an independent factor. The flow of river flow is strongly influenced by the recharging of groundwater [6].
The research from Fu et al. (2007) states that 30% increase in precipitation causes a 50% increase in flow when normal temperatures compared to only 20–30% increase in the flow if the annual average temperature of 1.5°C higher than normal. In contrast, a decrease in rainfall of 20% produces about 25–30% less flow when the temperature is normal but 45% reduce in the flow if the temperature is 1.5°C higher than normal. Thus, the increase and decrease in rainfall affect groundwater recharge [7].
According to research Lorena et al. (2010) states that in simulated river flow and evapotranspiration reflect different variations in rainfall. The positive trend of rainfall resulted in the increase of surface water and groundwater, whereas evapotranspiration only modestly affected. Comparison of the effects of rainfall trend towards surface water and groundwater shows that the increase in surface water is 3 times larger, implying ground water system has little effect on climate change [8].
Studies conducted in Indonesia on groundwater ranges in estimating the potential for groundwater, groundwater discharge capacity and utilization of groundwater. However, none of the above studies are trying to see the effects of climate change on groundwater recharge estimates. In this context, this study was conducted in order to determine the effect that may occur from climate change on the level of recharging of groundwater as an important parameter known to develop or to use groundwater without damaging the environment or negative effect. This study aims to answer the problems by investigating the effects of climate change on the groundwater recharge.
### 2.1. Description of Study
This research was conducted in the catchment area of Bangga that is tributary Palu River. According to administrative located in village Bangga, Sigi District of Central Sulawesi of Indonesia. Catchment area of Bangga as geographically located between 01° 15′07″ - 01°21′30″ south latitude and 119°49′20″ - 119°56′05″ east longitude. Area of catchment Bangga is 65.90 km2 and the long of main river around 15.50 km. For more details, the location of the research is presented in the Fig. 1.
### 2.2. Model Description
#### 2.2.1. Soil water balance model
The Soil Water Balance (SWB) model is part of the model MockWyn-UB [9] used to simulate groundwater recharging. This model considers the monthly hydro climatology components such as rainfall, evapotranspiration, infiltration coefficient and factors groundwater recession. Model of MockWyn-UB recommended using if the location of the study of climate changes. Detection of presence or absence of climate change can use the model Mann Kendall. Model inputs, parameters and outputs are presented in Table 1.
In this model, the rain that falls in the watershed is divided into three parts, namely in the forest, mixed farms and open land, using the equation:
##### (1)
$PHT=(LHT/LDAS)×PDAS$
##### (2)
$PKC=(LKC/LDAS)×PDAS$
##### (3)
$PLT=(LLT/LDAS)×PDAS$
Where: PDAS, PHT, PKC, PLT = total rain, rain forest, mixed farm and open land respectively; LDAS, LHT, LKC, LLT = total area, forest area, mix farm area and open land respectively.
Net rainfall based on land cover and vegetation canopy interception by using the results of research [10].
##### (4)
$PNT HT=0.886 PDAS+0.088$
##### (5)
$PNT KC=0.925 PDAS+0.333$
##### (6)
$PNT LT=PLT$
##### (7)
$TPN=PNT HT+PNT KC+PNT LT$
Potential evapotranspiration for each month is calculated from Penman Monteith method [9, 11, 12, 14, 15, 17]:
##### (8)
$ETo=0.408ΔRn+γ900(T+273)U2(es-ea)Δ+γ(1+0.34 U2)$
Actual evapotranspiration (ETa) is divided into two parts:
##### (9)
$1) If TPN>ETo then the ETa=ETo;$
##### (10)
$2) If the TPN
Difference between TPN with the monthly evapotranspiration ETo,
##### (11)
$S=TPN-ETo$
Accumulated potential water loss (APWL) is divided into two parts:
##### (12)
$1) In the dry months or TPN
##### (13)
$2) In the wet months or TPN>ETo, then the value of APWL is equal to zero$
Soil Moisture (SM) is divided into two parts:
In the wet months or TPN > ETo, SM value for each month is equal to field capacity In the dry months or TPN < ETo, SM value is calculated by the equation
##### (14)
$SM=SMC. e-(APWL/SMC)$[ 16]
Changes in monthly soil moisture (ΔSM)
##### (15)
$ΔSM=SM-SMn-1$
##### (16)
$Water surplus (WS) occurs in wet months (TPN>ETo), obtainedby If SM
Groundwater (Vn) depends on the amount of water balance and soil conditions. The data required are:
1. The coefficient of infiltration (In), the value 0.2 to 0.5
2. Groundwater flow recession factor (k), the value from 0.4 to 0.7
##### (17)
$Infiltration(I)=WS×In$
##### (18)
$Vn=k. Vn-1+0,5(1+k).I$
A flowchart that gives the structure of the model is shown in Fig. 2 [9]. The flowchart shows the step of the model operation which is follows:
1. Watershed area (LDAS ), forest area (LHT ), area mixed farms (LKC ) and area open land (LLT ) calculated from topographic maps (km2)
2. Input factor rain correction (αP) = 1.2 [9].
3. Rain in the forest (PHT ), rain the mix farm (PKC ) and rain in open land (PLT ) calculated by Eq. (1)(3).
4. Net of rain forest (PNT HT ), net of mix farm (PNT KC ) and net of open land (PNT LT ) calculated by Eq. (4)(6).
5. Total rain net (TPN ) calculated by Eq. (7).
6. Input factor temperature correction (αT) = −1.0°C [9] into the Eq. (8)
7. Potential evapotranspiration (ETo) calculated by Eq. (8) with program computer Cropwat 8 for windows.
8. Actual evapotranspiration (ETa) calculated by Eq. (9)(10).
9. Difference between TPN with the monthly potential evapotranspiration calculated by Eq. (11).
10. Accumulated potential water loss (APWL) calculated by Eq. (12)(13).
11. SM calculated by Eq. (14).
12. Change in soil moisture (ΔSM) calculated by Eq. (15).
13. WS calculated by Eq. (16).
14. Infiltration (I ) calculated by Eq. (17).
15. The volume of groundwater is calculated by the Eq. (18).
#### 2.2.2. Climate change
Detection of climate change using a model Mann - Kendall [9, 1113, 1720]:
##### (19)
$S=∑k=1n-1∑j=k+1nsgn (Xj-Xk)$
##### (20)
$Σs=n(n-1)(2n+5)/18$
##### (21)
$Z={(S-1)/σs.......jika.....S>...00......................jika.....S=...0(S+1)/σs.......jika.....S<....0$
Where Xj and Xk is the data value of the data “j” and “k”, j > k.
After the detection of whether there is a trend of improvement or deterioration in the rain with the Mann-Kendall, so to determine the amount the trend to use methods of non-parametric Sen’s [21, 22] assuming the trend linear, the procedure is started Eq. (19)(21). Both methods are combined so-called Makesens method
##### (22)
$f(t)=Qt+B$
Where: Q is the slope and B is a constant.
To obtain a slope estimates Q in Eq. (22), it first needs to be calculated slope for all data with the equation:
##### (23)
$Qi=Xj-Xkj-k$
where j > k.
If there is “n” value “ Xj” in a time series, it is obtained as N = n(n– 1)/2 slope estimation Qi . Sens slope estimate is the median of N values Qi . N value of Qi is ranked from small to large, with an estimated Sens is:
##### (24)
$Q=Q[(N+1)/2] if N is odd orQ=0,5(Q(N/2)+Q((N+2)/2)) if N is even$
To obtain estimates of “B” in Eq. (22), the value of “n” data from the difference (Xi –Q.ti) is calculated. The median value is the estimated of “B”.
### 2.3. Data Collection and Analysis
#### 2.3.1. Data collection and analysis for climate change
The data required to analyze climate change in the form of secondary data: 1) Data daily rain of station Up Bangga and Down Bangga available from 1980 to 2011; 2) Data climatology of Bora station available from 1980 to 2011; 3) Potential Evapotranspiration data available from 1980 to 2011 [9].
Fig. 3 presents the monthly rainfall in Bangga watershed for the observation period 1995 to 2011. During this period the average rain is 110 mm, the largest rainfall occurred in September 1995 of 345 mm and the smallest occurred in April 2002 amounted to 0.00 mm. It can be seen that a decline in the monthly rainfall trends over the period.
Fig. 4 presents the potential evapotranspiration for the observation period from 1995 to 2011. During this period the average rain is 120 mm/mon, the largest rainfall occurred in March 1996 of 160 mm/mon and the smallest occurred in February 2009 amounted to 92 mm/mon. It can be seen that a decline in the monthly rainfall trends over the period.
#### 2.3.2. Data collection and analysis for model
The data required to analyze the model in the form of secondary data and primary data. For secondary data are required: 1) Data monthly rain of station Up Bangga and Down Bangga available from 1995 to 2011); 2) Data temperature of Bora station available from 1995 to 2011); 3) Data Potential Evapotranspiration available from 1995 to 2011; 4) Maps satellite imagery; 5) Maps the earth in such a scale of 1:50,000 and 6) Soil type maps, scale 1:50,000. 7) The primary data obtained by directly sampling the soil in the research site to obtain the percentage fraction of land, SM content, soil bulk density and soil type conducted in accordance vegetation land cover (mixed garden and forest) to make a hole in the soil profile depth of the surface soil, topsoil and subsoil below (SM storage) [9].
### 3. Results and Discussion
The SWB model is part of the model MockWyn-UB which considers climate change in the form of rain correction factor (αP ) and temperature correction factor (αT) as input. Thus, before using this model needs to be analyzed in advance whether there is a climate change study site by using Mann-Kendall models. Furthermore, the projected changes are analyzed by the method of non-parametric Sen’s by Sutapa (2015) [9] there has been a climate change in Bangga watershed as mention in Table 2. Thus this model can be used SWB model calculation results are presented in Table 3.
Climate change is occurred if the values of Z test more or less than zero. Otherwise, if the value of Z test is equal to zero is not occurred climate change. A confidence value (α) used in the calculation of Mann-Kendall is 0.001; 0.01; 0.05 and 0.1. According to the table of normal standard “Z” the values are: Z0,001 = 3.292; Z0,01 = 2.576; Z0,05 = 1.96; Z0,1 = 1.645.
Signs of significance in the calculation of the Mann-Kendall categorized into five categories, namely: three star (***), two (**), one (*), the (+) and empty (..) which indicates the level of confidence (α) = 0.001; 0.01; 0.05 and 0.1 respectively. If there is no sign (blank) means a significant level (α) of more than 0.1 or can be said to be insignificant [9].
Fig. 5 presents the relationship between monthly rainfall (R) and potential evapotranspiration (ETo) with groundwater (GW) as a result of modeling taking into account climate change in year 1995. It can be seen that the potential evapotranspiration almost the same throughout the month, then the amount of ground water is only affected by the amount of monthly rainfall. This model may give good performances because of the variation of the ground water or linear follow the up and down of monthly rainfall. The mean effect of evapotranspiration (ETo) and rainfall (R) to be groundwater is 58% and 42%, where the influence of evapotraspiration ranged from 37% to 68%, while rainfall ranges between 32% to 54%.
Fig. 6 presents the results of the calculation model of groundwater in the period 1995 to 2011. At the beginning of the simulation in 1995 resulted in ground water is high enough then dropped dramatically in all months. In January (year 1995 to 2011) the value of the largest groundwater occurred in 2000, while in others nearly flat. In February and March the value of groundwater fluctuation but not too big. Period May–August (year 1995 to 2011) the largest fluctuation occurred in 2010. Finally, for the period September–December (year 1995 to 2011) the value of groundwater occurred in 1995, while for the others there is a fluctuation but not too big.
Fig. 7 presents variation of yearly groundwater for period 1995 to 2011. During this period the average groundwater is 147 mm/y, the largest groundwater is 676 mm/y in year 1995 and the smallest occurred in year 2006 amounted to 33 mm/y. It can be seen that there is a downward trend in groundwater from the beginning to the end of the simulation.
Fig. 8 and Fig. 9 present variation of yearly rainfall (R), evapotranspiration (ETo), groundwater (GW) and percentage of rainfall to be groundwater for period 1995 to 2011. During this period the average rainfall to be groundwater is 9.9%, the largest rainfall to be groundwater is 25.60% in year 1995 and the smallest occurred in year 2005 amounted to 3.24%. It can be seen that the fluctuation of the rain changes into groundwater.
From the results that have been presented can be seen that groundwater recharging period 1995 to 2011, in January decline of simulation start in 1995 until 1998, raised significantly in 2000 before finally almost the same until 2011. In February decreased very sharply from the beginning of the simulation until 1998, then slightly increased in 2000, 2007, 2009 and other years is almost the same. In March a very sharp decline from the beginning of the simulation until 1998 and then fluctuated until the end of the simulation (2011). In April there is a decrease until 1996, there is an increase in 1999, 2001, 2007, 2008 and 2009, while in others fluctuate. In May decline began early simulation and an increase in 1999, 2001 and 2007. In June decline began early simulation and an increase in 2000, 2007 and 2010. In July a decline from early simulation until 1997 and increase in 1998, 2008 and 2010. In August a decline from early simulation to 1997 and increased in 1998, 2007 and 2010. In September, October, November and December, there was a very sharp decline from the beginning of the simulation until 1996 and fluctuate for other years.
The annual simulation results indicate a trend decline in groundwater recharge from early 1995 until the end of the simulation in 2011. From 1998 to 2000, 2007, 2008, 2010 and 2011 were above the average annual groundwater for the period 1995 to 2011. The amount of groundwater is very linear with the amount of monthly rainfall is happening. It means that the higher the rainfall that occurred then recharging groundwater will be even greater, and vice versa.
Percentage rain to fill ground water is very small, ranging from 3% till 25%. This indicates that the rain that falls on the watershed Bangga to be a lot more runoff and evaporation. If observed Fig. 8, it is clear that in general the amount of annual rainfall is less than evaporation and no recharging groundwater. This is unlikely to happen. That is the amount of annual rainfall, evaporation annual and annual groundwater cannot be compared. What can be done is to compare a monthly basis in accordance with this simulation. It can be seen from Fig. 5 where the value is greater than the monthly rainfall evaporation and groundwater recharge occurs. This occurs and goes for the other years during the period 1995 to 2011.
### 4. Conclusions
The conclusion from this study is going on a downward trend in the annual groundwater recharge from early 1995 until the end of the simulation in 2011. For 1998 till 2000, 2007, 2008, 2010 and 2011 recharging groundwater is above the average annual groundwater for the simulation period (1995 to 2011). The amount of groundwater recharging linearly with the mount of monthly rainfall is happening. It means that the higher the rainfall that occurred then recharging groundwater will be even greater, and vice versa. Percentage of rain to fill ground water is very small, ranging from 3% to 25%. This indicates that the rain that falls on the Bangga watershed to be a lot more runoff and evaporation.
### Acknowledgments
The author thanks very much to the River Regional Department Office of Sulawesi III in Palu and Watershed Management Department Office of Palu Poso in Center Sulawesi Province of Indonesia which prepared the hydro-climatology data and land use map for supporting this research. The author would also like to thank the Faculty of Engineering, University of Tadulako, Palu, Central Sulawesi of Indonesia for providing technical support.
#### Nomenclature
B
Constant
ETo
Potential evapotranpiration, mm/d
ETa
Actual evapotranspiration, mm/d
es
Saturated water vapor pressure, kPa
ea
Actual water vapor pressure, kPa
I
Infiltration
In
Coefficient of infiltration
k
Groundwater flow recession factor
LDAS,
Total area in watershed, km2
LHT
Forest area, km2
LKC
Mix farm area, km2
LLT
Open land area, km2
n
The amount of observational data
PDAS
Total rain in watershed, mm
PHT
Rain in forest, mm
PKC
Rain in mixed farm, mm
PLT
Rain in open land, mm
PNT HT
Net rain forest, mm
PNT KC
Net rain mixed farm, mm
PNT LT
Net rain open land, mm
Q
Slope
Rn
S
Variants
TPN
Total net rain, mm
U2
Wind velocity at a height of 2 m above the ground, m/s
Vn
Volume of groundwater, mm/mon
Z
Statistical value
αP
Precipitation correction factor
αT
Temperature correction factor
σs
Root of Variants
Δ
The slope of the vapor pressure of water against temperature, kPa/°C
γ
psychrometric constant, kPa/°C
### References
1. The Government of the Republic of Indonesia. The Indonesian Government Regulation number 43 of 2008 on Groundwater. Indonesia: 2008.
2. Bisri MGroundwater. 1st edMalang, Indonesia: Univ. of Brawijaya Press (UB Press); 2012. p. 11–21.
3. WWF Indonesia and IPCC. Climate change in Indonesia, implications for humans and nature. Indonesia: 1999.
4. Rao A, Al-Wagdany AEffects of climate change in Wabah River basin. J Hydrol Eng. 1995;121:207–215.
5. Tung CP, Haith DAGlobal-warming effetcs on New York streamflows. J Water Resour Plann Manage. 1995;121:216–225.
6. Yates D, Strzepek KModeling the Nile basin under climate change. J Hydrol Eng. 1998;3:98–108.
7. Fu G, Barber M, Chen SImpacts of climate change on regional hydrological regims in the Spokane River watershed. J Hydrol Eng. 2007;12:452–461.
8. Lorena L, Leonardo VN, Enrique RV, Goffreda LLBasin-scale water resources assessment in Oklahoma under synthetic climate change scenario using a fully distributed hydrologic model. J Hydrol Eng. 2010;15:107–122.
9. Sutapa IWModeling discharge of Bangga watershed under climate change. Appl Mech Mater. 2015;776:133–138.
10. Dunne , Leopold Interception. Lecture 3A of Geography. IHE Delft; The Netherlands: 1978.
11. Sutapa IWApplication model Mann-Kendall and Sen’s (Makesens) for detecting climate change. Infrast J Civil Eng Univ Tadulako. 2014;4:31–40.
12. Sutapa IWLong-term trend climatology in Sigi, Central Sulawesi province. In : National Seminar on Civil Engineering; 28 February 2015; Narotama Univ., Surabaya. p. 267–277.
13. Sutapa IWApplication of non parametric test to detect trend rainfall in watershed Palu, Central Sulawesi, Indonesia. Int J Hydrol Sci Technol. 2016;6:238–253.
14. Hadisusanto NugrohoHydrology Application. 1st edMalang, Indonesia: Jogja Mediutama; 2011. p. 95–105.
15. Richard AGCrop evapotranspiration-guidelines for computing crop water requirement-FAO irrigation and drainage paper No. 56. Rome: Food Agriculture Organization of the United Nation; 1998. chapter 2:
16. Nasution , Syaifullah DjazimSpatial analysis of the North Coast regional drought index (North Coast) in West Java. J Air Indonesia. 2005;1:235–243.
17. Sutapa IWEffect of climate change modeling to discharge [dissertation]. Malang, Indonesia: Univ. of Brawijaya; 2013. 113–114.
18. Onoz B, Bayazit MThe power of statistical test for trend detection. Turkish J Eng Env Sci. 2002;27:247–251.
19. Wang S, Zhang Z, Sun G, et alLong-term streamflow response to climatic variability in the Loess Plateau, China. J Am Water Resour Assoc. 2008;44:1098–1107.
20. Hakan A, Savaş K, Osman STrend analysis of hydrometeorological parameters in climate regions of Turkey. In : BALWOIS 2010; Ohrid, Republic of Macedonia. 2010.
21. Salmi T, Määttä A, Anttila P, Ruoho-Airola T, Amnell TDetecting trends of annual of atmospheric pollutants by the Mann-Kendall test and Sen’s slope estimates. Helsingki: Finnish Meteorological Institute; 2002.
22. Deo RC, McAlpines CA, Syktus J, McGowan HA, Phinn QOn Australian heat waves: Time series analysis of extreme temperature event in Australia, 1950–2005. In : MODSIM 2007 International Congress on Modelling and Simulation; 10–13 December 2007; Canberra, Australia. The Modelling & Simulation Society of Australia and NZ Inc.; 2007. p. 626–636.
Map of location.
##### Fig. 2
Flowchart of soil water balance model.
##### Fig. 3
Variation of monthly rainfall data in year 1995–2011.
##### Fig. 4
Variation of monthly evapotranspiration data in year 1995–2011.
##### Fig. 5
Relationship between GW, R versus ETo in year 1995.
##### Fig. 6
Variation of monthly groundwater January-December.
##### Fig. 7
Variation of yearly groundwater in period 1995–2011.
##### Fig. 8
Variation of yearly rainfall (R), evapotranspiration (ETo), groundwater (GW).
##### Fig. 9
Variation of percentage rainfall (R) to be groundwater (GW).
##### Fig. 10
Relationship between groundwater (GW) versus rainfall (R) and potential evapotranpiration (ETo) base line 1995–2011 and projection to 2031.
