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<p>I have been looking all over and I can't find a detailed specification about the <a href="http://en.wikipedia.org/wiki/Mars_Science_Laboratory" rel="nofollow">Curiosity Rover</a>.
The length is about 10 feet but it would be cool to find out the dimensions of the wheels, clearance and so on.</p>
<p>Anybody know where I would find this information?</p>
| 3,457 |
<p>In <a href="http://astronomy.stackexchange.com/q/4/13">one of my questions</a> I asked about the career paths to become a professional astronomer. Now let's consider the next logical step. What if a person has gained his/her PhD in Astronomy and had an opportunity to do a couple of postdoc positions. The time comes for independent work but where
can the funding come from? What are the funding possibilities in different countries to continue the career as an independent researcher?</p>
<ul>
<li>Please provide links and general advice if there is for each funding agency</li>
<li>Keep it short and to the point</li>
</ul>
| 3,458 |
<p>I'm writing a paper and I used them as an example, but then reconsidered . . . maybe I'm not getting it right!
thanks!</p>
| 3,459 |
<p>If we set the <a href="http://en.wikipedia.org/wiki/Boltzmann_constant">Boltzmann constant</a> to $1$, then entropy would just be $\ln \Omega$, temperature would be measured in <a href="http://en.wikipedia.org/wiki/Joule">$\text{joules}$ ($\,\text{J}\,$)</a>, and average kinetic energy would be an integer times $\frac{T}{2}$. Why do we need separate units for temperature and energy?</p>
| 852 |
<p>I just read this article at <a href="https://medium.com/the-physics-arxiv-blog/7ef5eea6fd7a">https://medium.com/the-physics-arxiv-blog/7ef5eea6fd7a</a> about the work of a physicist called Bolotin, that states that P!=NP (from computer science) implies that large quantum mechanical objects are not possible.</p>
<p>The author starts by explaining the Schrodingers cat thought experiment. He then says "Nobody knows why we don’t observe these kinds of strange superpositions in the macroscopic world", which I find strange because you cannot observe a superposition, you either see that the cat is alive or dead.</p>
<p>But then he writes "For some reason, quantum mechanics just doesn’t work on that scale. And therein lies the mystery, one of the greatest in science." I thought that the correspondence principle actually explains nicely why quantum mechanics works very well on a large scale?</p>
<p>The main point of the article is that if you can have a large quantum mechanical system, you would get P=NP, which is believed to be not true by most computer scientists. But isn't that exactly what they try to do with quantum computers?</p>
| 3,460 |
<p>Consider a Pseudo-Riemannian Manifold with signature</p>
<p>$$ (\underbrace{+,\cdots,+}_p,\underbrace{-,\cdots,-}_q) $$</p>
<p>For any positive integers $p$ and $q$. Can this kind of manifold contain closed timelike curves (CTCs)? I know that if $p=1$, then we get a Lorentzian Manifold that can't contain CTCs, but I am interested in the cases where $p>1$.</p>
| 3,461 |
<p>Here is a picture of the usual vigorous Niagara Falls (in the winter).
<img src="http://i.stack.imgur.com/PgwrA.jpg" alt="enter image description here"></p>
<p>Here is the picture of Niagara Falls frozen in 1933 (in the very cold winter).
<img src="http://i.stack.imgur.com/FZB1W.jpg" alt="Niagara Falls frozen in 1933"></p>
<p>Here is the picture of Niagara Falls frozen in 1911 (in the extremely cold winter).
<img src="http://i.stack.imgur.com/lcaLv.jpg" alt="enter image description here"></p>
<p>Background knowledge: <a href="http://en.wikipedia.org/wiki/Melting_point" rel="nofollow">The Melting/Freezing point of the static water at 1 atm is 32°F (0.0°C or 273.15 K)</a>.</p>
<p>$\bullet$ I am curious to know under what conditions of the air pressure(atm), temperature, <a href="http://en.wikipedia.org/wiki/Freezing-point_depression" rel="nofollow">solute density in the water</a> <strong>would cause the Niagara fall frozen?</strong></p>
<p>Are other conditions like: winds, snowing or not, the strength of <a href="http://en.wikipedia.org/wiki/Van_der_Waals_force" rel="nofollow">Van der Waals force</a>, the moving speed/velocity of water and the pulling force of gravity (acceleration $g$), the height $H$ of waterfalls would be some important factors to predict <strong>the conditions to freeze the waterfalls</strong>?</p>
<p>Is there some simple qualitative relations or quantitative formulas describing the freezing point for waterfall? <strong>A $P$(pressure),$T$(temperature), etc, $\dots$ curve for the water-ice phase transition on the phase diagram?</strong> Perhaps the answer is very simple(?), but it is not entirely trivial to me.</p>
<p>ps. Happy winter/holiday times for those who reading the post, and especially those residing on the cold side of the north-hemisphere. Thanks for comments/replies!</p>
| 3,462 |
<p>Consider a triple-beam balance, like so:</p>
<p><img src="http://www.microscopesamerica.com/catalog/images/images/Large%20image%20Ohaus%20Triple%20Beam%20Balance%201610-00%20Cup%20Platform.png" alt="enter image description here"></p>
<p>An unknown mass is placed on the left pan, and the provided weights are moved on the right until the lever arm comes to rest at an exactly horizontal position, as indicated by a line on the rightmost tip of the lever arm. Often, when I'm using the device and I'm close to the equilibrium point, the lever arm will come to rest slightly higher or lower than horizontal, so I'll adjust the weights accordingly until the lines align perfectly.</p>
<p>However, I don't see why the arm must be perfectly horizontal to read an accurate value. Isn't a stationary lever -- no matter what the angle -- an indication that there is no net torque on the system? If so, wouldn't that demonstrate that the masses are balanced in an appropriate way?</p>
<p>In other words, why doesn't a lever which is stationary in a non-horizontal position remain stationary when placed horizontally (and vice versa), since the net torque is zero in both cases?</p>
| 3,463 |
<p>I recently read that the mass we deal with in Equation $F=Ma$ is called inertial mass and the mass we deal with in $F=Mg$ is gravitational mass. Suppose I am taking a same ball in a free fall and in accelerating it on the smooth horizontal plane the mass of ball should be mass. What is fundamental difference between the gravitational and inertial mass? I have also read that The inertial mass and the gravitational mass are equivalent and the general theory of relativity is based on this logic.So there must be something fundamental difference between definitions of inertial mass and gravitational mass otherwise how can a law be based on the equivalence between the above two. (I want to say that if they are the same than there must not be any logic related to their equivalence which can formulate the general theory of relativity because identicals are equivalent is a simple rule of maths no other thing can be concluded from it.) </p>
| 88 |
<p>When a clock is transported here and there into space and then brought to the same place it differs with the other clock.
When particles are accelerated with high speeds and then brought to rest their mass again gets back to its original rest mass.
