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<p>You age at a different rate depending on the force of gravity. Astronauts age fractions of fractions of fractions of a second less than earthlings.</p>
<p>If you took a sphere of equal masses, separated by space, then found the exact center of the gravitational pulls of all masses. How would time react?</p>
| 2,060 |
<p>I'm a non-engineer interested in the recent GP-B mission results: <a href="http://www.engadget.com/2011/05/06/nasa-concludes-gravity-probe-b-space-time-experiment-proves-e/#disqus_thread" rel="nofollow">http://www.engadget.com/2011/05/06/nasa-concludes-gravity-probe-b-space-time-experiment-proves-e/#disqus_thread</a> </p>
<p>Is it correct that this means that both the rotation of the earth and the size affect time? Would a result be that a person living on an earth-sized planet spinning faster make that person's perception of time relative to an earth-bound person different, and if so, in what way? Also, how massive must an object be and and how fast must an object rotate to have noticable effect? </p>
| 2,061 |
<p>In his book: <em>The Equation That Couldn't be Solved</em> Mario Livio explains the equivalence principle in laymen's terms. I took the statement on page 209: <em>The force of gravity and the force resulting from acceleration are in fact the same.</em> to mean that since our universe is accelerating, gravity is the result. In other words, we have gravity because we have acceleration. This made a lot of sense to me and seemed to explain things (like warps in space-time due to gravity of very large objects. They 'have' gravity because they are accelerating masses.</p>
<p>In my beginner's head I extended that conclusion to mean if the acceleration of the universe was zero we would have mass but no gravity. Based on what I've read since, this does not seem to be true...is it?</p>
| 2,062 |
<p>I have a circuit with capacitors on it:</p>
<p><img src="http://i.stack.imgur.com/h5N90.jpg" alt="circuit figure 1"></p>
<p>I am trying to figure out the charge on each capacitor.</p>
<p>The following is given: </p>
<p><img src="http://i.stack.imgur.com/YYXQN.jpg" alt="given"></p>
<p>i know that parallel capacitors follow the equation</p>
<p><img src="http://i.stack.imgur.com/MK9jh.png" alt="paraller"></p>
<p>and that capacitors in series behave according to this equation:
<img src="http://i.stack.imgur.com/9tmCe.png" alt="series"></p>
<p>I dont know how to use that knowledge to find out the charge of the capacitors.
I also dont know how to apply this to find the voltage across all the capacitors and the total voltage.</p>
<p>This is from a homework question but i want to find out the general concepts of calculating voltages and charges on any circuits. </p>
<p>Any help will be greatly appreciated. </p>
| 2,063 |
<p>Let's have Dirac spinor $\Psi (x)$. It transforms as $\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right)$ representation of the Lorentz group:
$$
\Psi = \begin{pmatrix} \psi_{a} \\ \kappa^{\dot {a}}\end{pmatrix}, \quad \Psi {'} = \hat {S}\Psi .
$$
Let's have spinor $\bar {\Psi} (x)$, which transforms also as $\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right)$, but as cospinor:
$$
\bar {\Psi} = \begin{pmatrix} \kappa^{a} & \psi_{\dot {a}}\end{pmatrix}, \quad \bar {\Psi}{'} = \bar {\Psi} \hat {S}^{-1}.
$$
How to show formally that
$$
\bar {\Psi}\Psi = inv?
$$
I mean that if $\Psi \bar {\Psi}$ refers to the direct product (correct it please, if I have done the mistake)
$$
\left[\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right) \right]\otimes \left[\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right) \right],
$$
what group operation corresponds to $\bar {\Psi} \Psi$?</p>
<p>This question is strongly connected with <a href="http://physics.stackexchange.com/questions/104688/transformation-law-for-spinor-functions-multiplication">this one</a>.</p>
| 2,064 |
<p>Say we had a particle moving in a frictionless funnel and was projected horizontally.</p>
<p>If we had some initial conditions for the energy E, then would these conditions be the same always?</p>
<p>For instance, in this particular question I got
$$ E = 1/2m\dot r^2 - mgz, $$ and we were given that $z = b\left ( \dfrac{b}{r} \right )^n$, and it was projected at the inner surface level $z = b$ horizontally with speed $U$. Using those initial conditions, I got $ E = 1/2m U^2 -mgb.$ However would I be correct in the stating that $$ 1/2m\dot r^2 - mgz = 1/2m U^2 -mgb?$$ I looked in the solutions and the lecturer wrote that they were equal, but does the energy not change of the particle?</p>
| 2,065 |
<p>Consider the situation when a cell of an unknown emf is being measured using a potentiometer. We slide the jockey so as to obtain the null point. Now, is there any current in the potentiometer wire at the null point? Since we know that there is no current in the arm containing the unknown cell, its terminals have acquired equal potentials,how is it possible that there is any current in the potentiometer wire when that is in parallel to that cell.</p>
<p>Potential difference across AB= Potential difference across CD?