##### Table 1
Summary of SWB Model, Inputs, Parameters, Outputs
(1) (2)
(a) Inputs Data
LDAS, LHT, LKC, LLT Total area, forest area, area mix farm and open land area respectively (km2)
PDAS Total monthly precipitation (mm)
ETo Monthly evapotranpiration (mm)
(b) Parameters
αP Precipitation correction factor
αT Temperature correction factor
In Coefficient of infiltration
k Groundwater flow recession factor
(c) Output
Vn Monthly groundwater volume (mm)
##### Table 2
Results of Mankendall Model
Uraian From To n Trend
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Temperature, T (°C)
Test Z, average daily 1980 2011 32 2.84
Pos and YS
Rainfall, R (mm/d)
Test Z, average daily −2.73
1993 2011 19 Neg and YS
Potential Evapotranspiration (ET, mm/d)
Test Z, average monthly −1.31
1980 2011 32 Neg and YS
[i] where:
Pos = Positif atau increasing trend
Neg = Negatif atau decreasing trend
YS = Yes significant
NS = No significant
NT = No trend
Zcal > Zα …………………… Yes significant (YS)
Zcal < Zα …………………… No significant (NS)
Z = 0 …………………… No trend (NT)
Table Z of normal standard
Z0,001=3,292 …………….. α = 0,1%
Z0,01=2,576 …………….. α = 1 %
Z0,05=1,96 ……………… α = 5%
Z0,1=1,645 ……………… α = 10%
##### Table 3
Estimates of Monthly Groundwater (mm)
No. Year Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Sum Average
1 1995 32.82 66.17 83.72 41.69 28.67 43.47 45.29 66.14 91.54 75.59 67.18 33.45 675.74 56.31
2 1996 16.06 19.24 7.70 6.13 6.10 10.59 4.24 8.82 6.37 2.55 1.02 0.41 88.22 7.35
3 1997 16.06 7.30 2.92 8.26 3.31 1.32 8.33 3.33 1.33 0.53 21.87 8.75 82.31 6.86
4 1998 15.06 6.02 2.41 15.75 6.30 13.59 56.67 22.67 17.53 7.01 2.80 1.12 166.92 13.91
5 1999 26.31 19.39 12.86 21.96 16.20 11.32 7.91 5.53 3.87 16.52 11.55 8.07 160.48 13.37
6 2000 54.18 21.67 8.67 3.47 1.39 31.39 12.55 5.02 2.01 0.80 0.32 0.13 141.60 11.80
7 2001 17.18 6.87 2.75 17.65 17.33 13.13 5.25 2.10 0.84 0.34 0.13 0.05 83.63 6.97
8 2002 15.06 6.02 2.41 0.96 0.39 17.12 6.85 2.74 1.10 0.44 10.42 4.17 67.67 5.64
9 2003 15.06 6.02 2.41 7.01 2.81 1.12 0.45 0.18 0.07 1.03 0.41 0.17 36.74 3.06
10 2004 15.06 6.02 2.41 2.40 8.76 3.50 1.40 0.56 0.22 0.09 0.04 0.01 40.47 3.37
11 2005 15.06 6.02 2.41 0.96 2.03 0.81 0.33 0.13 2.99 1.85 0.74 0.30 33.62 2.80
12 2006 15.06 6.02 2.41 5.58 2.23 0.89 0.36 0.14 0.06 0.02 0.01 0.00 32.79 2.73
13 2007 15.06 22.59 9.04 16.30 39.96 20.58 8.23 39.03 15.61 6.70 2.68 1.50 197.27 16.44
14 2008 15.06 6.02 10.23 20.97 17.63 17.44 44.34 17.74 7.09 2.84 1.14 0.45 160.94 13.41
15 2009 26.35 18.44 12.91 9.04 6.33 4.43 3.10 2.17 1.52 1.06 0.74 6.74 92.83 7.74
16 2010 15.06 6.02 2.41 0.96 0.39 53.11 65.35 76.29 30.52 12.21 4.88 1.95 9.15 22.43
17 2011 18.29 8.89 4.32 2.10 1.02 0.50 0.24 32.89 22.24 10.81 44.61 21.68 167.59 13.97
Average 20.04 13.99 10.12 10.66 9.46 14.37 15.94 16.79 12.05 8.26 10.03 5.23 146.94 12.24
TOOLS
Full text via DOI
E-Mail
Print
Share:
METRICS
0 Crossref
0 Scopus
2,014 View
|
2019-11-21 01:11:05
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 24, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4735413193702698, "perplexity": 3665.195693354677}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670643.58/warc/CC-MAIN-20191121000300-20191121024300-00206.warc.gz"}
|
https://en.wikipedia.org/wiki/Exponentially_equivalent_measures
|
Exponentially equivalent measures
In mathematics, exponential equivalence of measures is how two sequences or families of probability measures are “the same” from the point of view of large deviations theory.
Definition
Let (Md) be a metric space and consider two one-parameter families of probability measures on M, say (με)ε>0 and (νε)ε>0. These two families are said to be exponentially equivalent if there exist
• a one-parameter family of probability spaces ((Ω, ΣεPε))ε>0,
• two families of M-valued random variables (Yε)ε>0 and (Zε)ε>0,
such that
• for each ε > 0, the Pε-law (i.e. the push-forward measure) of Yε is με, and the Pε-law of Zε is νε,
• for each δ > 0, “Yε and Zε are further than δ apart” is a Σε-measurable event, i.e.
$\big\{ \omega \in \Omega \big| d(Y_{\varepsilon}(\omega), Z_{\varepsilon}(\omega)) > \delta \big\} \in \Sigma_{\varepsilon},$
• for each δ > 0,
$\limsup_{\varepsilon \downarrow 0} \varepsilon \log \mathbf{P}_{\varepsilon} \big[ d(Y_{\varepsilon}, Z_{\varepsilon}) > \delta \big] = - \infty.$
The two families of random variables (Yε)ε>0 and (Zε)ε>0 are also said to be exponentially equivalent.
Properties
The main use of exponential equivalence is that as far as large deviations principles are concerned, exponentially equivalent families of measures are indistinguishable. More precisely, if a large deviations principle holds for (με)ε>0 with good rate function I, and (με)ε>0 and (νε)ε>0 are exponentially equivalent, then the same large deviations principle holds for (νε)ε>0 with the same good rate function I.
References
• Dembo, Amir; Zeitouni, Ofer (1998). Large deviations techniques and applications. Applications of Mathematics (New York) 38 (Second edition ed.). New York: Springer-Verlag. pp. xvi+396. ISBN 0-387-98406-2. MR 1619036. (See section 4.2.2)
|
2015-05-05 15:37:50
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9285613298416138, "perplexity": 1840.3212641448697}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1430456773322.93/warc/CC-MAIN-20150501050613-00069-ip-10-235-10-82.ec2.internal.warc.gz"}
|
http://koreascience.or.kr/article/CFKO200111923005517.page
|
# Improvement of Permeability to Organic Solvent in Escherichia coli for a Toxicity Biosensor
• Bae, Hee-Kyung (Water Environment Research Center, Korea Institute of Science and Technology (KIST)) ;
• Shin, Pyong-Kyun (Water Environment Research Center, Korea Institute of Science and Technology (KIST)) ;
• Song, Bang-Ho (Department of Biology, Teachers College, Kyungpook National Universitiy)
• Published : 2001.06.01
• 49 5
#### Abstract
The outer membrane (OM) of gram-negative bacteria acts as an effective permeability barrier against noxious agents including several antibiotics and organic solvents, and lipopolysaccharide (LPS) is the key molecule for this function. Outer membrane modified mutants (Ml-166, M2-42, M3-21) of E. coli DH5$\alpha$/pBSl were selected through a mutation using EMS (ethyl-methane-sulfonate). Among the selected mutants, M3-21 was twice as sensitive as LumisTo $x^{ }$ to benzene and M2-41 was 8 times as sensitive as LumisTo $x^{ }$ to toluene. To identify the structural change in the membrane by mutation, the relative cell surface hydrophobicities and the absorption of the crystal violet to the organisms were measured. All the mutants absorbed more crystal violet than their parent and the absorption of crystal violet increased in cell walls as carbohydrate of lipopolysaccharide decreased. When the cell surface hydrophobicities of DH5/pBSl and its mutants were measured by the BATH, the hydrophobicities of mutants increased compared to their parent in several organic solvents. The difference of lipopolysaccharide between DH5/pBSl and its mutants was identified by various ways such as the SDS-PAGE gel, the screening of LPS molecular weights, the mass spectrometry, and MALDI-TOF.F.
|
2020-04-06 20:42:01
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.41523101925849915, "perplexity": 11036.571278646676}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371660550.75/warc/CC-MAIN-20200406200320-20200406230820-00076.warc.gz"}
|
https://nb.khanacademy.org/economics-finance-domain/macroeconomics/macroeconomics-income-inequality/the-2008-financial-crisis/v/cnn-understanding-the-crisis
|
# CNN: Understanding the crisis
## Video transkripsjon
Rick: Susan Liscovicz, thanks for that report. We'll be talking to you, obviously, throughout the course of this newscast. I want to bring somebody in now that I'm realy excited about having on this show. He's totally different from most of the people that we talk to. He's gained his fame on YouTube, of all places, where people to go to see how he explains things that many of us frankly don't understand, so let's start with the basics, Sal Khan, thanks so much for being with us from the Khan Academy, by the way. Basic question, explain to our viewers. Explain to the million people out there what is actually broken that we need to fix? Sal: Right, and right now, we have so many, you know, the stock market goes down every day, and people are having foreclosures, and banks are stopping lending and they're failing, so it's really hard for people to keep track of what is the problem. Are these just causes of the problem, or are they the actual problem? I think we need to take a step back, and I've drawn a little diagram, here. Rick: Let's put it up. Sal: Drew a little diagram here, and I think there's a basic fundamental question that we all have to ask, and what does the financial system do? They make a lot of money. Hopefully they add some value to our economy, and this diagram essentially explains what the financial system does, so in a capitalist economy, you have a lot of people with capital. That could be land. That could be gold. That could be, I don't know, animals. that could be cash, and what you need to do in order to make the economy grow is invest that capital into projects. Build factories, plow seeds, hire people. What the financial system does, the banks, and the markets, they take that capital and hopefully they allocate it on projects that actualy put people to work, hire people, grow our GDP. The problem right now is, as opposed to the financial system actually putting more capital into the real world, it's, for the most part, more concerned with self-preservation at the moment. They're kind of in survival mode, and because they're in survival mode, you don't have this flow of capital going from whoever has it to the financial system and then to the projects. You pretty much have everything just going back into the financial system because it's in essentially survival mode. Rick: So, I see the financial system in the middle. Why is that piece broken? Draw that one up for us. Sal: Sure, and here I'm going to attempt to give everyone a 5-minute MBA, but essentially, what I've drawn here is our balance sheets for some banks, and it might sound like a complicated term, but a balance sheet is, I think, something that most people are fairly familiar with when they talk about home equity. Rick: Mm-hmm. Sal: So, if I were to draw a balance sheet for someone's house, it would look something like this. So the asset would be your house, whatever it's worth, and then your liability, and I'll call that A for asset, and then your liability would be your mortgage. Your liability would be the loan. Rick: Mm-hmm. Now, if your house is worth $1 million, and your loan, let's say you owe 800,000 on your loan. Rick: Mm-hmm. Sal: Whatever's left over is your equity. Rick: Right. Sal: That's a balance sheet. That's a balance sheet that I think most people are fairly familiar with. For corporations, they have the exact same notion. And I think this is an important point to realize when people talk about the stock market and what does it mean for a company's stock to go to 0? When you buy a stock in a company, you're essentially, you're not buying a share in all of its assets. You're buying a share in its equity. So, if you look at, so I drew two banks here, and we'll talk a little bit more about what's broken, but a bank has assets, and a bank has liabilities. If you take the sum of the assets, and you subtract out the liabilities ... Rick: Mm-hmm. Sal: ... what's left over is essentially the equity. When you buy a stock in a bank or any company, you're essentially taking a share of that. So what's broken here is through the housing bubble, a lot of these banks, in order to facilitate this whole securitization, and I can go into more detail into what that is, they essentially accumulated these assets on their balance sheets. This has been, this has been the central focus of the entire bailout plan, and there's been a lot of talk about toxic CDOs and what are they worth, but I just want to show you what they do to a bank's balance sheet, and why it's so important that we figure out what these things are worth because you see here, I drew the, you can kind of view the height of this left-hand side as the assets. The height on the right-hand side is the liabilities. What's left over is the equity. If these toxic CDOs are worth nothing, if this disappeared, then all of a sudden, your assets are worth less than your liabilities. If your house is worth less than your mortgage, then you have no equity. In that situation, your stock would be worthless. In that situation, there's no reason why anyone should be lending to you, because you're essentially throwing good money after bad. Rick: So, essentially, the situation that the banks find themselves in is the same as the one that homeowners had found themselves in as a result of this. Sal: Pretty much. Homeowners, you have an asset that's worth less than your liabilities, what do you do? Do you try to get someone to bail you out and buy your asset for more than it's worth or do you just declare bankruptcy? Rick: Let's hold off right there. We're gonna come back in just a little bit, and we're gonna continue that conversation, and we're also gonna be bringing you something today that's an exclusive to this particular hour on CNN. The story of Curveball. Who is Curveball? He may just be the only person that the Bush Administration used as their source for going to war in Iraq, or certainly as much as anyone else. Was he right? No. Who is he? No one's known, up to now. An exclusive interview with Curveball. Stay with us. We'll be right back. (upbeat music) twitter.com/ricksanchezcnn So many of you, 23,000, I believe when we last checked are talking to us, watching this newscast as well as on MySpace and Facebook. In fact, there's somebody on MySpace right now. She's watching our show and she just wanted us to know this. She says, "My son is 17 years old "and excited about college. "I'm worried that we won't be able to get a loan "to send him now. "My husband's hours have been cut back from work "as well. "This absolutely affects all of America." And that's why we're doing what we're trying to do for you. Let's go back now and bring in Sal, if we can. Sal Khan. He explained to us moments ago what it is that is actually the problem. What it is that is broken. Now, let's take the next step. Let's try and understand because everyone seems to be wondering. I don't understand this bailout / rescue plan that the government's been in. What is it? What is the government actually doing? Explain that to us so we could understand it like a 5th grader. Sal: Right. The main problem, just to take the big picture, was that no one's lending to each other. What the government at least said that they wanted to do is try to get the lending started again. Their bailout plan was essentially, let's buy out these toxic CDOs, because if I'm, let's say I'm a good bank that didn't get involved in this. If Bank A is holding these toxic CDOs, I'm not sure if Bank A should be bankrupt, if the toxic CDOs are worth nothing, or maybe they are good for the money. There's a lot of uncertainty on what these are worth. The government, at least what they said they would do is if they went in and essentially did these reverse auctions and bought these assets, then that, all of a sudden, would give confidence in these banks. People wouldn't, there wouldn't be uncertainty as to whether you can loan them money and maybe that'll free up the credit, the whole system again. Rick: Will it work? Sal: Probably not. If you think about what's going to happen, psychologically, so Bank A, if you were to, let's say that, you know, if you were to just switch these toxic CDOs, and if you were to just turn that back into cash because the government just bought it, for exactly what Bank A says it's worth. If you just turn this into cash, the question is, is Bank A all of a sudden going to start lending? And there's a couple of things to think about because besides that toxic CDO, there were other things on its balance sheet, and frankly, a lot of these things are starting to starting to get a little bit toxic. No one's really talking about them right now because if you have a dead skunk in your house, you're not going to notice the milk's gone bad, so essentially, you have a lot of assets on your balance sheet right here, that if you're a prudent bank manager, you're going to see, you're going to say, "Boy, I should just keep that cash "because if these turn toxic "because the economy might be turning south "or for whatever reason, "I should just hold onto that cash "so that I don't go bankrupt." So everyone is in complete survival mode, even the good banks. If somehow you were to give them capital and Paulson's new version of the plan is that essentially, they'll inject equity, so buy stocks. If you buy stock, you'll essentially make this pie bigger and put some cash here, but even this guy, the good bank, it's not so clear to me that he'll actually start lending. Everyone is in survival mode. This isn't a time to make new loans and take on new risks. Rick: So it seems like the government went in, and they had decided, "You know what we're gonna do? "We're gonna go in there and help these guys out, "and as soon as we help them out, "everybody will realize that now they have "the government's backing, "so other people will be interested in helping them. "More people will give them their money." That hasn't worked. The people and other investors haven't bought into that. When we come back, I want to hear your idea of what, perhaps, the government should have done or could still do to rescue this. It's unique. It's different. You and I had talked about it earlier, but I want you to be able to share it with our viewers. Also, Curveball, the story that we've been telling you about. This is gonna be a CNN exclusive to this hour about one particular individual who no one has ever talked to before. He only speaks German and Arab. You'll hear his conversation here in the next 10 minutes. Stay with us. We'll be right back. (lively music) All right, so we understand now what actually is broken. We've got a sense of what the government tried to do to fix it, but it doesn't seem to be working because enough people / investors and regular folk don't seem to be buying into it. Let's bring in Sal Khan once again from the Khan Academy and talk about what may be done. Either A, would have been the original plan that would have worked a lot better, or something we could still do now, and that is ... Sal: So the big picture, just quick, going back to the big picture is that the financial system no one's lending through, right? Rick: Right. Sal: And every bailout plan so far, the government is just, it just keeps injecting money in either through loans, either through buying assets for maybe more than they're worth or now buying stock. and the logic here is maybe that'll start up the lending, but most probably any new money that you put into the system, to a large degree, is just going to go into survival mode or it's gonna go to essentially make up mistakes that have already been made. So, one idea, and this actually came from a friend of mine, Todd Plutsky, and I actually think it sounded crazy when he first said it, but it actually sounds like a really good idea is take that 700 billion, and remember, the problem isn't, you're not trying to save the financial system. You're trying to save the pipe that goes from the capital to the project. So why not take the 700 billion and capitalize a brand new financial system? Rick: Hmm. Sal: And one thing that Todd had pointed out, 700 billion. It's an astronomical amount of money. That's more than the book value of the equity, book equity is this piece right here on the balance sheet, than JP Morgan, Morgan Stanley, Goldman Sachs, Washington Mutual, Wachovia, and Bank of America combined. So the government could overnight create banks bigger than those banks, although I don't think they should concentrate it all in five or six banks. They should start up ... Rick: So in other words, you're saying why try and put money into something that is so absolutely messed up? If they're half dead, let 'em die. Create a new system all together. But wouldn't that be a problem too? Because then wouldn't those banks be owned by the government? Do we want that? Sal: No, and this is a solution. First of all, just to make a point clear, Paulson's current solution involves government ownership. This plan, what you could do is, the government could capitalize the banks, maybe 20 or 30 banks around the country with the$700 billion, and then each of those banks could have 300 million shares, and they could give one share to every American. Rick: New banks. New, totally new banks. Interesting. Sal Khan, thanks so much. We'll keep you on. We're gonna be talking about this throughout the hour, obviously. We're trying to get people to explain some of these things for us, and what's the market doing right now? Let's put that, let's make that as big as we can possibly can. Down 71, a lot better than we were looking at an hour ago. Maybe you've helped, Sal, in some ways, hopefully. All right. We're gonna come back to that in just a little bit, but obviously one of the key stories that we've been saving for you today, the story of Curveball, the man who seemed to have been the source, so the influence in the invasion of Iraq. As we expected, Sal Khan is a bit of a hit. Does things kind of differently, doesn't he? and, a lot of you have a lot of comments, a lot of questions for him, including this one. Let's go ahead and take a Twitter board, if we possibly can. This one comes in from sanchfan. Yeah, he watches this show a lot, like every day. He says, "Keep Khan talking! "Look, the market is going up!" Apparently, it had been. I hadn't even noticed that. I was so engaged in the conversation. Let's do this real quick. I want to just step away from this for a moment. Wolf Blitzer, thanks so much. Sal Khan standing by. Sal, tell us, if you could, I guess bring us back to basics. How are we being affected by what we're seeing in the news every day, the credit crunch and the situation with the market. Sal: Sure. If we don't unclog the credit system, we know that everyone's gonna unlever their debt, so we're gonna have deleverage. Rick: Mm. Sal: That's essentially everyone unwinding their loans. Deleverage, and this is, I think, a critical point because I think there's been a lot of misinformation out there. That actually contracts the money supply. So, a lot of people out there are worried about inflation, these deficits that lead to inflation, but when you have deleverage, and this is what happened in Japan. This is what happened in the '30s, the money supply contracts. That's our printing press. Leverage is our printing press in a fractional reserve system. That's going to lead to deflation. So, i think a lot of people are worried about are we going to see inflation, are we going to see deflation, and if we don't fix the problem, we're going to see what Japan saw, and we're going to see deflation, and so ... Rick: And deflation, deflation means what? Concretely, to me and you. Sal: Things get cheaper, the opposite of inflation, so unfortunately, maybe salaries will go down, but on the plus side, a lot of assets, you know, people buying a house, they'll get cheaper, even. Rick: Got it. Sal: I know that probably won't be a good thing in the short term. Rick: Yeah, exactly. Not with what the situation is right now. Thanks so much. Sal Khan, you've been a great guest. What an original way of being able to explain things. We're gonna be right back with the closing bell, which looks to be somewhere in the middle of ...
|
2019-02-18 03:19:30
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.35900649428367615, "perplexity": 1126.5213921840946}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247484020.33/warc/CC-MAIN-20190218013525-20190218035525-00634.warc.gz"}
|
https://math.stackexchange.com/questions/3148595/let-a-be-the-n-%C3%97-n-matrix-whose-i-j-entry-is-i-j-for-all-i-j-1-2
|
# Let A be the n × n matrix whose i, j entry is i + j for all i, j = 1, 2, . . . , n. What is the rank of A? [closed]
I tried finding the solution by assuming i and j are both n, but I'm not sure if this is the proper direction to go.