Why?</p>
<p>Answer allegorically please.</p>
| 3,464 |
<p>I was browsing through <em>Foundations of Space and Time,</em> a compilation of essays on various theories of quantum gravity. The following passage in the introduction intrigued me:</p>
<blockquote>
<p><em>Each compactification leads to a different vacuum state.... at least one state should describe our Universe in its entirety.... the enormous number (~10^500 at last count) of solutions, with no perturbative mechanism to select mechanism to select among them, leads some critics to question the predictive power of the theory..Even more worrying is that, while the theory is perturbatively finite order by order, the perturbation series does not seem to converge.</em></p>
</blockquote>
<p>I don't know anything about string theory and so I could not make head or tails this. All I know is that ~$10^{500}$ is a very large number. </p>
<ol>
<li><p>What exactly is a 'solution' in string theory? Is it a spacetime metric of some sort or the terms of a S-matrix of some sort? </p></li>
<li><p>Why are there so many 'solutions'? </p></li>
<li><p>I thought string theory was supposed to be finite, why do perturbative series still diverge?</p></li>
<li><p>Is there any experimental technique to limit the number of 'solutions'? </p></li>
<li><p>Will experimental techniques be able to pinpoint a solution within present day string theorists' lifetimes too? If not, how long will it take before we can experimentally probe these things? </p></li>
<li><p>Are string theorists completely relaxed about these issues? Or are they in anguish? </p></li>
</ol>
| 3,465 |
<p>I am an educator, and I am looking for a specific video. In the video, they ask some middle school students and some college graduates about why the moon has phases. Most of the students in both the groups get the answer wrong, saying that the phases of the moon happen because of the shadow of the Earth. I remember that they also interviewed the teacher of the middle school students, and she was really perplexed that her students didn't know the right answer.</p>
<p>Another thing I remember from the video was that the middle school kids were not sure of their answers, but the college graduates (who are at their graduation ceremony) were really sure.</p>
<p>I want to use that video for educational purposes, but I can't find it. Can you please help?</p>
| 3,466 |
<p>After reading much online I've decided to by a 10x50 porro prism Binoculars. The one I have in mind <a href="http://www.letsbuy.com/celestron-upclose-10x50-p-34652" rel="nofollow">http://www.letsbuy.com/celestron-upclose-10x50-p-34652</a> has BK7 prisms. The one I would like to buy is this: <a href="http://www.letsbuy.com/olympus-10x50-dps-i--p-34842" rel="nofollow">Olympus 10x50 DPS I</a> but it is never available and a tad bit more expensive. Is buying binoculars or telescopes online a bad thing as I've read you need to inspect them carefully before buying? Also will the prism difference (BK7 vs BAK4) make much of a difference for the 10x50 range?</p>
| 3,467 |
<p>How are interplanetary trajectories that involve gravity assist maneuvers found?</p>
<p>Examples:</p>
<ul>
<li>the MESSENGER spacecraft made flybys of Earth, Venus, and Mercury before getting into orbit</li>
<li>the Juno spacecraft will flyby Earth (returning after launch) before continuing to Jupiter</li>
<li>many more...</li>
</ul>
<p>Are these found using a brute force search of virtually all possible paths or is there a more direct method for finding these helpful alignments?</p>
<p>If I decided to leave for Neptune today (using gravity assist), how would I find a good route and are such tools available on the desktop?</p>
| 3,468 |
<p>Assuming that a person has understanding of theory of Lie groups, Lie algebras and basic quantum mechanics, what is the simplest route to gain a basic understanding of the SM of particle physics? Are there any particular books suited for people with this background?</p>
| 89 |
<p>I want to make a concave parabolic mirror that will have a focal point six inches below where I place it (on the edge of the sidewalk) so that it will melt the ice collecting in the gutter below and clear a path for the water to drain. I will probably be using shiny stainless steel for the reflector, so it won't be a fully reflective mirror. What would be the ideal size, and how do I calculate the curve needed to concentrate the heat to that focal point?</p>
<p>Edit: Here's a real life example of someone doing this unintentionally:
<a href="http://www.cnn.com/2013/09/03/world/europe/uk-london-building-melts-car/" rel="nofollow">Reflected light from London skyscraper melts car</a></p>
| 3,469 |
<p><img src="http://i.stack.imgur.com/QoSeL.png" alt="enter image description here"></p>
<p>Sorry for the ugly picture but it makes my question more understandable.
The $\Delta V$ from $A$ to $B$ is calculated by$$\int_A^B E \, \mathrm{d}r$$ where $r$ is the distance between $A$ and $B$. The curve line is the real path the charge take going from $A$ to $B$. Suppose that the green arrow is the $\vec E$ and the red arrow is $dl$ at some arbitrary point on the path the charge move.
From what I learned from class
$$\int_A^B E \, \mathrm{d}r = \int_A^B E \, \mathrm{d}l$$
This is the point I don't understand and make me reconsider what I learn about <a href="http://en.wikipedia.org/wiki/Work_%28electrical%29" rel="nofollow">work</a> in <a href="http://en.wikipedia.org/wiki/Work_%28physics%29" rel="nofollow">physics</a>. Can someone explain why $\int_A^B E \, \mathrm{d}r = \int_A^B E \, \mathrm{d}l$</p>
<p>Does the calculation of the work involve vector and $E\, \mathrm{d}l$ here is the dot product of $\vec E$ and $\vec {\mathrm{d}l}$ or it is calculated by projecting all the force exert to the object to the axis created by start point and end point?</p>
| 3,470 |
<p>In history there was an attempt to reach (+, +, +, +) by replacing "ct" with "ict", still employed today in form of the "Wick rotation". Wick rotation supposes that time is imaginary. I wonder if there is another way without need to have recourse to imaginary numbers.</p>
<p>Minkowski spacetime is based on the signature (-, +, +, +). In a Minkowski diagram we get the equation: $$ \delta t^2 - \delta x^2 = \tau^2 $$
Tau being the invariant spacetime interval or the proper time.</p>
<p>By replacing time with proper time on the y-axis of the Minkowski diagram, the equation would be
$$ \delta x^2 + \tau^2 = \delta t^2$$
In my new diagram this equation would describe a right-angled triangle, and the signature of (proper time, space, space, space) would be (+, +, +, +).</p>
<p><img src="http://i.stack.imgur.com/YVF18.jpg" alt="enter image description here"></p>
<p>I am aware of the fact that the signature (-, +, +, +) is necessary for the majority of physical calculations and applications (especially Lorentz transforms), and thus the (+, +, +, +) signature would absolutely not be practicable.( Edit: In contrast to some authors on the website about Euclidian spacetime mentioned in alemi’s comment below) </p>
<p>But I wonder if there are some few physical calculations/ applications where this signature is useful in physics (especially when studying the nature of time and of proper time).</p>
<p>Edit (drawing added): Both diagrams (time/space and proper time/space) are observer's views, even if, as it has been pointed out by John Rennie, dt is frame dependent and τ is not.</p>
| 3,471 |
<ol>
<li><p>In particle mechanics where the notion of rotation does not apply, particles are said to be in static equilibrium when the sum of the external forces acting on the particle of interest in all directions equals zero.</p></li>
<li><p>In rigid body mechanics where the notion of rotation applies (or at least the concept of invariant distances between points), rigid bodies are said to be in static equilibrium when (1) the sum of the external forces in all directions equals zero, and (2) the sum of the external moments in all directions equals zero.</p></li>
</ol>
<p>My question is: in rigid body mechanics, where does the equation on moments come from? Is it a principle similar to Newton's law in particle mechanics or can it be proven from other assumptions (like, for instance, the invariance of distances separating two points)</p>
| 3,472 |
<p>I am having a very basic problem understanding the idea of detailed balance, particularly in the context of the Ising model. Most references I have found contain the following phrase:</p>
<p>"In equilibrium, each elementary process must be equilibrated by its reverse process".</p>
<p>What does it mean for one process to equilibrate another?</p>
<p>In particular, I am trying to understand the section here: <a href="http://en.wikipedia.org/wiki/Ising_model#Algorithm_Specification" rel="nofollow">http://en.wikipedia.org/wiki/Ising_model#Algorithm_Specification</a> in which they state that the given form of selection probabilities is required by detailed balance, but I suspect that that will follow once I understand what detailed balance actually means.</p>
<p>Thanks!</p>
| 3,473 |
<p>I am working on a textbook problem of a grounded conductor inside a uniform electric field. The textbook states that "grounded" means potential = 0. In my opinion, "grounded" should mean "same potential as infinity". But in this case we can't set potential at infinity equal zero. So my question is, what is actually the meaning of "grounded". Am I right that it means "equal potential with infinity"? What does it mean by "grounded" in this question when the potential at infinity can't be set to zero?</p>
| 3,474 |
<p>Where does the energy of a light bulb come from?</p>
<p>Is it from the coil of wire, magnet in the generator, mechanical input to the generator or plug where the generator is connected to the wall?</p>
| 3,475 |
<p>To better explain my question, I will need to give a brief description of the configuration used in 2D MEMS switches.</p>
<p>So, the next figure shows a configuration of a 2D MEMS switch, a light beam arrives from a fiber to the input port, travels inside the switch until it meets a mirror that is in the standing state to reflect the beam to the output port.</p>