<img src="http://i.stack.imgur.com/YcUOl.gif" alt="enter image description here"></p>
| 2,066 |
<p>Most sources say that <a href="http://en.wikipedia.org/wiki/Wigner_quasiprobability_distribution" rel="nofollow">Wigner distribution</a> acts like a joint phase-space distribution in quantum mechanics and this is justified by the formula </p>
<p>$$\int_{\mathbb{R}^6}w(x,p)a(x,p)dxdp=<\psi|\hat{a}\psi>$$ as this allows us to compute the expectation value for the operator $\hat{a}$ corresponding to the physical quantity $a$. However, without the knowledge of this formula, how did Wigner come up with this definition? $$w(x,p)=\frac{1}{(2\pi)^3}\int_{\mathbb{R}^3}\psi(x-\frac{v}{2})\psi^*(x+\frac{v}{2})e^{iv.p}dv.$$ I would be greatly indebted for any mathematical motivation that anyone could provide. Similarly, I also wonder at the mathematical motivation for the Weyl quantization.</p>
| 2,067 |
<p>I am learning about calculating decay rates from quantum field theory amplitudes from <a href="http://www.damtp.cam.ac.uk/user/tong/qft/three.pdf" rel="nofollow">David Tong's lecture notes</a> (page 74 in the notes, 24 in document). However, I have some doubts:</p>
<ol>
<li><p>When he says <em>the first is to integrate over all possible momenta of the final particles,</em> he uses the measure $V \int \frac{d^{3} p_{i}}{(2 \pi)^{3}}$. I'm not sure why that factor of V is there. I have come up with some explanations, the best of which is that the delta function $\delta^{(4)} (p_{F}-p_{I})$, due to considering only a finite space does not actually multiply by infinity, but by V. However, I need several factors of V, which I don't see how to get.</p></li>
<li><p>The next question is about the meaning of the probability he calculates. He gets a Probability which is something finite times a $t$, so that it grows infinitely in time. I don't get how does he then get from this the half life, as I don't even know how to make sense of a diverging probability like this (normalizing it in the normal way I know, would just give all the probability way at infinity).</p></li>
</ol>
<p>So clearly, I must be missing some important things here.</p>
| 2,068 |
<p>How can one prove the Bohr-Sommerfeld quantization formula</p>
<p>$$ \oint p~dq ~=~2\pi n \hbar $$</p>
<p>from the WKB ansatz solution $$\Psi(x)~=~e^{iS(x)/ \hbar}$$ for the Schroedinger equation? </p>
<p>With $S$ the action of the particle defined by Hamilton-Jacobi equation</p>
<p>$$ \frac{\partial S}{\partial t}+ \frac{(\nabla S)^{2} }{2m}+V(x)~=~0 .$$</p>
| 2,069 |
<p>I'm trying to compute uncertainty for the density of the ball. </p>
<p>I measured its radius 6 times, so I was able to compute the stastistical uncertainty (we call it uncertainty type A, I don't know, if that's common used designation) and I knew accuracy of the micrometer, so I got standatd uncertainty (we call it Type B uncertainty) and made combined uncertainty, which is $4.3 \cdot 10^{-6}~m$.</p>
<p>Then I measured the weight of the ball, but only once and computed standard uncertainty (type B) as $0,29 \cdot 10^{-4}~g$.</p>
<p>Now I have this formula for the density: $\rho = \frac {m}{V} = \frac {m}{\frac 43 \pi r^3}$ and I need to find a way how to express uncertainty of this quantity.</p>
<p>Does anybody know how to do that?</p>
| 2,070 |
<p>Can anyone explain to me why most articles describe chromium as an acceptor in titanium dioxide? In TiO<sub>2</sub>, titanium has the charge state Ti$^{4+}$ and oxygen has the charge state O$^{2-}$. When Cr substitutes for Ti, it does so as Cr$^{3+}$. Now, at first glance, Cr has atomic number 24 and Ti 22. Cr therefore has two more valence electrons and is a donor. In TiO<sub>2</sub>, Cr$^{3+}$ actually has <em>three more</em> valence electrons than the Ti$^{4+}$ ([Ar]$3d^34s^0$ vs [Ar]$3d^04s^0$). It should therefore be a donor, right? The thing is, it forms a deep impurity level near the valence band. TiO<sub>2</sub> has an energy gap of around 3.2 eV, and the impurity state is about 1.0 eV from the valence band maximum. To me, that makes it a <strong>deep donor</strong>. For some reason, journals almost always describe it as an acceptor. Can someone help me make sense of this? </p>
<p>My understanding has always been simply this: more electrons than host $\Rightarrow$ donor, fewer electrons than host (more holes) $\Rightarrow$ acceptor. The position of the impurity level, to my (perhaps incorrect) knowledge, does not determine whether or not the impurity is actually a donor or acceptor, but rather whether it is a recombination center or trap. We can have localized states near the middle of the bandgap that are technically donors/acceptors but function as recombination centers, so I'm not sure what I'm missing here.</p>
| 2,071 |
<p>Suppose that it is given that the Riemann curvature tensor in a special kind of spacetime of dimension $d\geq2$ can be written as $$R_{abcd}=k(x^a)(g_{ac}g_{bd}-g_{ad}g_{bc})$$ where $x^a$ is a vector in the space.</p>
<p>What condition can I impose on $d$ to guarantee that $k(x^a)$ is a constant? I have been told that such a condition exists.</p>
<p>Not knowing what to do, I computed the contractions of the curvature tensor and got
$$R_{bd}=g^{ac}R_{abcd}=k(d-1)g_{bd}$$ and
$$R=g^{bd}g^{ac}R_{abcd}=kd(d-1)$$
But it probably is of no use?</p>
| 2,072 |
<p><img src="http://i.stack.imgur.com/DWU9K.jpg" alt="enter image description here"></p>