## closed as off-topic by Leucippus, Shailesh, Cesareo, Lee David Chung Lin, SaadMar 15 at 0:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Leucippus, Shailesh, Cesareo, Lee David Chung Lin, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
• It looks like you may have a fundamental misunderstanding of what the matrix looks like. Try writing out the matrix for, say, a $3 \times 3$ example and see if that helps you understand what's going on. – Robert Shore Mar 14 at 23:18
Hint: the first column is $$\mathbf v=(2,3,\dots,n)^T$$. Furthermore, the $$i$$th column is given by $$\mathbf v+(i-1)\mathbf 1$$ where $$\mathbf 1\in\mathbb R^n$$ is the vector whose entries are all $$1$$s. What does this tell you about the number of column vectors of the matrix that are linearly independent?
• Not that it changes the argument at all, but the entries in the first column are $2,3,....,n,n+1$. – Ned Mar 14 at 23:35
|
2019-03-22 15:02:06
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8235594034194946, "perplexity": 324.3193155501292}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202671.79/warc/CC-MAIN-20190322135230-20190322161230-00528.warc.gz"}
|
https://www.developer.here.com/documentation/android-premium/3.16/dev_guide/topics/map-disk-cache.html
|
HERE SDK for Android (Premium Edition)
SDK for Android Developer's Guide
# Map Disk Cache
HERE SDK allows you to set the map disk cache to another location such as an SD Card. Use MapSettings.setDiskCacheRootPath(String path) to set desired cache location. This call should occur before MapEngine initialization, otherwise default path will be used for disk cache. Default path for apps that have offline storage permission is Context.getExternalFilesDir(String), otherwise Context.getFilesDir(). For example, if you are modifying the application from the sample tutorial app, you can add the call in the BasicMapActivity.java file before mapFragment.init().
Note:
• If you are using an SD card, ensure the SD card is always present to avoid any unexpected behavior.
• You should only delete the map data cache when the app is in its early start-up stages, before any HERE SDK calls. Otherwise, map data corruption and unexpected app errors can occur.
• If you plan to support changing the storage location such as switching between internal storage and an SD card, be aware that this requires an app restart, as the storage location switch must be done before initializing MapEngine or AndroidXMapFragment.
• Starting from Android 10 (see Scoped storage), disk cache path must be set under application-specific file directory.
## Incompatibility with Older Versions
Starting from v3.4 HERE SDK is no longer compatible with pre-3.4 versions of the HERE SDK disk cache. Map data downloaded on pre-3.4 versions of HERE SDK cannot be used on v3.4 or later. This limitation can be avoided by upgrading the pre-3.4 cache using the DiskCacheUtility.migrate(String sourcePath, String destPath) method. This method takes the same path value as MapSettings.setDiskCacheRootPath(String), and it must be run before MapEngine is initialized for the first time.
Note: migrate(String, String) is marked as deprecated because it is offered temporarily to assist with the transition. It will be removed in a future release.
|
2021-12-06 21:14:14
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.28960639238357544, "perplexity": 4877.041882129673}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363312.79/warc/CC-MAIN-20211206194128-20211206224128-00043.warc.gz"}
|
https://www.physicsforums.com/threads/area-between-two-curves.122535/
|
# Area between two curves
Hi, im having a little trouble with this one, how do you do them when no boundaries are given?
$$y=sin \frac{\pi x}{2}$$
and
$$y=x$$
how do i find the boundaries?
Related Calculus and Beyond Homework Help News on Phys.org
EDIT:i know how to get the boundaries, but how do i know what numbers to integrate between?
arildno
Homework Helper
Gold Member
Dearly Missed
Eeh, what about the interval $0\leq{x}\leq{1}$??
why between 0 and 1
?..also..what is the integral of sin2x?
In an example in the book they find the points of intersection, is this how i do it?if so how would i find the points of intersection?
arildno
Homework Helper
Gold Member
Dearly Missed
What does the term "point of intersection" MEAN?
Answer that, and you automatically may set up an equation whose solutions are the points of intersection.
You may graph the function for clearer understanding .
k..well cancel everyhting i said before..we have the two equations..I want to isolate x..and then find where the two curves intersect...
making it $$\frac{2y}{sin \pi} = x$$ is wrong isnt it?
how do i isolate x?
If I can find the points where the two curves intersect, then I can integrate between these two points to find the Area
Yes, that is wrong.
You have two equations in y and x .
The point of intersection has to satisfy both the conditions .
Now can you form the equation ?
Hint : Try to get an equation in x alone .
|
2020-12-04 02:10:43
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8334130048751831, "perplexity": 783.4273558774513}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141733120.84/warc/CC-MAIN-20201204010410-20201204040410-00357.warc.gz"}
|
https://api-project-1022638073839.appspot.com/questions/how-do-you-use-partial-fraction-decomposition-to-decompose-the-fraction-to-integ-39
|
# How do you use partial fraction decomposition to decompose the fraction to integrate (6x)/((x-4) (x+4))?
Mar 28, 2018
The integral is $3 \ln | {x}^{2} - 16 | + C$
#### Explanation:
We have:
$\frac{A}{x - 4} + \frac{B}{x + 4} = \frac{6 x}{\left(x - 4\right) \left(x + 4\right)}$
$A \left(x + 4\right) + B \left(x - 4\right) = 6 x$
$A x + B x + 4 A - 4 B = 6 x$
Thus $\left\{\begin{matrix}A + B = 6 \\ 4 A - 4 B = 0\end{matrix}\right.$
Solving we get
$2 A = 6$
$A = 3$
$B = 3$
Therefore the integral becomes $\int \frac{3}{x - 4} + \frac{3}{x + 4} \mathrm{dx}$. This is readily integrated as $3 \ln | x - 4 | + 3 \ln | x + 4 | = 3 \ln | {x}^{2} - 16 | + C$#
Hopefully this helps!
|
2021-10-24 14:48:40
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9883977770805359, "perplexity": 2464.552582619905}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323586043.75/warc/CC-MAIN-20211024142824-20211024172824-00071.warc.gz"}
|
https://poker.stackexchange.com/questions/10498/poker-game-notation-pgn-as-in-chess
|
# Poker game notation PGN as in chess
Is there a notation for poker hands, as in PGN for chess? If not can you suggest one.
## 1 Answer
To my knowledge, there is no official notation quite like PGN for chess that is used for poker hands. However, there is multiple ways of sharing hands. If you are looking to show a specific hand to other players, you can use replay services like sharemypair.com that create a replay of the hand for others to watch. There is also a notation that is used by poker tracking software and poker sites to save hands that have already been played to a database locally, but this notation can vary across different services.
I think a universal notation would be a great idea, but as of now I don't think there is one.
|
2019-09-20 03:06:56
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5050355792045593, "perplexity": 853.5198025687104}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573827.2/warc/CC-MAIN-20190920030357-20190920052357-00086.warc.gz"}
|
http://googology.wikia.com/wiki/Fish_number_5
|
## FANDOM
10,219 Pages
Fish number 5 (F5) is a number defined by Japanese googologist Fish in 2003.[1] It is one of Fish numbers.
Fish function 5 uses m(n) map.
Definition and growth rate of Fish function 5, $$F_5(x)$$, is \begin{eqnarray*} F_5(x) & := & ((..((m(x)m(x-1))m(x-2))...m(2))m(1))(x) \\ F_5(x) & \approx & f_{\varepsilon_0}(x) \end{eqnarray*}
Then Fish number 5 is defined as: $F_5 := F_5^{63}(3)$
Therefore, Fish number 5 is greater than Fish number 3 and is approximately $$f_{\varepsilon_0+1}(63)$$.
## Approximations in other notations Edit
Fish number 5 is comparable to Grantethrathoth $$\approx f_{\epsilon_0+1}(100)$$.
Notation Approximation
BEAF $$\{63,63,2 (X \uparrow\uparrow X) 2\}$$[2]
Bird's array notation $$\{63,63,2 [1 \backslash 2] 2\}$$
Hyperfactorial array notation $$63![2,1,1,1,2]$$
Fast-growing hierarchy $$f_{\epsilon_0+1}(63)$$
Hardy hierarchy $$H_{\epsilon_0 \omega}(63)$$
Slow-growing hierarchy $$g_{\vartheta(\varepsilon_{\Omega+1}+1)}(63)$$
## Sources Edit
1. Fish, Googology in Japan - exploring large numbers (2013)
2. Using particular notation $$\{a,b (X \uparrow\uparrow X) 2\}$$ for $$X \uparrow\uparrow b \&\ a$$
|
2018-05-26 00:36:53
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9971703290939331, "perplexity": 12838.700055801917}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794867254.84/warc/CC-MAIN-20180525235049-20180526015049-00045.warc.gz"}
|
https://gateoverflow.in/204130/gate2018-55?show=252182
|
+14 votes
3.6k views
Consider a simple communication system where multiple nodes are connected by a shared broadcast medium (like Ethernet or wireless). The nodes in the system use the following carrier-sense the medium access protocol. A node that receives a packet to transmit will carrier-sense the medium for $5$ units of time. If the node does not detect any other transmission, it waits until this other transmission finishes, and then begins to carrier-sense for $5$ time units again. Once they start to transmit, nodes do not perform any collision detection and continue transmission even if a collision occurs. All transmissions last for $20$ units of time. Assume that the transmission signal travels at the speed of $10$ meters per unit time in the medium.
Assume that the system has two nodes $P$ and $Q$, located at a distance $d$ meters from each other. $P$ start transmitting a packet at time $t=0$ after successfully completing its carrier-sense phase. Node $Q$ has a packet to transmit at time $t=0$ and begins to carrier-sense the medium.
The maximum distance $d$ (in meters, rounded to the closest integer) that allows $Q$ to successfully avoid a collision between its proposed transmission and $P$’s ongoing transmission is _______.
asked
edited | 3.6k views
0
is it $150$?
+6
Question is incomplete.
"...If the node does not detect any other transmission in this duration, it starts transmitting its packet in the next time unit. If the node detects another transmission, it waits until this other transmission finishes and then begins
to carrier-sense for 5-time units again..."
## 5 Answers
+26 votes
Best answer
P starts transmission at $t=0$. If P's first bit reaches Q within Q's sensing window, then Q won't transmit and there shall be no collision.
Q senses carrier till $t=5$. At $t=6$ it starts its transmission.
If the first bit of P reaches Q by $t=5$, collision can be averted. Since signal speed is $10$ m/time (given), we can conclude that max distance between P and Q can be $50$ meters.
answered by (317 points)
edited
0
P SHOULD TAKE TIME T1 TO SEND DATA TO Q WHERE T1 IS GREATER THAN EQUAL TO T2 (WHERE T2 IS THE TIME WHEN P STOPPED TRASNMITTING".).
NOW T2=20.
THEREFOERE T1>=20.
FOR Q T1>=20 MEANS T1>=(5(SENSING)+15(TRANSMITTING)).
THEREFORE OUR MEDIUM LENGTH BETWEEN THE TWO SHOULD BE GREATER THAN 15 UNIT OF TIME OR 15*10=150 METRES.
D>=150
NOW THEY HAVE ASKED MAXIMUM. THEREFORE MEDIUM LENGTH SHOULD BE INFINITE. WHERE AM I DOING WRONG?
0
T2 is not 20. Its mentioned that P's transmission lasts for 20. That means time taken to transmit first bit to the last bit is 20 (Transmission Delay)
0
Sorry, my wrong. Its written we have only two nodes. I was taking case of more than 2 nodes
0
Sir, i somehow agree with your answer but as the time 5 sec has elapsed it has stopped carrier sensing, and even if the packet of P arrived it will ignore. So, should'nt it be less than 50? I answered 49 accordingly.
0
Here it is given that if node detect another transmissiion it will wait until other finishes it transmission
so q should start its transmission after 20s
+1
Here What does it mean by
All transmissions last for 20 units of time.
+8 votes
I can model my given scenario as given below
So, P is done with carrier sensing and begins transmitting at t=0. Now, Q will sense the medium for 5 units of time till t=4 and begin transmitting at t=5.
If untill this time unit, the first bit of P's data reaches Q, Q will refrain from sending its own data. If it doesn't reach Q at t=4, at t=5, Q will start sending it's own data.
This means we have only 5 units of time till which we can wait for the first bit of signal of P to reach Q. And we know that this is nothing but one-way propagation delay ($T_p$).
Given the distance is d meters and the velocity of the signal is 10 meters per time unit, $T_p=\frac{d}{10}$ units of time.
Now, as you can see in the image, the worst case till when we can wait for the P's signal is till 5 units of time.
So, my $T_p=5$ units of time
$5=\frac{d}{10}$
$d=50\,m$ Answer
answered by Boss (14.1k points)
+5 votes
P starts transmission (career sensing is already done) at t=0
P will take 20 unit time for completion of the transmission.
At t=0, Q starts channel sensing.
At t=5 it will send a packet. If the packet reaches before t=20 unit it will collied.
So to avoid a collision, distance between P and Q should be more than the 15-time unit.
Transmission speed is 10 meter per unit time. So for 15-time unit length would be 15*10 = 150
answered by Veteran (55.4k points)
0
i too got $150$
+5
Tx = 20 unit time
v = 10 m / unit time
First bit wil reach station B in 5 seconds not 15 seconds.
0
To avoid collision successfully shouldn't it be 150/2 = 75 ?
0
50? :/
0
I got 150 as answer, but there is another popular answer 50. I just don't know which one is right.
+2
Nothing like 'popular' :P
50 is correct :)
0
I think it would be 50 meter because let if we take 150 meter distance .than at t=0, p station start transmission and at same time q is sensing the medium for 5 unit time. at t=5, station q analyze that there is nothing in medium so it will start transmission at t=6. It continues transmission for next 20 unit time.
Now at t=15, p's data should arrive at q but q has also send the data so both get collied somewhere in medium.
0 votes
Here the vulnerable time is tp where tp is the propagation delay.
P and Q are separated by a distance of 'd' meters. When P starts transmission its first bit will take d/10 time units to reach the Q. If this first bit is caught by Q then Q can avoid collision.
Hence the first bit should reach Q within 5 time units so that Q detects it.
Therefore, d/10<=5 units
d<=50 metres
Hence maximum distance is 50 metres.
answered by Boss (14.7k points)
–1 vote
P SHOULD TAKE TIME T1 TO SEND DATA TO Q WHERE T1 IS GREATER THAN EQUAL TO T2 (WHERE T2 IS THE TIME WHEN P STOPPED TRASNMITTING".).
NOW T2=20.
THEREFOERE T1>=20.
FOR Q T1>=20 MEANS T1>=(5(SENSING)+15(TRANSMITTING)).
THEREFORE OUR MEDIUM LENGTH BETWEEN THE TWO SHOULD BE GREATER THAN 15 UNIT OF TIME OR 15*10=150 METRES.
D>=150
NOW THEY HAVE ASKED MAXIMUM. THEREFORE MEDIUM LENGTH SHOULD BE INFINITE. WHERE AM I DOING WRONG?
answered by (231 points)
0
Sorry, my wrong. Its written we have only two nodes. I was taking case of more than2 nodes
Answer:
+7 votes
3 answers
1
+11 votes
3 answers
2
+5 votes
3 answers
3
|
2018-10-22 17:12:41
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5851093530654907, "perplexity": 2126.6539941115984}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583515352.63/warc/CC-MAIN-20181022155502-20181022181002-00140.warc.gz"}
|
http://blog.math.toronto.edu/GraduateBlog/2012/06/04/departmental-phd-thesis-exam-jordan-watts/
|
## Departmental PhD Thesis Exam – Jordan Watts
Everyone welcome. Refreshments will be served in the Math
Lounge before the exam.
Friday, June 15, 2012, 11:10 a.m., in BA 6183, 40 St. George St.
PhD Candidate: Jordan Watts
Thesis Title: Diffeologies, Differential Spaces, and Symplectic Geometry
(http://www.math.toronto.edu/jwatts/storage/ut-thesis.pdf)
Thesis Abstract:
The notion of smoothness is well-understood on manifolds, but in
practice one often requires some kind of smooth structure on subsets
or quotients of manifolds as well. Many notions of a smooth structure
have been defined for more general spaces than manifolds to remedy
this issue; for example, diffeological, differential, and Frölicher
structures are defined on arbitrary sets.
We will focus on the application of diffeology to a problem in
symplectic geometry; in particular, to differential forms on a
symplectic quotient. A symplectic quotient coming from a Hamiltonian
action of a compact Lie group is generally not a manifold (it is a
symplectic stratified space). Sjamaar defines a differential form on
this space as an ordinary differential form on the open dense stratum
that lifts and extends to a form on the original symplectic manifold.
He proves a de Rham Theorem, Poincaré Lemma, and Stokes’ Theorem using
this de Rham complex of forms. But these forms are defined
extrinsically. The symplectic quotient does, however, obtain a
natural diffeological structure, and one can ask whether the Sjamaar
forms are isomorphic as a complex to something intrinsic to the
diffeology.
I will show, under certain conditions, that this is the case: the
complex defined by Sjamaar is isomorphic to the complex of
diffeological differential forms. A useful consequence of this is
that one may use the Sjamaar differential complex in conjunction with
reduction in stages.
|
2022-07-06 07:32:02
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8630867600440979, "perplexity": 2160.467786344486}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104668059.88/warc/CC-MAIN-20220706060502-20220706090502-00290.warc.gz"}
|
https://math.stackexchange.com/questions/1563866/understanding-proof-by-infinite-descent-fermats-last-theorem
|
Understanding proof by infinite descent, Fermat's Last Theorem.
See here. The question is as follows.
How do we see that there do not exist nonconstant, relatively prime, polynomials $a(t)$, $b(t)$, and $c(t) \in \mathbb{C}[t]$ such that$$a(t)^3 + b(t)^3 = c(t)^3?$$
There is the following answer, purportedly elementary.
We give a completely elementary proof simply working in the ring $\mathbb{C}[t]$ and using that it is a UFD. Suppose there are some solutions of$$a(t)^3 + b(t)^3 = c(t)^3$$in $\mathbb{C}[t]$. Choose a solution $(a(t), b(t), c(t))$ such that the maximum $m > 0$ of the degrees of $a$, $b$, $c$ is minimal possible among all solutions. Clearly, this choice ensures that $a(t)$, $b(t)$, $c(t)$ are coprime. Then we have$$a(t)^3 = c(t)^3 - b(t)^3 = (c(t) - b(t))(c(t) - \omega b(t)) (c(t) - \omega^2 b(t)),$$where $\omega$ is a third primitive root of unity. Now, we look at the factors $c(t) - b(t)$, $c(t) - \omega b(t)$. Suppose that they have a common factor $q(t)$. Considering their sum and difference, we conclude that $c(t)$ and $b(t)$ have a common factor too. Moreover, $q(t)$ is a factor of $a(t)$. Thus, $a$, $b$, $c$ are not relatively prime, which is a contradiction. Repeating the same game with other pairs of factors, we see that all three factors $c(t) - b(t)$, $c(t) - \omega b(t)$, $c(t) - \omega^2 b(t)$ are pairwise coprime. Therefore,$$c(t) - b(t) = d_1(t)^3,\text{ }c(t) - \omega b(t) = d_2(t)^3,\text{ }c(t) - \omega^2b(t) = d_3(t)^3,\text{ where }d_1,\,d_2,\,d_3 \in \mathbb{C}[t].$$Note that$$\omega^2 + \omega + 1 = 0.$$Multiplying the second equation by $\omega$ and the third equation by $\omega^2$ and adding all three, we arrive at $$d_1(t)^3 + \omega d_2(t)^3 + \omega^2d_3(t)^3 = 0.$$Choosing $\eta_1$ and $\eta_2$ as any third roots of $-\omega$ and $-\omega^2$, respectively, and letting$$a_1 = d_1^3,\text{ }b_1 = \eta_1d_2^3,\text{ }c_1 = \eta_2d_2^3,$$we get$$a_1^3 = b_1^3 + c_1^3.$$By construction, at least one of $a_1$, $b_1$, $c_1$ is a nonconstant polynomial, and the maximum of their degrees is smaller than that of $a$, $b$, $c$. This is a contradiction to the choice of $a$, $b$, $c$.
Unfortunately, I have never seen any abstract algebra before (I am only a student who has taken calculus), and therefore, do not really understand what is going on, aside from the fact infinite descent is being invoked. Can anyone explain the motivation/what is going on/the key steps/help me understand the proof? Thanks.
• This proof is quite long: you should tell us what are the specific steps you don't understand (for example, do you know what $\omega$ is? Do you know what is an UFD?), in order to better help you. – Crostul Dec 7 '15 at 8:13
• Yes, the answer is elementary. If you do not understand it, then you need to learn abstract algebra. It's not a big deal, many people on Earth don't know anything about algebra -- but you still need it to understand this proof. There's no way around it. You wouldn't expect someone who's never run in their life to be able to run a marathon, or someone who's never studied medicine to successfully perform surgery, right? – Najib Idrissi Dec 7 '15 at 20:27
• I'm really confused by the positive attention this question has received. The OP has verbatim copied a long answer from another question and asked us to explain it to them. There is no evidence of any effort (eg we don't even know where the OP first gets stuck); all we know is that they have no background in abstract algebra. Without more information, I don't think it is possible to answer this question adequately. – Mathmo123 Dec 9 '15 at 9:47
• This argument just uses simple arithmetic of polynomials (long division etc.) that is essentially high school algebra. The assertion that $\mathbb C[t]$ is a UFD appears, but this is just short hand for saying that polynomials have unique factorization into linear polynomials, which is again a piece of high school algebra. Did you try actually working through it? – tracing Dec 9 '15 at 12:21
• Just answered to your question with lots of details about the formal framework behind it and on it as well. – Olórin Dec 15 '15 at 22:18
The formal framework behind your concrete problem is the following : you want to show that "something sizeable" does not exist. "Something that does not exist" is coded by a set $X$ that we want to show to be empty. "Being sizeable" means that you have a "size function" $s : X \rightarrow \mathbf{N}$. If $X$ is not empty, as the set $s(X)$ is a non-empty part of $\mathbf{N}$, it has a smallest element $d_0$ reached for some $x_0\in X$ such that $s(x_0)=d_0$. Now, if by some argument, some "recipe", from $x_0$ you arrive to construct another element $x_1$ from $X$ such that $s(x_1) = d_1 < d_0$ then you will have contradicted the fact that $s(x_0)$ is supposed minimal, and show that $X$ is empty. The descent name of the argument comes from the fact that from $x_1$ you could, with the same recipe construct an $x_2$ such that $s(x_2) < s(x_1)$, and then an $x_3$ etc etc : by induction you construct a sequence $(x_n)_n$ such that the sequence $(s(x_n))_n$ strictly decreases : it "descends infinitely", hence the "descente infinie" french term given by Fermat himself to such kind of proofs.
