Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
What is a good introductory book on quantum mechanics? I'm really interested in quantum theory and would like to learn all that I can about it. I've followed a few tutorials and read a few books but none satisfied me completely. I'm looking for introductions for beginners which do not depend heavily on linear algebra or calculus, or which provide a soft introduction for the requisite mathematics as they go along.
What are good introductory guides to QM along these lines?
| OK.
First, you need a some comfort in Linear Algebra.
Go to the MIT Open Courseware site and watch the Linear Algebra lecture (videos) by Strang. These are great.
Next, watch the "Theorectical Minimum" videos by Leonard Susskind . They represent the theoretical minimum that you need to know about quantum mechanics. (i.e. the title of the video course is theoretical minimum, but it is in fact a course on quantum mechanics. Susskind is a great teacher and the videos are great. You can access them on itunes and You Tube. Search for Susskind lectures quantum mechanic from Stanford. They are just released (a few weeks ago)
Finally, the text you want is Principles of Quantum Mechanics by Shankar. He is also a great teacher. He does have some video lectures on general physics, but he does not have a video lecture on Quantum Mechanics. Nonetheless, his book is a great book for learning. It is about $70, but if you google around (with PDF in your google search) you may get lucky.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/33215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "103",
"answer_count": 19,
"answer_id": 10
} |
Must all symmetries have consequences? Must all symmetries have consequences?
We know that transnational invariance, for example, leads to momentum conservation, etc, cf. Noether's Theorem.
Is it possible for a theory or a model to have a symmetry of some kind with no physical consequences at all for that symmetry?
| If there is a continuous symmetry of the action, it is necessary to take the quotient by the symmetry--gauge fix it--when quantizing. One way to see this is to just consider perturbation theory. The quadratic coupling in the Lagrangian will be constant along the flow of the symmetry, so it will be have degenerate derivative in the tangent space directions along that flow and hence not be invertible. Thus, we won't be able to find a propagator.
Abstractly, perturbation theory works when the classical theory one is quantizing is integrable. When showing a 2n dimensional system is integrable, one must specify n Poisson-commuting integrals of motion. The continuous symmetry gives one by Noether's theorem, and n is the most possible in 2n dimensions, so one has to come up with n integrals of motion which combine linearly into the Noether charge of the symmetry. In other words, one eventually includes the symmetry in the solution regardless of whether he meant to.
I feel like there might be weirder things that happen nonperturbatively, but I am not sure. Certainly when one wants all observables to also obey the symmetry, one can cook up a BRST operator. The existence of this operator implies that some states in the "big" Hilbert space that doesn't take account of the symmetry necessarily have zero norm. Can one do this in all cases?
It is possible on the other hand to lose symmetries when one quantizes. Usually this is the fault of the regularization scheme, but in certain cases one can prove that no regularization scheme is invariant under the symmetry. An example is the 2+1 dimensional Chern Simons theory with Wilson loops. Even though the Lagrangian is fully topologically invariant, one needs to pick a framing of loops to regularize self-intersections. Similarly there are theories with manifestly conformal Lagrangians but which require breaking scale invariance in regularization.
Discrete symmetries are of course of a much different flavor.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/33242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Electrial Conductivity of Thin Metal Films What is the best way to find specific/electric conductivity which is dependent of very thin film thickness?
| Your problem is that the conductivity depends on the film structure. Back in the dawn of time (mid 1980s!) I spent a happy three years studying reactions of silver films, and one of the techniques used was to measure the resistance of the film. As the silver reacted the film thickness went down and the resistance went up.
However once the film thickness forms below around 400$\mathring{A}$ the film is no longer solid metal. As the thickness falls you get voids, then a reticulate structure and finally isolated islands of metal. You'll also find that annealing films can reduce the resistance by a factor of two. I found that if you control the film preparation very carefully you can get reproducible resistances, but it does take some work. For example the substrate cleanliness was very important and I had to plasma etch my glass substrates before growing the films. With careful preparation you can construct a resistivity vs thickness curve for your films (but note this curve will probably differ from researchers using different preparation methods).
Actually measuring the resistance is dead easy. I just used scored a track in the film then measured the resistance as a function of track length, and one linear regression later I had the resistance. You don't need any special techniques to get measurements more accurate than the natural variability in the films.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/33291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What's up in this SPAWAR video? Here is a video presentation of infrared recordings of anomalous heating in a deuterium palladium cell: ( youtube video) (see also this presentation if you want more detail, and have time). There are two papers referenced in the link, which give more detail. You can see with your own eyes sporadic localized bursts of energy, which show up as white flashes in the camera (I would like to encourage people to read the linked papers, which include thermal photos with a different color map, and also include piezoelectric detection of bursts)
Excluding cold fusion, what could possibly be causing this? Please try to account for the nature of the bursts--- the small radius, the qualitative amount of energy released, etc.
Just to be clear, I think the answer is nothing, that it is cold fusion, but I would like to see what explanations people come up with. I will also point out that this research group has been politically shut down recently.
| Without knowing anything about the experiment nor the camera, I would suggest that what is shown in the video is a combination of shot noise and aliasing due to a poor choice of gradient mapping. Note that the gradient bar at the bottom of the frame jumps from a fairly deep red (actually darker than precedent tones) to pure white in one increment - this may result in small fluctuations in temperature appearing much more visually significant than they are.
I had a go at my own gradient maps (in Photoshop) to show off this effect - I tried to go for something similar to their gradient mapping, which for all I know is hardwired into the camera:
Something to consider. I read an article recently about injudicious gradient mapping overemphasising tiny differences in geospatial and medical datasets (and also obscuring important distinctions in other scenarios). It's a good read and I'll link it when I find it again.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/33414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Understanding Tensors I don't seem to be able to visualize tensors. I am reading The Morgan Kauffman Game Physics Engine Development and he uses tensors to represent aerodynamics but he doesn't explain them so I am not really able to visualize them. Please explain in very simple ways. I just want to understand the basics.
| Following @peter4075, Fleische's book, A Student's Guide to Vectors and Tensors is a gold mine of clear, useful, practical explanation, by someone who clearly knows their topic. Highly recommended. I got it through my local library interlibrary loan program and am loathe to surrender it.
Good luck!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/33470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
References for nuclear masses, mass deficits, decay rates and modes Where can I find the base data for computing the energy release of nuclear decays and the spectra of the decay products?
My immediate need is to find the energy release by the beta decay of Thorium to Protactinum upon receiving a neutron:
$$\mathrm{Th}02 + n \to \mathrm{Th}03 \to \mathrm{Pa} 13 + e^- + \bar{\nu}$$
The estimated amount of energy released from Beta decay is roughly 1eV.
Minus the neutrino loss, how much kinetic energy is released as heat?
| In addition to the LBL interactive table of the isotopes that John mentions in the comments, I get a lot of use1 out of their associated radiation search tool and the TUNL Nuclear Data Evaluation Project site.
In general there are several major efforts to collect and collate nuclear data that go under the heading "evaluate nuclear data" and share the Evaluated Nuclear Data File format for reporting these things.I recently discovered the IAEA's online interface to their ENDFs.
1 Somewhere along the way some of my colleagues decided I was one of the guys to ask about source construction and feasibility even though I have only ever built one source (Two years to demonstrate feasibility, design and build plus six months to to analyze the data and write a paper which has generated a grand total of four (count them!) citations in the several years since I published it. ::sigh::), and now people ask me to suggest sources that might be useful for this or that.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/33520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Bound states in QCD: Why only bound states of 2 or 3 quarks and not more? Why when people/textbooks talk about strong interaction, they talk only about bound states of 2 or 3 quarks to form baryons and mesons?
Does the strong interaction allow bound states of more than 3 quarks?
If so, how is the stability of a bound state of more than 3 quarks studied?
| As a quick explanation: all bound states are color-neutral. The intuitive reason is that the strong interaction is so strong that it would pull any color-charged particles together. (Because the strong force increases with distance, you can't get around this by spreading out the charged particles, as you can with the EM interaction.)
Since there are 3 colors, you can either achieve a color-neutral state by combining one quark of each color, which gives you a baryon, or a quark and an antiquark of the same color (e.g. blue and antiblue), which gives you a meson. Any combination of more quarks or antiquarks that works out to being color-neutral, such as the hypothetical pentaquark, can be broken down into some combination of baryons and mesons, which means that such a particle would probably naturally decay in that way, if it could even exist (which there is no evidence for).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/33578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
Problems in the modern semiconductor/electronics technology? From what I have read, the problem with modern semiconductors/electronics seems to be quantum tunnelling and heat. The root of these problems is the size of the devices. The electrons are leaking out, and currents are causing active materials to melt.
How far have we become in this regard? Can we make our devices even smaller? What is being done to maintain advancements in computing power? What is the main research, particularly in quantum mechanics and in solid state physics, being done to compute faster using less energy and space?
| The main solution is to give up Silicon altogether since it is very inefficient, compared to some of the compound semiconductors(such as GaAs). However, GaAs will eventually reach the same limitations. In any case, you can't really make a transistor from half an atom, therefore there are some physical limits as to how small you can make a transistor and we're very close to that. Therefore, current focus is on cheap and widespread use of compound semiconductors.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/33629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is it safe to study from MIT and Berkeley course series, or they contain wrong information? After surveying most of the universities introductory physics courses, I found none is using Berkeley physics books or MIT physics books as textbooks. All are using Halliday, or Serway and the like.
What is the reason behind not using these books anymore (once they were the standard textbooks, may be before halliday I suppose) ? Do they contain old/outdated/wrong information or something?
Did Berkeley and MIT develop a new set of books to substitute those books?
| It is a mistake to assume that the books that are standardized are the ones that are superior, or have fewer mistakes. Generally, they have fewer typos, because they get more exposure, but the rate of typos and errors in most undergraduate books is roughly constant across the board.
If you find that a school is using a different set of books then the standard ones, and these are not written by local authors, so it isn't nepotism, this means somebody at the school put some thought into designing the curriculum. This is usually as sign that the books are superior.
So you should use the MIT physics books, as they are probably pretty good. Although to be honest, I think that nothing in the American physics system, with the possible unique exception of the Feynman's lectures, can compare to the Soviet era Russian books, in particular the Landau and Lifschitz series, which is superb in both scope and quality.
Purcell is a decent book, although it uses CGS, which takes a bit of getting used to. I don't know the other books. It is better to just ignore the majority opinion when considering the quality of books, since majority opinion tends to bury classics and elevate mediocrity. Popular opinion decides how true something is based on how familiar it sounds, and this is not a procedure which rewards originality or vision.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/33683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to charge an object with electricity I know this is a rather basic question, but how do you charge an object? Not a battery, an object. I'm guessing it involves static electricity, but I'm not sure. Some resources I've been reading talk about charging two objects with opposing voltages, and I am trying to figure out how you do it. I think you do it with DC current, but past that, I'm not sure.
Here is the paper I am talking about: http://www.avonhistory.org/school/gravitor.htm
| Are these not examples of static electricity charging?
*
*Combing dry hair with comb, the comb gets charged.
*A sweater worn for sometime...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/33758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Special Relativity Second Postulate That the speed of light is constant for all inertial frames is the second postulate of special relativity but this does not means that nothing can travel faster than light.
*
*so is it possible the point that nothing can travel faster than light was wrong?
| From a purely theoretical point of view, the Special Relativity (SR) is based on a space-time metric $$\eta=\begin{bmatrix}+&0&0&0\\0&−&0&0\\0&0&-&0\\0&0&0&-\end{bmatrix}$$
The most general transformation to preserve metric $\eta$ is global Poincaré group which is the limit of the de sitter group with sphere radius $R\rightarrow \infty$. There is an other type of de Sitter transformation with $R \rightarrow$ finite which also leads to a special relativity theory. Basically one plays with cphoton and c.
But, keep in mind that if it is possible that SR be a finite large $R$ de Sitter transformation, it has not been experimentally confirmed, and as far as we know we can use Einstein special relativity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/34024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
mathematical explanation for UV divergences and $ \delta ^{(n)}(0) $ is there any mathematical explanation for the UV divergences ??
i have read that in the framework of Epstein-Glser theory :D these UV divergences appear from the product of distributions
anyone does the numbers (divergences) of the form $ \delta ^{(n)}(0) $ have a meaning within the renormalization framework ?? since $ \delta ^{(n)}(0)=i^{n} \int_{-\infty}^{\infty}dx x^{n} $
| I think it's best to start from the works of Wilson and Kadanoff on the renormalization group of effective field theories with an ultraviolet regulator. They find the coefficients can be classified as relevant, marginal and irrelevant. UV divergences are what you get when you run the RG in the other direction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/34161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What law of electro-magnetics explains this? I took my son to a science museum where they had a solenoid oriented vertically with a plastic cylinder passing through the solenoid. An employee dropped an aluminum ring over the top of the cylinder when there was no current going through the solenoid. Then they turned on the current going through the solenoid and they aluminum ring went flying up and off the top of the solenoid. What law of electro-magnetics causes the force on the aluminum ring?
| This sounds like a Thompson's Jumping Ring setup.
Turning on the solenoid creates an increasing magnetic field B. Maxwell's (Faraday's Law) tells us
$$\frac{-\partial B}{\partial t}=\nabla \times E$$
So the increasing B field produces an azimuthal E field "curling" around the aluminum ring and since the aluminum is a conductor this will produce a current flowing around the ring, $I$.
Say the solenoid produces a B field in the positive vertical direction, z.
The electric field and current in the aluminum ring will then be in the -$\theta$ direction of a cylindrical r,$\theta$,z coordinate system, i.e.,
$$I=I_{\theta }<0$$
Finally the magnetic field produced by the solenoid will not be precisely vertical but will flare outward radially; that is there will be a positive component of B in the r direction.
Now we can see that the $I\times B$ force arising from the interaction between the solenoid field and the current in the aluminum ring is in the vertical direction $$F=I\times B=-I_{\theta }B_r\hat{z}$$ and since $I_{\theta }<0$, the force is in the vertical direction.
If the $I\times B$ is unfamiliar, it is just the summation of forces on a charges moving in an magnetic field.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/34265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Applying velocity Verlet algorithm I want to implement a simple particules system using the velocity form of the Verlet algorithm as integrator.