<p>Light is the yellow arrow. Observer is the black arrow. Observer is moving at a constant speed of v, w.r.t to a Galilean frame of reference. </p>
<p>Now from the point of view of the observer (O), how will the motion of the light ray look like? Will it bend away from him?</p>
<p>Looking for a good explanation. Thanks! </p>
| 2,073 |
<p>The representation $(\frac{1}{2},\frac{1}{2})$ of the Lorentz group correspond to a four- vector or a spin-one object. Right? Does it imply that any four-vector is <em>identical to</em> a spin-one object or any scalar is <em>identical to</em> a spin-0 object? This can't be correct, right? Because although $A^\mu$ is a four vector and a spin-one object at the same time (which is photon), there is no concept of spin associated with $p^\mu$ or $J^\mu$. I'm confused by terminologies of representation.</p>
<p>Edit- How can I show that $A^\mu$ represent a spin-1 object?</p>
| 2,074 |
<p>Allow me to preface this by stating that I am a high school student interested in physics and self-studying using a variety of resources, both on- and off-line, primarily GSU's HyperPhysics website, Halliday & Resnick's <em>Fundamentals of Physics</em>, Taylor's <em>Classical Mechanics</em>, and ultimately the Feynman lectures (mirrored by Caltech). Hopefully this gives somewhat of a feel of my level of physics understanding so as to avoid any answers that fly far above my head.</p>
<p>As I've understood from previous reading of electromagnetism (for example, in Halliday), a point charge is not affected by its own electromagnetic field. Unfortunately, as I recently read in the Feynman lecture on electromagnetism, <a href="http://www.feynmanlectures.caltech.edu/I_28.html#Ch28-S1-p16">this appears to not be so</a>:</p>
<blockquote>
<p>For those purists who know more (the professors who happen to be reading this), we should add that when we say that (<a href="http://www.feynmanlectures.caltech.edu/I_28.html#mjx-eqn-EqI283">28.3</a>) is a complete expression of the knowledge of electrodynamics, we are not being entirely accurate. There was a problem that was not quite solved at the end of the 19th century. When we try to calculate the field from all the charges <em>including the charge itself that we want the field to act on</em>, we get into trouble trying to find the distance, for example, of a charge from itself, and dividing something by that distance, which is zero. The problem of how to handle the part of this field which is generated by the very charge on which we want the field to act is not yet solved today. So we leave it there; we do not have a complete solution to that puzzle yet, and so we shall avoid the puzzle for as long as we can.</p>
</blockquote>
<p>At first I figured I must've misunderstood, but upon rereading, it's clear Feynman states the that electromagnetic field due to a point charge <strong>does</strong>, in fact, influence said charge; I inferred this "self-force" must be somewhat negligible for Halliday to assert otherwise. What stood out to me was that Feynman states this problem had <em>not</em> yet been solved.</p>
<p>I suppose my first major question is simply, <em>has this problem been solved yet</em>? After a bit of research I came across the <a href="http://en.wikipedia.org/wiki/Abraham%E2%80%93Lorentz_force">Abraham-Lorentz force</a> which appears to refer exactly to this "problem of self-force". As the article states the formula is entirely in the domain of classical physics and a quick Google search indicates it was derived by Abraham and Lorentz in 1903-4, why is it that Feynman state the problem was still unsolved in 1963? Has it been solved in the classical case but not in QED? </p>
<p>Lastly, despite the Wikipedia article <a href="http://en.wikipedia.org/wiki/Abraham%E2%80%93Lorentz_force#Background">somewhat addressing the topic</a>, <em>is this problem of self-force present with other forces</em> (e.g. gravity)? I believe it does state that standard renormalization methods fail in the case of GR and thus the problem is still present classically, though it does mention that non-classical theories of gravity purportedly solve the issue. Why is there not a similar Abraham-Lorentz-esque force possible in GR -- is there an underlying fundamental reason? Due to the relative weakness of gravity, can these self-force effects be ignored safely in practice?</p>
<p>I apologize for the long post size and appreciate any help I can receive. I only hope my post isn't too broad or vague!</p>
| 2,075 |
<p>I always hear people saying symmetry is beautiful, nature is symmetric intrinsically, physics and math show the inherent symmetry in nature <em>et cetera, et cetera</em>.</p>
<p>Today I learned that half of the quarks have +2/3 electric charge and other half have -1/3 magnitude electric charge. Is there any explanation for this? Why their net charge isn't zero?</p>
| 2,076 |
<p>In atomic and molecular physics we quite often encounter with electric dipole approximation. The dipole approximation we do when the wave-length of the type of electromagnetic radiation which induces, or is emitted during, transitions between different atomic energy levels is much larger than the typical size of a light atom. This is mostly the case. I have two questions regarding this:</p>
<p>1) Is there any case where we use quadrupole approximation or higher?</p>