Now, in the case of your concrete problem, can you find the right set $X$, and the right "size" function $s : X \rightarrow \mathbf{N}$ ? Can you then find the right recipe to get $x_1$ from $x_0$ ? (Note that the recipe is just a method giving you a $y\in X$ such that $s(y) < s(x)$, starting from a given $x\in X$.) I will answer to these question through the details I give below.
First, no need to know what a UFD (unique factorization domain) is : we are in the $\mathbf{C}[T]$ case and the only thing we will use is this : for any non constant polynomial $P\in\mathbf{C}[T]$ you have a unique (modulo the order of the factors) factorization $$P = a_d (T-\lambda_1^{\mu_{\lambda_1}})\ldots(T-\lambda_d^{\mu_{\lambda_d}}) = z \prod_{\lambda\in S} (T - \lambda^{\mu_{\lambda}})$$ where $S$ is the set of roots of $P$ and $\mu_{\lambda}$ is the multiplicity of the root $\lambda\in S$ (this factorization is called nice factorization from now on) where the $\lambda_i\mathbf{C}$ are the distinct roots of $P$, $a_d\in\mathbf{C}$ is $P$'s leading coefficient and $d$ is the degree of $P$. You already know this : it is called that the field $\mathbf{C}$ is algebraically closed.
I fixed once for all an integer $n\geq 3$. Now we note $X$ the set of triples $(a(T),b(T),c(T))\in\left(\mathbf{C}[T]\right)^3$ such that
• $a,b$ and $c$ are non constant and relatively prime : this means that you cannot find a $\lambda\in\mathbf{C}$ such that the polynomial $T-\lambda$ divides $a, b$ and $c$, that is, you cannot write $a(T) = (T-\lambda) a_0(T), b(T) = (T-\lambda) b_0(T)$ and $c(T) = (T-\lambda) c_0(T)$ for a triple $(a_0(T),b_0(T),c_0(T))\in\left(\mathbf{C}[T]\right)^3$ : that is the polynomials $a,b$ and $c$ have no common factor
• $a(t)^n + b(t)^n = c(t)^n$ (note that I replaced $3$ by my $n$, I will prove something more general than what you want to prove, and I think that the greater generality will make you understand more easily what is formal behind the proof and what is not.)
and we want to show that $X = \varnothing$. Note that the degree of the zero polynomial is $-\infty$ let's say by convention, and that for a triple $(a(T),b(T),c(T))\in X$, at least one polynomial is non-zero, as if all were zero, $a,b$ and $c$ wouldn't be relatively prime. All this to say that for a triple $(a,b,c)\in X$ we have $s((a,b,c)) := \max\left( \textrm{deg}(a), \textrm{deg}(b), \textrm{deg}(c) \right) \in \mathbf{N}$ : we have our size function !
We want to show that $X = \varnothing$, and as I wrote at the beginning, we suppose that $X \not= \varnothing$ and we therefore take a $(a,b,c)\in X$, and we saw at the beginning that we have the right to choose $(a,b,c)$ such that $s((a,b,c))$ is minimal.
Remark. If $(a,b,c)$ is a solution in $\mathbf{C}[T]^3$ such that $s((a,b,c))$ is minimal, then $a,b,c$ are de facto relatively prime : would they have a common non trivial (that is, of degree $>0$) factor $Q$, we could write $a = Q a_0$, $b=Q b_0$ and $c = Q c_0$ and plugging these relations in the equation $a^n + b^n = c^n$ and factoring out the $Q^n$ shows that $a_0^n + b_0^n = c_0^n$, so that $(a_0,b_0,c_0)$ is also a solution but with $s((a_0,b_0,c_0)) < s((a,b,c))$ which contradicts the minimality of $s((a,b,c))$ and shows that $a,b,c$ are relatively prime indeed
Let's proceed further. We rewrite $a(t)^n + b(t)^n = c(t)^n$ as $a(t)^n = c(t)^n - b(t)^n$. Now remark that would we have $n=2$, you would know how to factor $c(t)^n - b(t)^n$, as we have the remarkable identity $c^2 - b^2 = (c-b)(c+b)$. We look for such an identity for $c(t)^n - b(t)^n$ in the general case $n\geq 3$. I am sure you heard about $n$-th roots of unity : these are complex numbers $\omega$'s such that $\omega^n = 1$. You have to know that there are so-called primitive $n$-th roots of unity : those are $n$-th roots of unity $\omega$'s such that the list $1,\omega,\ldots,\omega^{n-1}$ give you all $n$-th roots of unity. (Note that there are indeed $n$ $n$-th roots of unity : can you see why ?) Fix such a primitive $n$-th roots of unity $\omega$. As the $\omega^i$ for $0\leq i < n$ are all roots of unity, that is, roots of the polynomial $T^n - 1$, we can write : $$T^n - 1 = \prod_{i=0}^{n-1} (T - \omega^i ).$$ This is a formal identity, so that you can replace $T$ by $\frac{c(t)}{b(t)}$ in it and you will then see that you have $$c(t)^n - b(t)^n = \prod_{i=0}^{n-1} (c(t) - \omega^i b(t))$$ which is the remarkable identity we were looking for.
Now, we can rewrite the equation $a(t)^n = c(t)^n - b(t)^n$ as $$a(t)^n = \prod_{i=0}^{n-1} (c(t) - \omega^i b(t)).$$ What do we see here ? A product (in the right-hand side) that is equal to the $n$-th power of something (of $a$, in the left-hand side.) We would like to be able to say that this implies that each term of the product is the $n$-th power of something. We are allowed to make this implication when the factors of the product is constituted by relatively prime polynomials. (This is by the way the same situation as in $\mathbf{Z}$.) I will detail this :
(1) The $c(t) - \omega^i b(t)$ form relatively prime polynomials. Indeed : suppose they aren't and let $Q$ a common non trivial (that is, of degree $>0$) factor $Q$. We have $c - b = Q R_1$ and $c - \omega b = Q R_2$ for polynomials $R_1$ and $R_2$. Solving the system of equations constituted by two previous equations will show that $Q$ divides $b$ and $c$, and as $a^n = c^n - b^n$, $Q$ divides also $a^n$, and therefore divides also $a$ (you see why ?). So $Q$ divides $a$, $b$ and $c$ : contradiction as $a$, $b$ and $c$ are relatively prime. So, we shown that $c(t) - \omega^i b(t)$ form relatively prime polynomials.
(2) Previous (1) and the equation $c(t)^n - b(t)^n = \prod_{i=0}^{n-1} (c(t) - \omega^i b(t))$ imply that all $c(t) - \omega^i b(t)$ are $n$-th powers. To see this, simply write nice factorizations each of the $c(t) - \omega^i b(t)$'s : $$c(t) - \omega^i b(t) = z_i \prod_{\lambda\in S_i} (T-\lambda^{\mu_{\lambda}})$$ where $S_i \subseteq \mathbf{C}$ for all $i$ such that $0\leq i < n$. As the $c(t) - \omega^i b(t)$'s are relatively prime we have $S_i \cap S_j$ for all $i,j$ such that $0\leq i,j <n$ and $i\not=j$. Now if you you note $a = z\prod_{\lambda\in S} (T-\lambda^{\mu_{\lambda}})$ the nice factorization of $a$ (where $S\subseteq \mathbf{C}$), we must have $S = \cup_{i=0}^{n-1} S_i$. Plugging all linear factorization described in this paragraph into the equation $a(t)^n = \prod_{i=0}^{n-1} (c(t) - \omega^i b(t))$ and regrouping in the left-hand side linear factors according to if they belong to the same $S_i$ and comparing with the right-hand side shows indeed that each $c(t) - \omega^i b(t)$ is the $n$-th power of some polynomial $D_i(T)$ : $c(t) - \omega^i b(t) = D_i(T)^n$ for $0\leq i < n$.
From now on I will diverge slightly from the proof you quote. As the $c(t) - \omega^i b(t)$'s are relatively prime, so are the $D_i (T)$'s (exercise : prove it). Looking at leading coefficients from $c$ and $b$ we see that at most one $D_i (T)$ can be constant. Now choose any triplet $(x,y,z)$ of distinct elements among the $D_i (T)$'s. (This is possible as $n\geq 3$.) As $x^n,y^n,z^n$ are in the sub-vector space of $\mathbf{C}[T]$ generated by $b$ and $c$ (remember what the definition of the $D_i (T)$'s is !) which as dimension (over $\mathbf{C}$) $\leq 2$, we have a non trivial relation $$\alpha x^n + \beta y^n = \gamma z^n$$ where $(\alpha,\beta,\gamma)\not=(0,0,0)$. But (as $\mathbf{C}$ is algebraically closed) as each complex number is a $n$-th power of some other complex number, we can write $\alpha = \alpha_O^n$, $\beta = \beta_0^n$ and $\gamma = \gamma_0^n$ and then write $x_0^n + y_0^n = z_0^n$ where $x_0 = \alpha_0 x$, $y_0 = \beta_0 y$ and $z_0 = \gamma_0 z$.
Now, $x_0, y_0, z_0)$ are relatively prime non constant polynomial satisfying the equation, and such that $s(x_0,y_0,z_0)) < s((a,b,c))$, which is the contradiction with the minimality of $s((a,b,c))$ we were looking for. ;-)
Remark. All of this is based on the fact that $\mathbf{C}[T]$ is an UFD. You could mimic the same proof over $\mathbf{Z}$ instead of $\mathbf{C}[T]$ and conclude falsely that you get a proof of Fermat's great theorem, as it is believed Fermat did himself. Why falsely ? Because there would be and error in the "over $\mathbf{Z}$" analogue of the point (2). First over $\mathbf{Z}$, this argument would have to take place inside the sub-ring $\mathbf{Z}[\omega]:=\{a+b\omega\;|\;a,b\in\mathbf{Z}\}$ of $\mathbf{C}$. Second, for the argument to be true, you would need $\mathbf{Z}[\omega]$ to be a UFD, which is false. It is believed that Fermat falsely thought the $\mathbf{Z}[\omega]$'s to be a UFD for $\omega$ primitive $n$-th root of $1$. I think you can find historical details on this in the historical notes from Algèbre, Bourbaki.
|
2021-04-14 22:48:57
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9397682547569275, "perplexity": 106.96335023477009}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038078900.34/warc/CC-MAIN-20210414215842-20210415005842-00309.warc.gz"}
|
https://www.projecteuclid.org/euclid.aaa/1393511912
|
## Abstract and Applied Analysis
### On the Domain of the Triangle $A\left(\lambda \right)$ on the Spaces of Null, Convergent, and Bounded Sequences
#### Abstract
We introduce the spaces of $A\left(\lambda \right)$-null, $A\left(\lambda \right)$-convergent, and $A\left(\lambda \right)$-bounded sequences. We examine some topological properties of the spaces and give some inclusion relations concerning these sequence spaces. Furthermore, we compute $\alpha$-, $\beta$-, and $\gamma$-duals of these spaces. Finally, we characterize some classes of matrix transformations from the spaces of $A\left(\lambda \right)$-bounded and $A\left(\lambda \right)$-convergent sequences to the spaces of bounded, almost convergent, almost null, and convergent sequences and present a Steinhaus type theorem.
#### Article information
Source
Abstr. Appl. Anal., Volume 2013 (2013), Article ID 476363, 9 pages.
Dates
First available in Project Euclid: 27 February 2014
https://projecteuclid.org/euclid.aaa/1393511912
Digital Object Identifier
doi:10.1155/2013/476363
Mathematical Reviews number (MathSciNet)
MR3064395
Zentralblatt MATH identifier
1303.40003
#### Citation
Braha, Naim L.; Başar, Feyzi. On the Domain of the Triangle $A\left(\lambda \right)$ on the Spaces of Null, Convergent, and Bounded Sequences. Abstr. Appl. Anal. 2013 (2013), Article ID 476363, 9 pages. doi:10.1155/2013/476363. https://projecteuclid.org/euclid.aaa/1393511912
#### References
• N. L. Braha, “A new class of sequences related to the ${l}_{p}$ spaces defined by sequences of Orlicz functions,” Journal of Inequalities and Applications, vol. 2011, Article ID 539745, 10 pages, 2011.
• M. Mursaleen and A. K. Noman, “On the spaces of $\lambda$-convergent and bounded sequences,” Thai Journal of Mathematics, vol. 8, no. 2, pp. 311–329, 2010.
• M. Stieglitz and H. Tietz, “Matrixtransformationen von folgenräumen. Eine ergebnisübersicht,” Mathematische Zeitschrift, vol. 154, no. 1, pp. 1–16, 1977.
• E. Malkowsky and E. Savaş, “Matrix transformations between sequence spaces of generalized weighted means,” Applied Mathematics and Computation, vol. 147, no. 2, pp. 333–345, 2004.
• N. L. Braha and T. Mansour, “On $\Lambda ^{2}$-strong convergence of numerical sequences and Fourier series,” Acta Mathematica Hungarica, 2013.
• A. Wilansky, Summability Through Functional Analysis, vol. 85 of North-Holland Mathematics Studies, Elsevier, New York, NY, USA, 1984.
• F. Başar, Summability Theory and Its Applications, Monographs, Bentham Science Publishers E-books, \.Istanbul, Turkey, 2012.
• G. G. Lorentz, “A contribution to the theory of divergent sequences,” Acta Mathematica, vol. 80, pp. 167–190, 1948.
• J. P. Duran, “Infinite matrices and almost-convergence,” Mathematische Zeitschrift, vol. 128, pp. 75–83, 1972.
• J. P. King, “Almost summable sequences,” Proceedings of the American Mathematical Society, vol. 17, pp. 1219–1225, 1966.
• F. Başar and B. Altay, “On the space of sequences of $p$-bounded variation and related matrix mappings,” Ukrainian Mathematical Journal, vol. 55, no. 1, pp. 136–147, 2003.
• B. Altay, F. Başar, and M. Mursaleen, “On the Euler sequence spaces which include the spaces ${\ell }_{p}$ and ${\ell }_{\infty }$\emphI,” Information Sciences, vol. 176, no. 10, pp. 1450–1462, 2006.
• P. N. Ng and P. Y. Lee, “Cesàro sequence spaces of non-absolute type,” Commentationes Mathematicae. Prace Matematyczne, vol. 20, no. 2, pp. 429–433, 1978.
• B. Altay and F. Başar, “On the paranormed Riesz sequence spaces of non-absolute type,” Southeast Asian Bulletin of Mathematics, vol. 26, no. 5, pp. 701–715, 2003.
• H. K\izmaz, “On certain sequence spaces,” Canadian Mathematical Bulletin, vol. 24, no. 2, pp. 169–176, 1981.
• B. Altay and F. Başar, “Some Euler sequence spaces of nonabsolute type,” Ukrainian Mathematical Journal , vol. 57, no. 1, pp. 1–17, 2005.
• M. Şengönül and F. Başar, “Some new Cesàro sequence spaces of non-absolute type which include the spaces ${c}_{0}$ and $c$,” Soochow Journal of Mathematics, vol. 31, no. 1, pp. 107–119, 2005.
• B. Altay and F. Başar, “Some paranormed Riesz sequence spaces of non-absolute type,” Southeast Asian Bulletin of Mathematics, vol. 30, no. 4, pp. 591–608, 2006.
• F. Başar and M. Kirişçi, “Almost convergence and generalized difference matrix,” Computers & Mathematics with Applications, vol. 61, no. 3, pp. 602–611, 2011.
• K. Kayaduman and K. Şengönül, “The spaces of Cesàro almost convergent sequences and core theorems,” Acta Mathematica Scientia B, vol. 32, no. 6, pp. 2265–2278, 2012.
• A. Sönmez, “Almost convergence and triple band matrix,” Mathematical and Computer Modelling, vol. 57, no. 9-10, pp. 2393–2402, 2013.
• M. Mursaleen and A. K. Noman, “On some new sequence spaces of non-absolute type related to the spaces ${\ell }_{p}$ and ${\ell }_{\infty }$\emphI,” Filomat, vol. 25, no. 2, pp. 33–51, 2011.
• M. Mursaleen and A. K. Noman, “On some new sequence spaces of non-absolute type related to the spaces ${\ell }_{p}$ and ${\ell }_{\infty }$II,” Mathematical Communications, vol. 16, no. 2, pp. 383–398, 2011.
• E. Malkowsky and V. Rakočević, “Measure of noncompactness of linear operators between spaces of sequences that are $(\overline{N},q)$ summable or bounded,” Czechoslovak Mathematical Journal, vol. 51, no. 3, pp. 505–522, 2001.
• A. Sönmez and F. Başar, “Generalized difference spaces of non-absolute type of convergent and null sequences,” Abstract and Applied Analysis, vol. 2012, Article ID 435076, 20 pages, 2012.
• M. Mursaleen and A. K. Noman, “On some new difference sequence spaces of non-absolute type,” Mathematical and Computer Modelling, vol. 52, no. 3-4, pp. 603–617, 2010.
• B. Altay and F. Başar, “The fine spectrum and the matrix domain of the difference operator $\Delta$ on the sequence space ${\ell }_{p}$, $0<p<1$,” Communications in Mathematical Analysis, vol. 2, no. 2, pp. 1–11, 2007.
|
2019-11-17 21:24:00
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 9, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4204241633415222, "perplexity": 2746.679482661774}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496669276.41/warc/CC-MAIN-20191117192728-20191117220728-00230.warc.gz"}
|
http://mathhelpforum.com/calculus/144142-stokes-theorem.html
|
# Math Help - Stokes Theorem
1. ## Stokes Theorem
Verify Stokes Theorem ∬(∇xF).N dA where surface S is the paraboloid z = 0.5(x^2 + y^2) bound by the plane z=2, C is its boundary, and the vector field F = 3yi - xzj + yzk
I had found (∇xF) = (z+x)i + (-z-3)k
r = [u, v, 0.5(u^2 + v^2)]
Therefore N= ru X rv = -ui -uj +k
Therefore (∇xF).N = [(z+x), 0, (-z-3)].[-x, -y, 1]
After that I substitute x = rcos(θ), y = rsin(θ), z = 0.5r^2
Thus ∫(0-2)∫(0-2pi) (∇xF).N r dθdr
But I cant seems to get the answer. Can anyone help? Help would greatly appreciated
2. [QUOTE=HeheZz;510745]Verify Stokes Theorem ∬(∇xF).N dA where surface S is the paraboloid z = 0.5(x^2 + y^2) bound by the plane z=2, C is its boundary, and the vector field F = 3yi - xzj + yzk
I had found (∇xF) = (z+x)i + (-z-3)k
r = [u, v, 0.5(u^2 + v^2)]
Therefore N= ru X rv = -ui -uj +k[/tex]
First this is wrong because this N is not a unit vector.
Therefore (∇xF).N = [(z+x), 0, (-z-3)].[-x, -y, 1]
After that I substitute x = rcos(θ), y = rsin(θ), z = 0.5r^2
Thus ∫(0-2)∫(0-2pi) (∇xF).N r dθdr
But I cant seems to get the answer. Can anyone help? Help would greatly appreciated
It seems to me simpler to use $u= r \cos(\theta)$, $v= r \sin(\theta)$ from the start: $\rho(r, \theta)= r \cos(\theta)\vec{i}+ r\sin(\theta)\vec{j}+ (1/2)r^2\vec{j}$ so that $\rho_r= \cos(\theta)\vec{i}+ \sin(\theta)\vec{j}+ r\vec{k}$ and $\rho_\theta= -r \sin(\theta)\vec{i}+ r \cos(\theta)\vec{j}$.
Now, $\rho_r\times\rho_\theta= -r^2 \cos(\theta)\vec{i}- r^2 \sin(\theta)\vec{j}+ r\vec{k}$.
There is no need to calculate "N" and "dS" separately since both would involve calculating the length of that vector which would then cancel. Instead $\vec{N}dS= d\vec{S}= (-r^2 \cos(\theta)\vec{i}- r^2 \sin(\theta)\vec{j}+ r\vec{k})dr d\theta$.
$\nabla F= (z+ x)\vec{i}+ (-z- 3)\vec{k}= ((1/2)r+ r \cos(\theta))\vec{i}- ((1/2)r- 3)\vec{j}$ so $\nabla F\cdot d\vec{S}= (-r^2((1/2)\cos(\theta)+ \cos^2(\theta)- ((1/2)r- 3)r^2 \sin(\theta))dr d\theta$.
The parts involving only $\sin(\theta)$ and $\cos(\theta)$ are easy- the integral with respect to $\theta$ is 0.