Initial conditions at $t=0$ for a given particule $p$:
*
*mass: $ m $
*position: $\overrightarrow x(t=0) = \overrightarrow x_0$
*velocity: $\overrightarrow v(t=0) = \overrightarrow v_0$
*forces applied to it: $ \overrightarrow F(t=0) = \overrightarrow F_0 $
Algorithm's recipe says:
*
*Calculate: $\overrightarrow v(t+\frac{1}{2}\Delta t) = \overrightarrow v(t) + \frac{1}{2}\overrightarrow a(t)\Delta t$
*Calculate: $\overrightarrow x(t+\Delta t) = \overrightarrow x(t) + \overrightarrow v(t+\frac{1}{2}\Delta t) \Delta t$
*Derive: $\overrightarrow a(t+\Delta t)$ from the interaction potential using $\overrightarrow x(t+\Delta t)$
*Calculate: $\overrightarrow v(t+\Delta t) = \overrightarrow v(t + \frac{1}{2}\Delta t) + \frac{1}{2}\overrightarrow a(t+\Delta t)\Delta t$
Let's apply it in order to find $\overrightarrow x(t=1)$ and $\overrightarrow v(t=1)$, so with $\Delta t = 1$:
*
*$\overrightarrow v(0+\frac{1}{2}1) = \overrightarrow v(0) + \frac{1}{2}\overrightarrow a(0)1 = \overrightarrow v_0 + \frac{1}{2}\frac{\overrightarrow F_0}{m}$ [OK using Newton's second law]
*$\overrightarrow x(0+1) = \overrightarrow x(0) + \overrightarrow v(0+\frac{1}{2}1) 1 = \overrightarrow x_0 + (1.)$ [OK using previous result (1.)]
*???
*$\overrightarrow v(0+1) = \overrightarrow v(0 + \frac{1}{2}1) + \frac{1}{2}\overrightarrow a(0+1)1 = (1.) + \frac{1}{2}(3.)$ [OK using (1.) and (3.)]
--
So I'm stuck with (3.)
I have to calculate $\overrightarrow a(0+1)$ but I don't know how...
I can't apply Newton's second law here since I don't know $\overrightarrow F(t=1)$.
Algorithm says "from the interaction potential using (1.)" but I don't understand what it means...
Can you help?
Thank you
| I think you all your steps are correct, I would suggest adding units though, otherwise adding $\vec{x}$ and $\vec{v}$ can be misleading.
For step 3. You can just calculate $a = F/m$, where $F$ can either be gravitation $F=m g$, so it does not depend on $x$, or for example for a spring dependent on $x$, so $F = -k x$, here you need the next position from step 2.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/34304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Tips on teaching Dimensional Analysis? What's a good way to explain dimensional analysis to a student?
Here's a simple question which this method would be useful:
Let's say a truck is moving with a speed of 18 m/s to a new speed of 13 m/s over a distance of 48 meters. How long did it take for the truck to break to its new speed, and what was the acceleration of the truck?
Edit
@TimGoodman and @Laar, Hmm. I was under the impression that you could solve such questions by analyzing their dimensional properties. For example, based on V1 and V0 in the example above I know the person decelerated 5m/s over a distance of 48 meters. If I want the time it took (first question) I can simply look at the units and see if I can eliminate all units except s (seconds). 48m/5m/s allows the meters to cancel out leaving 9.6 seconds.
Now I have a new peice of information and I can now solve the second question "What was the acceleration of the truck?" Since I'm solving for acceleration I know my final units will be m/s^2. I know the velocity of the deceleration was 5m/s and it took him 9.6s. Dividing 5m/s/9.6s = .52 m/s^2. My final units match and so by using dimensional analysis I've solved the problem.
Maybe I have a misunderstanding of dimensional analysis. Any suggestions?
| Since I could not figure out how to write a comment on your post, I will post it as an answer, even if it's just a link:
You can find some inspiring dimensional analysis examples here:
https://particlephd.wordpress.com/2008/12/08/dimensional-analysis-for-animals/ (see also the comments)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/34357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Hawking radiation and black hole entropy Is black hole entropy, computed by means of quantum field theory on curved spacetime, the entropy of matter degrees of freedom i.e. non-gravitational dofs? What is one actually counting?
| There are a multiplicity of ways of deriving the Hawking formula for black hole entropy. Some techniques, like Bekenstein's argument, do equate the entropy of matter falling into the hole with the entropy of the hole. Some actually count gravitational microstates in various quantum gravity schema. The result $S\propto A$ seems to be quite generic in all of these approaches, however, at least to the first order of approximation in $\hbar$ (I believe the LQG result has $\mathbb{O}(\hbar^{2})$ corrections to the Hawking formula).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/34432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How much does electromagnetic radiation contribute to dark matter? EM radiation has a relativistic mass (see for instance,
Does a photon exert a gravitational pull?), and therefore exerts a gravitational pull.
Intuitively it makes sense to include EM radiation itself in the galactic mass used to calculate rotation curves, but I've never actually seen that done before...
So: if we were to sum up all the electromagnetic radiation present in a galaxy, what fraction of the dark matter would it account for?
| I found it surprisingly hard to find an authoritative statement of the density of the CMB. According to this article it's about $5 \times 10^{-34}\mathrm{g\ cm}^{-3}$, and since the critical density is somewhere in the range $10^{-30}$ to $10^{-29}\mathrm{g\ cm}^{-3}$ photons don't make a significant contribution.
Photons wouldn't be dark of course. If there were enough photons to make a significant contribution to the mass/energy of the universe we'd see them, just as we can see the CMB.
Response to comment: oops yes, I didn't read your question properly - sorry!
Anyhow, my comment that photons aren't dark matter still applies, but it's easy to make an estimate of the gravitational contribution of the EM radiation in e.g. the Solar System. The Sun converts about $4 \times 10^9$ kg of matter to energy every second. Since it weighs about $2 \times 10^{30}$ kg every second it loses about $2 \times 10^{-19}$% of it's mass every second.
If you're prepared to assume the photon density in the Solar System is dominated by the Sun's output (which seems plausible) and take the size of the Solar System to be Neptune's orbit, i.e. $1.5 \times 10^4$ light seconds then the mass/energy of photons in the Solar System is $3 \times 10^{-15}$% of the Sun's mass. So it's utterly insignificant.
The reason photons make a much lower contribution to the Solar System than to the universe as a whole is because mass is much more concentrated in the Solar System than in the universe as a whole.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/34516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
The observation of a non-SM resonance at 38 MeV Was reported here. Of course if this is real it is very exciting. It leads me to the question: given that it took so long to find this resonance at a meager 38 MeV, is it possible that all SUSY particles are hiding down in the MeV or KeV range (or lower)?
| The paper has been retracted.
Due to non ordinariness of the obtained results (standing out of The Standard Model) and at the request of co-authors the first version of the article is withdrawn for further verification and more detailed description of the experiment and data analysis. The second version is being prepared.
(from the replaced abstract on the arXiv)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/34698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Why isn't Hawking radiation frozen on the boundary, like in-falling matter? From the perspective of a far-away observer, matter falling into a black hole never crosses the boundary. Why doesn't a basic symmetry argument prove that Hawking radiation is therefore also frozen on the boundary, and therefore not observable? Wouldn't the hawking radiation have to have started its journey before the formation of the black hole? Furthermore, wouldn't the radiation be infinitely red-shifted?
| Evaporating black holes behave qualitatively differently from static ones if you allow an infinite amount of time to pass. In particular, there is no event horizon for an evaporating BH, only an apparent horizon. Furthermore, if you stack the apparent horizons of a shrinking black hole plus time, the resultant surface is two-way transversible.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/34812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Why does heterodyne laser Doppler vibrometry require a modulating frequency shift? On the wikipedia article (and other texts such as Optical Inspections of Microsystems) for laser Doppler vibrometry, it states that a modulating frequency must be added such that the detector can measure the interference signal with frequency $f_b + f_d$. Why couldn't you remove the modulating frequency $f_b$ and interfere the two beams with frequencies $f_0$ and $f_0+f_d$ to produce a signal with frequency $f_d$ at the detector? I haven't been able to find any reasoning on the subject.
My first idea was that the Doppler frequency might fall inside the laser's spectral linewidth and thus not be resolvable, but for a stabilized low-power CW laser (linewidth on the order of KHz) and a typical $f_d$ in the tens of MHz range I don't see this being an issue.
| The 'heterodyne' is based in the same principle used by the 'nonio' also called 'vernier' invented by the portuguese Pedro Nunes to increase the precision of measures of lengths. The principle is widely used also in telecomunications .
A visualization.
edit add:
If you ommit the Bragg Cell the Photo Detector, should have to work in the optical range (modulated with $f_d$ kHz). With the Bragg Cell the signal $f_d$ will be mounted on $f_b$ (Mhz),a more electronic adequate band and the noise will be reduced. This autobalanced photoreceiver seems to simplify the design.
A sensible expanation is here: Mapping optical frequencies to electronic frequencies allows sensitive measurements
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/34871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What is the significance of action? What is the physical interpretation of
$$ \int_{t_1}^{t_2} (T -V) dt $$
where, $T$ is Kinetic Energy and $V$ is potential energy.
How does it give trajectory?
| The only real physical interpretation of this quantity is in quantum mechanics. This is the phase of a contribution from a path that goes from $t_1$ to $t_2$ along the path $x(t)$. The two terms then are relatively clear, when you exponentiate this to make a phase and make time a lattice:
$$ e^{i \int_{t_1}^{t_2} (T - V)} = \prod_{t_1<t<t_2} e^{i {m(x(t+\epsilon)-x(t))^2\over 2\epsilon}}e^{-i \epsilon V(x(t))} $$
Where the product is over all t's between $t_1$ and $t_2$ in $\epsilon$ size steps.
The first term gives you the phase for free particle propagation from $x(t)$ to $x(t+\epsilon)$. The second term gives an extra phase rotation for the potential energy at the position $x(t)$. The two phases add up, and you add the phases over all the paths to get the total quantum propagation.
The classical path is then the place where the phase is stationary, so that the paths tend to add together with the same phase, rather than cancel due to interference. This is the place where a first-order variation in the path makes no change in the action.
To find this, you can shift $x(t)$ to $x(t)+\delta x(t)$, and find the leading variation
$$ e^S \int_t ( m \dot{x} {d\over dt}(\delta x) - V'(x) \delta x) dt $$
Integrating the first term by parts, you find
$$ m\ddot{x} = - V'(x) $$
or that the particle obeys Newton's law. The same derivation works in the path integral formalism to demonstrate Ehrenfest's theorem, and the Heisenberg equation of motion. This is because the path integral is invariant under shifts of the integration variable $x(t)$ by a constant amount $\delta x(t)$, even if that constant is different from time to time.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/34946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Can double entanglement preserve correlations? We have 2 EPR experiments running in parallel, with Alice having one leg of each (a1,a2) and Bob the other leg of each (b1,b2). Thus (a1,b1) are anticorrelated, as are (a2,b2). Thus also (a1,a2) are uncorrelated as are (b1,b2). Now Alice locally entangles (a1,a2), and Bob measures b1 and b2. After repeating the entire experiment (including setting up the initial entanglements) many times, does Bob see consistent correlation or anticorrelation between his measured b1 and b2?
How Alice accomplishes this final entanglement (a1,a2) is either via entanglement swapping, or via the method described in Yurke and Stoler, Phys. Rev. A46, 2229 (1992): "Bell’s-inequality experiments using independent-particle sources".
Put another way, it's clear that b1 and b2 will show no correlation if Alice does not entangle (a1,a2), since the two EPR experiments are independent. Will this situation change for Bob as a result of Alice's entanglement of (a1,a2)?
| It's clear from no signalling--- by entangling $a_1$ and $a_2$, Alice uses local operators which necessarily commute with the spin operators on $b_1$ and $b_2$, so the reduced density matrix for $b_1$ and $b_2$ stays completely random. It makes no difference what method Alice uses, unless it involves mucking around with Bob's electrons.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/34996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why is current not 0 in a regular resistor - battery circuit immediately after you closed a circuit? In regular open circuits with either a capacitor or inductor element, (when capacitor is uncharged) with a battery, when a switch is closed to complete the circuit the current is said to be 0 because current doesn't jump immediately.
But in a circuit with just resistors, as soon as a switch is closed the current isn't 0?
Example is this question from 2008 AP Physics C Exam
http://apcentral.collegeboard.com/apc/public/repository/ap08_physics_c_em_frq.pdf
http://www.collegeboard.com/prod_downloads/ap/students/physics/ap08_physics_c_e&m_sgs_rev.pdf
Go to Question 2 for details.
| This is really a footnote to user1631's answer: even in the absence of any inductance the current obviously can't change instantly because no signal can propagate faster than the speed of light. In typical circuits the increase in current propagates somewhere between $0.1c$ and $c$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/35042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Black-holes are in which state of matter? Wikipedia says,
A black hole grows by absorbing everything nearby, during its life-cycle. By absorbing other stars, objects, and by merging with other black-holes, they could form supermassive Black-holes
*
*When two black-holes come to merge, don't they rotate with an increasing angular velocity as they come closer and closer (how does it from a neutron star? I mean, who's powerful?)
And it also says,
Inside of the event horizon, all paths bring the particle closer to the center of the black hole.
*
*What happens to the objects that are absorbed into a black-hole? Which state are they really are now? They would've already been plasma during their accretion spin. Would they be on the surface (deposited), or would they still be attracted and moved towards the center? If so, then the surface of black-hole couldn't be a solid.
| I would imagine it would be a solid. Gas = atoms that are connected loosely fluid = connected more than gas but still loose solid = compact and held together I would imagine that since the gravity well and escape velocity being so high that not even light can escape, that it has traits of a very dense solid. Hence Singularity would be a very compact solid, since whenever a solid is compressed beyond a point of no return singularity is formed. There is no need for heat to act as a catalyst then plasma is omitted, it wouldn't be gaseous or a liquid. Unless you can compress gas/plasma without it becoming a liquid. So maybe a super compressed liquid can be the core. Now I think that, my theory can be wrong, so consider this empirical observation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/35150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
The Hermiticity of the Laplacian (and other operators) Is the Laplacian operator, $\nabla^{2}$, a Hermitian operator?