<p>2) In the case of transition in molecules (for eg. large organic molecules or polymers) the size of the molecule is larger than the EM radiation. This case how we choose the approximation?</p>
| 2,077 |
<p>This <a href="http://www.youtube.com/watch?v=1TKSfAkWWN0">Veritasium video</a> explains how electromagnets can be explained by special relativity, and how the magnetic field surrounding a current-carrying wire can also be viewed as an electric field, if your frame of reference is moving with respect to the wire.</p>
<p>The example they use is a positively-charged cat, moving along a current-carrying wire in the same direction as the electron drift:</p>
<p><img src="http://i.stack.imgur.com/HHUiP.png" alt="positively charged cat next to a current-carrying wire"></p>
<p>If you view this from the rest frame of the cat, then the electron's drift velocity is zero, while the protons are moving to the left. Because the protons are moving, length contraction makes it look like (to the cat) there are more of them, giving the wire a net <em>positive</em> charge, <em>repelling</em> the cat.</p>
<p>This makes sense and is all kinds of elegant and intuitive. It explains electromagnets in a way that depends on only three simple concepts:</p>
<ol>
<li>motion is relative</li>
<li>things contract in their direction of apparent motion</li>
<li>opposite charges attract, like charges repel</li>
</ol>
<p>Groovy. Now back up in the video. Derek says:</p>
<blockquote>
<p>Now the number of protons is equal to the number of negative electrons, so overall the wire is neutral. So if there were a positively-charged cat nearby, it would experience no force from the wire at all. And even if there were a current in the wire, the electrons would just be drifting in one direction, but the density of positive and negative charges would still be the same, and so the wire would be neutral, so no force on the kitty.</p>
</blockquote>
<p><img src="http://i.stack.imgur.com/Gb796.png" alt="Derek standing next to a current-carrying wire"></p>
<p>Wait...<em>what</em>? Why is it that in the cat's frame, the protons are moving, are contracted, and the wire is charged, but in Derek's frame, the electrons are moving, <em>but are not contracted</em>, and the wire is still neutral?</p>
<p>How can you say "well, length contraction creates charge imbalances, allowing magnetic forces to be explained as electrical ones if you choose the right reference frame", but simultaneously say "but length contraction doesn't happen sometimes"? That's not elegant at all. Is there an elegant, intuitive<sup>1</sup> explanation?</p>
<p><sup>1: meaning, I've seen the math on Wikipedia and it's over my head. There is also <a href="http://physics.stackexchange.com/questions/82821/current-in-wire-special-relativity-magnetism">current in wire + special relativity = magnetism</a> where the answer to my question seems to be "the Lorentz force". OK, but that negates the elegance of the explanation above, with only three simple axioms. Are they not sufficient? If so, why?</sup></p>
| 2,078 |
<p>I'm reading through Kardar's Statistical Mechanics of Particles; in the section 1.5 he says:</p>
<blockquote>
<p>A reversible process is one that can be run backward in time by simply reversing its inputs and outputs. It is the thermodynamic equivalent of frictionless motion in mechanics. <strong>Since time reversibility implies equilibrium</strong>, a reversible transformation must be quasi-static, but the reverse is not necessarily true.</p>
</blockquote>
<p>Why time reversibility implies equilibrium ?</p>
| 39 |
<p>I came up last night with a talk given by <a href="http://www.sjcrothers.plasmaresources.com/index.html" rel="nofollow">Stephen J. Crothers</a> in which he claims that black holes and the Big Bang have no basis in general relativity. But is he really true? How legitimate are his claims?</p>
<p>Here is the talk on YouTube, "The Non-existence of the Black Hole and the Failure of General Relativity": <a href="http://www.youtube.com/watch?v=Q185InpONK4" rel="nofollow">Part One</a>, <a href="http://www.youtube.com/watch?v=CHZ5O0jTH8A" rel="nofollow">Part Two</a>.</p>
| 2,079 |
<p>Just like classical potential, it's stated that equilibrium is obtained when the corresponding thermodynamics potential reaches the minimum.</p>
<p>Explicitly, according to <a href="http://en.wikipedia.org/wiki/Thermodynamic_potential" rel="nofollow">Wikipedia</a>,
in particular: </p>
<blockquote>
<p>When the entropy (S ) and "external parameters" (e.g. volume) of a closed system are held constant, the internal energy (U ) decreases and reaches a minimum value at equilibrium.</p>
<p>When the temperature (T ) and external parameters of a closed system are held constant, the Helmholtz free energy (F ) decreases and reaches a minimum value at equilibrium.</p>
<p>When the pressure (p) and external parameters of a closed system are held constant, the enthalpy (H ) decreases and reaches a minimum value at equilibrium.</p>
<p>When the temperature (T ), pressure (p) and external parameters of a closed system are held constant, the Gibbs free energy (G ) decreases and reaches a minimum value at equilibrium.</p>
</blockquote>
<p>There is something confuse me here. Take the first claim for example. If we have the formula: $dU=TdS-pdV$ then when $S$ and $V$ are held constant, is it not clear that $dU=0$ for whatever changes occurring? Furthermore, if there are $two$ fixed parameter, is it not true that the state of system is fixed? (assuming there is a state formula $f(T,p,V)=0$).</p>