That leaves only $-\int_{r= 0}^2 \int_{\theta= 0}^{2\pi} r^2 \cos^2(\theta) d\theta dr$.
By the way, notice the negative sign in front of that. Doing a vector integration over a surface involves a choice of orientation. I chose to orient with "z component upward". Choosing z component downard will reverse the sign. Of course, in using Stoke's theorem, you also have to choose an orientation for the path integral around the boundary- which direction you will integrate around the boundary. An orientation for one implies a corresponding orientation for the other.
|
2015-08-28 10:14:30
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9401752948760986, "perplexity": 2559.9193095788096}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644062760.2/warc/CC-MAIN-20150827025422-00249-ip-10-171-96-226.ec2.internal.warc.gz"}
|
https://samacheerkalviguru.com/samacheer-kalvi-12th-chemistry-solutions-chapter-4/
|
Are you searching for the Samacheer Kalvi 12th Chemistry Chapter Wise Solutions PDF? Then, get your Samacheer Kalvi 12th Chapter Wise Solutions PDF for free on our website. Students can Download Chemistry Chapter 4 Transition and Inner Transition Elements Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Chemistry Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 4 Transition and Inner Transition Elements
All concepts are explained in an easy way on our website. So, students can easily learn and practice Tamilnadu State Board 12th Chemistry Chapter 4 Transition and Inner Transition Elements Question and Answers. You can enjoy the digitized learning with the help of the Tamilnadu State Board Chemistry Online Material.
### Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements Text Book Evalution
12th Chemistry Chapter 4 Book Back Answers Questions 1.
Sc ( Z=21) is a transition element but Zinc (z=30) is not because ……………..
(a) both Sc3+ and Zn2+ ions are colourless and form white compounds.
(b) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled
(c) last electron as assumed to be added to 4s level in case of zinc
(d) both Sc and Zn do not exhibit variable oxidation states
(c) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled
12th Chemistry Unit 4 Book Back Answers Question 2.
Which of the following d block element has half filled penultimate d sub shell as well as half filled valence sub shell?
(a) Cr
(b) Pd
(c) Pt
(d) none of these
(a) Cr
Hint: Cr ⇒ [Ar]3d5 4s1
12th Chemistry Lesson 4 Book Back Answers Question 3.
Among the transition metals of 3d series, the one that has highest negative (M2+/ M) standard electrode potential is ……………..
(a) Ti
(b) Cu
(c) Mn
(d) Zn
(a) Ti
12th Chemistry Chapter 4 Question 4.
Which one of the following ions has the same number of unpaired electrons as present in V3+?
(a) Ti3+
(b) Fe3+
(c) Ni2+
(d) Cr3+
(c) Ni2+
Transition And Inner Transition Elements Class 12 Question 5.
The magnetic moment of Mn2+ ion is ……………..
(a) 5.92BM
(b) 2.80BM
(c) 8.95BM
(d) 3.90BM
(a) 5.92BM
Hint: Mn2+ ⇒ 3d contains 5 unpaired electrons
n = 5
= $$\sqrt { n(n+2) } BM$$
= $$\sqrt { 5(5+2) }$$ = $$\sqrt { 35 }$$ = 5.92 BM
Samacheer Kalvi 12th Chemistry Question 6.
Which of the following compounds is colourless?
(a) Fe3+
(6) Ti4+
(c) CO2+
(d) Ni2
(b) Ti4+
Hint: Ti4+ contains no unpaired electrons in d orbital, hence no d-d transition.
12th Chemistry Samacheer Kalvi Question 7.
The catalytic behaviour of transition metals and their compounds is ascribed mainly due to ……………..
(a) their magnetic behaviour
(b) their unfilled d orbitals
(c) their ability to adopt variable oxidation states
(d) their chemical reactivity
(c) their ability to adopt variable oxidation states
Samacheer Kalvi 12th Chemistry Solutions Question 8.
The correct order of increasing oxidizing power in the series ……………..
(a) $${ VO }_{ 2 }^{ + }$$ < $${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }$$ < $$Mn{ O }_{ 4 }^{ – }$$
(b) $${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }$$ < $${ VO }_{ 2 }^{ + }$$ < $$Mn{ O }_{ 4 }^{ – }$$
(c) $${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }$$ < $$Mn{ O }_{ 4 }^{ – }$$ < $${ VO }_{ 2 }^{ + }$$
(d) $$Mn{ O }_{ 4 }^{ – }$$ < $${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }$$ < $${ VO }_{ 2 }^{ + }$$
(a) $${ VO }_{ 2 }^{ + }$$ < $${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }$$ < $$Mn{ O }_{ 4 }^{ – }$$
Hint: greater the oxidation state, higher is the oxidising power.
12th Chemistry Solutions Samacheer Kalvi Question 9.
The alloy of copper that contain Zinc is ……………..
(a) Monel metal
(b) Bronze
(c) bell metal
(d) brass
(d) brass
Class 12 Chemistry Samacheer Kalvi Question 10.
Which of the following does not give oxygen on heating?
(a) K2Cr2O7
(b) (NH4)2Cr2O7
(c) KClO3
(d) Zn(ClO3)2
(b) (NH4)2Cr2O7
Hint:
12 Chemistry Samacheer Kalvi Question 11.
In acid medium, potassium permanganate oxidizes oxalic acid to ……………..
(a) Oxalate
(b) Carbon dioxide
(c) acetate
(d) acetic acid
(b) Carbon dioxide
Hint: $${ 5(COO) }_{ 2 }^{ 2- }$$ + $${ 2MnO }_{ 4 }^{ – }$$ + 16H+ $$\underrightarrow { \triangle }$$ 2Mn2+ + 10CO2+$$\uparrow$$ + 8H2O
12th Chemistry 4th Lesson Question 12.
Which of the following statements is not true?
(a) on passing H2S, through acidified K2Cr2O7 solution, a milky colour is observed.
(b) Na2Cr2O7 is preferred over K2Cr2O7 in volumetric analysis
(c) K2Cr2O7 solution in acidic medium is orange in colour
(d) K2Cr2O7 solution becomes yellow on increasing the pH beyond 7
(b) Na2Cr2O7 is preferred over K2Cr2O7 in volumetric analysis
Transition And Inner Transition Elements Class 12 Exercise Question 13.
Permanganate ion changes to in acidic medium ……………..
(a) $${ MnO }_{ 4 }^{ 2- }$$
(b) Mn2+
(c) Mn3+
(d) MnO2
(b) Mn2+
Hint: $${ MnO }_{ 4 }^{ – }$$ + 8H+ + 5e → Mn2+ + 4H2O
Samacheer Kalvi Chemistry 12th Question 14.
A white crystalline salt (A) react with dilute HCl to liberate a suffocating gas (B) and also forms a yellow precipitate . The gas (B) turns potassium dichromate acidified with dil H2SO4 to a green coloured solution(C). A,B and C are respectively ……………..
(a) Na2SO3, SO2, Cr2(SO4)3
(b) Na2S2O3, SO2, Cr2(SO4)3
(c) Na2S, SO2, Cr2(SO4)3
(d) Na2SO4, SO2, Cr2(SO4)3
(a) Na2SO3, SO2, Cr2(SO4)3
Class 12 Chemistry Chapter 4 Notes Question 15.
$${ MnO }_{ 4 }^{ – }$$ react with Br in alkaline PH to give ……………..
(a) $${ BrO }_{ 3 }^{ – }$$, MnO2
(b) Br3, $${ MnO }_{ 4 }^{ 2- }$$
(c) Br3, MnO3
(d) BrO, $${ MnO }_{ 4 }^{ 2- }$$
(a) $${ BrO }_{ 3 }^{ – }$$, MnO2
Hint: $${ 2MnO }_{ 4 }^{ – }$$ + Br- + H2O → 2OH- + 2MnO2 + $${ BrO }_{ 3 }^{ – }$$
Class 12 Chemistry Chapter 4 Question 16.
How many moles of I2 are liberated when 1 mole of potassium dichromate react with potassium iodide?
(a) 1
(b) 2
(c) 3
(d) 4
(c) 3
Hint: K2Cr2O7 + 6KI + 7H2SO4 → 4K2SO4 + Cr2 (SO4)3 + 7H2O + 3I2
Chapter 4 Chemistry Class 12 Notes Question 17.
The number of moles of acidified KMnO4 required to oxidize 1 mole of ferrous oxalate(FeC2O4) is …………..
(a) 5
(b) 3
(c) 0.6
(d) 1.5
(c) 0.6
Question 18.
When a brown compound of Mn (A) ids treated with HCl, it gives a gas (B). The gas (B) taken in excess reacts with NH3 to give an explosive compound (C). The compound A, B and C are ……………
(a) MnO2, Cl2, NCl3
(b) MnO, Cl2, NH4Cl
(c) Mn3O4, Cl2, NCl3
(d) MnO3, Cl2, NCl2
(a) MnO2, Cl2, NCl3
Question 19.
Which one of the following statements related to lanthanons is incorrect?
(a) Europium shows +2 oxidation state.
(b) The basicity decreases as the ionic radius decreases from Pr to Lu.
(c) All the lanthanons are much more reactive than aluminium.
(d) Ce4+ solutions are widely used as oxidising agents in volumetric analysis.
(c) All the lanthanons are much more reactive than aluminium.
Hint: As we move from La to Lu , their metallic behaviour because almost similar to that of aluminium.
Question 20.
Which of the following lanthanoid ions is diamagnetic?
(a) Eu2+
(b) Yb2+
(c) Ce2+
(d) Sm2+
(6) Yb2+
Hint: Yb2+ – 4f14 – no unpaired electrons – diamagnetic
Question 21.
Which of the following oxidation states is most common among the lanthanoids?
(a) 4
(b) 2
(c) 5
(d) 3
(d) 3
Question 22.
Assertion: Ce4+ is used as an oxidizing agent in volumetric analysis.
Reason: Ce4+ has the tendency of attaining +3 oxidation state.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false. .
(d) Both assertion and reason are false.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
Question 23.
The most common oxidation state of actinoids is ……………
(a) +2
(b) +3
(c) +4
(d) +6
(c) +4
Question 24.
The actinoid elements which show the highest oxidation state of +7 are ……………
(a) Np, Pu, Am
(b) U, Fm, Th
(c) U, Th, Md
(d) Es, No, Lr
(a) Np, Pu, Am
Question 25.
Which one of the following is not correct?
(a) La(OH)2 is less basic than Lu(OH)3
(b) In lanthanoid series ionic radius of Ln3+ ions decreases
(c) La is actually an element of transition metal series rather than lanthanide series
(d) Atomic radii of Zr and Hf are same because of lanthanide contraction
(a) La(OH)2 is less basic than Lu(OH)3
Question 1.
What are transition metals? Give four examples.
1. Transition metal is an element whose atom has an incomplete d sub shell or which can give rise to cations with an incomplete d sub shell.
2. They occupy the central position of the periodic table, between s-block and p-block elements.
3. Their properties are transitional between highly reactive metals of s-block and elements of p-block which are mostly non metals.
4. Example – Iron, Copper, Tungsten, Titanium.
Question 2.
Explain the oxidation states of 3d series elements.
1. The first transition metal Scandium exhibits only +3 oxidation state, but all other transition elements exhibit variable oxidation states by losing electrons from (n-l)d orbital and ns orbital as the energy difference between them is very small.
2. At the beginning of the 3d series, +3 oxidation state is stable but towards the end +2 oxidation state becomes stable.
3. The number of oxidation states increases with the number of electrons available, and it decreases as the number of paired electrons increases. For example, in 3d series, first element Sc has only one oxidation state +3; the middle element Mn has six different oxidation states from +2 to +7. The last element Cu shows +1 and +2 oxidation states only.
4. Mn2+ (3d5) is more stable than Mn4+ (3d3) is due to half filled stable configuration.
Question 3.
What are inner transition elements?
1. The inner transition elements have two series of elements.
• Lanthanoids
• Actinoids
2. Lanthanoid series consists of 14 elements from Cerium (58Ce) to Lutetium (71Lu) following Lanthanum (57La). These elements are characterised by the preferential filling of 4f orbitals. .
3. Actinoids consists of 14 elements from Thorium (90Th) to Lawrencium (103Lr) following Actinium (89Ac). These elements are characterised by the preferential filling of 5f orbital.
Question 4.
Justify the position of lanthanides and actinides in the periodic table.
1. In sixth period after lanthanum, the electrons are preferentially filled in inner 4f sub shell and these 14 elements following lanthanum show similar chemical properties. Therefore these elements are grouped together and placed at the bottom of the periodic table. This position can be justified as follows.
• Lanthanoids have general electronic configuration [Xe] 4f2-14 5d0-1 6s2
• The common oxidation state of lanthanoids is +3
• All these elements have similar physical and chemical properties.
2. Similarly the fourteen elements following actinium resemble in their physical and chemical properties.
3. If we place these elements after Lanthanum in the periodic table below 4d series and actinides below 5d series, the properties of the elements belongs to a group would be different and it would affect the proper structure of the periodic table.
4. Hence a separate position is provided to the inner transition elements at the bottom of the periodic table.
Question 5.
What are actinoides? Give three examples.
1. The fourteen elements following actinium ,i.e., from thorium (Th) to lawrentium (Lr) are called actinoids.
2. All the actinoids are radioactive and most of them have short half lives.
3. The heavier members being extremely unstable and not of natural occurrence. They are produced synthetically by the artificial transformation of naturally occuring elements by nuclear reactions.
4. Example – Thorium, Uranium, Plutonium, Californium.
Question 6.
Why Gd3+ is colourless?
Gd – Electronic Configuration: [Xe] 4f7 5d1 6s2
Gd3+ – Electronic Configuration: [Xe] 4f7
In Gd3+, no electrons are there in outer d-orbitals. d-d transition is not possible. So it is colourless.
Question 7.
Explain why compounds of Cu2+ are coloured but those of Zn2+ are colourless.
Cu (Z = 29) Electronic configuration is [Ar] 3d10 4s1
Cu2+: Electronic configuration is [Ar] 3d9.
In Cu2+, promotion of electrons take place in outer d-orbital by the absorption of light form visible region involves d-d transition. Due to this Cu2+ compounds are coloured. Where in Zn2+ electronic configuration is [Ar]3d10. It has completely filled d-orbital. So there is no chance of d – d transition. So Zn2+ compounds are colourless.
Question 8.
Describe the preparation of potassium dichromate.
Preparation of potassium dichromate:
1. Potassium dichromate is prepared from chromite ore. The ore FeO. Cr2O3 is concentrated by gravity separation process.
2. The concentrated ore is mixed with excess sodium carbonate and lime and roasted in a reverbratory furnace.
4FeCr2O4 + 8Na2CO3 + 7O2 $$\underrightarrow { 900-{ 1000 }^{ 0 }C }$$ 8 Na2CrO4 + 2Fe2O3 + 8CO2$$\uparrow$$
3. The roasted mass is treated with water to separate soluble sodium chromate from insoluble iron oxide. The yellow solution of sodium chromate is treated with concentrated sulphuric acid which converts sodium chromate into sodium dichromate.
4. The above solution is concentrated to remove less soluble sodium sulphate. The resulting solution is filtered and concentrated. It is cooled to get the crystals of Na2SO2.2H2O.
5. The saturated solution of sodium dichromate in water is mixed with KCl and then concentrated to get crystals of NaCl. It is filtered while hot and the filtrate is cooled to obtain K2Cr2O7 crystals.
Question 9.
What is lanthanide contraction and what are the effects of lanthanide contraction?
As we move across 4f series, the atomic and ionic radii of lanthanoids show gradual decrease with increase in atomic number. This decrease in ionic size is called lanthanoid contraction.
Effects (or) Consequences of lanthanoid contraction:
1. Basicity differences: As we move from Ce3+ to Lu3+ , the basic character of Ln3+ ions decrease. Due to the decrease in the size of Ln3+ ions, the ionic character of Ln – OH bond decreases (covalent character increases) which results in the decrease in the basicity.
2. Similarities among lanthanoids – In the complete f-series only 10 pm decrease in atomic radii and 20 pm decrease in ionic radii is observed. Because of this very small change in radii of lanthanoids, their chemical properties are quite similar.
The elements of second and third transition series resemble each other more closely than the elements of first and second transition series due to lanthanoid contraction. For example,
• 4d series – Zr – Atomic radius 145 pm
• 5d series – Hf – Atomic radius 144 pm
Question 10.
Complete the following
Question 11.
What are interstitial compounds?
1. An interstitial compound or alloy is a compound that is formed when small atoms like hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice.
2. They are usually non-stoichiometric compounds.
3. Transition metals form a number of interstitial compounds such as TiC, ZrH1.92, Mn4N etc.
4. The elements that occupy the metal lattice provide them new properties.
• They are hard and show electrical and thermal conductivity
• They have high melting points higher than those of pure metals
• Transition metal hydrides are used as powerful reducing agents
• Metallic carbides are chemically inert.
Question 12.
Calculate the number of unpaired electrons in Ti3+, Mn2+ and calculate the spin only magnetic moment.
Ti3+:
Ti (Z = 22). Electronic configuration [Ar] 3d2 4s2
Ti3+ – Electronic configuration [Ar] 3d1
So, the number of unpaired electrons in Ti3+ is equal to 1.
Spin only magnetic moment of Ti3+ = $$\sqrt { 1(1+2) }$$ = $$\sqrt { 3 }$$ = 1.73µB
Mn2+:
Mn (Z = 25). Electronic configuration [Ar] 3d5 4s2
Mn2+ – Electronic configuration [Ar] 3d5
Mn2+ has 5 unpaired electrons.
Spin only magnetic moment of Mn2+ = $$\sqrt { 5(5+2) }$$ = $$\sqrt { 35 }$$ = 5.91µB
Question 13.
Write the electronic configuration of Ce4+ and CO2+.
Ce (Z = 58) → Ce4+4e
Ce4+ – Is2 – 2s22p63s23p64s23d104p6 5s24d10 5p6
CO2+ – Is22s22p63s23p64s23d5.
Question 14.
Explain briefly how +2 states becomes more and more stable in the first half of the first row transition elements with increasing atomic number.
First transition series.
Question 15.
Which is more stable? Fe3+ or Fe2+ – explain.
Fe (Z = 26)
Fe → Fe2+ + 2e
Fe → Fe3+ + 3e
Fe2+ [Number of electrons 24]
Electronic configuration = [Ar]3d6
Fe3+ [Number of electrons 23]
Electronic configuration = [Ar]3d5
Among Fe3+ and Fe2+, Fe3+ is more stable due to half filled d-orbital. This can be explained by Aufbau principle. Half filled and completely filled d-orbitals are more stable than partially filled d-orbitals. So Fe3+ is more stable than Fe2+.
Question 16.
Explain the variation in E0M2+/M3+3d series.
1. In transition series, as we move down from Ti to Zn, the standard reduction potential E0M2+/M3 value is approaching towards less negative value and copper has a positive reduction potential, i.e. elemental copper is more stable than Cu2+.
2. E0M2+/M value for manganese and zinc are more negative than regular trend. It is due to extra stability arises due to the half filled d5 configuration in Mn2+ and completely filled d10 configuration in Zn2+.
3. The standard electrode potential for the M3+ / M2+ half cell gives the relative stability between M3+ and M2+.
4. The high reduction potential of Mn3+ / Mn2+ indicates Mn2+ is more stable than Mn3+.
5. For Fe3+ / Fe2+ the reduction potential is 0.77 V, and this low value indicates that both Fe3+ and Fe2+ can exist under normal condition.
6. Mn3+ has a 3d2 configuration while that of Mn2+ is 3d5. The extra stability associated with a half filled d sub-shell makes the reduction of Mn3+ very feasible [E° = +1.51 V]
Question 17.
Compare lanthanides and actinides.
Lanthanoids:
1. Differentiating electron enters in 4f orbital
2. Binding energy of 4f orbitals are higher
3. They show less tendency to form complexes
4. Most of the lanthanoids are colourless
5. They do not form oxo cations
6. Besides +3 oxidation states lanthanoids show +2 and +4 oxidation states in few cases.
Actinoids:
1. Differentiating electron enters in 5f orbital
2. Binding energy of 5f orbitals are lower
3. They show greater tendency to form complexes
4. ost of the actinoids are coloured.
E.g : U3+ (red), U4+ (green), $${ UO }_{ 2 }^{ 2+ }$$(yellow)
5. They do form oxo cations such as $${ UO }_{ 2 }^{ 2+ }$$ $${ NpO }_{ 2 }^{ 2+ }$$ etc.
6. Besides +3 oxidation states actinoids show higher oxidation states such as +4, +5, +6 and +7.
Question 18.
Explain why Cr2+ is strongly reducing while Mn3+ is strongly oxidizing.
Cr2+ is strong reducing while Mn3+ is strongly oxidising.
E0Cr3+/Cr2+ is -0.41 V
Cr2+ + 2e → Cr E° = – 0.91V.
If the standard electrode potential E° of a metal is large and negative, the metal is a powerful reducing agent because it loses electrons easily.
Mn → Mn3+ + 3e
Mn3+ + e → Mn2+
Mn3+ [Ar]3d4
E° = + 1.51 V
If the standard electrode potential E° of a metal is large and positive, the metal is a powerful oxidising agent because it gains electrons easily.
Question 19.
Compare the ionization enthalpies of first series of the transition elements. Ionization enthalpies of first transition series:
1. Ionization energy of transition element is intermediate between those of s and p block elements.
2. As we move from left to right in a transition metal series, the ionization enthalpy increases as expected. This is due to increase in nuclear charge corresponding to the filling of d electrons.