Alternatively: is the matrix representation of the Laplacian Hermitian?
i.e.
$$\langle \nabla^{2} x | y \rangle = \langle x | \nabla^{2} y \rangle$$
I believe that $\nabla^{2}$ is Hermitian (if it was not, then the Hamiltonian in the time-independent Schroedinger equation would not be Hermitian), but I do not know how one would demonstrate that this is the case.
More broadly, how would one determine whether a general operator is Hermitian?
One could calculate every element in a matrix representation of the operator to see whether the matrix is equal to it's conjugate transpose, but this would neither efficient or general.
It is my understanding that Hermiticity is a property that does not depend on the matrix representation of the operator. I feel that there should be a general way to test the Hermiticity of an operator without evaluating matrix elements in a particular matrix representation.
Apologies if this question is poorly posed. I am not sure if I need to be more specific with the definitions of "Hermitian" and "Laplacian". Feel free to request clarification.
| In general, one needs to write down the integrals for $\langle\phi|\Delta\psi\rangle$ and $\langle\Delta\phi|\psi\rangle$ and transform them into each other using integration by parts.
Hermiticity does not depend on the basis (matrix representation) used.
But it depends on the boundary conditions imposed, as one needs to ensure that integration by parts does not generate nonhermitian boundary terms.
With the boundary conditions usually used in quantum mechanics (square integrability in $R^n$), it is a self-adjoint operator, and in particular Hermitian.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/35210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Where can I find hamiltonians + lagrangians? Where would you say I can start learning about Hamiltonians, Lagrangians ... Jacobians? and the like?
I was trying to read Ibach and Luth - Solid State Physics, and suddenly
(suddenly a Hamiltonian pops up. and then a wave equation and then $H_{aa}\ and\ H_{ab}$?
| Well, for solid state physics, you should know quantum mechanics first, which will teach you about Hamiltonians. So pick up any standard undergrad quantum text. Lagrangians don't come into undergrad QM too much, unless you happen to get an introduction to path integrals, so if you want to learn about those I'd suggest a classical mechanics text (again, any standard undergrad book will do).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/35274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why does observation collapse the wave function? In one of the first lectures on QM we are always taught about Young's experiment and how particles behave either as waves or as particles depending on whether or not they are being observed. I want to know what about observation causes this change?
| Towards a better picture of the duality
On Young's double-slit experiment the wave-particle duality (one by one photon) is more a problem of "picture of the model" than a philosophic one: see Y. Couder interpretation, by your self (!),
Youtube Couder experiments
The quantum particle HAVE a location, there are only a limitation in choose a good pictoric model when you is constrained by "wave or particle" picture options: Couder demonstrates that a good picture,
of an "intermediary wave/particle object" model, exists!
Imagine a "localizable object" that haven't a well-defined boundary, but have a well-defined distance-limit (~lambda) to interact with obstacles (other objects).
On this little video you see the objects one by one, changing (or not) the rectilinear trajectory by the "oscilating interaction" with the (double-slit) obstacle, not by the after-obstacle-screen as "observer".
There are an article online about the experiment.
PS: of course, if a observation constitutes measurement before the screen, it will be interfere on result, changing the interference pattern at the screen.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/35328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 2
} |
Why does the aligning of magnetic dipoles in a material cause its heat capacity to decrease? This is with regards to adiabatic magnetisation.
| If the magnetic dipoles in a material are ordered, the material has a lower entropy because there are many fewer ways how the spins may be oriented if most of them (or all of them!) are required to be aligned.
Such an alignment also reduces the heat capacity because before the dipoles got aligned, the orientation (direction) of each dipole was a degree of freedom that was storing something like $O(kT)$ of energy where $k$ is Boltzmann's constant and $T$ is the temperature in kelvins.
However, when the dipoles are kept aligned, the direction is no longer variable so the heat capacity from each dipole decreases by $O(k)$ or so, or $O(R)$ if expressed per mole.
It's a similar difference as the heat capacity of monoatomic vs diatomic gas, which carry $3kT/2$ vs $5kT/2$ per molecule, respectively. ($3/2$ is from 3 linear momenta and the extra $2/2$ is from the longitude and latitude remembering the rotational motion and/or direction of the diatomic molecule.) Note that the diatomic gas has a greater energy per molecule which scales with $T$, and therefore also steeper dependence of the energy on the temperature (the capacity is $5R/2$) because the energy may also be stored in the random rotations of the diatomic molecule that don't exist for the monoatomic molecule. The random chaotic thermal rotation of the dipoles is analogous to the random rotation of the diatomic gas molecules and it becomes banned or indistinguishable for monoatomic gas as well as the magnetic material with aligned spins which is why the heat capacity decreases analogously to the decrease from $5R/2$ to $3R/2$ in the transition from a diatomic molecule to a monoatomic one.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/35391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Gravitational inverse-square law I was looking at the gravitational inverse-square law:
$$
F_G = G \frac{Mm}{r^2}
$$
This law comes from some experimental data? Why it is an exact inverse-square law? Could it be
$$
F_G = G \frac{Mm}{r^{2.00000000000000001}}
$$
or there is a mathematical method to find exactly this law?
| To answer your question of whether or not there is experimental data, here is one of what I am sure are many papers regarding the classical definition of the gravitational force: http://www.physics.uci.edu/~glab/papers/HoskinsPaper.pdf
Nothing in physics is ever exact, but considered an approximation to the reality of the event under consideration. To test whether or not this approximation is valid, experimental data must be obtained and the standard deviation between the expected and measured results must be analysed. If a measured and expected quantity agree to within some deviation of each other over numerous iterations, the approximation is considered "good enough" to be true.
The gravitational inverse square law is a classical approximation, not taking into account relativistic effects. Even so, the standard deviation of the gravitational inverse square law is a close enough approximation that it will in most cases suffice as an accurate description of the system under study.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/35457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Numerical renormalization is there a numerical algorithm (Numerical methods) to get 'renormalization' ?? i mean you have the action $ S[ \phi] = \int d^{4}x L (\phi , \partial _{\mu} \phi ) $ for a given theory and you want to get it renormalized without using perturbation theory .. can this be made with numerical methods ??
or on the other side let us suppose we have a divergent integral in perturbative renormalization and you want to evaluate it with numerical methods to extract some finite part of it.
| The answer to this question is a resounding YES. Lattice field theorists do their computations entirely numerically. As a result, they must resort to numerical (and hence, nonperturbative) renormalization (by extrapolating down the lattice spacing).
They would not deal with counterterms, but rather deal directly with the various $Z$ factors appearing in the renormalized Lagrangian.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/35521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Do we live in a world with 4 or more dimension? A NOVA show have told the audience that we are live in 3 dimensional world, the world we lived in is compose by 3 element: the energy, matter, space. By the time Einstein have invented the 4-dimensional model including the time element, why no one claim that we actually live in 4-dimensional world
| Victor,
This is a good question. Probably one of the greatest tragedies surrounding modern science in popular literature is the portrayal of dimensions and our understanding of them.
Many popular shows are inconsistent in when they talk about dimensions, in some cases they refer to the 3 spatial and 1 time dimension that we commonly observe as separate things, and in some cases they talk about the 4-dimensional spacetime that is used in modern physics on a routine basis without any real discussion about what they mean.
Spacetime is a combination of the 3 space dimensions and 1 time dimension, e.g. 3+1 dimensions which equals 4 dimensions.
In the early days of relativistic mechanics, in order to represent the 3 spatial dimensions and 1 time dimension as one entity, they would give the time dimension an imaginary value. So the 4 dimensional spacetime would be represented as:
$$x + y + z + it$$
This covenient, because when one takes the inner product (sometimes called the dot product) of the representation of space time one gets:
$$x^2 + y^2 + z^2 - t^2$$
This sort of representation is convenient because one can begin to understand the relationship between space and time in terms of the Pythagorean theorem.
This is the simplest example of how one can begin to relate the nature of spacetime to geometry. The broader theory of how geometry of spacetime changes as you move in spacetime is Einstein's Theory of General Relativity.
General Relativity assumes that there are 3 dimensions of space and 1 dimension of time that make up the 4-dimensional spacetime that programs sometimes reference.
There are other theories that propose there are more than 3 space dimensions of space and 1 dimension of time. However, these are very speculative still.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/35644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Increased mass from signals traveling close to the speed of light As you travel close to the speed of light, it is to my understanding you gain mass. Does this also apply when the brain sends electrical signals to the muscles? Do the signals (that are traveling at the speed of light) cause the body to weigh more?
| The signals travelling from your brain along your nerves travel at much much less than the speed of light. The maximum speed is about 100 m/sec.
In any case what travels down the nerve is a change in the sodium and potassium ion concentration not some mass, so there isn't anything to gain relativistic mass.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/35762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
A problem of missing energy when charging a second capacitor A capacitor is charged. It is then connected to an identical uncharged capacitor using superconducting wires. Each capacitor has 1/2 the charge as the original, so 1/4 the energy - so we only have 1/2 the energy we started with. What happened?
my first thoughts were that the difference in energy is due to heat produced in the wire. It may be heat, or it may be that this is needed to keep equilibrium.
| Indeed, there will be at least some losses in the superconducting wires: first, as far as I know, losses in superconductors only vanish for zero frequency, second, initial high current can exceed the critical current of the superconductor.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/35843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
the difference between the operators between $\delta$ and $d$ In classical mechanics, when talking about the principle of virtual work, what is difference between $\delta r$ and $dr$? e.g. $W=\int \overrightarrow{F} \cdot \delta \overrightarrow{r} $ and $W=\int \overrightarrow{F} \cdot d \overrightarrow{r} $ .
Why can one exchange the place of $d\delta$ and $d$ in derivative calculation? e.g. $d\delta r=\delta d r$?
| They are not different, they are the same, but "dr" is ossified notation, meaning the integration differential, and physicists often think about infinitesimal increments, so they use different letters to indicate smallness. You sort it out from context.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/35908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Complete set of observables in classical mechanics I'm reading "Symplectic geometry and geometric quantization" by Matthias Blau and he introduces a complete set of observables for the classical case:
The functions $q^k$ and $p_l$ form a complete set of observables in the sense that any function which Poisson commutes (has vanishing Poisson brackets) with all of them is a constant.
I wonder why is it so? That is why do we call it a complete set of observables? As I understand it means the functions satisfying the condition above form coordinates on a symplectic manifold, but I don't see how.
| Blau's definition is a classical analog of the Schur's lemma. The
reasoning behind this definition is the requirement that under a faithful
quantization map which carries functions on the phase space to operators
on some Hilbert space, the representation of the algebra of the quantum
observables is irreducible. The irreducibility requirement has a physical
origin as irreducible representations correspond to "single" quantum
systems and if a representation is reducible, then it can be reduced to
independent subsystems. Of course due to the Groenwold-Van Hove theorem,
in general, no such quantization map exists. We usually give
up faithfulness for the sake of irreducibility.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/36017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Sinusoidal Wave Displacement Function I am learning about waves (intro course) and as I was studying Wave Functions, I got a little confused.
The book claims that the wave function of a sinusoidal wave moving in the $+x$ direction is $y(x,t) = A\cos(kx - wt)$.
However, I see a drawing of the wave and they always seem to be $\cos$ graphs. Are sinusoidal waves always cosine graphs? Or can they be sine? If I ever see a sine wave, then does that mean that this is merely a pulse/wave travelling once and not oscillating in a periodic motion?
Sorry for the basic beginners question.
| Since
$$\cos\left(x-\frac{\pi}{2}\right)=\sin x,$$
using $\cos$ or $\sin$ does not matter, it depends on the choice of initial conditions.
In addition, in general, there will be a initial phase $\phi$, so sinusoidal wave is written like
$$ y(x,t)=A \cos(kx-\omega t+\phi).$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/36114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
How does the Sun's magnetic field continue to exist at such high temperatures? The temperature at the surface of the Sun is apparently well above 5000 C; I'm assuming the layers beneath the surface may be even hotter.
At school, we learned that heating a metal beyond a certain temperature, specific to each metal, would demagnetize the magnet.
How does the Sun's magnetic field continue to exist at such high temperatures?
| If the sun's internal plasma was at rest (the sun would have to stop rotating and other factors would need to occur), then I believe the magnetic field would dissipate and dissolve, essentially being 'demagnetized'.
However, because the star is rotating, and different layers of it at varying rates, the churning of the plasma (which is charged) generates the magnetic field as it moves past other charged plasma.
From Wikipedia:
A stellar magnetic field is a magnetic field generated by the motion of conductive plasma inside a star. This motion is created through convection, which is a form of energy transport involving the physical movement of material. A localized magnetic field exerts a force on the plasma, effectively increasing the pressure without a comparable gain in density. As a result, the magnetized region rises relative to the remainder of the plasma, until it reaches the star's photosphere. This creates starspots on the surface, and the related phenomenon of coronal loops
Also, a central part you're missing in the relationship to metals on earth is that you're not reaching a plasma state with the metal. Above a certain temperature, a solid metal will reconfigure itself into random alignment (no aligned charges), but if you go too far and make it a plasma, it's essentially all charged nuclei. This just isn't discussed because a majority of people don't tend to make metals into plasmas on a very routine basis.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/36182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Reason for the Gaussian wave packet spreading I have recently read how the Gaussian wave packet spreads while propagating.
see:
http://en.wikipedia.org/wiki/Wave_packet#Gaussian_wavepackets_in_quantum_mechanics
Though I understand the mathematics I don't understand the physical explanation behind it. Can you please explain?
| The explanation is really very simple to understand intuitively, and very beautiful.
Imagine that a particle an uncertainity in its velocity $v$ of $\delta v$. Suppose at $t=0$ we have $x=x_{0}$. After $t=T$, the location of the particle will be given by the range $(x_{0}+Tv-T\delta v,x_{0}+Tv+T\delta v)$, because we dont know the exact velocity the particle started with. It evident that a probability of finding a particle has changed from being localised in the beginning to being diffused after some time: this is wave packet spreading.
Note that the range (wave packet size) increases with time monotonically, this means that even if we started with a diffused particle density for general case, it will just become much more diffused by extension of the arguement.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/36430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
What are distinguishable and indistinguishable particles in statistical mechanics? What are distinguishable and indistinguishable particles in statistical mechanics? While learning different distributions in statistical mechanics I came across this doubt; Maxwell-Boltzmann distribution is used for solving distinguishable particle and Fermi-Dirac, Bose-Einstein for indistinguishable particles. What is the significance of these two terms in these distributions?
| On the deepest level, particles are indistinguishable if and only if they have the same quantum numbers (mass, spin, and charges).