<p><img src="http://i.stack.imgur.com/bSdBJ.jpg" alt="2D MEMS Switch"></p>
<p>The output beam from the fiber is wide, therefore, they collimate it using a collimator at the input. This collimator is picked such that the beam waist is minimum at the mirror of the worst case scenario (i.e., longest path).</p>
<p>This path is shown in the next figure.</p>
<p><img src="http://i.stack.imgur.com/QVZHy.jpg" alt="Analysis of the beam Gaussian divergence"></p>
<p>During the analysis of the power loss, there is a loss due to the imperfection in the reflection of the mirror, which ranges from 1% to 3%, and there is a loss due to the Gaussian beam divergence and the existence of the mirror 1d in the beam's path. According to the papers, "The optical signal loss due to Gaussian-beam divergence for a mirror of radius R at distance z is":</p>
<p><img src="http://i.stack.imgur.com/hCE7I.jpg" alt="loss due to Gaussian beam divergence"></p>
<p>Usually, they maintain a ratio R/W(z) of 1.5 to 2.</p>
<p>Note that the beam propagates in the free space from the input port to the mirror and from the mirror to the output port. This is because mirrors other than the mirror 1d are in the sleeping state (i.e., they are out of the beam's path)</p>
<p>My question is:</p>
<p>If we assume a case in which the beam propagates in a similar configuration, however, at the positions of the mirrors from 1a to 1c, there are switching elements with the same positions and dimensions made of glass such that the beam propagates through them. </p>
<p>What would be the impact of these elements from 1a to 1c on the loss due to the Gaussian beam divergence? Would it be the summation of the losses using the provided equation by at different distances, in other words</p>
<p>$L_{Gauss(at\ 1a)}+L_{Gauss(at\ 1b)}+L_{Gauss(at\ 1c)}+L_{Gauss(at\ 1 d)}+...$</p>
<p>If not, then why? and how can I calculate it? </p>
<p>Reference:
CY Li et al., "Using 2x2 switching modules to build large 2-D MEMS optical switches"</p>
| 3,476 |
<p>According to Newton's law of action-reaction, there is a reaction force for the action force. He did not say when the reaction appears, whether immediately or with a delay.</p>
<p>Could you tell me the truth?</p>
<p>If the reaction appear immediately, does it disobey the relativistic principle that the changes must not be faster than the speed of light.</p>
| 3,477 |
<p>Two particles $A, B$ are travelling along parallel straight paths. At some point, the velocity of $A$ exceeds that of $B$. Does this <em>necessarily</em> mean that the acceleration of $A$ is greater than the acceleration of $B$?</p>
<p>If you look at the $v - t$ graph of the two particles, the lines would intersect. Probably, starting off, the velocity of $B$ would be greater, but since the slope of the velocity of $A$ would be greater it would intersect with the graph of $B$ and exceed it. I couldn't think of any other situation. So, my conclusion was that the acceleration has to be greater. But my textbook says otherwise. How come?</p>
<p>EDIT: This is question 13 from chapter 2 in Resnick halliday physics. </p>
<p>To clarify: the problem does NOT assume that initally A's velocity was lower than B's. (See comments)</p>
| 3,478 |
<p>This is a bit of an odd question. I'm not a physicist, so bear with me if I say something wrong.</p>
<p>Lets say you have some sort of quantum event where matter is in a superposition. Standing next to you is another scientist waiting to observe the results (and, in theory, collapse the suposition). You go to get a cup of coffee while your fellow scientist stays in the room. Your fellow scientist observes the result of the experiment while you are out of the room but does not tell you what the result was. There are three possible options here for when you return to the room:</p>
<ol>
<li>The matter is still in a superposition for both you and the other scientist.</li>
<li>The matter is no longer in a superposition for either of you (even though you have not observed the event and have no knowledge of what happened).</li>
<li>The matter is in a superposition only to you.</li>
</ol>
<p>So which option is it? If the answer is option 2, would that mean that we as humans have some kind of "superpower" forcing quantum superpositions to collapse as soon as we see them?</p>
| 3,479 |
<p>In the <a href="http://en.wikipedia.org/wiki/Standard_Model">standard model</a> of particle physics, there are three generations of quarks (up/down, strange/charm, and top/bottom), along with three generations of leptons (electron, muon, and tau). All of these particles have been observed experimentally, and we don't seem to have seen anything new along these lines. A priori, this doesn't eliminate the possibility of a fourth generation, but the physicists I've spoken to do not think additional generations are likely.</p>
<blockquote>
<p><strong>Question:</strong> What sort of theoretical or experimental reasons do we have for this limitation?</p>
</blockquote>
<p>One reason I heard from my officemate is that we haven't seen new neutrinos. Neutrinos seem to be light enough that if another generation's neutrino is too heavy to be detected, then the corresponding quarks would be massive enough that new physics might interfere with their existence. This suggests the question: is there a general rule relating neutrino masses to quark masses, or would an exceptionally heavy neutrino just look bizarre but otherwise be okay with our current state of knowledge?</p>
<p>Another reason I've heard involves the Yukawa coupling between quarks and the Higgs field. Apparently, if quark masses get much beyond the top quark mass, the coupling gets strong enough that QCD fails to accurately describe the resulting theory. My wild guess is that this really means perturbative expansions in Feynman diagrams don't even pretend to converge, but that it may not necessarily eliminate alternative techniques like lattice QCD (about which I know nothing).</p>
<p>Additional reasons would be greatly appreciated, and any words or references (the more mathy the better) that would help to illuminate the previous paragraphs would be nice.</p>
| 959 |
<p>How is the following classical optics phenomenon explained in quantum electrodynamics?</p>
<ul>
<li>Color</li>
</ul>
<p>According to Schroedinger's model of the atom, only particular colors are emitted depending on the type of atom and the state of its electrons. Then, how it is possible that the same type of matter changes color in relation to temperature? For example, water is transparent with a light blue tint, but snow is white (which means, pretty much, that photons have no particular color in the visible spectrum).</p>
<p>Split by request: See the other part of this question <a href="http://physics.stackexchange.com/questions/2041/how-are-classical-optics-phenomena-explained-in-qed-snells-law">here</a>.</p>
| 3,480 |
<blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="http://physics.stackexchange.com/questions/2023/is-it-safe-to-use-any-wireless-device-during-a-lightning-storm">Is it safe to use any wireless device during a lightning storm?</a> </p>
</blockquote>
<p>We often hear the tale that a person gets stuck by lighting when he is unfortunate enough to use a cell phone outdoor.</p>
<p>But why is this so?</p>
| 90 |
<p>From <a href="http://en.wikipedia.org/wiki/Kinetic_theory" rel="nofollow">Wikipedia entry on Kinetic Theory</a></p>
<blockquote>
<p>The temperature of an ideal monatomic gas is a measure of the average kinetic energy of its atoms. </p>
</blockquote>
<p>Now if I remove all the particles from the box shown below will the temperature be zero?</p>
<p><img src="http://i.stack.imgur.com/3E0Ak.gif" alt="alt text"></p>
| 3,481 |
<p>I have a ball attached to a spring and the spring is attached to a wall. There is no gravity for simplicity. In the rest RF the oscillating ball energy is conserved: T + U = const. In a moving RF it is not conserved. I would like to see the shortest answer to the question "Why?".</p>
<p>| <-- --><br>
|/\/\/\/\/\/\/<strong>O</strong><br>
|</p>
<p>=====> $V_{RF}-->$ </p>
<p><strong>Edit:</strong> For those who has doubt - I choose the moving RF velocity equal to the maximum ball velocity, for example. Then the ball with its maximum velocity in the rest RF looks as still and does not have any potential energy in the moving RF (the maximum velocity is attained at the equilibrium position where no force acts). So T = 0 and U = 0. </p>
<p>At the farthest right position the ball looks as moving with $-V_{RF}$ and having a potential energy $kx^2/2$, both energy terms are positive ($v^2 > 0, x^2 > 0$). So now the total energy of the ball is positive and thus is not conserved. It oscillates from zero to some maximum value. Why?</p>
| 3,482 |
<p>Is 89MHZ station emitting photons of 89MHZ frequency? (I mean $\nu$ in $E=h\nu$).</p>
| 3,483 |
<p>A photon emitted from a receding source (Doppler redshift) has less energy when detected at an observer's location. Please explain the energy loss from the perspective of energy conservation.</p>
| 3,484 |
<p>Is there any explanation on a qualitative level why we can see in the observed magnitude vs. redshift z plot that the universe is expanding accelerated? See for example here:
<a href="http://hyperphysics.phy-astr.gsu.edu/hbase/astro/univacc.html" rel="nofollow">http://hyperphysics.phy-astr.gsu.edu/hbase/astro/univacc.html</a></p>
<p>The observed magnitude is, if the absolute magnitude is well know as it is the case for Supernova 1a, a measure for the distance. The redshift z is a measure for how much the object hast moved relative to us., i.e. also some kind of distance indicator. My problem is that I can't get together how we now conclude that we have an accelerated expansion, besides the best fit has $\Omega_{\Lambda}\neq 0$. Any help or link would be awesome.</p>
| 3,485 |
<p>A firecracker explodes at the origin of an inertial reference frame. Then, 2.0 microseconds later, a second firecracker explodes 300m away. Astronauts in a passing rocket measure the distance between the explosions to be 200m. According to the astronauts, how much time elapses between the two explosions?