3. The increase in first ionisation enthalpy with increase in atomic number along a particular series is not regular. The added electron enters (n-l)d orbital and the inner electrons act as a shield and decrease the effect of nuclear charge on valence ‘ns’ electrons. Therefore, it leads to variation in the ionization energy values.
Question 20.
Actinoid contraction is greater from element to element than the lanthanoid contraction, why?
1. Actinoid contraction is greater from element to element than lanthanoid contraction. The 5f orbitals in Actinoids have a very poorer shielding effect than 4f orbitals in lanthanoids.
2. Thus, the effective nuclear charge experienced by electron in valence shells in case of actinoids is much more than that experienced by lanthanoids.
3. In actinoids, electrons are shielded by 5d, 4f, 4d and 3d whereas in lanthanoids, electrons are shielded by 4d, 4f only.
4. Hence, the size contraction in actinoids is greater as compared to that in lanthanoids.
Question 21.
Out of LU(OH)3 and La(OH)3 which is more basic and why?
1. As we move from Ce3+ to Lu3+, the basic character of Lu3+ ions decreases.
2. Due to the decrease in the size of Lu3+ ions, the ionic character of Lu – OH bond decreases, covalent character increases which results in the decrease in the basicity.
3. Hence, La(OH)3 is more basic than Lu(OH)3.
Question 22.
Why europium (II) is more stable than Cerium (II)?
Eu (Z = 63) – Electronic configuration – [Xe] 4f7 5d° 6s2
Eu2+ – Electronic configuration Electronic 6s1
Ce (Z = 58) – configuration – [Xe] 4f2+ 5d° 6s2+
Ce2+ – Electronic confluration – [Xe] 4f2 5d°
According to Aufbau principle, half filled and completely filled d (or) f orbitals are more stable than partially filled f orbitals.
Hence Eu2+ [Xe] 4f7 5d° is more stable than Ce2+ [Xe] 4f2 5d°
Question 23.
Why do zirconium and Hafnium exhibit similar properties?
1. The element of second and third transition series resemble each other more closely than the elements of first and second transition series due to lanthanoid contraction.
2. e.g., Zr – 4d series -Atomic radius 145 pm Hf – 5d series – Atomic radius 144 pm
3. The radii are very similar even though the number of electrons increases.
4. Zr and Hf have very similar chemical behaviour, having closely similar radii and electronic configuration.
5. Radius dependent properties such as lattice energy, solvation energy are similar.
6. Thus lanthanides contraction leads to formation of pair of elements and those known as chemical twins, e.g., Zr – Hf
Question 24.
Which is stronger reducing agent Cr2+ or Fe2+ ?
Cr2+ and Fe2+
Cr (Z = 24) – Electronic configuration – [Ar] 3d5 4s1
Cr2+ Electronic configuration – [Ar] 3d4 4s0
Fe (Z = 26) – Electronic configuration – [Ar] 3d6 4s2
Fe2+ Electronic confimiration – [Ar] 3d6 4s0
If standard electrode potential (E°) of a metal is large and negative, the metal is a powerful reducing agent
Cr2+ + 2e → Cr
Fe2++2e → Fe
E° = – 0.91V
E°= – 0.44V
By comparing the above equation, Cr2+ is a powerful reducing agent.
Question 25.
The E0M2+/M value for copper is positive. Suggest a possible reason for this.
1. Copper has a positive reduction potential. Elemental copper is more stable than Cu2+.
2. Copper having positive sign for electrode potential merely means that copper can undergo reduction at faster rate than reduction of hydrogen.
3. The electron giving reaction (oxidation) of copper is slower than that of hydrogen. It is determined from the result of S.H.E (Standard Hydrogen Electrode) potential value experiment.
Question 26.
Predict which of the following will be coloured in aqueous solution Ti2+ , V3+, Sc4+, Cu+, SC3+, Fe3+, Ni2+ and CO3+
Among Ti2+ , V3+, Sc4+, Cu+, Sc3+, Fe3+, Ni2+ and CO3+ in aqueous solution state.
Ti (Z = 22) – Ti2+ – Electronic configuration is [Ar] 3d2
V (Z = 23) – V3+ – Electronic configuration is [Ar] 3d2
Sc (Z = 21) – SC4+ – Electronic configuration is [Ar] 1s2 2s2 2p6 3s2 3p5
Cu (Z = 29) – Cu+ – Electronic configuration is [Ar] 3d10
Sc (Z = 21) – SC3+ – Electronic configuration is [Ar] 3d°4s°
Fe (Z = 26) – Fe3+ – Electronic configuration is [Ar] 3d5
Ni (Z = 28) – Ni2+ – Electronic configuration is [Ar] 3d5
CO (Z = 27) – CO3+ – Electronic configuration is [Ar] 3d6
A transition metal ion is coloured if it has one or more unpaired electrons in (n-l)d orbital, i.e. 3d orbitals in the case of first transition series, when such species are exposed to visible radiation, d – d transition take place and the species are coloured.
Question 27.
Describe the variable oxidation state of 3d series elements.
1. The first transition metal Scandium exhibits only +3 oxidation state, but all other transition elements exhibit variable oxidation states by loosing electrons from (n-l)d orbital and ns orbital as the energy difference between them is very small.
2. At the beginning of the 3d series, +3 oxidation state is stable but towards the end +2 oxidation state becomes stable.
3. The number of oxidation states increases with the number of electrons available, and it decreases as the number of paired electrons increases. For example, in the 3d series, first element Sc has only one oxidation state +3 the middle element Mn has six different oxidation states from +2 to +7. The last element Cu shows +1 and +2 oxidation states only.
4. Mn2+ (3d5) is more stable than Mn4+ (3d3) is due to half filled stable configuration.
Question 28.
Which metal in the 3d series exhibits +1 oxidation state most frequently and why?
1. The first transition metal copper exhibits only +1 oxidation state.
2. It is unique in 3d series having a stable +1 oxidation state.
Cu (Z = 29) Electronic configuration is [Ar] 3d10 4s2
3. So copper element only can have +1 oxidation state.
Question 29.
Why first ionization enthalpy of chromium is lower than that of zinc?
The first ionization enthalpy of chromium is lower than that of zinc. Cr (Z = 24) Electronic configuration [Ar] 3d5 4s1. In the case of Cr, first electron has to be removed easily from 4s orbital to attain the more stable half filled configuration. So Cr has lower ionization enthalpy. But in the case of Zinc (Z = 30), electronic configuration [Ar] 3d10 4s2. The first electron has to be removed from the most stable fully filled electronic configuration becomes difficult and it requires more energy.
Question 30.
Transition metals show high melting points why?
1. All the transition metals are hard.
2. Most of them are hexagonal close packed, cubic close packed (or) body centered cubic which are characteristics of true metals.
3. The maximum melting point at about the middle of transition metal series indicates that d5 configuration of favourable for strong inter atomic attraction.
4. Due to the strong metallic bonds, atoms of the transition elements are closely packed and held together. This leads to high melting point and boiling point.
Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements Evaluate yourself
Question 1.
Compare the stability of Ni4+ and Pt4+ from their ionisation enthalpy values
Pt4+ compounds are stable than Ni4+ compounds because the energy needed to remove 4 electrons in Pt is less than that of Ni.
Question 2.
Why iron is more stable in +3 oxidation state than in +2 and the reverse is true for Manganese?
Fe (Z = 26). Electronic configuration [Ar] 3d4s2
Fe → Fe3+ + 3e
Fe3+ Electronic configuration p° [Ar]3d5.
If’d’ orbital is half filled, it is more stable than . Fe2+ where it is [Ar]3d6.
Mn (Z = 25). Electronic configuration [Ar]3d5 4s2
Mn → Mn2+ + 2e . By the loss of 2e, Mn2+ is more stable due to half filled configuration. Mn → Mn3+ + 3e.
Mn3+ Electronic configuration [Ar]3d4 4s°.
Among this Fe3+ is more stable than Fe2+ and the Mn2+ is more stable than Mn3+.
### Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements Additional Questions
Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements 1 Mark Questions and Answers
Question 1.
The elements whose atom has incomplete d sub-shell are called …………..
(a) s-block element
(b) Alkali metals
(c) transition elements
(d) Representative elements
(c) transition elements
2. Which one of the following is the other name of d-block elements?
(a) Chalcogens
(b) Halogens
(c) Inner-transition elements
(d) Transition elements
(d) Transition elements
Question 3.
Which metals play an important role in the development of human civilization?
(a) Al and Mg
(b) Na and K
(c) Fe and Cu
(d) Mn and Ni
(c) Fe and Cu
Question 4.
Which transition element is used in light bulb filaments?
(a) Al
(b) Ni
(c) W
(d) Fe
(c) W
Question 5.
Which metal is used in manufacturing artificial joints?
(a) Molybdenum
(6) Titanium
(c) Tungsten
(d) Iron
(b) Titanium
Question 6.
Which transition metal is applied in the manufacturing of boiler plants?
(a) Iron
(b) Copper
(c) Aluminium
(d) Molybdenum
(d) Molybdenum
Question 7.
Identify the transition metal present in Hemoglobin …………..
(a) Cobalt
(b) Iron
(c) Manganese
(d) Copper
(b) Iron
Question 8.
Which of the following transition metal is present in Vitamin B12?
(a) Cobalt
(b) Platinum
(c) Copper
(d) Iron
(a) Cobalt
Question 9.
The metal cobalt is present in
(a) Vitamin-A
(b) Vitamin-B
(c) Vitamin-B12
(d) Vitamin-B6
(c) Vitamin-B12
Question 10.
Consider the following statements.
(i) Transition metals occupy group-3 to group-12 of the modem periodic table.
(ii) Representative elements occupy group-3 to group-12 of the modem periodic table.
(iii) Except group-11 elements of all transition metals are hard.
(iv) d-block elements are mostly non-metals.
Which of the above statements is/are incorrect?
(a) (ii) and (iv)
(b) (i) and (iii)
(c) (iii) only
(d) (i) only
(a) (ii) and (iv)
Question 11.
Consider the following statements.
(i) d-block elements composed of 3d series, Sc to Zn (4th period).
(ii) 4d series composed of Y to Cd.
(iii) 5d series composed of La, Hf to Mercury.
(iv) d-block elements composed of 4d series Y to Cd.
Which of the above statements is/are incorrect
(a) (i) and (iv)
(b) (i), (ii) and (iii)
(c) (iii) and (iv)
(d) (iv) only
Ans.
(b) (i), (ii) and (iii)
Question 12.
Which of the following is the correct electronic configuration of Sc (Z = 21)?
(a) [Ar] 3d3
(b) [Ar] 3d’ 4s2
(c) [Ar] 3d2 4s1
(d) [Ar] 4s2 4p’
(b) [Ar] 3d’ 4s2
Question 13.
The correct electronic configuration of Cr is …………..
(a) [Ar] 3d4 4s2
(b) [Ar] 3d5
(c) [Ar] 3d5 4s1
(d) [Ar] 3d6
(c) [Ar] 3d5 4s1
Question 14.
Which of the following is the correct electronic configuration of copper?
(a) [Ar] 3d5 4s1
(b) [Ar] 3d10 4s1
(c) [Ar] 3d9 4s2
(d) [Ar] 3d8 4s2 4p1
(b) [Ar] 3d10 4s1
Question 15.
Which one of the following is the general electronic configuration of transition elements?
(a) [Noble gas] ns2 np6
(b) [Noble gas] ( n – 2 ) f1-14(n-l)d1-10 ns2
(c) [Noble gas] ( n – 1 ) d1-10 (n-l)f1-14 ns2
(d) [Noble gas] ( n – 1 ) d1-10 ns2
(d) [Noble gas] ( n – 1 ) d1-10 ns2
Question 16.
Which of the following d-block elements has the highest electrical conductivity at room temperature?
(a) Copper
(b) Silver
(c) Aluminium
(d) Tungsten
(b) Silver
Question 17.
Consider the following statements.
(i) The melting point decreases from Scandium to Vanadium in 3d series.
(ii) In 3d transition series, atomic radius decreases from Sc to V and upto copper atomic radius nearly remains the same.
(iii) As we move down in 3d transition series, atomic radius increases.
Which of the above statements is/are incorrect?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (i), (ii) and (iii)
(a) (i) only
Question 18.
Which of the following transition element exhibit only +3 oxidation state?
(a) Cu
(b) Sc
(c) Mn
(d) Cr
(b) Sc
Question 19.
Which one of the following transition element has maximum oxidation states?
(a) Manganese
(b) Copper
(c) Scandium
(d) Titanium
(a) Manganese
Question 20.
Consider the following statements.
(i) In 3d series, the middle element Mn has +2 to +7 oxidation states.
(ii) The oxidation state of Ru and Os is +8.
(iii) Scandium has six different oxidation states.
Which of the above statements is/are not correct?
(a) (i) and (ii)
(b) (ii) only
(c) (i) only
(d) (iii) only
(d) (iii) only
Question 21.
Which one of the following elements show high positive electrode potential?
(a) Ti+
(b) Mn2+
(c) CO2+
(d) Cr2+
Ans.wer:
(c) CO2+
Question 22.
Which one of the following elements show high negative electrode potential?
(a) Copper
(b) Manganese
(c) Cobalt
(d) Zinc
(d) Zinc
Question 23.
Which one of the following is diamagnetic in nature?
(a) Ti3+
(b) Cu2+
(c) Zn2+
(d) V3+
(c) Zn2+
Question 24.
Which one of the following is paramagnetic in nature?
(a) Sc3+
(b) Ti4+
(c) V5+
(d) Cu2+
(d) Cu2+
Question 25.
Which of the following pair has maximum number of unpaired electrons?
(a) Mn2+, Fe3+
(b) CO3+, Fe2+
(c) Cr3+, Mn4+
(d) Ti2+, V3+
(a) Mn2+, Fe3+
Question 26.
Which of the following pair has d10 electrons?
(a) Ti3+, V4+
(b) CO3+, Fe2+
(c) Cu+ , Zn2+
(d) Mn2+, Fe3+
(c) Cu+, Zn2+
Question 27.
Which of the following is used as a catalyst in the manufacture of sulphuric acid form SO3.
(a) V3O5
(b) Rh-Ir
(c) Ni
(d) Fe
(a) V5O5
Question 28.
Which one of the following is Zeigler – Natta catalyst?
(a) CO2(CO)8
(b) Rh/Ir complex
(c) TiCl4 + Al(C2H5)3
(d) Fe / Mo
(c) TiCl4 + Al(C2H5)3
Question 29.
Which one of the following is used as a catalyst in the polymerisation of propylene?
(a) V2O5
(b) Pt
(c) TiCl4 + Al(C2H5)3
(d) Fe / Mo
(c)TiCl4 + Al(C2H5)3
Question 30.
Consider the following statements.
(i) Transition metal hydrides are used as powerful oxidising agents.
(ii) Metallic carbides are chemically active.
(iii) Interstitial compounds are hard and show electrical and thermal conductivity.
Which of the above statements is/are incorrect?
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) only
(d) (i) only
(a) (i) and (ii)
Question 31.
Which one of the following oxide is covalent?
(a) Cr2O3
(b) CrO
(c) Mn2O7
(d) Na2O
(c) Mn2O7
Question 32.
Which one of the following oxide is amphoteric in nature?
(a) CrO
(b) Cr2O3
(c) Mn2O7
(d) MnO
(b) Cr2O3
Question 33.
The oxidation state of Chromium in $${ CrO }_{ 4 }^{ 2- }$$ and in $${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }$$ are …………..
(a) +3, +6
(b) +7, +4
(c) +6, +6
(d) +4, +6
(c) +6, +6
Question 34.
Which one of the following is used to identify chloride ion in inorganic qualitative analysis?
(a) Barium chloride test
(b) Chromyl chloride test
(c) Brown ring test
(d) Ammonium molybdate test
(b) Chromyl chloride test
Question 35.
Which one of the following is the formula of chromyl chloride?
(a) CrOCl2
(b) CrCl3
(c) CrO2 Cl2
(d) CrCl
(c) CrO2 Cl2
Question 36.
Which ore is used to prepare potassium permanganate?
(a) Pyrolusite
(b) Chromite
(c) Argentite
(d) Cuprite
(a) Pyrolusite
Question 37.
Which one of the following geometry is possesed by permanganate ion?
(a) Pyramidal
(b) Tetrahedral
(c) Octahedral
(d) linear
(b) Tetrahedral
Question 38.
The hybridisation state of Mn+7 is permanganate ion is …………..
(a) sp2 hybridisation
(b) dsp2 hybridisation
(c) d2sp3 hybridisation
(d) sp3 hybridisation
(d) sp3 hybridisation
Question 39.
Which one of the following is known as Baeyer’s reagent?
(a) Cold dilute alkaline KMnO4
(b) Chromyl Chloride
(c) Acidified potassium dichromate
(d) Acidified potassium manganate
(a) Cold dilute alkaline KMnO4
Question 40.
Which reagent is used in the conversion of ethylene into ethylene glycol?
(a) Chromyl chloride
(b) Zeigler – Natta catalyst
(c) Cold dilute alkaline KMnO4
(d) Acidified K4Cr2O7
(c) Cold dilute alkaline KMnO4
Question 41.
Baeyer’s reagent is used to detect unsaturation in an organic compound.
(a) Chloride ion
(b) unsaturated organic compound
(c) Sulphate ion
(d) Chromate ion
(b) unsaturated organic compound
Question 42.
What is the equivalent weight of KMnO4 in acid medium?
(a) 158
(b) 52.67
(c) 31.6
(d) 392
(c) 31.6
Question 43.
What is the equivalent weight of KMnO4 in basic medium?
(a) 158
(b) 52.67
(c) 392
(d) 31.6
(a) 158
Question 44.
Which one of the following is used for the estimation of ferrous salts, oxalates, hydrogen peroxide and iodides?
(a) K2MnO4
(ft) KMnO4
(c) K2Cr2O7
(d) CrO2 Cl2
(b) KMnO4
Question 45.
Which of the following is the general electronic configuration of lanthanoids?
(a) [Xe] 4f7 3d1-10 5s2
(b) [Xe] 4f1-14 3d10 6s2
(c) [Xe] 5f2-14 4d10 6s2
(d) [Xe] 4f2-14 5d1-10 6s2
(d) [Xe] 4f2-14 5d1-10 6s2
Question 46.
The most common oxidation state of Lanthanoids is …………..
(a) + 4
(b) + 3
(c) + 6
(d) +5
(b) + 3
Question 47.
The expected electronic configuration of lanthanum (La) (Z = 57) is …………..
(a) [Xe] 4f1 5d° 6s2
(b) [Xe] 4f0 5d1 6s2
(c) [Xe] 4f3
(d) [Xe] 4f0 5d3
(a) [Xe] 4f1 5d° 6s2
Question 48.
The actual electronic configuration of La (Z = 57) is …………..
(a) [Xe] 4f1 5d° 6s2
(b) [Xe] 4f3
(c) [Xe] 4f0 5d1 6s2
(d) [Xe] 4f0 5d3
(c) [Xe] 4f0 5d1 6s2
Question 49.
Which of the following lanthanoids have half-filled 4f orbital?
(a) Gd
(b) Tb
(c) Lu
(d) La
(a) Gd
Question 50.
Which one of the following lanthanoids have completely filled 4f orbital?
(a) Gd and Eu
(b) La and Ce
(c) Yb and Lu
(d) Pr and Pm
(c) Yb and Lu
Question 51.
Which one of the following is the main cause of lanthanoid contraction?
(a) Poor shielding effect of 5f sub-shell
(b) More shielding effect of 4f sub-shell
(c) Poor shielding effect of 4f sub-shell
(d) More shielding effect of 5f sub-shell
(c) Poor shielding effect of 4f sub-shell
Question 52.
Which of the following pair has more or less same atomic radius due to lanthanide contraction?
(a) Ti and V
(b) Fm and Md
(c) No and Lr
(d) Zr and Hf
(d) Zr and Hf
Question 53.
Consider the following statement.
(i) All the actinoids are non radioactive.
(ii) Neptunium and other heavier elements are produced by artificial transformation of naturally occurring elements by nuclear reactions.
(iii) Most of the actinoids have long half lives.
Which of the above statements is/are not correct.
(a) (i) only
(b) (i) and (ii)
(c) (ii) and (iii)
(d) (i) and (iii)
(d) (i) and (iii)
Question 54.
The general valence shell electronic configuration of actinoids is …………..
(a) [Xe] 4f2-14 5d0-2 6s2
(b) [Rn] 4f2-14 5d0-2 6s2
(c) [Rn] 5f2-14 6d0-1 7s2
(d) [Rn] 4f0-7 5d0-1 s2
(c) [Rn] 5f2-14 6d0-2 7s2
Question 55.
Which pair of actinoids show +2 oxidation state?
(a) Am and Th
(b) Pa and U
(c) Pu and Cm
(d) No and Lr
(a) Am and Th
Question 56.
Neptunium and Plutonium exhibit the maximum oxidation state as …………..
(a) + 8
(b) + 7
(c) + 6
(d) + 4
(b) + 7
Question 57.
Consider the following statement.
(i) Most of the actinoids are coloured.
(ii) Actinoids show greater tendency to form complexes.
(iii) Most of the actinoids are non-radioactive.
Which of the above statements is/are correct.
(a) (i) only
(b) (i) and (iii)
(c) (i) and (ii)
(d) (ii) and (iii)
(c) (i) and (ii)
Question 58.
Consider the following statement.
(i) Lanthanoids do not form oxo cations.