However, in statistical mechanics one often studies effective theories where there are additional means of distinguishing particles. Two important examples:
*
*In modeling molecular fluids, two atoms on the same molecule are distinguishable if and only if there is no molecular symmetry interchanging the two atoms, and two atoms in different molecules are distinguishable if and only if there is no congruent matching of the two molecules such that the two atoms correspond to each other.
*In modeling the solid state, one typically assumes that the atoms are confined to lattice sites, and that each site is occupied at most once.. In this case, the position in the lattice is a distinguishable label, which makes all atoms distinguishable.
The computational relevance of the distinction is that permutations of (in)distinguishable particles (don't) count towards the weighting factor.
For an expanded discussion see my article at PhysicsForums.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/37556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 1
} |
Why is the partition function called ''partition function''? The partition function plays a central role in statistical mechanics.
But why is it called ''partition function''?
| First, recall what a partition is. A partition of a set $X$ is a way to write $X$ as a disjoint union of subsets: $X=\coprod_i X_i$, $X_i\cap X_j=\emptyset$ for $i\neq j$. When the elements of the set $X$ are considered undistinguishable, what matters are the cardinals of the set only, and we have a partition of an integer number, $n=n_1+\ldots+n_k$. For numbers, the name "partition function" denotes the number of ways in which the number $n$ can be written like this. It is different than the "partition function" in statistical mechanics, but both refer to partitions.
In statistical mechanics, a partition describes how $n$ particles are distributed among $k$ energy levels. Probably the "partition function" is named so (indeed a bit uninspired), because it is a function associated to the way particles are partitioned among energy levels. An interesting explanation of this can be found in "The Partition Function: If That’s What It Is Why Don’t They Say So!". But I don't know a historical account of this.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/37617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
"answer_count": 1,
"answer_id": 0
} |
What is a virtual photon pair? When describing a black hole evaporation in the hawking black body radiation it is usually said that is due to a virtual photon pair, is it this what happens? And what is virtual photon pair, does the photon has anti particle or it is its own ?
I am not looking for a deep theory, just the general picture.
| As Anna says, while it's commonly stated that Hawking radiation is due to one member of a particle anti-particle pair falling through the event horizon, this is nothing more than a metaphor and it's not actually what happens.
I actually answered this question is some detail in Black holes and positive/negative-energy particles, but I suspect this answer may be overly complex. The phenomenon is related to the curvature of spacetime. Have a look at my answer and if you want to explore any of the points I discuss post another question and I'll help if I can.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/37697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Misaligned Mirror on Michelson Inferometer If one of the outer mirrors on a Michelson interferometer was to be misaligned by a small angle of theta, what would be the shape of the interference pattern in the detector plane? What would happen to this pattern as the other mirror moves?
| Misaligning one of the end mirrors will produce a set of vertical or horizontal fringes at the detector plane (depending on the misalignment of the mirror).
The number of fringes is proportional to the misalignment angle of the mirror and inversely proportional to the wavelength of the light. When first setting up the alignment of the interferometer, this effect can be used by adjusting the alignment until the number of fringes is reduced.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/37801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How do I figure out the probability of finding a particle between two barriers?
Given a delta function $\alpha\delta(x+a)$ and an infinite energy potential barrier at $[0,\infty)$, calculate the scattered state, calculate the probability of reflection as a function of $\alpha$, momentum of the packet and energy. Also calculate the probability of finding the particle between the two barriers.
I start by setting up the standard equations for the wave function:
$$\begin{align}\psi_I &= Ae^{ikx}+Be^{-ikx} &&\text{when } x<-a, \\
\psi_{II} &= Ce^{ikx}+De^{-ikx} &&\text{when } -a<x<0\end{align}$$
The requirement for continuity at $x=-a$ means
$$Ae^{-ika}+Be^{ika}=Ce^{-ika}+De^{ika}$$
Then the requirement for specific discontinuity of the derivative at $x=-a$ gives
$$ik(-Ce^{-ika}+De^{ika}+Ae^{-ika}-Be^{ika}) = -\frac{2m\alpha}{\hbar^2}(Ae^{-ika}+Be^{ika})$$
At this point I set $A = 1$ (for a single wave packet) and set $D=0$ to calculate reflection and transmission probabilities. After a great deal of algebra I arrive at
$$\begin{align}B &= \frac{\gamma e^{-ika}}{-\gamma e^{ika} - 2ike^{ika}} & C &= \frac{2e^{-ika}}{\gamma e^{-ika} - 2ike^{-ika}}\end{align}$$
(where $\gamma = -\frac{2m\alpha}{\hbar^2}$) and so reflection prob. $R=\frac{\gamma^2}{\gamma^2+4}$ and transmission prob. $T=\frac{4}{\gamma^2+4}$.
Here's where I run into the trouble of figuring out the probability of finding the particle between the 2 barriers. Since the barrier at $0$ is infinite the only leak could be over the delta function barrier at $-a$. Would I want to use the previous conditions but this time set $A=1$ and $C=D$ due to the total reflection of the barrier at $0$ and then calculate $D^*D$?
| Hints to the question(v5):
*
*OP correctly imposes two conditions because of the delta function potential at $x=-a$, but OP should also impose the boundary condition $\psi(x\!=\!0)=0$ because of the infinite potential barrier at $x\geq 0$.
*There is zero probability of transmission because of the infinite potential barrier at $x\geq 0$. (Recall that transmission would imply that the particle could be found at $x\to \infty$, which is impossible.)
*Hence there is a 100 percent probability of reflection, cf. the unitarity of the $S$-matrix. See also this Phys.SE answer.
*As OP writes, away from the two obstacles, one has simply a free solution to the time-independent Schrödinger equation, namely a linear combination of the two oscillatory exponentials $e^{\pm ikx}$. This solution is non-normalizable over a non-compact interval $x\in ]-\infty,0]$.
*To make the wave function normalizable, let us truncate space for $x< -K$, where $K>0$ is a very large constant. So now $x\in [-K,0]$. One may then define and calculate the probability $P(-a \leq x\leq 0)$ of finding the particle between the two barriers via the usual probabilistic interpretation of the square of the wave function.
*If we now let the truncation parameter $K\to \infty$, then we can deduce without calculation that this probability $P(-a \leq x\leq 0)\to 0$ goes to zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/37857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Total momentum of the Universe What is the total momentum of the whole Universe in reference to the point in space where the Big Bang took place?
According to my reasoning (and a bit elementary knowledge) it should be exactly equal to 0 since the 'explosion' and scattering of the matter throught the space would not change the total momentum in any way.
| There is no point in space where the big bang took place. It happened everywhere, simultaneously. Centered on Earth, since everything is moving away from us with a uniform velocity (or is stationary with respect to the CMB, if you prefer), the net momentum of the universe is approximately zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/37957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is Heisenberg's matrix mechanics or Schrödinger's wave mechanics more preferred? Which quantum mechanics formulation is popular: Schrödinger's wave mechanics or Heisenberg's matrix mechanics? I find this extremely confusing: Some post-quantum mechanics textbooks seem to prefer wave mechanics version, while quantum mechanics textbooks themselves seem to prefer matrix mechanics more (as most formulations are given in matrix mechanics formulation.)
So, which one is more preferred?
Add: also, how is generalized matrix mechanics different from matrix mechanics?
| If you do a "physics for non-physicists" course (in my case as part of a Physical Chemistry degree) you'll almost certainly be taught the Shrodinger equation because it requires less mathematical sophistication to use. For most chemical applications it's also a lot easier to use.
I'm not sure if anyone uses matrix mechanics these days. I get the impression it's regarded as a worthy but slightly clumsy step along the way to quantum field theory, but that may simply mean I mix with the wrong type of scientist :-)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/38070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
2D Ghost CFT and two-point functions For some reason I am suddenly confused over something which should be quit elementary.
In two-dimensional CFT's the two-point functions of quasi-primary fields are fixed by global $SL(2,\mathbb C)/\mathbb Z_2$ invariance to have the form
$$\langle \phi_i(z)\phi_j(w)\rangle = \frac{d_{ij}}{(z-w)^{2h_i}}\delta_{h_i,h_j}.$$
So a necessary requirement for a non-vanishing two-point function is $h_i = h_j$. Now consider the Ghost System which contains the two primary fields $b(z)$ and $c(z)$ with the OPE's
$$T(z)b(w)\sim \frac{\lambda}{(z-w)^2}b(w) + \frac 1{z-w}\partial b(w),$$
$$T(z)c(w)\sim \frac{1-\lambda}{(z-w)^2}c(w) + \frac 1{z-w}\partial c(w).$$
These primary fields clearly don't have the same conformal weight for generic $\lambda$, $h_b\neq h_c$. However their two-point function is
$$\langle c(z)b(w)\rangle = \frac 1{z-w}.$$
Why isn't this forced to be zero? Am I missing something very trivial, or are there any subtleties here?
| 1) Everything OP writes(v1) above his last equation is correct. The $bc$ OPE reads
$$ {\cal R}c(z)b(w) ~\sim~ \frac 1{z-w} ,$$
where ${\cal R}$ denotes radial ordering.
2) To calculate the two-point function
$$\langle c(z)b(w)\rangle $$
(which as OP writes must vanish if the conformal dimensions for $b$ and $c$ are different) is more subtle due to the presence of the ghost number anomaly, i.e. the vacuum should be prepared with certain modes of the $bc$ system, see e.g. Polchinski, String Theory, Vol. 1, Sections 2.5-2.7.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/38215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Why is the solar noon time different every day? If you check the local time for solar noon is different every day. Why is it so? Is it because Earth doesn't make a complete rotation in exactly 24 hours?
The following is an example of the solar noon differences (also sunrise and sunset), computed by the Python Astral module for the city of Guayaquil
| Solar noon becomes earlier and then back again more than once a year; it moves 18 minutes earlier from about Feb. 10th until May, then 12 minutes later by July , then back 23 minutes earlier by late October/early November, then forwards again by half an hour by Feb tenth. So the biggest change in Solar noon occurs between early November and mid February by 31 minutes. I hope that has answered questions, but it is also a question, because I don't know why direction of solar noon time changes 4 times a year; two per direction, and also why the largest difference is in between early November and mid February. If anybody can answer that I'd appreciate it. Also, what are the 4 days user 11266 is referring to the end of his answer?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/38270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 4
} |
Magnitude of New Comet C/2012 S1 (ISON) A new comet (magnitude 18.8) has been discovered beyond the orbit of Jupiter.
Comet ISON will get within 0.012 AU of the Sun by the end of November 2013 and ~0.4 AU from of Earth early in January 2014. It may reach very welcome negative magnitudes at the end of November 2013. Forecasts of comet magnitudes have been disappointing in the past. It's clear that comet magnitudes depend on the quantity of dust and ices that are ablated by the Sun's energy, and therefore nucleus size and distance from the Sun and Earth.
My question is, to what extent is a comet's magnitude dependent on the ratio of nucleus dust to ice and on the type of ices?
| OR maybe comets aren't "dirty snowballs", rather highly charged asteroids projected into space, as a result of an impact of charged plasma to an interstellar planetary body. The tail composed of plasma and not dirt and ice.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/38342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to make something charged using electricity? If I had a piece of metal and i wanted it to be negatively charged.
How can I do that?
| There are many ways to charge a piece of metal, but they tend to be variations on the principle used by a Van de Graaff generator. When you run two materials together you will usually transfer electrons from one to the other. Which way the electrons go depends on where the two surfaces are in the triboelectric series. Rubbing materials directly on the metal tends not to work as the charge simply flows off again. Typically you charge an intermediary object then touch that to the metal to charge it.
Alternatively find a radioactive beta emitter and point it at the metal then wait a long time.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/38397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Newton 2nd Law: Does vertical force (mass) affect the horizontal acceleration? I learnt before that if 2 forces are perpendicular to each other, they should not affect each other. However in a recent experiment setup (asked in another question):
I believe the theoratical equation by newton 2nd law is
$$\begin{aligned}
F_{horizontal} &= F_{vertical} \\
m_{cart}a_{cart} &= mg \\
a_{cart} &= \frac{mg}{m_{cart}} \\
\end{aligned}$$
Am I right so far? If so, this seem to imply that $m_{cart}$ (vertical force) is somehow affecting acceleration (horizontal)? Why is that?
| What the diagram doesn't show is the force on the pulley:
It's the vector sum of this force and the force due to the weight that gives a horizontal force on the cart.
The tension in the string must be constant, because if it varied along the string the string would strtech or contract until the tension was constant, so $F$ is the tension in the string times $\sqrt{2}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/38448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What's driving the bucket up? Just saw this cool video from Plymouth University, which I actually found through Matthen's blog.
They fill a plastic bottle with liquid nitrogen, screw the cap on, drop it in a bucket full of warm water, cover it with ping pong balls, and when the heated, expanding gas bursts the bottle, the balls go jumping all over the place...
If you scroll the video to 3:50, you can see that the bucket jumps in the air, seemingly at the same time as the ping pong balls. Of course the force of the explosion is not pushing it up, but down. So what exactly is making it rise over 1 m off the ground?
I can only think of two possible explanations:
*
*Elastic recoil from the material of the bucket. This seems pretty unlikely to me, as in my experience plastic buckets don't bounce much.
*Pressure difference: the explosion drives everything out of the bucket, leaving a partial vacuum behind, so the atmospheric pressure outside the bucket pushes it up before air rushes back in to equalize the pressure. This seems more likely, but I would had thought that there would have been a longer delay.
Am I leaving any other possible explanation? Anybody willing to tackle a back of the envelope calculation of the height or delay for either of these?
| I'm with Dmitry on this one. Note that before the bottle explodes there are some ping pong balls lying on the floor next to the bin, and when the bottle explodes those ping pong balls fly up (though not as much as the bin).
The explosion deforms the floor below the bin, and when the floor rebounds it flings the bin into the air. The spilled ping pong balls alongside the bin also get thrown up, but because they are a little way away from the bin the deformation of the floor is less and they get thrown up less.