Okay. My textbook answers this question using spacetime interval consistency which is simple.
I don't understand why the time dilation formula does not work:
$$Δt = \frac {Δτ}{ \sqrt{1-\frac{v^2}{c^2}}}$$
Where v is simply the ratio between 300m and 2 microseconds.
Δτ=?
Δt=2x10^-6 s</p>
| 3,486 |
<p>As we know that matter-antimatter asymmetry is one of unsolved problems in physics. One possible solution to this problem is given as baryogenesis which produce asymmetry in rate of creation between matter and antimatter particles. But doesn't alternate solution like "different regions of space with different type of particles " holds more ground..</p>
<p>Means in one region matter particle dominates (the region we live ) and another region anti matter particle will dominate. When particle-antiparticle pair created from energy (at that energy is so much high that particle- antiparticle pair could be created from energy according to $E=mc^2$) before they meet and annihilate with each other they also has equal chance of meeting with same type of particle in neighborhood, because after some time of big bang the four force were united and gravity was as powerful as remaining three forces. So now in 50% of all pairs destroyed by annihilation and other 50% clumps together with same type of particle means matter with matter and antimatter with antimatter and then inflationary epoch throw these clumps from each other at very large distance so that they can not meet each other and have chance of annihilation. And we now can not see these regions because of accelerated expansion of universe because of dark energy. </p>
<p>Doesn't this hypothesis is more or at least equally valid with baryogeneis hypothesis.</p>
| 3,487 |
<p>I am doing some research to see if there is a simple equation (or an equation that can be simplified) to estimate the volume of a drop of liquid falling due to gravity based upon the approximate flow rate, and perhaps the viscosity/surface tension of the fluid.</p>
<p>Basically this is a drip chamber fluid set for delivering medications. Clinicians routinely calculate fluid flow based upon the number of drops per minute assuming a fixed drop size. However since drop size varies with flow rate. Using a fixed volume size has some amount of inaccuracy. I would like to see if I can be more accurate in determining actual flow rate.</p>
| 3,488 |
<p>I am reading Page 435, <em>General Relativity</em> by Wald.</p>
<p>Let $T^*\subset V^*$ be a subspace of the dual tangent space of a manifold, $W\subset V$ be the subspace of the tangent space annihilated by $T^*$, i.e. for every $\omega\in T^*$ and every $Y\in W$ we have $\omega_aY^a=0$.</p>
<p>Now the author says the <a href="http://en.wikipedia.org/wiki/Frobenius_theorem_%28differential_topology%29" rel="nofollow">Frobenius's theorem</a> implies
$$
2Y^aZ^b\nabla_{[b}\omega_{a]}=0
$$
for every $Y,Z\in W$ and every $\omega\in T^*$.</p>
<p>Then the author claims it holds if and only if</p>
<blockquote>
<p>$$ \nabla_{[b}\omega_{a]}=\sum_{\alpha=1}^{n-m}\mu^\alpha_{[a}
v^\alpha_{b]} $$ where each $v^\alpha$ is an arbitrary one-form and
each $\mu^\alpha\in T^*$</p>
</blockquote>
<p>Can anyone explain to me why?</p>
<hr>
<p>Also, the author later says </p>
<blockquote>
<p>$\nabla_{[a}\xi_{b]}=\xi_{[a}v_{b]}$ for some one-form $v$ if and only
if $\xi_{[a}\nabla_b\xi_{c]}=0$.</p>
</blockquote>
<p>I can work out the "=>" part, but don't know why "<=" holds. Could anyone help?</p>
| 3,489 |
<p>Visible light emitted or reflected from the objects around us provides information about the world. </p>
<p>If I sit in a dark room, and see the bright room outside, I am able to see all the objects of that bright room. In other words, it mean light emitted or reflected from those objects of the bright room are entering into the dark room. Shouldn't that make my room bright? But, my room is dark and I am able to see those objects of bright room, both are contrary? </p>
<p>Edit: If we consider another situation like the one in the picture below, lets assume that we move as far from the street light, that even in the presence of street light the place remains as dark as it was before. I believe even then we can see the street light, won't we? </p>
<p>If we agree that we still can see the lamp, it mean that we are able to see the lamp, without the ray from lamp hitting our eye? Won't it be contrary?
<img src="http://i.stack.imgur.com/ChAQ6.jpg" alt="enter image description here">
Edit: Lets consider a laser beam in the dark room (in the picture room is not totally dark, but assume a dark room). Look the laser in the same angle as shown in the picture, as laser remains narrow over long distance, I hope we can assume light is not spreading in other directions. But I hope we can see it even though light is not spreading and reaching me, how is it possible?, I hope this made my question more clearer. Sorry, I am unable to understand the concept, if this question is silly, pardon me.</p>
<p><img src="http://i.stack.imgur.com/uHxJx.jpg" alt="enter image description here"></p>
<p>Sometimes I might have misunderstood the concept, if so, pardon me and explain.</p>
| 3,490 |
<p>In a discussion concerning:</p>
<p><a href="http://physics.stackexchange.com/questions/87332/physical-meaning-of-non-trivial-solutions-of-vacuum-einsteins-field-equations">Physical meaning of non-trivial solutions of vacuum Einstein's field equations</a></p>
<p>there were a number of answers claiming that the flatness of the Ricci space (Rµv=0) does not necessarily entail the flatness of the Riemannian spacetime. This is explained by the fact that besides the Ricci tensor, the Riemann curvature tensor depends also on the Weyl tensor said to govern the propagation of gravitational radiation. And although Rµv=0, the Weyl tensor can still be non-zero.</p>
<p>That being so, it seems that there are some gravitational forces not included in Einstein's field equations, as the Ricci tensor (Rµv) and the Einstein's tensor (Gµv) are said to be trace-reversed (meaning that when one of them vanishes, the other does too). So, if Rµv=0 does not exclude the presence of gravitation - gravitational radiation can still be found through Weyl tensor - then (since Gµv=0) Einstein's gravitational field equations must be incomplete?</p>
<p>Yet there immediately arises another - and in my opinion, more important - question:</p>
<p><strong>What is the source of this gravitational radiation, since Rµv=0 means the whole universe is void of matter?</strong></p>
| 3,491 |
<p>Why we don't use <a href="http://www.ask.com/question/how-do-permanent-magnet-generators-work" rel="nofollow">this</a> in our houses to generate the electricity? Isn't the electricity generated by the magnets strong enough? Why do we have to pay for electricity if we can just create it by pushing the magnets attached to a rotor of the generator?</p>
| 3,492 |
<p><a href="http://en.wikipedia.org/wiki/Principle_of_relativity" rel="nofollow">Wikipedia</a> has this quote:</p>
<blockquote>
<p><em>Special principle of relativity:</em> If a system of coordinates K is
chosen so that, in relation to it, physical laws hold good in their
simplest form, the same laws hold good in relation to any other system
of coordinates K' moving in uniform translation relatively to K. —
Albert Einstein: The Foundation of the General Theory of Relativity,
Part A, §1</p>
</blockquote>
<p>Does this simply mean that any sound theory expressed in K, should be able to withstand a transfer to another system Z and still hold "true"? Or is there more to "uniform translation relatively to K"?</p>
| 3,493 |
<p>The area of a circle is $\pi r^2$ if you increase $r$ the area will increase by the square so if this area was of energy and you increase the area it is dispersed you would expect its energy to weaken by the square. Is this the intuition behind gravity weakening be the square of the distance?</p>
| 91 |
<p>For a discrete symmetry: At the minimum value of the potential, $V$, in the Lagrangian density, why do we take $\phi= \langle v\rangle + \eta$? Aren't we deliberately breaking the symmetry? If we don't do this, the symmetry is intact.