(ii) Most of the lanthanoids are colourless.
(iii) Binding energy of 4f orbitals are lower.
Which of the above statement is/are not correct.
(a) (i) and (ii)
(b) (iii) only
(c) (i) and (iii)
(d) (i), (ii) and (iii)
(b) (iii) only
Question 59.
Which one of the following is more basic in nature?
(a) La(OH)3
(b) Ce(OH)3
(c) Gd(OH)3
(d) Lu(OH)3
(a) La(OH)3
Question 60.
Which one of the following is less basic in nature?
(a) La(OH)3
(b) Gd(OH)3
(c) Lu(OH)3
(d) Ce(OH)3
(c) LU(OH)3
II. Fill in the blanks.
1. Transition elements occupy the central position of the periodic table between …………. elements.
2. Except …………. elements, all transition metals are hard and have very high melting point.
3. The metal present in Vitamin – B12 is ………….
4. …………. metal is used in manufacture of artificial joints.
5. The extra stability of Cr and Cu is due to …………. of electrons and exchange energy.
6. Of all the known elements …………. has the highest electrical conductivity at room temperature.
7. The maximum melting point at about the middle of transition metal series indicates that …………. configuration is favourable for strong attraction.
8. The atomic radius of 5d elements and 4d elements are nearly same due to ………….
9. Ni (II) compounds are thermodynamically …………. than Pt (II) compounds.
10. The first transition metal …………. exhibits only +3 oxidation state.
11. The middle transition element …………. has six different oxidation states.
12. The oxidation state of Ru and Os is ………….
13. …………. is unique in 3d series having a stable +1 oxidation state.
14. The substance which is oxidised is a …………. agent and the one which is reduced is an …………. agent.
15. The oxidising and reducing power of an element is measured in terms of ………….
16. If the E° of a metal is large and negative, the metal is a ………….
17. The species with all paired electrons exhibit ………….
18. The magnetic moment of an ion is given by ………….
19. Many industrial processes use …………. or their as catalyst.
20. In the catalytic hydrogenation of an alkene …………. is used as catalyst.
21. In the preparation of acetic acid from acetaldehyde the catalyst used in ………….
22. The catalyst used in the hydroformylation of olefins is ………….
23. …………. catalyst is used in polymerization of propylene.
24. Metallic carbides are chemically ………….
25. Except Scandium all other 3d series transition elements form …………. metal oxides.
26. Cr2O3 …………. is and CrO is …………. in nature.
27. Mn2O7 dissolves in water to give ………….
28. On heating potassium dichromate, it decomposes to give …………. and molecular oxygen.
29. Potassium dichromate is a powerful …………. agent in acidic medium.
30. ………….is used in leather tanneries for chrome tanning.
31. Potassium dichromate is used in quantitative analysis for the estimation of …………. and ………….
32. Permanganate ion has …………. geometry in which Mn+7 is …………. hybridised.
33. Cold dilute alkaline KMnO4 is known as ………….
34. …………. is used for the treatment of skin infections and fungal infections of the foot.
35. Baeyer’s reagent is used for detecting …………. in an organic compounds.
36. The equivalent weight of KMnO4 in neutral medium is ………….
37. The common oxidation state of lanthanoids is ………….
38. Due to the decrease in the size of Ln3+ ions, the ionic character of Ln – OH bond decreases which results in the ………….
39. All the actinoids are …………. and most of them have …………. half lives.
40. ………….do not form oxo cations.
1. sandpblock
2. group-II
3. cobalt
4. Titanium
5. symmetrical distribution
6. silver
7. d5, inter atomic
8. lanthanoid contraction
9. more stable
10. Scandium
11. Manganese
12. + 8
13. Copper
14. reducing, oxidising
15. Standard electrode potential
16. powerful reducing agent
17. diamagnetism
18. μ=$$g\sqrt { S(S+1) }$$μB
19. transition metals, compounds
20. Nickel
21. Rh/Ir complex
22. CO2(CO)8
23. Zeigler – Natta (or) TiCl4 + Al(C2H5)3
24. inert
25. ionic
26. amphoteric, basic
27. permanganic acid (HMnO4)
28. Chromium (III) oxide – Cr2O3
29. Potassium dichrom.ate
30. iron compounds, iodides
31. tetrahedral,sp3
32. Baeyer’s reagent
33. Potassium permanganate
34. unsaturation
35. 52.67
36. + 3
37. decrease in the basicity
39. Lanthanoids
III. Match the following using the code given below.
Question 1.
A. Tungsten – 1. Development of human civilization
B. Titanium – 2. Light bulb filament
C. Molybdenum – 3. Artificial joint
D. Copper – 4. Boiler plants
(a) 2, 3, 4, 1
Question 2.
A. Iron – 1. Artificial joints
B. Platinum – 2. Hemoglobin
C. Cobalt – 3. Catalysis
D. Titanium – 4. Vitamin – B12
(b) 2, 3, 4, 1
Question 3.
A. Sc to Zn – 1. 5d series
B. Y to Cd – 2. Actinoids
C. LatoHg – 3. 3dseries
D. Ac to Lr – 4. 4d series
(a) 3, 4, 1, 2
Question 4.
A. Cr – 1. [Ar] 3d10 4s2
B. Cu – 2. [Ar] 3d5 4s1
C. Zn – 3. [Ar]3d1 4s2
D. Sc – 4. [Ar] 3d10 4s1
(c) 2, 4, 1, 3
Question 5.
A. Scandium – 1. + 7
B. Manganese – 2. + 2
C. Copper – 3.+3
D. Titanium – 4. + I
(a) 3, 1, 4, 2
Question 6.
A. Sc3+ – 1. 3d1
B. Ti3 – 2. 3d0
C. Mn2+ – 3. 3d10
D. Zn2+ – 4. 3d5
(b) 2, 1, 4, 3
IV. Assertion and Reason.
Question 1.
Assertion (A) – Cr and Cu having [Ar] 3d5 4s1 and [Ar] 3d10 4s1 are more stable.
Reason (R) – The extra stability of elements Cr and Cu is due to symmetrical distribution of electrons and exchange energy.
(a) Both (A) and (R) are correct and (R) explains (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
(a) Both (A) and (R) are correct and (R) explains (A).
Question 2.
Assertion (A) – In 3d transition elements, the expected decrease in atomic radius is observed from Sc to V, thereafter upto Cu, the atomic radius nearly remains the same.
Reason (R) – As we move from Sc to V, the added 3d electrons only partially shield the increased nuclear charge but upto Cu, the extra electron added to 3d sub-shell repel the 4s electrons and the slight increase in nuclear charge operated in opposite direction and it leads to constancy in atomic radii.
(a) Both (A) and (R) are correct and (R) explains (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
(a) Both (A) and (R) are correct and (R) explains (A).
Question 3.
Assertion (A) – In transition metal series, the ionization enthalpy increases.
Reason (R) – This is due to increase in nuclear charge corresponding to the filling of d electrons.
(a) Both (A) and (R) are correct and (R) explains (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
(a) Both (A) and (R) are correct and (R) explains (A).
Question 4.
Assertion (A) – Ni (II) compounds are thermodynamically more stable than Pt (II) compounds.
Reason (R) – The energy required to form Ni2- is less than that of Pt2+.
(a) Both (A) and (R) are correct and (R) explains (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
(a) Both (A) and (R) are correct and (R) explains (A).
Question 5.
Assertion (A) – Except Scandium all 3d series, transition elements exhibit variable oxidation states.
Reason (R) – By loosing electrons from (n – l)d orbital and ns orbital as the energy difference between them is very small.
(a) Both (A) and (R) are correct and (R) explains (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
(a) Both (A) and (R) are correct and (R) explains (A).
Question 6.
Assertion (A) – Mn2+ is more stable than Mn4+.
Reason (R) – Mn2+(3d5) is more stable than Mn4+(3d3) is due to extra stability of half-filled electronic configuration.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A)
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
Question 7.
Assertion (A) – Copper is unique in 3d series having a stable +1 oxidation state.
Reason (R) – Copper is prone to disproportionate to the +2 and 0 oxidation states.
(a) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(b) Both (A) and (R) are correct and (R) explains (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
(b) Both (A) and (R) are correct and (R) explains (A).
Question 8.
Assertion (A) – Transition metals form large number of complexes.
Reason (R) – Transition metals are small and highly charged and they have vacant low energy orbitals to accept an electron pair donated by other groups.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A)
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is wrong
(a) Both (A) and (R) are correct and (R) is the correct explanation
V. Find the odd one out.
Question 1.
(a) Sc
(b) Ti
(c) Y
(d) Cr
(c) Y
Reason: Yttrium be1ons to 4d series whereas others are 3d series
Question 2.
(a)Ru
(b)Rh
(c)Pd
(d) Pt
(d) Pt
Reason: Pt belongs to 5d series whereas others are 4d series.
Question 3.
(a) Th
(b) La
(c) Ce
(d) Lu
(a) Th
Reason: Th belongs to actinoids whereas others are lanthanoids.
Question 4.
(a) La
(b) Pr
(c) Am
(d) Lu
(c) Am
Reason: Am belongs to actinoids whereas others are lanthanoids.
Question 5.
(a) Ce
(b) Th
(c) U
(d) Pu
(a) Ce
Reason: Ce belongs to lanthanoids whereas others are actinoids.
Question 6.
(a) Ac
(b) U
(c) Pa
(4) Np
(a) Ac
Reason: Except Actinium all the remaining elements are synthetically prepared.
Question 7.
(a) Ti2+
(b) V2+
(c) Zn2+
(d) Cu2+
(d) Cu2+
Reason: It has positive reduction potential whereas others have negative reduction potential.
Question 8.
(a) CO3+
(b) Cr3+
(c) V3+
(4) Ti3+
(a) CO3+
Reason: It has positive reduction potential whereas others have negative reduction potential.
Question 9.
(a) Mn3+
(b) Fe3+
(c) Cr3+
(d) CO3+
(c) Cr3+
Reason: It has negative reduction potential whereas others have positive reduction potential.
Question 10.
(a) SC3+
(b) Ti4+
(c) V5+
(d) Cu2+
(d) Cu2+
Reason: It is paramagnetic whereas others are diamagnetic.
Question 11.
(a) Cr3+
(b) Mn4+
(c) V2+
(d) Zn2+
(d) Zn2+
Reason: It is diamagnetic whereas others are paramagnetic
VI. Find Out the correct pair.
Question 1.
(a) $${ CrO }_{ 4 }^{ – }$$ and Cr2$${ O }_{ 7 }^{ 2- }$$
(b) $${ MnO }_{ 4 }^{ – }$$ and $${ MnO }_{ 4 }^{ 2- }$$
(c) H2CrO4 and HMnO4
(d) Cr2O3 and Mn2O7
(a) $${ CrO }_{ 4 }^{ – }$$ and Cr2$${ O }_{ 7 }^{ 2- }$$
In this pair Cr has +6 oxidation states whereas in others the metal has different oxidation state.
Question 2.
(a) Zn, Cu
(b) Hf, Zr
(c) Ag , Au
(d) Ti, Cu
(b) Hf, Zr
This pair has same atomic radius whereas others have different atomic radius.
Question 3.
(a) Ru , Os
(b) Mn , Cu
(c) Sc , Cu
(d) Ni, Co
(a) Ru , Os
Both has +8 as oxidation state whereas others have different oxidation state.
Question 4.
(a) Ti2+, CO2+
(b) Cr2+, Mn3+
(c) Fe2+ and CO3+
(d) CO3+ and Cu2+
(d) CO3+ and Cu2+
Both have positive electrode potential whereas others have different value.
Question 5.
(a) Cu2+, Zn2+
(b) CO3+, Cr3+
(c) Ti3+, V3+
(d) Mn3+, Cr3+
(c) Ti3+, V3+
have negative electrode potential whereas others have different values.
VII. Find out the incorrect pair.
Question 1.
(a) Sc3+, Ti4+
(b) Ti3+, Ti2+
(c) Cr2+, Mn3+
(d) Cu+, Zn2+
(b) Ti3+, Ti2+
have d1 and d1 configuration whereas others have same configuration.
Question 2.
(a) Sc and Zn
(b) Y and Cd
(c) Ag and Au
(d) Na and K
(d) Na and K. They belong to alkaline metals whereas others are d-block elements.
Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements 2 Mark Questions and Answers
Question 1.
d-block elements are called transition elements. Justify this statement.
1. d-block elements occupy the central position of the periodic table, between s and p block elements.
2. Their properties are transitional between highly reactive metals of s-block and elements of p-block which are mostly non-metals. That is why d-block elements are called transitional elements.
Question 2.
How many series are in d-bloclc elements? What are they?
• There are 4 series in d-block e1ements They are,
• 3d series – 4th period – Scandium to Zinc
• 4d series – 5th period – Yttrium to Cadmium
• 5d series – 6th period – Lanthanum to Mercury
• 6d series – 7th period – Actinium to Californium
Question 3.
Zn, Cd, Hg belong to d-block elements even though they do not have partially filled d-orbitals. Give reason.
1. Zn, Cd, Hg belong to d-block elements even though they do not have partially filled d-orbitals either in their elemental state or in their normal oxidation states.
2. However they are treated as transition elements, because their properties are an extension of the properties of the respective transition elements.
Question 4.
Applying Aufbau principle, write down the electronic configuration of Sc (Z = 21) and Zn (Z = 30).
1. According to Aufbau principle, the electron first fills the 4s orbital before 3d orbital.
2. Sc (Z = 21) Is2 2s2 2p6 3s2 3p6 4s2 3d1
• Zn (Z = 30) Is2 2s2 2p6 3s2 3p6 4s2 3d10
Question 5.
At the end of 3d series, d-orbitals of Zinc contain 10 electrons in which the repulsive interaction between the electrons is more than the effective nuclear charge and hence the orbitals slightly expand and atomic radius slightly increases.
Question 6.
Write a not about oxidation state of 3d series.
1. The number of oxidation states increases with the number of electrons available, and it decreases as the number of paired electrons increases.
2. Hence, the first and last elements show less number of oxidation states and the middle elements with more number of oxidation states.
3. For example, the first element Sc has only one oxidation state +3, the middle element Mn has six different oxidation states from +2 to +7. The last element Cu shows +1 and +2 oxidation states only.
Question 7.
Mn2+ is more stable than Mn4+. Why?
1. The relative stability of different oxidation states of 3d metals is correlated with the extra stability of half-filled and fully filled electronic configurations.
2. Example – Mn2+ (3d5) is more stable than Mn4+ (3d3)
Question 8.
Ru and Os have highest oxidation state in which compounds? Explain with example.
1. Ru and Os have +8 as the highest oxidation state.
2. The highest oxidation state of 4d and 5d elements are found in their compounds with the higher electronegative elements like O, F and Cl. For example: RuO4, OsO4
Question 9.
Copper is unique in 3d series. Prove this statement.
Copper is unique in 3d series having a stable +1 oxidation state. It is prone to disproportionate to the +2 and 0 oxidation states.
Question 10.
Define – Standard electrode potential.
Standard electrode potential is the value of the standard emf of a cell in which molecular hydrogen under standard pressure (latm) and temperature (273K) is oxidised to solvated protons at the electrode.
Question 11.
Which metal is used to reduce Cr3+ ion? Why?
A stable Cr3+ ion, strong reducing agent which has high negative value for reduction potential like metallic zinc (E° = – 0.76 V) is required. Metallic zinc is a powerful reducing agent due to its large negative values.
Question 12.
Sc3+, Ti4+, V5+ are diamagnetic. Give reason.
1. Sc3+, Ti4+, V5+ have d° electronic configuration, n = 0
2. µ = $$\sqrt { 0(0+2) }$$ = 0 µB So they are diamagnetic.
Question 13.
Calculate the magnetic moment of Ti3+ and V4+.
Ti (Z = 22) Ti3+ 3d1
V (Z = 23) V4+ 3d1
µ = $$\sqrt { 1(1+2) }$$ = $$\sqrt { 35 }$$ = 1.73 µB. So they are paramagnetic.
Question 14.
Cr3+, Mn4+, V2+ are paramagnetic. Calculate their magnetic moment values.
Cr3+, Mn4+, V2+ Configuration is d3. Due to 3 unpaired electrons, they are paramagnetic. µ = $$\sqrt { 3(3+2) }$$ = $$\sqrt { 35 }$$ = 3.87 µB
Question 15.
Mn2+, Fe3+ have high magnetic moment. Prove it.
1. Mn2+, Fe3+ configuration is d5.
2. µ = $$\sqrt { 5(5+2) }$$ = $$\sqrt { 35 }$$ = 5.916 µB
Among 3d series, Mn2+, Fe3+ have high magnetic moment as 5.916 µB .
Question 16.
How many unpaired electrons are present in CO3+, Fe2+? Calculate their magnetic moment.
CO (Z = 27) CO3+ [Ar] 3d6
Fe (Z = 26) Fe2+ [Ar] 3d6
The number of unpaired electrons are 4 as follows:
Their magnetic moment is µ = $$\sqrt { 4(4+2) }$$ = $$\sqrt { 24 }$$ = 4.89 µB
Question 17.
Calculate the magnetic moment and the number of unpaired electrons in Cu2+.
Cu (Z = 29) Electronic configuration [Ar] 3d10 4s1
Cu2+ Electronic configuration [Ar] 3d9
The number of unpaired electrons
is 1.
Magnetic moment µ = $$\sqrt { 1(1+2) }$$ = $$\sqrt { 3 }$$ = 1.732 µB
Question 18.
Cu+, Zn2+ are diamagnetic. Prove it.
Cu+, Zn2+ electronic configuration [Ar] 3d10
The number of unpaired electron is 0.
µ = $$\sqrt { 0(0+2) }$$ = 0 µB. Cu+, Zn2+ are diamagnetic.
Question 19.
Most of the transition metals act as catalyst. Justify this statement.
1. Many industrial processes use transition metals or their compounds as catalysts. Transition metal has energetically available d orbitals that can accept electrons from reactant molecule or metal can form bond with reactant molecule using its ‘d’ electrons.
2. For example, in the catalytic hydrogenation of an alkene, the alkene bonds to an active site by using its n electrons with an empty d orbital of the catalyst.
Question 20.
Explain the catalytic hydrogenation of alkene to alkane with equation.
The σ bond in the hydrogen molecule breaks, and each hydrogen atom forms a bond with a d electron on an atom in the catalyst Nickel. The two hydrogen atoms then bond with the partially broken π -bond in the alkene to form an alkane.
Question 21.
Which catalyst is used in the hydroformylation of olefins? Give equation
Question 22.
Which catalyst is used in the conversion of acetaldehyde to acetic acid? Give equation
Question 23.
What is Zeigler -Natta catalyst? In which reaction it is used? Give equation.
A mixture of TiCl4 and trialkyl aluminium is Zeigler – Natta catalyst. It is used in the polymerization
Question 24.
d-block elements readily form complexes. Give reason.
1. Transition elements (d-block elements) have a tendency to form coordination compounds (complexes) with a species that has an ability to donate an electron pair to form a coordinate
2. Transition metal ions are small and highly charged and they have vacant low energy orbitals to accept an electron pair donated by other groups. Due to these properties, transition metals form large number of complexes.
3. Examples – [Fe(CN)6]4-, [CO(NH3)6]3+
Question 25.
Prove that acidified potassium dichromate is a powerful oxidising agent.
K2Cr2O7 act as power oxidising agent in acidic medium. In the presence of H+ ions, the oxidation state of Cr froms Cr6+ is changed to Cr3+
Cr2$${ O }_{ 7 }^{ 2- }$$ + 14H+ + 6 → 2Cr3+ + 6Fe3+ + 7H2O
Example:
Acidified K2Cr2O7 oxidises Ferrous salts to Ferric salts.
Cr2$${ O }_{ 7 }^{ 2- }$$ + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O.
Question 26.
What are the uses of potassium dichromate?
K2Cr2O7 is used
1. as a strong oxidising agent
2. in dyeing and printing
3. in leather tanneries for chrome plating
4. in quantitative analysis for the estimation of iron compounds and iodides
Question 27.
Draw and explain about the structure of permanganate ion.
Permanganate ion has tetrahedral geometry in which the central Mn7+ is sp3 hybridised.
Question 28.
Explain the action of heat on potassium permanganate.
Question 30.
What happens when thiosuiphate ion is treated with permanganate ion?
Permanganate ion oxidises thiosulphate into sulphate.
$${ 8MnO }_{ 4 }^{ – }$$ + 3S2$${ O }_{ 3 }^{ 2- }$$ → $${ 6SO }_{ 4 }^{ 2- }$$ + 8MnO2 + 2OH
Question 31.
What is Baeyer’s reagent? Where it is used?
1. Cold dilute alkaline KMnO4 is known as Baeyer’s reagent. It is used to oxidise alkene into diols.
2. For example, ethylene can be converted into ethylene glycol and this reaction is used as a test for unsaturation.
Question 32.
Acidified KMnO4 is a very strong oxidising agent. Prove it.
1. In the presence of dilute sulphuric acid, potassium permanganate acts as a very strong oxidising agent. Permanganate ion is converted into Mn2+ ion.
$${ MnO }_{ 4 }^{ – }$$ + 8H+ + 5e → Mn2+ + 4H2O
2. Permanganate oxidises ferrous salt to ferric salt.
$${ 2MnO }_{ 4 }^{ – }$$ + 10Fe2+ + 16H+ → 2Mn2+ + 10Fe3+ + 8H2O
Question 33.
KMnO4 does not act as oxidising agent in the presence of HCl. Why?