I wish we had done experiments like that at school :-)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/38494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Determining the center of mass of a cone I'm having some trouble with a simple classical mechanics problem, where I need to calculate the center of mass of a cone whose base radius is $a$ and height $h$..!
I know the required equation. But, I think that I may be making a mistake either with my integral bounds or that $r$ at the last..!
$$z_{cm} = \frac{1}{M}\int_0^h \int_0^{2\pi} \int_0^{a(1-z/h)} r \cdot r \:dr d\phi dz$$
'Cause, once I work this out, I obtain $a \over 2$ instead of $h \over 4$...!
Could someone help me?
| The volume element is $ (dr)*(rd \phi)*(dz) $. Hence, the extra r in your integrand should be eliminated.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/38624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
What does $\psi_j(r_i)$ mean? I have a mean-field Hamiltonian for N electrons. The mean-field potential felt by electron $i$ at position ${\bf r}_i$ is given by
$V^{(i)}_{int}({\bf r}_i)=\sum_{j\ne i}|\psi_j({\bf r}_i)|^2$
I can understand why this is the case. However, I need to clear up my understanding of the term $\psi_j({\bf r}_i)$. Is it simply the wavefunction of the $j^{th}$ electron at position ${\bf r}_i$?
| The key idea in the mean field approach is taking into account the contribution, to the potential(the cause; see below) at each point ${\bf r}$, of the complete electronic configuration. By electronic configuration, we mean the probability density distribution of each of the one electron states $|\psi_j({\bf r}^')|^2 $. So the cause of the $j^{th}$ electron(at ${\bf r}^'$) "on" `the' electron at ${\bf r}$ is given by
$u({\bf r}) = -e \frac{|\psi_j({\bf r}^')|^2}{|{\bf r} - {\bf r}^'|}$.
Hence the total cause would be integrating over all the possible positions ${\bf r}^'$ of the $j^{th}$ electron and summing over all the states $j$ including 'the' electron which feels the effect. This total cause is used then to deduce the wavefunction of `the' electron using Schr\:odinger equation (the effect).
This may be confusing as we are including the wavefunction of the interested electron (the effect) in calculating what could be the possible cause "on" it to have that effect. This is the crux of the mean field approach where effect feedbacks the cause until we reach at a 'self-consistent' description of the cause-effect process! (self-consistent description is the physics lexicon used for this process.)
Some caveats : While integrating over all possible ${\bf r}^'$ we encounter a singularity when ${\bf r}^' = {\bf r}$. This is dealt with what is called "self-interaction" term and correction. I dont have good understanding of this.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/38709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What does the term liquid mean in condensed matter physics? In condensed matter physics, people always say quantum liquid or spin liquid. What does liquid mean?
| gas = particles are so little packed that they can easily move.
liquid = particles are fairly dense packed but can move over long distances.
solid = particles are so densely packed that they are confined to small vibrations araound an equilibrium position (site), and larger moves (site changes) are quite rare. In many cases, the sites form a periodic lattice; then we have a crystal.
Between these phases are so-called phase transitions, that provide (in the thermodynamic limit) a clear dividing line.
The prefix ''quantum'' just means that the motion of the particles cannot be described by classical mechanics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/38800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Is a volumetric rate frame-invariant in general relativity? Imagine that I have a radioactive material with a long half life. The atoms in this material decay at a certain rate $R$. The rate is the decay constant times the number density $R = \lambda N $. It has dimensionality:
$$ \left( \frac{ \text{decays} }{m^3 s} \right) $$
Imagine that the material is on board a spaceship traveling at some significant fraction of the speed of light. Length is contracted and time is dilated.
$$ \Delta t' = \Delta t \gamma = \frac{\Delta t}{\sqrt{1-v^2/c^2}} $$
$$ L'=\frac{L}{\gamma}=L\sqrt{1-v^{2}/c^{2}} $$
The volumetric decay rate according to the lab reference frame is found by correcting for both the increased density (due to length contraction) and the decreased decay constant (due to time dilation).
$$ N' = L' A = \frac{N}{\gamma} $$
$$ R' = \frac{N'}{N} \frac{\Delta t'}{\Delta t} R = R$$
It's the same volumetric decay rate! Amusingly, the $Q$ value of the decay would be greater, but that's aside the point.
Question:
What if the material was put in a large gravity well? If you use the coordinates from outside the gravity well, would you obtain this same result?
| I don't know whether it applies to all physically possible metrics, but the volumetric decay rate you define does stay constant in a Schwarzschild metric. Well, it does if the box is small compared to the curvature i.e. the time dilation etc is constant thoughout the box. I would need to think more about what happens if the box is very large.
Anyhow the Scharzschild metric is:
$$ ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1}dr^2 + r^2 d\Omega^2 $$
The time dilation is easy, as we see time moving more slowly for the box by a factor of $(1 - 2M/r)^{1/2}$. I had to think a bit about length contraction, but I think this is a sensible way to define it:
The Schwarzschild radial co-ordinate $r$ is defined as the radius of a circle with circumference $2\pi r$. So we can take a shell with circumference $2\pi r$ and another with circumference $2\pi (r + dr)$ and that defines our ruler of length $dr$. But the observer standing alongside the box would measure a different radial distance between the shells. Specifically they would measure the distance to be $dr/(1 - 2M/r)^{1/2}$. This distance is bigger than the observer at infinity measures, and therefore this means the shell observer's ruler is shorter than ours by a factor of $(1 - 2M/r)^{1/2}$. This factor is exactly the same as the time dilation factor, which means the time dilation and length contraction balance out, and the volumetric decay rate stays the same.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/38986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Resolution of twin paradox using Lorentz velocity addition In the following lecture, starting at minute 29:00 and going further, the professor resolves the Twin Paradox using Lorentz velocity addition. I have a question about this:
Isn't the figure given below (taken from a slide in the lecture, at time 37:50) referring to the case where there is an independent object moving relative to the inertial frames S and S'? In the example of the twins that he discusses, there is no object independent of the frames of reference. We simply have the twins, each attached to her respective reference frame. Why, then, does he use the formula for Lorentz velocity addition? What object is he referring to?
Link for the lecture: http://www.youtube.com/watch?v=4A5EQaXhCTw
| Briefly skimming the video, it seems like he's trying to resolve the paradox by having the twins meet up again somewhere else than their initial position (see 35:55). Basically, the traveling twin keeps going, and then the stationary twin tries to overtake her, starting at a later time. The velocity addition rule is used to find how fast the lingering twin needs to go to catch up so they both meet at a certain point.
The twin paradox is what happens when people don't realize there are three reference frames in play - the stationary one, the outgoing one, and the ingoing one. By treating the latter two as the same, one gets nonsensical results. The usual resolution comes from carefully drawing a couple spacetime diagrams and making sure whatever quantity you're considering is well defined.
This video tries to eliminate the problem in a way I've never seen before. It may be novel, but I've also never seen the twin paradox require so much cumbersome arithmetic. Also, there are still three different frames in this case - the stationary one, the outgoing one, and the even faster outgoing one. I'm doubtful whatever resolution comes from this will result in a better intuition for SR.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/39117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Schrödinger and thermodynamics I heard that Schrödinger pointed out that (classical/statistical) thermodynamics is impaired by logical inconsistencies and conceptual ambiguities.
I am not sure why he said this and what he is talking about. Can anyone point some direction to study what he said?
| The following source discusses some inconsistency of quantum statistical mechanics indicated by Schrödinger (see the reference there): http://jvr.freewebpage.org/TableOfContents/Volume5/Issue2/Beretta4BdQProceedings%5B1%5D.pdf
The explanation is long and cannot be outlined here. I cannot be sure that Schrödinger did not discuss other issues with thermodynamics or statistical mechanics, so I cannot be sure this answer is useful for you.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/39168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What is predicted to happen for electron beams in the Stern-Gerlach experiment? The Stern–Gerlach experiment has been carried out for silver and hydrogen atoms, with the result that the beams are deflected discretely rather than continuously by an inhomogenous magnetic field. What is theoretically predicted to happen for electron beams?
| Electron beams cannot be split by a stern Gerlach apparatus, because the spin splitting and the orbital splitting cannot be practically separated. The orbital splitting in a constant magnetic field is exactly of the same magnitude as the spin splitting, meaning that the spin anti-aligned electron in a given Landau level is more or less precisely degenerate with the spin aligned electron in the previous Landau level. This means that you can't separate the velocity deflection of the electron from the spin deflection.
This is why Stern Gerlach experiments are only done on atomic beams. There is no simple practical known way to correct for this.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/39301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Optics Paradox? Imagine we have two lens, one convex and one concave, spaced in such a way that the convex lens is before the concave lens. Now each lens has its own focus length and both are spaced such that the concave lens focus is to the right of the convex's. Furthermore, imagine that an arrow with a real length (in the $+y$-direction) is placed at the focus of the convex lens.
It seems that the rays that are incident on the convex lens will be redirected so as to emerge parallel to the optical axis. The image will then appear to be coming from infinite to the concave lens. Now these rays, once going through the concave lens, will diverge with the rays appearing to come from the concave lens focal point.
The question is: what is the magnification of the arrow by the optical system? In other words, can the arrow have a magnification and if so, how?
| I suspect you're worrying about the appearance of infinities in the equations. For example the equation for the concave lens with the object at the focal point $f$ gives:
$$ \frac{1}{f} + \frac{1}{v} = \frac{1}{f} $$
and therefore $v$ = infinity. This leaves you wondering how to calculate the magnification for the second step. If you want to do this rigorously put the object at $f + \delta$, where $\delta$ is some small distance that you'll eventually set to zero. The equation for the first (concave) lens now gives:
$$ \frac{1}{f + \delta} + \frac{1}{v} = \frac{1}{f} $$
so:
$$ \frac{1}{v} = \frac{1}{f} - \frac{1}{f + \delta} $$
The image from the concave lens become the object for the convex lens, so take this expression for $1/v$ and put it in the convex lens equation as $1/u$:
$$ \frac{1}{f} - \frac{1}{f + \delta} + \frac{1}{v} = \frac{1}{-F} $$
where $F$ is the focal length of the convex lens. A quick rearrangement gives:
$$ \frac{1}{v} = \frac{1}{f + \delta} - \frac{1}{f} - \frac{1}{F} $$
and now set $\delta$ to zero and we get:
$$ \frac{1}{v} = - \frac{1}{F} $$
so the image is a virtual image at a distance $F$, which is exactly what you'd expect since parallel rays form a virtual image at the focal point. The magnification is just the distance of the final (virtual) image divided by the distance of the original object, i.e. $F/f$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/39378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Proof of equality of the integral and differential form of Maxwell's equation Just curious, can anyone show how the integral and differential form of Maxwell's equation is equivalent? (While it is conceptually obvious, I am thinking rigorous mathematical proof may be useful in some occasions..)
| Well, as the people said in the comments, the Theorems of Green, Stokes and Gauss will do the job, and are about as mathematically rigorous as you could hope for here!
The two different sets of formula follow directly.
I don't want to write all four of them out, you should be able to do them yourself, but for example, let's consider the Gauss Law.
Starting with the integral form, we have (ignoring physical constants)
$$ \int_{\partial \Omega} \vec{E} . d\vec{S} = \int_{\Omega} \rho\space dV$$
Then by Gauss, we have
$$ \int_{V} \mbox{div} \vec{F} \space dV = \int_{S} \vec{F} .d \vec{S} $$
Hence, we can replace
$$ \int_{\partial \Omega} \vec{E} . d\vec{S} \rightarrow \int_{\Omega} \mbox{div} \vec{E} \space dV $$
to give
$$ \int_{\Omega} \mbox{div} \vec{E} \space dV = \int_{\Omega} \rho\space dV $$
or dropping the integrals,
$$ \mbox{div} \vec{E} = \rho\space $$
which is the differential form.
You should try to derive the other three. This may be helpful in showing you where to start, and where you want to get to.
As for proofs of Green's, Stoke's and Guass Theorems, I recall learning them for some maths exams some years ago, but I wouldn't know where to begin now! Look at any differential geometry course or book and they should be somewhere early on. I can assure you though that the mathematicians have rigirous proofs for them, so we do not need to be shy in using the results of the theorems!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/39419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How does this problems are solved (modeling/simulation)? Can somebody guide me in what to read and learn in order to be able to solve or understand how to solve the following types of problems:
*
*The modeling/simulation of the bullet, shot into the water container, at the bottom of which there is a glass cup
*Automobile crash into the wall/another automobile.
I want to be able to construct the correct mathematical models of such problems. And then to be able to solve them numerically and get the simulation. Of course depending on the problem I will need supercomputer's power, but that is another question.
| Learn about Finite Element Methods. For crash dynamics Smooth Particle Dynamics is good.
Computer codes
*
*LS-DYNA
*OpenFOAM
*DEAL.II
Book
Physics of Shock Waves and High-Temperature Hydrodynamic Phenomena- Ya. B. Zel'dovich & Yu. P. Raizer
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/39541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why do physicists think that the electron is an elementary particle? When we first discovered the proton and neutron, I'm sure scientists didn't think that it was made up of quark arrangements, but then we figured they could be and experiments proved that they were.
So, what is it about the electron that leads us to believe that it isn't a composite particle? What evidence do we have to suggest that it it isn't?
| Believe you me, people have devoted a lot of time to coming up with composite models of the electron, without much to show for it. For example, see the preon.
High energy scattering experiments have shown that the charge radius of the electron is very small, and yet the rest mass of the electron is also very small. It's difficult (though not impossible) to achieve both in a composite model.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/39590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
} |
Calculating force required to stop bungee jumper Given that:
*
*bungee jumper weighs 700N
*jumps off from a height of 36m
*needs to stop safely at 32m (4m above ground)
*unstretched length of bungee cord is 25m
Whats the force required to stop the jumper (4m above ground)
First what equation do I use?
$F = ma$? But even if $a = 0$ $v$ may not equals 0 (still moving)
$W = F \Delta x$? Can I say if $\Delta x = 0$ object is not moving? Even then, I don't know the work ...
I tried doing:
$-32 = \frac{1}{2} (-9.8) t^2$
$t = 2.556s$
Then I'm stuck ... I know $t$ but I cant seem to use any other equations... $v_f, v_i =0 $
| As others here have pointed out, the force of the bungee cord would vary, increasing as it is stretched.