On the other hand, if we replace $\phi$ by $\phi= \langle v\rangle + \eta$, even when not in a ground state, then also the symmetry will be broken in $\eta$ field. Please explain. </p>
| 3,494 |
<p>There's a guide here on how to seal a weather balloon after filling:
<a href="http://ukhas.org.uk/guides:sealing_the_balloon" rel="nofollow">http://ukhas.org.uk/guides:sealing_the_balloon</a>
<a href="http://ukhas.org.uk/_media/guides:neck8.jpg" rel="nofollow">http://ukhas.org.uk/_media/guides:neck8.jpg</a></p>
<p>Won't the static electricity from the duct tape cause the hydrogen gas to ignite (there's always a little air (oxygen) in the balloon before filling, plus, a very little amount can leak from the filler)?</p>
| 3,495 |
<p>I was always told that to find whether or not a vector field is <a href="http://en.wikipedia.org/wiki/Conservative_vector_field" rel="nofollow">conservative</a>, see if the curl is zero.</p>
<p>I have now been told that just because the curl is zero does not necessarily mean it is conservative. However, if it is non-zero it is definitely not-conservative. </p>
<p>IE: $\vec{F}=\frac{K}{r^3}\hat{r}$ (Spherical coordinates, $K$ is constant)</p>
<p>To show this is conservative I would go ahead and take the curl. It will be zero - but that is not definitive proof it is conservative? How would I show it is? </p>
| 3,496 |
<p>So, in doing some numerical computations in QFT, I've run into the following Wigner <a href="http://mathworld.wolfram.com/Wigner6j-Symbol.html" rel="nofollow">6j-Symbol</a>: </p>
<p>$
\left\{
\begin{array}{ccc}
x & J_1 & J_2 \\
\frac{N}{2} & \frac{N}{2} & \frac{N}{2} \\
\end{array}
\right\}
$</p>
<p>In the regime where $x \ll J_1,J_2,N$ and $J_1 \approx J_2 \approx N$, and $N$ is large. I would like to know if there is an asymptotic formula for such a symbol, or if one can be derived. Using symmetries we can get</p>
<p>$
\left\{
\begin{array}{ccc}
x & \frac{1}{2} \left(J_1+J_2\right) & \frac{1}{2} \left(J_1+J_2\right) \\
\frac{N}{2} & \frac{1}{2} \left(N+J_1-J_2\right) & \frac{1}{2} \left(N-J_1+J_2\right) \\
\end{array}
\right\}
$</p>
<p>Perhaps this could help, I'm really not sure.</p>
| 3,497 |
<p>The Legendre transformation defines the helmholtz free energy (at least according to my lectures) as:</p>
<p>$F(T,V,N)=E-TS$</p>
<p>It also says to start with</p>
<p>$E(S,V,N)$ and $T=\frac{\partial{E}}{\partial{S}}$</p>
<p>My question is I have a relation for entropy (I won't give it because I want this to be more of a concept based question), in terms of the standard $S(U,N,V)$ where U is the energy, I can re-arrange for energy easily. </p>
<p>Obviously the E and T to put into the equation are pretty standard it quite clearly states to use the function for each but the ENTROPY, do I just write the variable S next to the partial of E wrt S or do I use the relation I have for Entropy?</p>
| 3,498 |
<p>How does one solve the tensor differential equation for the relativistic motion of a partilcle of charge $e$ and mass $m$, with 4-momentum $p^a$ and electromagnetic field tensor $F_{ab}$ of a constant magetic field $\vec B$ perpendicular to the plane of motion. $$\frac{dp^a}{d\tau}=\frac{e}{m}F^a{}_bp^b$$
?
Let the the initial condition be $$p^a=(E_0 ,\vec 0)$$</p>
<p>I can see that the differential equation resembles that of a SHM equation or a cosh, sinh one if it's a scalar equation. However, I don't know how to deal with a tensor equation. Could anyone please explain? Thank you.</p>
| 3,499 |
<p>We need to know two of the following three to calculate the third: redshifted color, baseline color, and velocity. The velocity is related to the difference between the redshifted color and the objects baseline color. How do we know the baseline color of distant objects to know the amount of redshifting?</p>
| 3,500 |
<p>If a radio tunes to a specific frequency, where does the excess energy go? If one continues to hit the resonant frequency, shouldn't the wire begin to melt at some point from too much energy?</p>
| 3,501 |
<p>Isn't a magnetic field basically an electric field originating from a line charge, but as seen in a boosted frame? I mean, that's where the relativistic explanation comes from right?</p>
| 3,502 |
<p>The question is in the title.</p>
<p>If it is possible, what are some examples of gapped systems--either quantum field theories or condensed matter systems--which exhibit some kind of anomaly when coupled to a metric with curvature or placed on a spacetime with non-trivial topology?</p>
| 3,503 |
<blockquote>
<p>A positive point charge $Q$ is kept eccentrically inside a neutral conducting shell. An external uniform field E is applied. Then:</p>
<p>a) Force on Q due to E is zero</p>
<p>b) Net force on Q is zero</p>
<p>c) Net force acting on Q and conducting shell considered as a system is zero</p>
<p>d) Net force on the shell due to E is zero</p>
</blockquote>
<p>My try:</p>
<p>Since the shell is neutral, thus net force on it by E must also be zero.