<p>It seems like I'm really misunderstanding some concepts of the thermodynamics potential. Can someone give me a clear explanation for my confusion?</p>
| 912 |
<p>I'm building an airplane (Super Baby Great Lakes) and I'm wondering something about <a href="http://en.wikipedia.org/wiki/Airfoil" rel="nofollow">airfoils</a>. In particular (this plane is fabric covered), I'm wondering about the lifting forces on the main wings. I've read something about it being very important that the fabric adheres very well on the top of the wing to the ribs so that the fabric doesn't separate when lift is generated.</p>
<p>My question is this: how much lift is generated by direct pressure of the slipstream against the bottom of the wing because of high angle of attack vs. how much "sucking" force is generated due to low pressure on the top of the wing? Is the vacuum on the top of the wing simply a lack of atmospheric pressure, or is it genuinely a sucking force, like a powerful vacuum cleaner which could actually tear the sheet out of a notebook, for example?</p>
<p>Thanks,
Jay</p>
| 2,080 |
<p>Consider a system of $n$ undistinguishable particles moving in $d$-dimensional Euclidean space $E^d$. The configuration space is $M=((E^d)^n \setminus \Delta)/S_n$ where $\Delta$ is the diagonal (subspace where at least 2 particles have coincidental positions) and $S_n$ is the group permuting the particles</p>
<p>Quantization of this system yields superselection sectors corresponding to unitary irreducible representations of $\pi_1(M)$: $S_n$ for $d > 2$, $B_n$ for $d = 2$. The trivial representation yields bosonic statistics, the sign representations yield fermionic statistics. For $d > 2$ there are no other 1-dimensional representations. For $d = 2$ there are other 1-dimensional representations in which switching two particles generates an arbitrary phase. These yield anyonic statistics.</p>
<p>What about higher dimensional irreducible representations? These correspond to <a href="http://en.wikipedia.org/wiki/Parastatistics" rel="nofollow">parastatistics</a>. It is said that for $d > 2$ we can safely ignore them because in some sense they are equivalent to ordinary bosons/fermions. However for $d = 2$ this is not the case. Why?</p>
<blockquote>
<p>Why is parastatistics redundant for $d > 2$ but not for $d = 2$?</p>
</blockquote>
| 2,081 |
<p>Is there an entropy that one can use for the Wigner quasi-probability distribution?
(In the sense of a phase-space probability distribution, not - just von Neumann entropy.)</p>
<p>One cannot simply use $\int - W(q,p) \ln\left[ W(q,p) \right] dq dp$, as the Wigner function is not positively defined. </p>
<p>The motivation behind the question is the following:</p>
<p>A paper I. Białynicki-Birula, J. Mycielski, <a href="http://dx.doi.org/10.1007/BF01608825">Uncertainty relations for information entropy in wave mechanics</a> (Comm. Math. Phys. 1975) (or <a href="http://www.cft.edu.pl/~birula/publ/Uncertainty.pdf">here</a>) contains a derivation of an uncertainty principle based on an information entropy:
$$-\int |\psi(q)|^2 \ln\left[|\psi(q)|^2\right]dq-\int |\tilde{\psi}(p)|^2 \ln\left[|\tilde{\psi}(p)|^2\right]dp\geq1+\ln\pi.$$
One of the consequences of the above relation is the Heisenberg's uncertainty principle. However, the entropic version works also in more general settings (e.g. a ring and the relation of position - angular momentum uncertainty).</p>
<p>As $|\psi(q)|^2=\int W(q,p)dp$ and $|\tilde{\psi}(p)|^2=\int W(q,p)dq$ and in the separable case (i.e. a gaussian wave function) the Winger function is just a product of the probabilities in position an in momentum, it is tempting to search for an entropy-like functional fulfilling the following relation:
$$1+\ln\pi\leq\text{some_entropy}\left[ W\right]\leq -\int |\psi(q)|^2 \ln\left[|\psi(q)|^2\right]dq-\int |\tilde{\psi}(p)|^2 \ln\left[|\tilde{\psi}(p)|^2\right]dp.$$</p>
| 2,082 |
<p>The photon field is the non chiral piece of SU(2)xU(1), independently of symmetry breaking or not, isn't it? </p>
<p>But before symmetry breaking, each gauge boson has only a chiral gaugino as superpartner. Is it still possible, <strong>and correct</strong>, to arrange two of them in order to form an electrically charged "Dirac" fermion, able to couple to the photon field? </p>
<p>Probably this is a textbook question, but all the textbooks I have seed do first break susy, then electroweak, then this start point is never seen. </p>
| 2,083 |
<p>Single particle states in quantum field theory appear as discrete components in the spectrum of the Poincare group's action on the state space (i.e. in the decomposition of the Hilbert space of quantum states into irreducible representations of the Poincare group). Classification of irreducible unitary representations of the Poincare group leads to the notions of mass and spin.</p>
<p>Now, suppose we have a conformal QFT and are doing the same trick with the conformal group. Which irreducible representations do we have?</p>
<p>We still have the massless particles (at least I'm pretty sure we do although I don't immediately see the action of special conformal transformations). However, all representations for a given spin s and any mass m > 0 combine into a single irreducible representation.</p>
<blockquote>
<p>What sort of physical object corresponds to this representation?