HCl cannot be used for making acidified KMnO4 as oxidising agent, since it reacts with KMnO4 as follows.
$${ 2MnO }_{ 4 }^{ – }$$ + 10Cl + 16H+ → 2Mn2+ + 5Cl2 + 8H2O
Question 34.
HNO3 cannot be used as an acid medium along with KMnO4. Why?
HNO3 cannot be used since it is a good oxidising agent and it reacts with reducing agents in the reaction.
Question 35.
Among HCl, HNO3 and H2SO4, which is the suitable medium for KMnO4 in oxidising reaction?
HCl and HNO3 cannot be used. HCl react with KMnO4. HNO3 is itself a good oxidising agent. However, H2SO4 is found to be most suitable since it does not react with potassium permanganate.
Question 36.
Explain about the causes of lanthanide contraction.
1. As we move from one element to another in 4f series ( Ce to Lu) the nuclear charge increases by one unit and an additional electron is added into the same inner 4f sub-shell.
2. 4f sub-shell have a diffused shapes and therefore the shielding effect of 4f electrons are relatively poor. Hence, with increase of nuclear charge, the valence shell is pulled slightly towards nucleus.
3. As a result, the effective nuclear charge experienced by the 4f elelctoms increases and the size of Ln3+ ions decreases.
Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements 3 Mark Questions and Answers
Question 1.
Cr and Cu are more stable. Give reason.
Answwer:
1. The electronic configuration of Cr and Cu are [Ar] 3d3 4s1 and [Ar] 3d10 4s1 respectively. The extra stability of half filled and fully filled d orbitals, as already explained in XI STD, is due to symmetrical distribution of electrons and exchange energy.
2. The extra stability of half filled and fully filled d orbitals is due to symmetrical distribution of electrons and exchange energy.
3. When the d orbitals are considered together, they will constitute a sphere. So the half filled and fully filled configuration leads to complete symmetrical distribution of electron density.
4. On the other hand, an unsymmetrical distribution of electron density will result in building up of a potential difference. To decrease this and to achieve a tension free state with lower energy, a symmetrical distribution is preferred.
Question 2.
Explain about the metallic behaviour of d-block elements.
1. All the transition elements are metals. They are good conductors of heat and electricity. Of all the known elements, silver has the highest electrical conductivity at room temperature.
2. Most of the transition elements are hexagonal close packed, cubic close packed or body centered cubic which are the characteristics of true metals.
Question 3.
Explain about the variation of melting point among the transition metal series.
1. As we move from left to right along the transition metal series, melting point first increases as the number of unpaired d electrons available for metallic bonding increases, reach a maximum value and then decreases, as the d electron pairs up and become less available for bonding.
2. For example, in the first series the melting point increases from Scandium to a maximum of 2183 K for Vanadium, which is close to 2180K for chromium.
3. Manganese in 3d series and has low melting point. The maximum melting point at about the middle of transition metal series indicates that d5 configuration is favorable for strong interatomic attraction.
Question 4.
Explain about the variation of atomic radius along a period of 3d series.
1. In general, atomic radius decreases along a period. But for the 3d transition elements, the expected decrease in atomic radius is observed from Sc to V , thereafter upto Cu the atomic radius nearly remains the same.
2. As we move from Sc to Zn in 3d series, the extra electrons are added to the 3d orbitals, the added 3d electrons only partially shield the increased nuclear charge and hence the effective nuclear charge increases slightly.
3. However, the extra electrons added to the 3d sub shell strongly repel the 4s electrons and these two forces are operated in opposite direction and as they tend to balance each other, it leads to constancy in atomic radii.
Question 5.
Ni (II) compounds are more stable than Pt (II) compounds. Give reason.
1. The ionisation enthalpy values can be used to predict the thermodynamic stability of their compounds. .
2. For Nickel I.E1 + I.E2 = 737+ 1753
= 2490 kJ mol-1
For platinum, I.E1 + I.E2 = 864+ 1791
= 2655 kJ mol-1
Since, the energy required to form Ni2+ is less than that of Pt2+ , Ni(II) compounds are thermodynamically more stable than Pt(II) compounds.
Question 6.
Compare the reduction potentials of Mn3+ / Mn2+ and Fe3+ / Fe2+ .
1. Mn3+ + e → Mn2+
E° = + 1.51V
Fe3+ + e → Fe2+
E° = + 0.77V
2. The high reduction potential of Mn3+ / Mn2+ indicates Mn2+ is more stable than Mn3+. For Fe3+ / Fe2+ the reduction potential is 0.77V, this low value indicates that both Fe3+ and Fe2+ can exist under normal conditions.
3. The drop from Mn to Fe is due to the electronic structure of the ions concerned. Mn3+ has 3d4 configuration while that of Mn2+ is 3d5. The extra stability associated with a half filled d sub-shell makes the reduction of Mn3+ very feasible.
Question 7.
How alloys are formed in d-biock elements?
1. An alloy is formed by blending a metal with one or more other elements. The elements may be metals or non-metals or both.
2. The bulk metal is named as solvent, and the other elements in smaller portion is called solute.
3. According to Hume – Rothery rule to form an alloy, the difference between the atomic radii of the solvent and solute is less than 15%. Both the solvent and solute must have the same crystal structure and valence and their electro negativity difference must be close to zero.
4. Since their atomic sizes are similar and one metal atom can be easily replaced by another metal atom from its crystal lattice to form an alloy. The alloys are hard and have high melting points. Examples – Gold – copper alloy.
Question 8.
What are interstitial compounds? Give their properties.
An interstitial compound or alloy is a compound that is formed when small atoms like hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice. They are usually non-stoichiometric compounds, e.g., TiC, ZrH1.92, Mn4N.
1. They are hard and show electrical and thermal conductivity.
2. They have high melting points.
3. Transition metal hydrides are used as powerful reducing agents.
4. Metallic carbides are chemically inert.
Question 9.
Explain the action of heat on potassium dichromate.
Question10.
Draw and explain about the structure of chromate and dichromate ion.
1. Both chromate and dichromate ions are oxo anions of chromium and they are moderately strong oxidising agents.
2. In both structures, chromium is in +6 oxidation state.
3. In an aqueous solution, chromate and dichromate ions can be inter convertible, and in an alkaline solution, chromate ion is predominant, whereas dichromate ion becomes predominant in acidic solutions. .
Question 11.
Explain the action of acidified K2Cr2O7 – with
1. Iodide
2. Sulphide
1. Acidified K2Cr2O7 oxidises iodide ions to iodine.
Cr2$${ O }_{ 7 }^{ 2- }$$ + 6I + 14H+ → 2Cr3+ +3I2 + 7H2O
2. Acidified K2Cr2O7 oxidises Sulphide ions to Sulphur.
Cr2$${ O }_{ 7 }^{ 2- }$$ + 3S2- + 14H+ → 2Cr3+ + 3S + 7H2O
Question 12.
Explain the action of acidified K2Cr2O7 with
1. Sulphur dioxide
2. Alcohols.
1. Acidified K2Cr2O7 oxidises SO2 to sulphate ion.
Cr2$${ O }_{ 7 }^{ 2- }$$ + 3SO2 + 2H+ -> 2Cr3+ + $${ 3SO }_{ 4 }^{ 2- }$$ + H2O
2. Acidified K2Cr2O7 oxidises alcohol to acid. .
Question 13.
1. When potassium dichromate is heated with any chloride salt in the presence of Conc.H2SO4, orange red vapours of chromyl chloride (CrO2Cl2) is evolved. This reaction is used to confirm the presence of chloride ion in inorganic qualitative analysis.
2. The chromyl chloride vapours are dissolved in sodium hydroxide solution and then acidified with acetic acid and treated with lead acetate. A yellow precipitate of lead chromate is obtained.
Question 14.
Explain the action of Conc.H2SO4 on potassium permanganate.
1. On treating with cold conc.H2SO4, it decomposes to form manganese heptoxide, which subsequently decomposes explosively.
2. With hot Conc.H2SO4, potassium permanganate give MnSO4 [Manganese(II) sulphate]
Question 15.
What are the uses of Potassium permanganate?
Potassium permanganate is used,
1. as a strong oxidising agent.
2. for the treatment of various skin infections and fungal infections of foot.
3. used in water treatment industries to remove iron and hydrogen sulphide from well water.
4. as a Baeyer’s reagent for detecting unsaturation in an organic compound.
5. in quantitative analysis for the estimation of ferrous salts, oxalates, hydrogen peroxide (H2O2) and iodides.
Question 16.
Calculate the equivalent weight of KMnO4 in
1. Acidic medium
2. Basic medium
3. Neutral medium
1. Equivalent weight of KMnO4 in acidic medium
2. Equivalent weight of KMnO4 in basic medium
3. Equivalent weight of KMnO4 in neutral medium
Question 17.
Explain about the oxidation state of actinoids.
1. The most common oxidation state of actinoids is +3.
2. In addition to that actinoids show variable oxidation states such as +2, +3, +4, +5, +6 and +7.
3. The elements Americium (Am) and Thorium (Th) show +2 oxidation state in some compounds.
4. Th, Pa, U, Np, Pu and Am show +5 oxidation states.
5. Np and Pu exhibit +7 oxidation state.
Question 18.
Write the electronic configuration of
1. Ac (Z = 89)
2. Am (Z = 95)
3. Lr (Z = 103)
1. Ac (Z = 89): [Rn] 5f0 6d1 7s2
2. Am (Z = 95) : [Rn] 5f7 6d0 7s2
3. Lr (Z = 103): [Rn] 5f14 6d1 7s2
Samacheer Kalvi 12th Chemistry Transition and Inner Transition Elements 5 Mark Questions and Answers
Question 1.
Explain about the magnetic properties of transition elements.
1. Most of the compounds of transition elements are paramagnetic.
2. Materials with no elementary magnetic dipoles are diamagnetic. In other words a species with all paired electrons exhibits diamagnetism.
3. Paramagnetic solids having unpaired electrons possess magnetic dipoles which are isolated from one another.
4. Ferromagnetic materials have domain structure and in each domain the magnetic dipoles are arranged. But the spin dipoles of the adjacent domains are randomly oriented. Some transition elements or ions with unpaired d electrons show ferromagnetism.
5. 3d transition metal ions in paramagnetic solids often have a magnetic dipole moments corresponding to the electron spin contribution only. So the magnetic moment of the ion is given by
µ = $$g\sqrt { S(S+1) }$$ µB
Where g = 2, S is the total spin quantum number of the electrons.
µB = Bohr Magneton.
For an ion with ‘n’ unpaired electrons S = $$\frac { n }{ 2 }$$
Therefore the spin only magnetic moment is µ = $$2\sqrt { \left( \frac { n }{ 2 } \right) \left( \frac { n }{ 2 } +1 \right) }$$µB
µ = $$2\sqrt { \left( \frac { n\left( n+2 \right) }{ 4 } \right) }$$µB
µ = $$2\sqrt { n(n+2) }$$µB
Question 2.
How will you prepare potassium permanganate from pyrolusite ore?
Potassium permanganate is prepared from pyrolusite (MnO2) ore. The preparation involves the following steps.
1. Conversion of MnO2 to potassium manganate:
Powdered ore is fused with KOH in the presence of air or oxidising agents like KNO3. A green coloured potassium manganate is formed.
2. Oxidation of potassium manganate to potassium permanganate:
Potassium manganate can be oxidised in two ways, either by chemical oxidation or electrolytic oxidation.
3. Chemical oxidation – In this method potassium manganate is treated with ozone (O3) or chlorine to get potassium permanganate.
$${ 2MnO }_{ 4 }^{ 2- }$$ + O3 + H2O → $${ 2MnO }_{ 4 }^{ 2- }$$ + 2OH + O2
$${ 2MnO }_{ 4 }^{ 2- }$$ + Cl2 → $${ 2MnO }_{ 4 }^{ 2- }$$ + 2Cl
4. Electrolytic oxidation: In this method aqueous solution of potassium manganate is electrolyzed in the presence of little alkali.
K2MnO4 $$\equiv$$ 2K+ + $${ 2MnO }_{ 4 }^{ 2- }$$
H2O $$\equiv$$ H+ + OH
Manganate ions are converted into permanganate ions at anode.
The purple coloured solution is concentrated by evaporation and forms crystals of potassium permanganate on cooling.
Question 3.
Explain acidified KMnO4 is a powerful oxidising agent with 5 examples.
1. In the presence of dilute sulphuric acid, potassium permanganate acts as a very strong oxidising agent. Permanganate ions is converted into Mn2+ ion.
$${ MnO }_{ 4 }^{ – }$$ + 8H+ +5e → Mn2+ + 4H2O
Example:
1. Potassium permanganate oxidises ferrous salts to ferric salts.
2. Potassium permanganate oxidises iodide ions to iodine.
$${ 2MnO }_{ 4 }^{ – }$$ + 10I + 16H+ → 2Mn2+ + 5I2 + 8H2O
3. Potassium permanganate oxidises sulphide ion to sulphur.
$${ 2MnO }_{ 4 }^{ – }$$ + 5S2- + 16H+ → 2Mn2+ + 5S + 8H2O
4. Potassium permanganate oxidises oxalic acid to CO2.
$${ 2MnO }_{ 4 }^{ – }$$ + 5(COO)2- + 16H+ → 2Mn2+ + 10CO2 + 8H2O
5. Potassium permanganate oxidises alcohols to aldehyde.
2KMnO4 + 3H2SO4 + 5CH3CH2OH → 2K2SO4 + 2MnSO4 + 5CH3CHO + 8H2O
Question 4.
Explain about the oxidation state of Lanthanoids.
1. The common oxidation state of lanthanoids is +3. In addition to that some of the lanthanoids also show either +2 or +4 oxidation states.
2. Gd3+ and Lu3+ ions have extra stability, it is due to half filled and completely filled f-orbitals.
3. Cerium and terbium attain 4f7 and 4f14 configurations respectively in the +4 oxidation states.
4. Eu2+ and Yb2+ ions have exactly half filled and completely filled f orbitals.
5. Lu shows only +3 oxidation state.
6. Ce, Pr, Nd, Tb and Dy exhibit +3 and +4 oxidation states.
7. Nd, Sm, Eu, Tm, Yb exhibit +2 oxidation states also.
Common Errors:
Atomic number of d-block elements may get confused.
Rectifications:
Atomic number of d-block elements in group wise jumped as 18, 18, 18, 32.
e.g. Sc (Z = 21)
Y (Z = 39)
La (Z = 57)
Ac (Z = 89)
Hope you love the Samacheer Kalvi 12th Chemistry Chapter Wise Material. Clearly understand the deep concept of Chemistry learning with the help of Tamilnadu State Board 12th Chemistry Chapter 4 Transition and Inner Transition Elements Questions and AnswersPDF. Refer your friends to and bookmark our website for instant updates. Also, keep in touch with us using the comment section.
|
2022-08-12 15:27:17
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6300960779190063, "perplexity": 12021.827114714291}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571719.48/warc/CC-MAIN-20220812140019-20220812170019-00044.warc.gz"}
|
https://www.shaalaa.com/question-bank-solutions/volume-cuboid-a-rectangular-tank-80-m-long-25-m-broad-water-flows-it-through-pipe-whose-cross-section-25-cm2_38105
|
Share
Books Shortlist
# Solution for A Rectangular Tank is 80 M Long and 25 M Broad. Water Flows into It Through a Pipe Whose Cross-section is 25 Cm2, - CBSE Class 9 - Mathematics
ConceptVolume of a Cuboid
#### Question
A rectangular tank is 80 m long and 25 m broad. Water flows into it through a pipe whose cross-section is 25 cm2, at the rate of 16 km per hour. How much the level of the water rises in the tank in 45 minutes.
#### Solution
Let the level of water be risen by h cm.
Then,
Volume of water in the tank = 8000xx2500xxhcm^2
Area of cross – section of the pipe =25cm^2
Water coming out of the pipe forms a cuboid of base area 25cm^2 and length equal to the distance travelled in 45 minutes with the speed 16km/hour.
i.e., length=16000xx100xx45/60cm
∴Volume of water coming out of pipe in 45 minutes
=25xx16000xx100(45/60)
Now, volume of water in the tank = volume of water coming out of the pipe in 45 minutes
⇒8000xx2500xxh=16000xx100xx45/60xx25
⇒h=(16000xx100xx45xx25)/(8000xx2500xx60)cm=1.5cm.
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Solution for Mathematics for Class 9 by R D Sharma (2018-19 Session) (2018 to Current)
Chapter 18: Surface Areas and Volume of a Cuboid and Cube
Q: 23 | Page no. 31
Solution A Rectangular Tank is 80 M Long and 25 M Broad. Water Flows into It Through a Pipe Whose Cross-section is 25 Cm2, Concept: Volume of a Cuboid.
S
|
2019-04-18 15:24:04
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3223114609718323, "perplexity": 1203.6908692402938}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578517682.16/warc/CC-MAIN-20190418141430-20190418163430-00273.warc.gz"}
|
http://en.wikipedia.org/wiki/Kendall's_W
|
# Kendall's W
Kendall's W (also known as Kendall's coefficient of concordance) is a non-parametric statistic. It is a normalization of the statistic of the Friedman test, and can be used for assessing agreement among raters. Kendall's W ranges from 0 (no agreement) to 1 (complete agreement).
Suppose, for instance, that a number of people have been asked to rank a list of political concerns, from most important to least important. Kendall's W can be calculated from these data. If the test statistic W is 1, then all the survey respondents have been unanimous, and each respondent has assigned the same order to the list of concerns. If W is 0, then there is no overall trend of agreement among the respondents, and their responses may be regarded as essentially random. Intermediate values of W indicate a greater or lesser degree of unanimity among the various responses.
While tests using the standard Pearson correlation coefficient assume normally distributed values and compare two sequences of outcomes at a time, Kendall's W makes no assumptions regarding the nature of the probability distribution and can handle any number of distinct outcomes.
W is linearly related to the mean value of the Spearman's rank correlation coefficients between all pairs of the rankings over which it is calculated.
## Definition
Suppose that object i is given the rank ri,j by judge number j, where there are in total n objects and m judges. Then the total rank given to object i is
$R_i=\sum_{j=1}^m r_{i,j} ,$
and the mean value of these total ranks is
$\bar R= \frac{1}{2} m(n+1).$
The sum of squared deviations, S, is defined as
$S=\sum_{i=1}^n (R_i- \bar R)^2 ,$
and then Kendall's W is defined as[1]
$W=\frac{12 S}{m^2(n^3-n)}.$
If the test statistic W is 1, then all the judges or survey respondents have been unanimous, and each judge or respondent has assigned the same order to the list of objects or concerns. If W is 0, then there is no overall trend of agreement among the respondents, and their responses may be regarded as essentially random. Intermediate values of W indicate a greater or lesser degree of unanimity among the various judges or respondents.
Legendre[2] discusses a variant of the W statistic which accommodates ties in the rankings and also describes methods of making significance tests based on W.
## Correction for ties
When tied values occur, they are each given the average of the ranks that would have been given had no ties occurred. For example, the data set {80,76,34,80,73,80} has values of 80 tied for 4th, 5th, and 6th place; since the mean of {4,5,6} = 5, ranks would be assigned to the raw data values as follows: {5,3,1,5,2,5}.
The effect of ties is to reduce the value of W; however, this effect is small unless there are a large number of ties. To correct for ties, assign ranks to tied values as above and compute the correction factors
$T_j=\sum_{i=1}^{g_j} (t_i^3-t_i),$
where ti is the number of tied ranks in the ith group of tied ranks, (where a group is a set of values having constant (tied) rank,) and gj is the number of groups of ties in the set of ranks (ranging from 1 to n) for judge j. Thus, Tj is the correction factor required for the set of ranks for judge j, i.e. the jth set of ranks. Note that if there are no tied ranks for judge j, Tj equals 0.
With the correction for ties, the formula for W becomes
$W=\frac{12\sum_{i=1}^n (R_i^2)-3m^2n(n+1)^2}{m^2n(n^2-1)-m\sum_{j=1}^m (T_j)},$
where Ri is the sum of the ranks for object i, and $\sum_{j=1}^m (T_j)$ is the sum of the values of Tj over all m sets of ranks.[3]
## Notes
1. ^ Dodge (2003): see "concordance, coefficient of"
2. ^ Legendre (2005)
3. ^ Siegel & Castellan (1988, p. 266)
## References
• Kendall, M. G.; Babington Smith, B. (Sep 1939). "The Problem of m Rankings". The Annals of Mathematical Statistics 10 (3): 275–287. doi:10.1214/aoms/1177732186. JSTOR 2235668.
• Corder, G.W., Foreman, D.I. (2009).Nonparametric Statistics for Non-Statisticians: A Step-by-Step Approach Wiley, ISBN 978-0-470-45461-9
• Dodge, Y (2003) The Oxford Dictionary of Statistical Terms, OUP. ISBN 0-19-920613-9
• Legendre, P (2005) Species Associations: The Kendall Coefficient of Concordance Revisited. Journal of Agricultural, Biological and Environmental Statistics, 10(2), 226–245. [1]
• Siegel, Sidney; N. John Castellan, Jr. (1988). Nonparametric Statistics for the Behavioral Sciences (2nd ed.). New York: McGraw-Hill. p. 266. ISBN 0-07-057357-3.
|
2014-07-23 15:41:21
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8800498843193054, "perplexity": 1424.0986488623573}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997879037.61/warc/CC-MAIN-20140722025759-00248-ip-10-33-131-23.ec2.internal.warc.gz"}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.