Also since Q is electrostatically shielded(by placing it inside a shell), thus net force on it must also be zero.</p>
<p>But the answer I'm getting(as you might have guessed) is wrong(that's why I've asked this question).</p>
<p>Please point out the flaw in my reasoning.</p>
| 3,504 |
<p>In traditional mirrors, some of the input light is absorbed by atoms in the mirrors surface and are 'lost' as heat, degrading the quality of the reflected image.</p>
<p>Could this loss be compensated by an array of "powered elements" arranged to reflect nearby photons repelling them with some sort of electo-magnetic force, to completely prevent loss? </p>
<p><em>This would realize the <a href="http://en.wikipedia.org/wiki/Perfect_mirror" rel="nofollow">perfect mirror</a>, reflecting all light back.</em></p>
<p><img src="http://i.stack.imgur.com/OD5FI.png" alt="enter image description here"></p>
| 3,505 |
<p>A Wick rotation is a transformation that allows to change from a Lorentzian manifold to a Riemaniann manifold. In the cases when this is possible, is the Levi-Civita connection of the Riemaniann manifold the same connection as the initial Lorentzian manifold? </p>
| 3,506 |
<p><a href="http://en.wikipedia.org/wiki/Photon#Wave.E2.80.93particle_duality_and_uncertainty_principles">Wikipedia</a> claims the following:</p>
<blockquote>
<p><em>More generally, the normal concept of a Schrödinger probability wave function cannot be applied to photons. Being massless, they cannot be localized without being destroyed; technically, photons cannot have a position eigenstate and, thus, the normal Heisenberg uncertainty principle does not pertain to photons.</em></p>
</blockquote>
<p>Edit:</p>
<p>We can localize electrons to arbitrarily high precision, but can we do the same for photons? Several sources say "no." <a href="https://docs.google.com/viewer?a=v&q=cache%3adjf1u49p9WcJ%3abayes.wustl.edu/etj/articles/review.extended.prob.pdf%20&hl=en&gl=us&pid=bl&srcid=ADGEESg4snrOkd4hCQkvjZxcHcZFR-O5x7nZ90plCgW6lKNsP7BlAocqZBE5pso4GW6sxNTklwQv5h1aKScNbAcrSmH6SV3rSf1STPn8Jht5NlmRXmw4wWPSgtaQ70TCJ_3NhJJfxFXg&sig=AHIEtbT2CFFI2L0yuWk5FM-NAn07vZEwaw">See eq. 3.49</a> for an argument that says, in so many words, that if we could localize photons then we could define a current density which doesn't exist. (Or something like that, I'll admit I don't fully understand.)</p>
<p>It's the above question that I'd like clarification on.</p>
| 3,507 |
<p>What is the physical interpretation of
$$ \int_{t_1}^{t_2} (T -V) dt $$
where, $T$ is Kinetic Energy and $V$ is potential energy.<br>
How does it give trajectory?</p>
| 3,508 |
<p>Is there an upper limit to the angular momentum of a rotating (Kerr) black hole? </p>
| 3,509 |
<p>In other words, if a photon is emitted from source, is it possible to change its course en route either by introducing a gravitational lensing or some sort to change the road it travels (spacetime) before it reaches it's source?</p>
| 3,510 |
<p>When we canonically quantize the scalar field in QFT, we use a Lorentz invariant integration measure given by
$$\widetilde{dk} \equiv \frac{d^3k}{(2\pi)^3 2\omega(\textbf{k})}.$$</p>
<p>How can I show that it is Lorentz invariant?</p>
| 3,511 |
<p>I am reading some articles about the ionosphere and I am a little bit confused about the terms <em>mean ionospheric height</em> and <em>effective height of the ionosphere</em>. Are these the same thing?</p>
<hr>
<p>I would refer to two texts:
<a href="http://www.ips.gov.au/IPSHosted/INAG/web-65/2004/es-heights.pdf" rel="nofollow">Variations of yearly mean values of effective heights for the ionospheric sporadic E-layer by V.F.Petrukhin, E.A.Ponomarev, V.D.Kokourov, and N.A.Sutyrin</a> </p>
<blockquote>
<p>Each type was analysed separately by calculating the yearly mean
values of effective heights h’Es. It was found that the effective
heights of the sporadic EsC, EsL and EsF vary within 105 - 132 km
region with a typical period of a ... </p>
<p>...During 1967-1969 the average height for this sporadic layer type
decreased relative to the mean level by about 20 km; for a long time
afterwards (about 17 years) these values remained low, and it was not
until 1985-1987 that they returned to the long-term mean level.</p>
</blockquote>
<p>and
<a href="http://www.dtic.mil/cgi-bin/GetTRDoc?AD=ADA018862" rel="nofollow">Klobuchar, J. A. (1975). Klobuchar - First-Order Time delay Degrees (pp. 1–24). Massachusetts: Air Force Cambridge Research Laboratories.</a></p>
<blockquote>
<p>... The TEC must be found at the geographic point where the ray path
intersects the mean ionospheric height, rather than at the user
location. This point is taken here at a mean height of 350
kilometers. ...</p>
</blockquote>
<hr>
<p>I know that the mean height is used when one would like to express at which height a signal would pierce the ionosphere if we would approximate the ionosphere with one thin layer. And I also know that this is not the height of the electron density maximum.</p>
<p><strong>The question is then, how it is (how are they) calculated?</strong> </p>
| 3,512 |
<p>What occurs in atomic scale that cause the photon to be reflected?<br>
In other words, what is the reason for photons to change its direction and why material can reflect certain wavelengths and absorb the rest? What are the atomic properties that causes that behaviour?</p>
| 92 |
<p>While reviewing some basic field theory, I once again encountered the Bianchi identity (in the context of electromagnetism). It can be written as
$$\partial_{[\lambda}\partial_{[\mu}A_{\nu]]}=0$$
Here, $A_\nu$ is of course the electromagnetic potential. This formula is immediately reminiscent of the Jacobi identity:
$$[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0$$</p>
<p>This is even clearer in general relativity, where we have
$$\nabla_{[\lambda}R_{\rho\sigma]\mu\nu}$$
which we can rewrite, remembering the definition of the Riemann tensor in terms of the commutator of covariant derivatives, as
$$[[\nabla_\lambda,\nabla_\rho],\nabla_\sigma]+[[\nabla_\rho,\nabla_\sigma],\nabla_\lambda]+[[\nabla_\sigma,\nabla_\lambda],\nabla_\rho]=0 $$
This all looks like there should be some profound connection here, but I'm incapable of pinpointing it. Maybe one of the experts here can make this more precise? I'd love to get to know more about this. Any comments are much appreciated. I'd also be grateful if someone could suggest (more) appropriate tags to use.</p>
| 3,513 |
<p>How would you determine what initial velocity $v(r)$ of an object would have to be at some arbitrary point in the vicinity of a planet such that the object will end up on the surface of the planet with a zero velocity with respect to the point on the surface of the planet where it lands? Assume the planet is rotating and the object is traveling by the planet and one could somehow alter it's velocity at a point with an impulse. Hence what kind of initial velocity does the object need to be given so that it will eventually be "drug" to the surface of the planet in a way that it lands. One way to think of it, is to imagine it spiraling in and when it reaches the surface it's velocity is equal to the rotational velocity of the surface and it's verical velocity w.r.t the surface is zero as well. (Assuming no atmosphere on the planet)</p>
| 3,514 |
<p>Would the effect of gravity on me change if I were to dig a very deep hole and stand in it? If so, how would it change? Am I more likely to be pulled downwards, or pulled towards the edges of the hole? If there would be no change, why not?</p>
| 50 |
<p>Consider a hollow cylinder of different outer radius and inner radius and two different temperatures are maintained at the outer and inner surfaces such that inner temperature is higher. Because of this heat will flow radially outward.</p>
<p>Now I don't understand why we need to use integration to find rate of heat flow that is, why isn't the temperature gradient constant? Why can't we directly apply heat conduction formula?</p>
| 3,515 |
<p>What is the significance of $8\pi/3$ in the first <a href="http://en.wikipedia.org/wiki/Friedmann_equations" rel="nofollow">Friedmann Equation</a>, and in the question concerning the time independence of the Hubble Constant?</p>
<p>Is it the 'same' $8\pi/3$ that appears in the total cross section formula of Thomson Scattering ?</p>
| 3,516 |
<p>Part of the definition of the concept of force is that if particle $1$ exerts a force $F_1$ on particle $3$ and particle $2$ exerts a force $F_2$ on particle $3$, the total force on particle $3$ is $F_1+F_2$.</p>
<p>But, is the principle of superposition deducible from Newton's laws or is it an additional assumption? If so, is it always valid? Is this fact linked to the non-existence of three-body forces (or do such forces exist?) or to some kind of linearity in the laws of mechanics or to some kind of fundamental symmetry? </p>
| 3,517 |
<p>Why do photons have zero chemical potential and what is its the physical significance?</p>
<p>From what I know the chemical potential could be interpreted as the energy per unit particle that is put into a (thermodynamic) system... but surely photons also carry energy?</p>
| 93 |
<p>I am quite new to thermodynamics and statistical mechanics so this might be an easy question:</p>
<p>In thermodynamics you get a bunch of thermodynamics potentials, so as for example enthalpy, internal energy, gibbs energy, helmholtz energy and so on. Now my idea was, that you use them if the natural variables related to this quantity are constant. But I do have a question of interpretation: It is always true that if you have some process that the difference of the thermodynamic potentials, whose natural variables are left unchanged is equal to the difference in total energy?</p>
| 3,518 |
<p>An exercise from Goldstein (9.31-3rd Ed) asks to show that for a one-dimensional harmonic oscillator $u(q,p,t)$ is a constant of motion where
$$
u(q,p,t)=\ln(p+im\omega q)-i\omega t
$$
and $\omega=(k/m)^{1/2}$. The demonstration is easy but the physical significance of the constant of motion is not so clear to me. Indeed I can show that $u$ can be rewritten like:
$$
u(q,p,t)=i\phi+\ln(m\omega A)
$$
where $\phi$ is the phase and $A$ the amplitude of the vibration of the oscillator. I can also demonstrate that $m\omega A=\sqrt{2mE}$, where $E$ is the total energy of the oscillator. But there is any further significance of $u$ that I'm missing?</p>
| 3,519 |
<p>I know it doesn't really make sense if looking at the photon from the wave point of view, but is there any law of physics which prohibits a photon from stopping completely? Thanks.</p>
| 94 |
<p>As far as I understand it (which admittedly isn't very far), surface tension forces are made up by the tension-attractive forces of molecules at the liquid-gas/vacuum interface, such as those between hydrogen-bond-capable molecules in water.</p>
<p>But what happens when this liquid surface is in motion? I would assume that, given that there is no acceleration at, the surface tension doesn't really change much since they would be in some sort of "stasis", and the movement would only be that of all the molecules acting in concert. But say, if the molecules were not moving uniformly (but in a single direction), say the molecules in one part of the liquid body accelerates away from the other part, in some "pulling" motion. </p>
<p>What happens to the surface tension here? </p>
| 3,520 |
<p>When you observe or measure a process in classical physics it almost never really alters the experiment. For example, if you have an Carnot engine and measure the volume and pressure of a gas in some a cylinder while the process is running you can do it without disturbing the process, meaning that if you "ran the laws of physics backwards" or ran the process backwards in time for the same amount of time you ran it forwards it will end up at the same place it started at.</p>
<p>However, in Quantum Mechanics if you in any way observe a particle during an experiment it can ruin the whole experiment. For example:</p>
<p>If you sent a particle through a ring and then stopped the process and ran it backwards without ever observing it, the particle would go back to where it came from but if you observed the particle in any way, (even after the experiment) it would not matter whether or not you looks at it forward or backward in time, because the probability that particle's path might fluctuate to the left is the same for either case. How does this work?</p>
| 3,521 |
<p>Does the Hamiltonian always translate to the energy of a system? What about in QM? So by the Schrodinger equation, is it true then that $i\hbar{\partial\over\partial t}|\psi\rangle=H|\psi\rangle$ means that $i\hbar{\partial\over\partial t}$ is also an energy operator? How can we interpret this?