Is it possible to construct a scattering theory for such objects?
Is it possible to define unstable objects of this sort?</p>
</blockquote>
| 2,084 |
<p>In all thermodynamics texts that I have seen, expressions such as $\operatorname{ln}T$ and $\operatorname{ln}S$ are used, where $T$ is temperature and $S$ is entropy, and also with other thermodynamic quantities such as volume $V$ etc. But I have always thought that this is incorrect because the arguments $x$ in expessions such as $\operatorname{ln}x$ and $e^x$ ought to be dimensionless. Indeed at undergraduate level I always tried to rewrite these expressions in the form $\operatorname{ln}\frac{T}{T_0}$. So is it correct to use expressions such as $\operatorname{ln}T$ at some level?</p>
| 2,085 |
<p>Consider a point mass $x$ (like for example the earth in space) and let $A$ and $B$ be two sets of point masses which each hold the point mass $x$ in equilibrium, meaning the acceleration induced by $A$ on $x$ is zero, and the same for $B$. Now if I put the point masses of both $A$ and $B$ in space then (by superposition) the accelaration exerted by all the point masses in $A$ and $B$ together would still be zero on $x$. This seems quite obvious, but in <a href="http://philpapers.org/rec/KRAFMO" rel="nofollow">this article</a> on p. 485 something different seems to be suggested, to quote</p>
<blockquote>
<p>If we restrict to partial configurations that do not $a$ include the weight of the sphere, then the obvious superposition law fails. In fact let $\hat E$ denote the subset of partial configurations, excluding the weight of the sphere, for which the sphere has no observed acceleration.</p>
</blockquote>
<p>Then it is claimed that $\hat E$ is not closed under superposition (i.e. addition of force configurations)? Does I understand something wrong?</p>
| 2,086 |
<p>I am working through the derivation of an adiabatic process of an ideal gas $pV^{\gamma}$ and I can't see how to go from one step to the next. Here is my derivation so far which I understand:</p>
<p>$$dE=dQ+dW$$
$$dW=-pdV$$
$$dQ=0$$
$$dE=C_VdT$$</p>
<p>therefore</p>
<p>$$C_VdT=-pdV$$</p>
<p>differentiate the ideal gas equation $pV=Nk_BT$</p>
<p>$$pdV+Vdp=Nk_BdT$$</p>
<p>rearrange for $dT$ and substitute into the 1st law:</p>
<p>$$\frac{C_V}{Nk_B}(pdV+Vdp)=-pdV$$.</p>
<p>The next part is what I am stuck with I can't see how the next line works specifically how to go from $\frac{C_V}{C_p-C_V}=\frac{1}{\gamma -1}$</p>
<p>using the fact that $C_p-C_V=Nk_B$ and $\gamma = \frac{C_p}{C_V}$ it can be written</p>
<p>$$\frac{C_v}{Nk_B}=\frac{C_V}{C_p-C_V}=\frac{1}{\gamma -1}$$.</p>
<p>If this could be explained to me, I suspect it is some form of algebraic rearrangement that I am not comfortable with that is hindering me.</p>
| 2,087 |
<p>Can you explain what happens when a particle and its antiparticle are created. Do they whiz away from each other at the speed of light or what? I suppose that they don't because otherwise they would never meet and annihilate each other, but then, if I had just been created with an antiparticle I would do all I could to stay away from him/her. On the other hand, for the sadistic/suicidal type, they might actually be attracted to one another.</p>
<p>[the question is serious, even though it's written light-heartedly. Please explain.]</p>
| 2,088 |
<p>In bosonic string theory, to obtain the photon as the first excited state, the ground state must have a negative mass (tachyon). By applying $1 + 2 + 3 + \cdots = -1/12$, it can be shown (in a simplified way...) that a total of 26 spacetime dimensions are needed to obtain such a tachyonic ground state.</p>
<p>Going from bosonic string theory to superstring theory, the number of spacetime dimensions is 10 and the mass of the ground state is 0, such that the tachyons are removed trom the spectrum of the theory.</p>
<p>Can somebody explain to me in some detail and step by step how the number of dimensions is reduced (or fixed) to 10 and how the tachyons are removed (or avoided) in superstring theory? </p>
| 2,089 |
<p>The determinant of a two-qubit (4 x 4) density matrix lies between 0 and (1/2)^8. (A pure state has determinant zero, and the fully mixed [classical] state, determinant (1/2)^8.)</p>
<p>The determinant of the partial transpose (transpose in place the four 2 x 2 blocks)
of such a matrix (nonnegative values
indicating separability) lies between -(1/2)^4 and (1/2)^8. (The minimum is achieved by a Bell state and the maximum, again by the fully mixed state.)</p>