So your question is not well posed. If this is an actual homework problem I would guess that you misread it and you are actually being asked to find the force constant of the bungee cord (assuming, as I will below, that it obeys Hooke's law). Or perhaps you want the maximum force on the jumper due to the bungee cord (which would be when it is stretched the most). Here is how you would get those...
Ignoring air drag, the only forces on the jumper are due to the spring (bungee cord) and gravity, both of which are conservative forces, so you have:
$$ \frac{1}{2} m v_{f}^{2} + \frac{1}{2} k x_{f}^{2} + m g y_{f} = \frac{1}{2} m v_{i}^{2} + \frac{1}{2} k x_{i}^{2} + m g y_{i} $$
Going from the start of the fall of the jumper to when the stretch of the spring is maximum, the initial speed and final speed of the jumper are zero, and the initial x is zero (spring is unstretched at the start), while you can choose the final y to be zero, leaving:
$$ \frac{1}{2} k x_{f}^{2} = m g y_{i} $$
Which can be simply interpreted as saying that the initial gravitational potential energy of the jumper-earth system ends up stored in the spring as elastic potential energy. From this equation you can get $ k $. You can then go on to find the force due to the spring on the jumper when it is stretched by any amount using $ F_{x} = - k x $, and in particular, the maximum force due to the spring.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/39622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Differences between classical, analytical, rational and theoretical mechanics Can you explain me what are the differences between the four following subjects?
*
*analytical mechanics
*rational mechanics
*classical mechanics
*theoretical mechanics
| Analytical mechanics is a branch of classical mechanics that is not vectorial mechanics (original Newton's work). Analytical mechanics uses two scalar properties of motion, the kinetic and potential energies, instead of vector forces, to analyse the motion. Analytical mechanics includes Lagrangian mechanics, Hamiltonian mechanics, Routhian mechanics...
Theoretical mechanics is a branch of mechanics which employs mathematical models and abstractions of physics to rationalize, explain and predict mechanical phenomena. This is in contrast to experimental mechanics, which uses experimental tools to probe these phenomena.
Rational mechanics is a branch of theoretical mechanics characterized by a purely axiomatic approach, where some few axioms are selected and then the rest of the theory logically derived as theorems and corollaries. This branch is usually more mathematically oriented than others.
Classical mechanics is that branch of mechanics that ignores quantum effects. Classical mechanics can be either relativistic or non-relativistic, although in older literature classical mechanics often means pre-relativistic classical mechanics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/39677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
What are some of the best books on complex systems and emergence? I'm rather interested in getting my feet wet at the interface of complex systems and emergence. Can anybody give me references to some good books on these topics? I'm looking for very introductory technical books.
| An Introduction to Complex Systems is a recently published (2019) book by Tranquillo that has been very well reviewed in the Nov. 2019 Physics Today issue:
The text provides a useful overview of complex systems, with enough detail to allow a reader unfamiliar with the topic to understand the basics. The book stands out for its comprehensiveness and approachability. It will be particularly useful as a text for introductory physics courses.
[...]
Tranquillo has written a thorough textbook that gives a useful introduction to complex adaptive systems as a whole field. It can also serve as a quick reference for seasoned practitioners who need a refresher on a particular subject.
Which also points out that:
math is well-integrated into the text, a reader could skim over the equations and focus instead on the prose without losing much. Tranquillo’s presentation will allow math-phobes to be slowly exposed to equations—alongside good explanations—without being forced to read through proofs line by line; thus the book is a useful text for mixed-population undergraduate courses.
Which seems to make it fit very well into the "very introductory technical books" category.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/39712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 5,
"answer_id": 2
} |
From how high could have Felix Baumgartner jumped without disintegrating like a shooting star? Today Felix Baumgartner jumped from 39 kilometres high and reached the earth safely.
Just considering friction, from how high can a human jump?
I expect that from a certain height, he would have reached a speed so high that he would have burnt when entering dense atmosphere?
Some meteoroids disintegrate when they reach the atmosphere, other don't... so I am not sure what would happen for a 100kg human+spacesuit.
Maybe he could jump from arbitrarily high? (notwithstanding digestion/food/drinks, gravity from other stars/planets, lifespan, cold)
| He "only" flew at the maximum speed of 370 m/s or so which is much less than the speed of the meteoroids – the latter hit the Earth by speeds between 11,000 and 70,000 m/s. So he was about 2 orders of magnitude slower. The friction is correspondingly lower for Baumgartner.
Note that even if he jumped from "infinity", he would only reach the escape velocity which is 11,200 m/s for the Earth, just like the slowest meteoroids. I guess that a good enough (and cooled) suit inspired by NASA rockets might be capable of protecting a human against such relative speeds even though for generic surfaces, they would almost certainly start to burn at the surface.
However, it wouldn't be pleasant to slow down from such speeds in the atmosphere. ;-) You see that if you uniformly slow down from 10 km/s to 0 km/s while flying through 10 km of the atmosphere, the penetration through the atmosphere takes about 2 seconds. However, getting from 10 km/s to 0 km/s in two seconds means that the deceleration is 5000 m/s/s or 500 $g$. I guess that not even he could survive that. ;-)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/40829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
Why are synthetic elements unstable? So far 20 synthetic elements have been synthesized. All are unstable, decaying with half-lives between years and milliseconds.
Why is that?
| The artificial elements are artificial because they're rapidly radioactive, and not regenerated through decay. 81 of the first 83 elements, as well as #90 (thorium) and 92 (uranium) can practically be considered stable for most purposes. The exceptions are technetium (43) and promethium (61). Some other elements are rapidly radioactive, such as polonium (84), radon (86) and radium (88), but occur in nature as decay products of heavier elements such as thorium and uranium. The elements #43, 61, 85 (astatine), 87 (francium), 93 (neptunium) and 94 (plutonium) are also found in nature, as they occur in very minor decay branches of elements 90 and 92. Also, elements #1 (hydrogen)-94 either formed in the Big Bang or in stars. Heavier elements are produced artificially. See http://ryanmarciniak.com/archives/1627; https://en.wikipedia.org/wiki/Transuranium_element; and https://en.wikipedia.org/wiki/Thorium.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/40894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Distribution of charge on a hollow metal sphere
A hollow metal sphere is electrically neutral (no excess
charge). A small amount of negative charge is suddenly
placed at one point P on this metal sphere. If we check on
this excess negative charge a few seconds later we will find
one of the following possibilities:
(a) All of the excess charge remains right around P.
(b) The excess charge has distributed itself evenly over the
outside surface of the sphere.
(c) The excess charge is evenly distributed over the inside
and outside surface.
(d) Most of the charge is still at point P, but some will have
spread over the sphere.
(e) There will be no excess charge left.
Which one is correct and why?
I guess it is some kind of electrostatic induction - phenomena going on. Am I right? I understand that excess charge is distributed over hollow sphere and that negative and positive charges are distributed opposite sides, but don't know which one positive or negative go to inside surface.
| B, Since the sphere has no charge the negative charge would distribute evenly across the surface as like charges repel the push themselves away from each other.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/40993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
What are the reasons for leaving the dissipative energy term out of the Hamiltonian when writing the Lyapunov function? I have a problem with one of my study questions for an oral exam:
The Hamiltonian of a nonlinear mechanical system, i.e. the sum of the kinetic and potential energies, is often used as a Lyapunov function for controlling the position and velocity of the system. Consider a damped single degree-of-freedom system, $m\ddot{x}+c\dot{x}+kx=0$, where $m$ is the mass, $c$ is the velocity-proportional damping and $k$ is the stiffness. A candidate Lyapunov function is the Hamiltonian $V=\frac{1}{2}m\dot{x}^2+\frac{1}{2}kx^2$. What are the reasons for leaving out the dissipative energy term when writing the Lyapunov function?
The only thing what comes into my mind for this question is, that a dissipative energy term in the Lyapunov function would have a "-" sign and the Lyapunov function would thus not be positive definite anymore. Is that correct?
| I am not quite sure what "dissipative energy term" means, but I do know that you can't add anything proportional to $\dot{x}$.
To see why, just take a point close to the $(x, \dot{x})=(0,0)$ point. In a neighbourhood of this point the $\dot{x}$ term will dominate over $\dot{x}^2$ and either the point $(0,\epsilon)$ or $(0,-\epsilon)$ would give a negative value.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/41034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How do I integrate $\frac{1}{\Psi}\frac{\partial \Psi}{\partial x} = Cx$ How do I integrate the following?
$$\frac{1}{\Psi}\frac{\partial \Psi}{\partial x} = Cx$$
where $C$ is a constant.
I'm supposed to get a Gaussian function out of the above by integrating but don't know how to proceed.
| Sorry, I can't comment in the right place due to low rep.
@Killercam, you never need to ''treat this [the partial derivative] as an ordinary derivative''. Doing so ignores the possibility of other variables, and doesn't find the most general solution.
The only change in Killercam's derivation, is that $\kappa$ should be a function of any variables which are held constant during the partial derivative, $\partial_x$.
For example, consider the function on $\mathbb{R}^3$
$$ \Psi(x,y,z) = f(y,z)\,e^{\frac{1}{2}Cx^2}. $$
If the partial derivative holds $y$ and $z$ constant, we find that $\partial_x\Psi = Cx\Psi$. Exactly as required.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/41099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why does bad smell follow people (assuming they are not the source)? When you are sitting in a room where there is a source of bad smell, such as somebody smoking or some other source of bad smell, it is often a solution to simply move to another spot where bad smell is not present. Assuming you are not actually the source of the smell, this will work for a while until you notice the smell has somehow migrated to exactly the spot where you are now sitting. Frustrating.
This got me thinking about the fluid mechanics of this problem. Treat bad smell as a gas that is (perhaps continuously) emitted at a certain fixed source. One explanation could be that human breathes and perhaps creates a pressure differential that causes the smell to move around. Is there any truth to this? Please provide a reasoned argument with reference to the relevant thermodynamic and/or fluid quantities in answering the question. Theoretical explanation is desired, but extra kudos if you know of an experiment.
| From a fluid dynamics standpoint, as a body moves through a fluid, a small region of fluid is dragged along with it. This is what forms the boundary layer. In the near-body region, odor will be dragged along with the body.
Likewise, behind a moving person is a turbulent wake and a low pressure region. The low pressure reason will "suck" the odor along with the body, and the turbulence will mix the odor into the air which will also help distribute it.
Turns out there is an experiment, in this paper, that looks at the effect of a stationary body and a moving body (as in human body) in a room with stratified contaminants. The principles discussed therein are along the lines of your question.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/41149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
What is the difference between a spinor and a vector or a tensor? Why do we call a 1/2 spin particle satisfying the Dirac equation a spinor, and not a vector or a tensor?
| A four-vector transforms under the Lorentz-Group $SO(3,1)$, i.e. a "standard Lorentz transformation". The Lorentz transformation for a spinor is under $SU(2)\times SU(2)$ (to be exact the representation $2 \times \bar 2$) which is locally isomorph to $SO(3,1)$ but not the same. To get a better understanding, you could read chapter two here and meditate a bit about it. It took me a while to get everything sorted in my head. (In fact, I doubt everything is sorted yet, but at least I start to get the grasp of it.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/41211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 4,
"answer_id": 1
} |
Does infinite energy imply Infinite mass (and vice versa)? If some kind of source was able to supply an infinite amount of energy, does that imply that it also must have an infinite mass? Is the contrary also true?
| It is very dangerous to talk about infinity in physics, especially when talking special relativity.
To your question: Yes the object would need an infinite amount of mass.
E.G. Take a battery that would have an finite amount of energy inside.
Then you would have
$M_{\text{Total mass of the battery}}=M_{0,(\text{the usual Mass of an uncharged battery})}+\frac{E_{\text{Amount of Energy the Battery can supply}}}{c^2}$
Now If you set $\lim E\rightarrow \infty$ of course $M_{\text{Total mass of the battery}}$ also diverges
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/41287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Will the siphoning effect help a system pump water upwards if the water's entry and exit points are at the same height? I am looking to pump water from a pool up to a roof for solar heating (black plastic tubing) and then back into the pool with the original source water. Does the gravitational force of the water flowing back down the pipe into the pool assist the pump and therefore decrease the pumps required strength?
I have been told that the strength of the pump needed would have to be the same regardless of the exit point's height. i.e. If the water was being pumped into a tank on the roof instead of flowing back to the pool.
| "I have been told that the strength of the pump needed would have to be the same regardless of the exit point's height."
This is false. You would want to bring the pipe exit to the same inlet height as the pump or you are adding required head. As long as it's not over ~33 feet (from pump center line to the highest point in the pipe), you should be fine. That being said, the pipe, bends, fittings and solar collector will all add length in the form of equivalent length due to frictional losses. This equivalent length will also increase the required head. One other thing you may want to consider is the outlet size of the pump and the inlet size of the solar collector.
*
*Don't go over ~33 feet
*Outlet height = Inlet height
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/41331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
The earth's magnetic field This might sound like a silly question. Is it possible for the earth's magnetic field to actually destroy or harm earth? (implosion, crushing etc.)
| I'd say a big NO for that...
Perhaps, the real truth is that "Without the magnetic field of Earth, we'd face a lot of awful things than just Greenhouse effect & Global warming..."
The magnetic field protects Earth from most of the charged particles (based on Lorentz force) in Cosmic rays and Coronal mass ejections (Not all are deflected away. It just reduces the effect and doesn't prevent it completely). It would've been better if it also protects us from neutrinos..! But, we aren't so lucky that they're neutral...
This paper would be useful for your question...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/41397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Free falling objects My teacher and I are in the middle of an argument because she says that if you were to drop two objects at the same time and the same height, but with different initial velocities, both of them would hit the ground at the same time. She also said this was proven by Galileo's unregistered experiment when he let two objects fall from the Leaning Tower of Pisa. So I told her that she was right, they would fall at the same time only if both initial velocities had been the same, and that Galileo's experiment was mainly to prove that mass has nothing to do with the timing registered on two objects falling from the same height.
Could you please help me out, using 'pure' physics to prove my teacher wrong? I've already searched the internet, but I haven't found something that would convince her because it's what I already have explained to her, but she's in denial and I'm thinking she just doesn't want to accept that she's wrong.