Thanks.</p>
| 3,522 |
<p>Suppose a dynamical system of one variable $x$ with discrete time-steps. I've seen in some papers a type of graph in which $x(n+1)$ is plotted versus $x(n)$.
My questions are :
1/ Can this be considered as the phase portrait of the system ?
2/ Does this method has a specific name ?
3/ Has there been any studies with regard to the topology of this space ?</p>
<p>Thanks for your help!</p>
| 3,523 |
<p>Why do water of water based liquids like sweat cool objects? The most clear example I have is the sweat on the skin. I learned that it absorbed heat because it evaporates, but this is something strange. I also read that by putting a little water on a can under the sun will cool the inside of the can, for the same reason as sweat. Why is this?
Why does a liquid take away the heat, while it isn't even at the boiling point? The body is only 37 degrees Celcius, why does sweat "want" to evaporate already? I do know that liquids take energy to change its phase without adding to the temperature, but this is only at boiling point.</p>
| 95 |
<p>I've always been confused by this very VERY basic and important fact about two-dimensional CFTs. I hope I can get a satisfactory explanation here. In a classical CFT, the generators of the conformal transformation satisfy the Witt algebra
$$[ \ell_m, \ell_n ] = (m-n)\ell_{m+n}.$$
In the quantum theory, the same generators satisfy a different algebra
$$[ \hat{\ell}_m, \hat{\ell}_n ] = (m-n) \hat{\ell}_{m+n} + \frac{\hbar c}{12} (m^3-m)\delta_{n+m,0}.$$
Why is this?</p>
<p>How come we don't see similar things for other algebras? For example, why isn't the Poincare algebra modified when going from a classical to quantum theory?</p>
<p>Please, try to be as descriptive as possible in your answer. </p>
| 3,524 |
<p>I came to this thought experiment as I was pondering good teaching examples of stable and unstable systems. It occurred to me that stable systems are really quite abundant. For a shoot-from-the-hip example, the speed of a car is stable about a given speed given a constant rate of fuel injection, since any perturbation will disappear over time. To be perfectly illustrative, here is a ball on a hill. Note that the forces are <em>balanced</em> in both these cases, but <em>stability</em> is different.</p>
<p><img src="http://i.stack.imgur.com/u8ss8.png" alt="ball on a hill"></p>
<p>Now, the above illustrates the concepts well enough, but I like to present something that's slightly more non-trivial and try to get people to exercise real physical understanding.</p>
<p>Consider:</p>
<p>A submersible with a compressible cavity. A model sufficient for this will be that of a weight attached to a balloon. Saying it's an airtight bag instead of a balloon might be more straightforward, as it avoids the contribution to pressure from the balloon elasticity itself.</p>
<p><img src="http://i.stack.imgur.com/kdZAS.png" alt="submarine"></p>
<p>I want to make the <em>qualitative</em> argument that given the buoyancy and gravitational forces are balanced at some underwater point (obviously below the surface and above the ocean floor), this point is unstable.</p>
<p>The logic is that the balloon will increase in volume and decrease in density as it gets closer to the surface, where the pressure is lesser. Correspondingly, increasing depth and pressure contracts the balloon. That means that upward movement causes a net upward force and moving down causes a net downward force. Unstable.</p>
<p><img src="http://i.stack.imgur.com/i4wzs.png" alt="buoyancy water"></p>
<p><strong>Several questions</strong></p>
<ul>
<li>Does this imply that any submersible (below the surface and above the floor) must actively maintain its depth level? Is the condition associated with this claim that the submersible be more compressible than the liquid? What about a compressible atmosphere?</li>
<li>Are there passive control systems that could maintain a depth setpoint?</li>
<li>Would anyone like to address the problem and questions with equations, derivatives, and all that good stuff?</li>
</ul>
| 3,525 |
<p>In <a href="http://www.newscientist.com/article/mg21929333.200-death-by-higgs-rids-cosmos-of-space-brain-threat.html" rel="nofollow">this article</a>, they mention that the <a href="http://en.wikipedia.org/wiki/Boltzmann_brain" rel="nofollow">Boltzmann Brain</a> paradox is supported by 'the best cosmological models.' However, it gives no references, and I haven't found any by a Google search.</p>
<p>I was wondering if someone here might have some links to where I can read more about this?</p>
| 3,526 |
<p>How is Base emitter junction and collector emitter junction biased? How do we determine the value of potential difference between emitter and collector required to be maintained in order to determine input characteristic of transistor? Explanations covering nitty gritty are most welcome.</p>
| 3,527 |
<p>What is the reason for some writing Faraday's Induction Law as $$ \nabla \times E= -\frac{1}{c}\frac{\partial B}{\partial t} $$ versus $$ \nabla \times E= -\frac{\partial B}{\partial t} ?$$</p>
| 3,528 |
<p>The first law of thermodynamics says energy cannot be created or destroyed. But we can collide photons to form electrons and positrons. Does this means that law does not apply in these microscopic scales?</p>
<p>And we can create mass from energy in the above process.but is it possible to make atoms that way? Like converting an electron to proton as a step in it? That would be like real energy- mass conversion.</p>
| 96 |
<p>Since mass can be given to particles via the interaction with the Higgs Field could there be a "Charger Field" that supplies particles with charge? Possibly this would require two different "charger bosons" one for + and one for -.</p>
| 3,529 |
<p>Do you know where can I find simple diagram (solid, liquid, gas) for for oxygen?</p>
<p>EDIT: I made a mistake. I wanted oxygen instead of water. sorry. </p>
| 3,530 |
<p>I have read several times by different supposedly knowledgeable authors that conscious observation affects quantum experiments. </p>
<p>I think these authors are confused, by which I mean they forget that "observing" is not passive. </p>
<p>To take the double slit experiment for example, observing means that there is at least one photon (used for measuring) interacting with the photon(s) passing thru the slit(s). </p>
<p>My understanding is that this interaction between the photons traveling thru the slits and the photons used for measuring cause the change in the experiment, ie. whether someone is there "observing" the experiment is irrelevant.</p>
<p>Is my understanding correct?</p>
| 3,531 |
<blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="http://physics.stackexchange.com/questions/21753/why-do-electron-and-proton-have-the-same-but-opposite-electric-charge">Why do electron and proton have the same but opposite electric charge?</a> </p>
</blockquote>
<p>Doesn't it seem very curious that one is an elementary particle and the other a subatomic particle yet the charge is exact same.. </p>
| 97 |
<p>I want to know whether decreasing the weight of the frame of a car will increase its horsepower. From what I understand horsepower is a measure of the car's ability to transport load, and decreasing the frame's weight will increase this ability. I tried to search this on internet and all I got was that it depends on the horsepower of the engine not the car itself. It is very confusing. If my assumption that the car's weight has an effect on horsepower is right, then what is the possible way for me to calculate it?</p>
| 3,532 |
<p>I was wondering if there was something equivalent to the property of being wet with water, but with air instead. For example, if I drop water on my shirt, I'll notice by its appearance and feel that it is wet, so in a sense its properties were changed by being exposed to water.