<p>What is the range (upper and lower limits) of the difference of these two determinants?</p>
| 2,090 |
<p>I cannot seem to prove that the derivative of the duel tensor = 0.</p>
<p>$$ \frac{1}{2}\partial_{\alpha}\epsilon^{\alpha \beta \gamma \delta} F_{\gamma \delta} = 0. $$</p>
<p>Writing this out I get (for some fixed $\alpha$ and $\beta$),</p>
<p>$$ \partial_{\alpha} (\partial_{\gamma}A_{\delta} - \partial_{\delta}A_{\gamma}). $$</p>
<p>From here I get stuck.</p>
<p>Any ideas?</p>
| 2,091 |
<p>Firstly I can say that I would love us to come up with a sustainable fusion solution. However with the latest estimates being 2050 at the earliest for an effective fusion solution and the planets energy needs growing by the year should we not be looking at alternatives?</p>
<p>I understand that in the EU alone we are spending billions of Euro's on an experimental fusion reactor.</p>
<p>We sit on top of a thin crust which is in turn on top of a mantle with abundant heat. Should we not be investing serious money into tapping into this heat source as a long term, sustainable solution to the planet's energy needs?</p>
<p>How far could we get if the money used for fusion was diverted to geothermal drilling and research?</p>
| 2,092 |
<p>Is the opening of the NOVA program on PBS a Calabi-Yau space?</p>
| 2,093 |
<p>I have a doubt about the electric field lines for a continuous distribution. Well, if there's only some point charges, I know that the field starts on the positive ones (or at infinite), ends at negative ones (or at infinity) and the number of lines is proportional to the number of charges. That's fine, but what are the "instructions" to draw the field lines when the distribution is continous ?</p>
<p>I've seem some places where they draw the lines just imagining that the continuous distribution is the same as a finite distribution of small charges $dq$, but I'm not sure that this is the way to do it.</p>
<p>Can someone point out how do we work with this ?</p>
<p>Thanks in advance!</p>
| 2,094 |
<p>I have a problem that asks for the minimum energy of a wave that we will use to see a particle of size $.1\text{ nm}$. I understand that I can not see a $.1\text{ nm}$ particle with any wave length larger than $.1\text{ nm}$. I thought this would be easy, and I would use De Broglie's relation of electron waves,</p>
<p>$$f=\frac{E}{h}\quad\text{or}\quad E=fh=\frac{hc}{λ}$$</p>
<p>Using this I get $12400\text{ eV}$... this is the wrong answer. </p>
<p>What the book says to do is use an equation "wavelength associated with a particle of mass $M$." It is:</p>
<p>$$λ=\frac{hc}{\sqrt{2mc^2K}}$$</p>
<p>OR for my specific case:</p>
<p>$$λ=\frac{1.226}{\sqrt{K}}\text{ nm}$$</p>
<p>This second equation, if I'm correct, is getting the kinetic energy of the wavelength, not the total energy.</p>
<p>I do not understand what I should be looking for in problems asking for energy of wavelengths to distinguish the use of the first equation I presented vs. the second one. Any enlightenment on this area would be appreciated.</p>
| 2,095 |
<p>In <a href="http://en.wikipedia.org/wiki/Aharonov%E2%80%93Bohm_effect" rel="nofollow">Aharonov-Bohm effect</a>, how to derive that the wave function of a electric charge $q$ acquires a phase shift $\phi=\frac{q}{\hbar}\int \mathbf{A} \cdot d\mathbf{x}$ after travelling in the non-zero magnetic vector potential $\mathbf{A}$?</p>
| 2,096 |
<p>Which of two objects at the same tempreature can cause more intense burns when you touch it: the one with the greater specific heat capacity or the one with the smaller specific heat capacity and why?</p>
| 2,097 |
<p>Suppose one inertial observer measures a rod at rest w.r.t. him and another observer is moving w.r.t. rod. We then say that length will be shorter for moving observer but at the instants the first observer is measuring the length, the second observer doesn't even get the length of the rod, he just gets distance between two points in space after Lorentz Transformations because simultaneity is a relative concept. So how is it a length contraction in literal sense? Isn't it a misnomer ?</p>
| 2,098 |
<p>I know the famous beta function of asymptotic free, but that seems describe the running coupling beyond confinement/QCD scale so that a perturbative analysis can apply. But how coupling runs below that scale? Any comment or references are greatly appreciated.</p>
| 2,099 |