Assume no drag/air resistance.
| If we are throwing two objects directly to the ground you are right.
So from our kinematic equations:
$$V_f = V_i + at$$
I would ask your teacher. What happens to the $V_f$ if $V_i=0$? Then Follow it up with what would $V_f$ be if $V_i$ was very large?
The initial velocity DOES have an effect here.
HOWEVER: Make sure that you are not misinterpreting the question! If you are throwing two objects in something that is sort of parabolic arc (i.e. a projectile) and you threw them at the same time, the time it would take for each one to hit the ground would be THE SAME.
This is because at their peak the objects both only have one thing accelerating them (gravity) and because at their peak their V = 0! This is very un-intuitive and you should look into exploring it!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/41444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Minkowski diagram and time dilation After i figured out how to show length contraction in this topic. I tried to use a similar way to show time dilation in Minkowski diagram. Time dilation means that time interval between two events is the shortest in the frame in which those two events happen in same place. We call this frame its rest frame.
According to just said I set up my problem so that frame $x,ct$ is my rest frame as edges of a time interval $\Delta t$ happen on a same vertical line. This means that time interval $\Delta t$ should be the shortest time interval compared to time intervals in any other frames. Including frame $x',ct'$.
But when i draw time interval $\Delta t'$ it is obvious that it is shorter than interval $\Delta t$. So why do i get time contraction instead time dilation?
Marks $1'$ and $2'$ on $ct'$ axis are further appart than marks $1$ and $2$ on $ct$ axis (i even draw a parabola to show this). This is why $\Delta t = 1$ and $\Delta t' < 1'$.
| We want to find the (coordinate) elapsed time between two events.
Let those two events be the origin and $ct=1, x=0$. Clearly, these two events are co-located in the unprimed coordinate system and so, the unprimed coordinate elapsed time is equal to the proper elapsed time between these two events.
To find the coordinate elapsed time in the primed coordinate system, draw a line, parallel with the $x'$ axis, through the 2nd event. This is a line of simultaneity in the primed coordinate system. In other words, all the events along that line occur at the same time according to the clocks moving with the primed system.
So, the intersection of this line with the $ct'$ axis gives the primed coordinate time of the 2nd event. You shall see that the primed elapsed time is larger.
In the image above, $\gamma = 1.25$.
For the $x'$ axis, $ct'=0$. For the line parallel to the $x'$ axis through the 2nd event, $ct'=\gamma=1.25$.
So we have that $\Delta t' = \gamma \Delta t > \Delta t$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/41586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why do green lasers appear brighter and stronger than red and blue lasers? This is mostly for my own personal illumination, and isn't directly related to any school or work projects. I just picked up a trio of laser pointers (red, green, and blue), and I notice that when I project them, the red and the blue appear to be dimmer to my eye than the green one. I had a fleeting suspicion that, perhaps this is an effect of blue and red being at the periphery of the visual light scale, but I honestly have no idea if this is the case or if it's just my eyes playing tricks on me. All three lasers have the same nominal strength, in this case.
| Its because human eyes are more sensitive to green light than other colors. Here's a reference:
http://www.physicsclassroom.com/class/light/u12l2b.cfm
I also heard from a friend, although I can't find a reference, that traffic lights display red at a higher intensity so that it appears just as "bright" as green light does.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/41624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
Why do physicists believe that particles are pointlike? String theory gives physicists reason to believe that particles are 1-dimensional strings because the theory has a purpose - unifying gravity with the gauge theories.
So why is it that it's popular belief that particles are 0-dimensional points? Was there ever a proposed theory of them being like this? And why?
What reason do physicists have to believe that particles are 0-dimensional points as opposed to 1-dimensional strings?
| I had been preparing an answer to the question made duplicate .
there were more questions than in the question above, so I am answering here:
Elementary particles are like mathematical points?
In the standard model of physics they are assumed so.
Does make sense in quantum mechanics and standard model think this way?
This is the table of elementary particles of the standard model of particle physics.
All matter is a composite of these particles, and yes, they are modeled as point particles. Yes, the mathematical model of the standard model has been validated over and over again, and its quantum mechanics based predictions are fulfilled, as recently as the discovery of the Higgs.
Is is true that two elementary particles are indistinguible?
No, this is wrong as a general statement. Different types of particles (electrons, quarks ...) are characterized by different quantum numbers and are distinguishable.
The same type of elementary particles are experimentally indistinguishable, two electrons are interchangeable, except by their quantum numbers in specific boundary conditions. In general one cannot attach an identity card on an elementary particle.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/41676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 7,
"answer_id": 5
} |
Is it possible to travel at precisely the speed of sound? I've been talking to a friend, and he said that it's impossible to travel at exactly the same speed as the speed of sound is. He argued that it's only possible to break through the sound barrier using enough acceleration, but it's impossible to maintain speed exactly equal to that of sound. Is it true? And if it's true, why?
| No, this is not true. Unlike the speed of light, for example, there isn't anything particularly special about the speed of sound that would prevent you from traveling exactly at it. There isn't really anything else I can explain about it without knowing what reason your friend gave for making his argument.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/41721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
Why do birds sitting on electric wires not get shocked? When we touch electric wires, we get shocked. Why don't birds sitting on electric wires not get shocked?
| A simple googling would've provided you an answer. We won't get a shock if we fly or when we aren't grounded... Because, Current flows only in closed circuits (Maybe in Plasma "as an open"). A bird sitting in the wire doesn't form a complete circuit in order for the current to flow. In other words, Birds have their feet in the same wire (It also has high resistance, Now, that's another problem).
I've seen many days, birds (like "crows" in our country) get shocks. Sometimes birds too touch the Phase and neutral (or Earthing) wires thereby creating a largest potential difference, get toasted & fall down. Not all birds are clever.
Imagine: Take a circuit with battery. Connect one terminal of the battery to a resistance and a galvanometer. Leave the other terminal open. If it shows deflection, then you'd solve your question and you'd definitely get a Nobel...
Have a look over this overview of the topic...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/41770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
} |
Why do we need more power to do a job fast? Let's assume we have two identical electric trains. One has a big electric motor (high power) and the other has small motor (low power). Let us assume the electric motors are of the same brand and the power of the motor is directly proportional to the size. Now, the two trains used their max power to travel between to stations and the powerful train arrives quicker. So can we relate time (the saved time) to the size (mass) of the motor or if we neglect the mass variance of the motors, to the mass of additional coal used at the power station? If this so, we can now measure time in kg's or lb's of coal?
| In physics, the average power $P$ is defined as the amount of work done per unit time interval, i.e., $P = \frac{\Delta W}{\Delta t}$. So, a greater $P$ implies a smaller $\Delta t$ for the same amount of work. This answers the first part of your question. As regards whether you can measure time in terms of kg of coal, you could. Assume that the power output of the train engine scales linearly with the mass of the coal. You could define 3 minutes as the time it takes for the train to burn up, say, 100 kg of coal. Then, any other time interval can be measured in terms of the mass of coal. So, 6 minutes is the time required to burn up 200 kg of coal. Whereas this may be useful to the driver of the train, it is not a very useful definition to the rest of us because it is not at all clear what the efficiency of the engine is, the mass of the train is, how he accelerates and decelerates, etc.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/41845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What are the fields in this problem? In problem 3 of chapter 2 of Landau Lifshitz "Mechanics," I don't understand the meaning of the fields as defined in the following statement:
Which components of momentum and angular momentum are conserved in motion in the following fields?
(a) the field of an infinite homogeneous plane,
(b) that of an infinite homogeneous cylinder,
(c) that of an infinite homogeneous prism,
(d) that of two points
etc.
I just don't get what he's trying to say. Does he mean an electric field produced by a uniformly charged infinite plane, or an uniformly charged cylinder, etc.?
| Yes, that is what Landau and Lify are getting at. I don't really see another interpretation. I mean, are momentum and angular momentum conserved under the following applied potential fields? I think that is a reasonable interpretation. Haven't looked at Landau and co. Mechanics in a bit, but this seems reasonable.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/41925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Equation of the saddle-like surface with constant negative curvature? What is the equation for the saddle-like 2d surface (embeded in 3d Euclidean space with cartesian coordinates x, y and z) with constant negative curvature frequently used to illustrate open universe (for example in the following image is taken from Wikipedia)?
Flat universe (zero curvature) is illustrated by a plane, with equation say
$z = 0$
Closed universe (constant positive curvature) is illustrated by a surface of sphere, say
$x^2 + y^2 + z^2 = r$
| I don't think you can embed a surface of constant negative curvature in Euclidean space. However you can embed it in Minkowski space. See http://en.wikipedia.org/wiki/De_Sitter_space for details. If you have Minkowski space defined by:
$$ ds^2 = -dx_0^2 + dx_1^2 + dx_2^2 + dx_3^2 $$
then the corresponding surface is:
$$ -x_0^2 + x_1^2 + x_2^2 + x_3^2 = a^2$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/43056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Cause of buoyant force? Can you explain to me what causes the buoyant force? Is this a result of a density gradient, or is it like a normal force with solid objects?
| One can think of buoyancy as the same force that lets you stand on the ground without sinking into it. When you are standing on the ground, the earth's gravity exerts a downward force equal to your weight, and the ground offers a normal reaction of equal magnitude in the upward direction which balances out your weight. Now, think of the water in a filled tumbler, and visualize the water in upper half as you, and the water in lower half of the glass as the ground. The weight of upper half of the water is supported by the normal reaction, or in this case, the pressure difference created by the rest of the fluid. Using this analogy, it is very simple to understand the Archimedes principle too. The buoyancy force exerted on any immersed body is equal to the weight of the fluid displaced by it, because this weight was initially balanced out by the pressure difference created by the fluid beneath it. Fundamentally, it is the intermolecular forces and the lattice structure which provides solids and fluids a degree of resistance to compression. This resistance is exhibited as the compressional stress in solids and as pressure in fluids.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/43087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
What is the difference between weight and mass? What is the difference between the weight of an object and the mass of an object?
| To elaborate on John Rennie's answer - As he said, the mass is the inertial mass of the body, which isn't the same as the weight.
The weight is typically defined in context to a gravitational field.
No doubt you know that the acceleration due to gravity on the earth's surface is $9.8 m/s^{2}$ (on average). So if your mass on earth is say, 5 kilograms, your weight on earth would be $9.8 \times 5 = 49 N$. So the weight that your weighing machine registers is your weight, not your mass. You'd have to divide by the acceleration due to gravity at that point to calculate the mass.
So basically your weight basically measures the force that you exert on the weighing scale, not the actual mass of your body. Which is why if you stand on a weighing scale in free fall it'll register zero, since you aren't exerting any additional force on it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/43195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 11,
"answer_id": 3
} |
Intuition for multiple temporal dimensions It’s easy, relatively speaking, to develop an intuition for higher spatial dimensions, usually by induction on familiar lower-dimensional spaces. But I’m having difficulty envisioning a universe with multiple dimensions of time. Even if such a thing may not be real or possible, it seems like a good intellectual exercise. Can anyone offer an illustrative example?
| Theories can be formulated in any number of space-time dimensions. Examples of multidimensional theories:
*
*F-theory, 12d theory with two times.
*S-theory, 13d theory with three times.
*Kalitzin's relativity with r-times or even infinite-dimensional times (the latter much less developed) was studied.
Indeed, I think to remember some people tried to understand quantum mechanics in a multitemporal set-up.
In philosophy, some people studied three-dimensional time theories. These times are called time, hyparxis and eternity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/43322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 4,
"answer_id": 1
} |
Why do diamonds shine? I have always wondered why diamonds shine. Can anyone tell me why?
| Diamond is one of the hardest material. We know that it's an allotrope of carbon. A diamond (crystalline in nature) has a three dimensional arrangement of carbon atoms linked to each other by strong covalent bonds. What you've shown a round brilliant cut diamond.
Actually, the secret that's rattling inside a diamond is refraction, total internal reflection (not to be confused with ordinary reflection) & dispersion. The refractive index of diamond is pretty high (2.417) and is also dispersive (coefficient is 0.044). Due to this fact, diamond is an important application in optics.
Consider an ideal cut diamond. I explain according to the figure below. When the light is incident at an angle $1$, it refracts inside and travels through the lattice. At the surface which separates air & diamond media, the incident angle $2$ is very well above the critical angle ($c_a$) and simultaneously ($3$ & $4$) the reflection takes place at different surfaces of the diamond. Finally, the light refracts out.
The first one shows the mechanism of internal dispersive reflection. The second figure shows the reflections inside ideal cut, deep and shallow cut diamonds.
Note: For total internal reflection to take place, light must travel from an optically denser medium to a relatively rarer medium. Also, the incident angle should be far high above thee critical angle.
There are youtube goodies regarding the topic...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/43361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Wheatstone bridge galvanometer error We had to measure the resistance of $R_x$, we balanced the Wheatstone bridge and did calculations. My question is: we didn't include galvanometer error into calculations. Why is that? I read that it's very precise, but that doesn't seem like a good enough explanation in exact sciences :/
Edit: The precision is not the case as I was told, I need to go into more detail.
When I measured, I set a value on the potentiometer, then I adjusted the adjustable resistor, so that the galvanometer would show zero. Is the error somehow compensated?
ɛ - electromotive force.
K - switch.
R - adjustable resistor.
C - potentiometer.
| What it means by very precise is that the error from the galvanometer is significantly less then the error within your system. The Wheatstone Bridge calculations do not require serious precision; the galvanometer are the least of your concerns. IF you were using this device in some type of research environment that required highly calibrated and precise equipment, this device may need to be considered if critical to the experiment (it probably would not be).
The error of the galvonometer may be on the order of $\pm .0001$ or smaller while the rest of your error may be more worthy of mention around $\pm .1$ or $\pm.01$ or something to that effect. I would not worry about it, but if you are writing a lab report it would be a good idea to write about the possibility of it being considered; if the device tells you the error associated with it (on a sticker on the device or in the manuel), I would write it down.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/43437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Minimal Maxwell's Demon I would like to understand where the waste heat is generated in the Maxwell's demon problem. To this end I've come up with the simplest scenario I can think of. If my scenario is workable I am hoping someone will be kind enough to apply the right math to it to show what is going on and why it does not work to provide a "free lunch."