So I'm wondering if similarly, by being exposed to ambient air, my shirt is somehow being changed, i.e. if it was in a vacuum would it feel or appear different than when it's exposed to air?</p>
| 3,533 |
<p>When electric charges of equal magnitude and sign are released on a regular sphere (and assume that they stick to the surface of the sphere, but they are free to move along its surface), what is the shape of the figure made by the charges as vertexes when they come to a state of equilibrium? </p>
<p>Case 1 - Only one charge is there:</p>
<p>Already in equilibrium as there are no other charges.</p>
<p>Case 2 - Two charges:</p>
<p>Two charges are on opposite points of one diameter of the sphere.</p>
<p>Case 3 - Three charges:</p>
<p>Three charges make a shape of an equilateral triangle.</p>
<p>Case 4 - Four charges:</p>
<p>A regular tetrahedron comes up when they reach the state of equilibrium.</p>
<p>While extrapolating these cases to higher number of charges, one roadblock comes up: how to adapt a more-than-three-dimensional figure to the three-dimensional sphere?</p>
<p>Any hint taking to the correct answer or better approach to the problem is welcome.</p>
| 3,534 |
<p>What does the term <em>limb of the earth</em> (see <a href="http://physics.stackexchange.com/q/25038/5291">this question</a>, for example) or <em>atmospheric limb</em> mean? The phrase strikes me as very odd, since earth is nearly spherical. Do other planets with atmospheres also have a limb?</p>
| 3,535 |
<blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="http://physics.stackexchange.com/questions/6068/list-of-good-classical-physics-books">List of good classical physics books</a> </p>
</blockquote>
<p>I'm having a hard time understanding force in my highschool AP physics class, i looked through youtube but without luck. Could anyone link me some tutorial's that might help? The current notes we're working on: <a href="http://www.myteacherpages.com/webpages/NBolken/file_viewer.cfm?secFile=5534" rel="nofollow">http://www.myteacherpages.com/webpages/NBolken/file_viewer.cfm?secFile=5534</a></p>
| 98 |
<p>In flat space-time the electric potential energy between two charges is $\frac{k Q_1 Q_2}{r_{12}}$, where $Q$'s are charges and $r_{12}$ is the distance between them. What would happen if the two charges are placed in a strongly curved space-time, which makes "distance" and "duration" measures different from place to place? Will the two feel the same strength of electric force/potential energy?</p>
| 3,536 |
<p>Some Soviet space stations reportedly <a href="http://en.wikipedia.org/wiki/Almaz" rel="nofollow">had anti-aircraft cannons installed</a>. Could such a cannon hit the firing space station accidentally on a subsequent orbit? The muzzle velocity of <a href="http://en.wikipedia.org/wiki/Nudelman-Rikhter_NR-23" rel="nofollow">the cannon</a> is under 700 m/s, significantly slower than orbital velocity so the projectiles should have similar orbits to the station.</p>
| 3,537 |
<p>Outside a narrow charged stream (say, a beam of ions or electrons) is the same as observing a current through a conducting wire - there is a circular magnetic field around it.</p>
<p>What would happen inside a charged stream (for example, inside a conducting wire or inside a solar flare)? I have a feeling that symmetry will rule that there is no magnetic field, but I am not sure.</p>
| 3,538 |
<ul>
<li><p>is there any difference between these Differential $dx^2$ and $(dx)^2$!?</p></li>
<li><p>what is relation between them?</p></li>
</ul>
| 3,539 |
<p>If the universe is just a Matrix- like simulation, how could we ever know? Physicist Silas Beane of the University of Bonn, Germany, thinks he has the answer!. His paper <a href="http://arxiv.org/abs/1210.1847" rel="nofollow">“Constraints on the Universe as a Numerical Simulation</a>” has been submitted to the journal Physical Review D !.</p>
<p><a href="http://www.slate.com/articles/health_and_science/new_scientist/2012/12/simulated_universe_testing_the_laws_of_physics_to_determine_whether_we_re.html?original_referrer=http%3A%2F%2Fmath.stackexchange.com%2Fusers%2F42480%2Fneo" rel="nofollow">Are We Living in a Simulated Universe?</a></p>
<p><strong>Remark:</strong>
This person claimed that the whole world is a great simulation, and the whole world started with a simulated big bang.
and every things are results of spontaneous program self-organization and automorphisms.
The big bang happened in a kind of supercomputer And now we're inside of it.
the Big Bang occurred approximately 13.75 billion years to our eyes, But perhaps within less than a second for simulator.</p>
<ul>
<li>If the universe is just a Matrix- like simulation, how could we ever know?</li>
</ul>
| 3,540 |
<p>The example I'm trying to understand is:</p>
<p>$ \hat{S}_{x} \begin{pmatrix}
\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}
\end{pmatrix} = 1/2 \begin{pmatrix}
\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}
\end{pmatrix} $</p>
<p>My interpretation of this is that the vector shows you the probabilities of a particle being spin up or spin down if you square them.</p>
<p>And I've been told that $ \hat{S}_{x} $ gives you the spin as an eigenvalue, but how? Since its 50:50 of getting -1/2 and 1/2. $ \hat{S}_{x} $ has only given you one of them.</p>
<p>Is it that $ \hat{S}_{x} $ only measures the magnitude of spin in the x direction?</p>
| 3,541 |
<p><strong>Heads up</strong>: This question has never been asked (here) before the way I will ask it here, so let's shed some light on it a bit.</p>
<p><strong>Prelude and anecdote</strong>(can be skipped): The other day I was walking home, and I passed a stop sign. I looked at it and asked myself, "<strong>What kind of atoms make up this macro-sized metal-like object that visible light enables my eyes to see, which are also atoms I don't know?</strong>"</p>
<p>So I thought to myself ... <em>I really should go to physics.stackexchange.com to ask this.</em>
They have so many great people there who volunteer their time and help without any immediate financial gain, and there is some very smart people there who are open-minded, and know that repetition is key to success.</p>
<p><strong>The question:</strong></p>
<p>So here I am ... my question is how do you know what materials are composed of? </p>
<p>I ask this because:</p>
<p>1.There are no "grab and go" ways to quickly check out what atoms something is made of. How would I, as a curious person that I am who's into science in all respects, determine what the world is around me?</p>
<p>2.The tools needed to do this don't come cheap ... how could I, by using cheaper tools or techniques, determine what kind of atoms, molecules/chemical compounds, determine which exact atoms make up the things visible light enables me to see around me?</p>
<p>I would like to look at something and go beyond just "It's plastic-like", but actually dig down to which monomer, polymer, or molecular structure it is, and how I can break it down to the individual atoms.</p>
<p>Please ... I ask this with all respect, all seriousness, and all desire to want to further know the materials, compounds, and structures of the objects around me.</p>
| 3,542 |
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