My scenario is as follows:
Take a 1 dimensional container of length l. Start with an evenly distributed population of particles; slow red ones and fast blue ones, traveling at +v or -v, and +2v or -2v respectively.
The demon sits at the boundary at l/2.
He lets slow red particles at +v and fast blue particles at -2v through and keeps the opening between the two halves of the container closed otherwise.
The idea of using a one dimensional container is to keep the math as simple as possible. In order for this to work the particles have to be able to go right through one another without interacting.
I am imagining that if one can work out what happens in this simplest case then it should be possible to extend the scenario to a 2-dimensional container in which the line forming the barrier between the two halves of the container is made up of discrete copies of the Maxwell's Demon in my 1-dimensional scenario.
| The dimensionality is not relevant. The commonly-believed "solution" of the Maxwell's demon puzzle utilizes the fact that the demon must store information as measurements are made. To bring the demon back to its initial state, that memory must be erased. According to Landauer's theorem, erasure of each bit of information generates at least $k \rm{ln}(2)$ J/K of entropy in the environment via a heat process. Erasure is dissipative. You can read about this in various places, including Leff & Rex, "Maxwell's Demon 2: Entropy, Classical and Quantum Information, Computing (Institute of Physics, Bristol, 2003) and von Baeyer, "Maxwell's Demon: Why Warmth Disperses and Time Passes"(Random House, 1998). C.H. Bennett also had a nice article in Scientific American magazine: C. H. Bennett, "Demons, engines and the second law," Sci. Am. Vol. 257, 108-16 (1987).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/43504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Projectile motion along the earth Suppose a projectile is launched from the Earth's surface with initial velocity $v_0$ well below speed of light and initial angle $\theta_0$ with respect to the vertical line perpendicular to the Earth's surface. Omitting Earth's rotation, but knowing that Earth is not flat (as in the real world), what is the maximum height of the projectile with respect to center of the Earth? (Suppose air resistance is negligible.)
Is it possible to solve this problem using only the material in the first $8$ chapters of Fundamentals of Physics; D. Holliday, R. Resnick & J. Walker? I'm eager to see a various number of solutions!
| I do not have the book you mentioned, but I could find something in my notes from last year's Classical Mechanics lecture.
Firstly, if your projectile reaches escape velocity, then it will of course travel away from the Earth forever, which answers your question with $\infty$. When does this happen? The potential energy of the projectile is given by:
$V(r) = -\frac{GMm}{r}$
We will escape the Earth's potential, if we have enough energy to reach the potential at $r=\infty$, which obviously corresponds to $V=0$. So we need at least the potential at the Earth's radius in kinetic energy:
$K = \frac{1}{2}mv_0^2$
Therefore
$v_{escape}=\sqrt{\frac{2GM}{R_E}}$
Where $G$ is the gravitational constant, $M$ the mass of the Earth, $m$ the mass of your projectile and $R_E$ the radius of the Earth (which is the current distance of your projectile from the Earth's center). So if the projectile's velocity is higher than this we have an unbound orbit, and there is no maximum height. If it is less than this we have a bound orbit, whose trajectory we need to figure out.
Now gravitation is a radial inverse-square law force of the form $\textbf{F}=\frac{k}{r^2}\textbf{r}$ where $\textbf{r}$ is a unit vector pointing radially outwards. Since it is attractive $k<0$ (in fact $k=-GMm$). For these forces a bound orbit will be elliptical, with the center of the Earth at one of the ellipse's foci. The equations I could turn up let you determine the eccentricity $e$ and the semi-latus rectum $h$, which define the ellipse, from the angular momentum and the total energy of the projectile.
$E = K + V = \frac{1}{2}mv_0^2-\frac{GMm}{R_E} \\
\textbf{L} = m\textbf{R}_E\times\textbf{v}_0=mR_Ev_0\sin(\theta_0) \\
h = -\frac{L^2}{mk} = \frac{L^2}{GMm^2} \\
e = \sqrt{1+\frac{2EL^2}{mk^2}}=\sqrt{1+\frac{2EL^2}{G^2M^2m^3}}$
Using $e$ and $h$ we can figure out $r_A$ and $r_B$, the distance from the Earth's center to the "perigee" and "apogee" (the points that are nearest and farthest from the Earth's center, respectively). We do this using standard ellipses formulae:
$r_A = \frac{h}{1+e} \\
r_B = \frac{h}{1-e} \\$
Since $e$ is between $0$ and $1$, we obviously have $r_B > r_A$ as expected. So the largest height above the Earth's surface would then be $r_B - R_E$. We would be done here... if there was not the possibility that the projectile hits the Earth's surface before reaching the apogee (for example because $\theta_0>\pi/2$). I am honestly at a loss when it comes to figuring out whether the projectile reaches the apogee, the perigee or both, but I have a feeling that if it reaches any of the two, it would be the apogee as required. But I cannot figure it out properly right now. When I do, I will edit the answer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/43571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Definition of Fine-Tuning I've looked in and out the forum, and found no precise definition of the meaning of fine-tuning in physics.
QUESTION
Is it possible to give a precise definition of fine-tuning?
Of course, I guess most of us understand the empirical meaning of the phrase... but it seem so ethereal, that's the reason behind my question.
| All we can do precisely is give a probability for some physical quantity to have its observed value. For example (subject to various assumptions!) the probability of the cosmological constant having it's observed value is around 1 in $10^{120}$. Since this is absurdly low we say it's fine tuned.
But where you draw the line between fine tuned and not fined tuned is a matter of debate. Most of us wouldn't consider a 10% probability fine tuned, but what about 1% or 0.1%? Particle physics required a $5\sigma$ probability to be considered proof, and this is about 1 in 3.5 million and this is about 0.00003%, so that seems like a reasonable lower bound for not fine tuned. However I'd guess most people would consider considerably higher probabilities than this as evidence of fine tuning.
The point is that the probability of an observed value can be calculated precisely, but whether this corresponds to fine tuning is a matter of personal opinion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/43632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Current in series resistors and voltage drop in parallel resistors When we have resistors in series, the current through all the resistors is same and the voltage drop (or simply voltage) at each resistor is different.
Question 1: It is fine that voltage drop (potential drop) across each resistor is different because each resistor offers different resistance (suppose). but how is the current through each resistor same? If we have resistors of different resistance, shouldn't the current be different through each resistor?
Similarly, when we have resistors in parallel, the current through each resistor is different but the voltage drop at each resistor is same.
Question 2: Current through each resistor is different because resistance of each resistor is different (suppose). but how is the voltage drop across each resistor same here? Shouldn't the voltage drop at each resistor be different because each resistor offers different resistance?
| To be clear, current is charge passing through a certain area per unit time. This does not imply a second parameter in the denominator of the formula for current $\frac{dq}{dt}$; just a guideline for how to measure $dq$. The larger the cross sectional area, the larger the perceived current will be. This is why resistance is seen to decrease for larger cross sectional areas when selecting wire gauges.
Secondly, it is true that different drift velocities along a circuit would be impossible as the electric field would vary in the areas wherever charge density is different and force an equalization of the rate of flow.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/43782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
The definition of entropy in quantum mechanics I have seen entropy with several different definitions. Like Von Neumann entropy and Rényi entropy, etc.
So I am curious why there are so many different definitions in quantum mechanics while only one in classical mechanics named after Boltzmann?
| All the quantum entropies that you cite have a classical analogue. E.g. the Von Neumann entropy $\langle S \rangle = -k_B \mathrm{Tr} (\hat{\rho} \ln \hat{\rho})$ is the quantum version of the Gibbs entropy $\langle S_\mathrm{cl} \rangle = -k_B \int \mathrm{d}p \mathrm{d}x (\rho \ln \rho)$ used in classical statistical mechanics. The Boltzmann entropy is a special case of it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/43816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If wave packets spread, why don't objects disappear? If you have an electron moving in empty space, it will be represented by a wave packet. But packets can spread over time, that is, their width increases, with it's uncertainty in position increasing. Now, if I throw a basketball, why doesn't the basketball's packet spread as well? Wouldn't that cause its uncertainty in position to increase so much to the point it disappears?
EDIT: I realize I wasn't clear what I meant by disappear. Basically, suppose the wave packet is spread over the entire Solar System. Your field of vision covers only an extremely tiny part of the Solar System. Therefore, the probability that you will find the basketball that you threw in your field of vision is very small.
| Spreading the wave packet does not mean spreading and disappearing the electron.
If a basketball has initially uncertainty $\Delta V$ of its velocity, then with time the ball position uncertainty will grow as $\Delta V\cdot t$.. With time this uncertainty gets so large that the ball disappears from your sight (I mean you will not find it where you expect it to be without knowing the initial velocity spread).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/43860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to compute the expectation value $\langle x^2 \rangle$ in quantum mechanics? $$\langle x^2 \rangle = \int_{-\infty}^\infty x^2 |\psi(x)|^2 \text d x$$
What is the meaning of $|\psi(x)|^2$? Does that just mean one has to multiply the wave function with itself?
| About your other question, the meaning of $|\psi(x)|^2$ is that of a density of probability, with $[|\psi(x)|^2 \mathrm{d}x]$ giving the probability that the particle is found between $x$ and $x + \mathrm{d}x$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/44147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Is physics rigorous in the mathematical sense? I am a student studying Mathematics with no prior knowledge of Physics whatsoever except for very simple equations. I would like to ask, due to my experience with Mathematics:
Is there a set of axioms to which it adheres? In Mathematics, we have given sets of axioms, and we build up equations from these sets.
How does one come up with seemingly simple equations that describe physical processes in nature? I mean, it's not like you can see an apple falling and intuitively come up with an equation for motion... Is there something to build up hypotheses from, and how are they proven, if the only way of verifying the truth is to do it experimentally? Is Physics rigorous?
| Since you are a student of mathematics with little knowledge of physics, I strongly urge you to take a few courses in modern physics before you finish your mathematics education (General Relativity and Quantum Mechanics the very least). This is a must if you are specializing in geometry/topology.
Having said that, the "axioms" of the two disciplines are not the same kind, and therefore the "rigor" of proofs is different.
Physical "axioms" are those that are invariant in every experiment. E.g., the "axiom" of conservation of energy; or that the speed of light cannot be exceeded, and so on. Newton's laws were once "axioms" but they had to be modified in extreme conditions. But that does not mean Newton's laws are false: they are still true to the world we normally experience.
Mathematical axioms can be "abstract" but not based on any experiment: parallel lines do not intersect is an ancient axiom. Whether that is true or not in our universe is not important to a mathematician.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/44196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 10,
"answer_id": 5
} |
Poincare Patch covers half of the hyperboloid of AdS We start with the general case of $AdS_{p+2}$ i.e AdS space in $p+2$ dimension.
\begin{equation}
X_{0}^{2}+X_{p+2}^{2}-\sum_{i=1}^{p+1}X_{i}^{2} = R^2
\end{equation}
This space has an isometry $SO(2,p+1)$ and is homogeneous and isotropic. The Poincare Patch is given by
\begin{equation}
ds^2 = R^{2}\left(\frac{du^2}{u^2}+u^2(-dt^2 +d\mathbf{x}^{2})\right)
\end{equation}
According to Equation (2.27) of the article http://arxiv.org/abs/hep-th/9905111, The second metric covers only half of the hyperboloid. Firstly, how do I show this. Secondly, when I go to the asymptotic limit (small radial distance), should the topology of the two spaces be different?
| It can be shown by conformal compactification of the spacetime, i.e. using coordinates which allow you to draw a penrose diagram. After you have done so, you can analytically continue the geometry and discover the other half (see chapter 2 of this) . Regarding differences in topology: I don't see any reason for this to be the case.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/44335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Photon energy - momentum in matter $E = h\nu$ and $P = h\nu/c$ in vacuum.
If a photon enters water, its frequency $\nu$ doesn't change.
What are its energy and momentum: $h\nu$ and $h\nu/c$ ?
Since part of its energy and momentum have been transferred to water, it should be less.
If water's refractive index is $n$, are the energy and momentum equal to $h\nu/n$ and $h\nu/c/n$ ?
| It's a non-trivial problem, which also involves how you define a photon in a medium - as a interacting particle and treating excitation of medium separately, or as a "dressed particle", including the interaction.
From Abraham–Minkowski controversy Wikipedia page:
The Abraham–Minkowski controversy is a physics debate concerning electromagnetic momentum within dielectric media.
[...]
*
*The Minkowski version:
$$p=\frac {n h \nu}{c}$$
*The Abraham version:
$$p=\frac {h \nu}{n c}$$
[...]
A 2010 study suggested that both equations are correct, with the Abraham version being the kinetic momentum and the Minkowski version being the canonical momentum, and claims to explain the contradicting experimental results using this interpretation
*
*Stephen M. Barnett, Resolution of the Abraham-Minkowski Dilemma, Phys. Rev. Lett. 104, 070401 (2010), free pdf here
Look also (Google Scholar?) at "electromagnetic momentum in a medium" or "electromagnetic energy in a medium", as essentially its related to a classical problem.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/44509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Could we have assumed the speed of light to be different in different reference frames? Ptolemy's model of universe assumes that our earth is the static center of universe and everything else move relative to it (ref: The grand design ch:3). This model would give us a consistent picture of universe the only complication would be that the trajectories of other heavenly bodies would be fairly complicated with our earth on center. So assuming any center doesn't contradict fundamental property of nature.
Similarly could we have assumed that the speed of light too was not the same on different frames and had similar properties as speed of sound (for example) which is different on different frame and still not contradict fundamental property of nature only giving some other complicated (may be) equations describing nature?
| The given answer is in fact not complete. While aether construction is trivially knocked down, yet another construction must be used to get rid of the remaining theory in which this set up works, that is light takes some convenient reference frame relative to its emitter or collector.
The problem must be solved (again) in two dimensions. The calculation for the angle that light is observed at due to being in a different frame as its source (this is known as Aberration of light) yields the right answer only using relativistic velocity addition.
There are a couple of choices for "convenient reference frame". One is taking the reference frame of the emitter so that all others must add the emitter's velocity. I can't find a name for this one anywhere despite it being the most obvious choice to me. The other is taking the reference frame of the observer (known as aether dragging). The main point is neither work.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/44568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.