<p>Different authors seem to attach different levels of importance to keeping track of the exact tensor valences of various physical quantities. In the strict-Catholic-school-nun camp, we have Burke 1980, which emphasizes that you don't always have a metric available, so it may not always be possible to raise and lower indices at will. Burke makes firm pronouncements, e.g., that force is a covector (I recapped his argument <a href="http://physics.stackexchange.com/a/62564/4552">here</a>). At the permissive end of the spectrum, Rindler 1997 has a disclaimer early on in the book that he doesn't want to worry about distinguishing upper and lower indices, and won't do so until some later point in the book. Sometimes it feels a little strained to try to maintain such distinctions, especially in relativity, which we don't even know how to formulate without a nondegenerate metric. E.g., Burke argues that momentum is really a covector, because you can get it by differentiation of the Lagrangian with respect to $\dot{x}$. But then a perfectly natural index-gymnastics expression like $p^i=m v^i$ becomes something wrong and naughty.</p>
<p>I find this particularly confusing when it comes to higher-rank tensors and questions about which form of a tensor is the one that corresponds to actual measurements. Measurements with rulers measure $\Delta x^i$, not $\Delta x_i$, which is essentially a definition that breaks the otherwise perfect symmetry between vectors and covectors. But for me, at least, it gets a lot more muddled when we're talking about something like the stress-energy tensor. For example, in <a href="http://physics.stackexchange.com/questions/77374/equilibrium-for-a-rope-hanging-in-a-schwarzschild-spacetime">this question</a>, I was working through a calculation in Brown 2012 in which he essentially writes down $T^\mu_\nu=\operatorname{diag}(\rho,P,0,0)$ for the stress-energy tensor of a rope hanging in a Schwarzschild spacetime. It's not obvious to me that this corresponds better with measurements than writing down the same r.h.s. but with $T^{\mu\nu}$ or $T_{\mu\nu}$ on the left. Misner 1973 has a nice little summary of this sort of thing on p. 131, with, e.g., a rule stating that $T^\mu_\nu v_\mu v^\nu$ is to be interpreted as the density of mass-energy seen by an observer with four-velocity $v$. Most, but not all, of their rules are, like this one, expressed as scalars. This is very attractive, because we have identities such as $a^ib_i=a_ib^i$, which means that it makes absolutely no difference whether we discuss an object like $a^i$ or its dual $a_i$, and we never have to discuss which form of a tensor matches up with measurements, because our measurements are scalars.</p>
<p>Is this approach, of reducing every measurement to a scalar, universally workable in GR? Is it universally desirable? Is it valid philosophically to say that all measurements are ultimately measurements of scalars?</p>
<p>A few examples:</p>
<p>Some quantities like mass are defined as scalars, so we're good.</p>
<p>Mass-energy is $p^i v_i=p_i v^i$, where $v$ is the velocity vector of the observer.</p>
<p>In the case of a Killing vector, there is no way to reduce it to a scalar, but a Killing vector isn't really something you can measure directly, so maybe that's OK...?</p>
<p>Relations like $\nabla_i T^{ij}=0$ and $\nabla_i \xi_j+\nabla_j \xi_i=0$ could be reduced to scalars, e.g., $v_j\nabla_i T^{ij}=0$, but there is no real need to do so, because we're saying a tensor is zero, and a zero tensor is zero regardless of how you raise or lower its indices.</p>
<p>I would be particularly interested in answers that spelled out how one should reason about examples like the hanging rope. In the treatment on p. 131 of Misner, they give $T_{\mu\nu}=(\rho+P)v_\mu v_\nu+P g_{\mu\nu}$ for a perfect fluid; this is not scalar-ized, and in fact appears to contradict Brown's use of $T^\mu_\nu$.</p>
<p><em>Update</em></p>
<p>After chewing this over with Cristi Stoica and Trimok, I think I've understood the issue about $T^\mu_\nu$ versus $T_{\mu\nu}$ better. Contrary to what I said above, the expression $T_{\mu\nu}=(\rho+P)v_\mu v_\nu+P g_{\mu\nu}$ (with $-+++$ signature) for a perfect fluid <em>is</em> really scalar-ized, in the sense that $\rho$ and $P$ have no tensor indices, so they're notated as scalars. This makes sense, because $\rho$ and $P$ are defined by reference to a particular frame of reference, the rest frame of the fluid. This is exactly analogous to the way in which we define the scalar proper time by reference to the rest frame of a clock.</p>
<p>Now suppose we have coordinates in which the metric is diagonal, e.g., the Schwarzschild metric written in Schwarzschild coordinates. Let $g_{\mu\nu}=\operatorname{diag}(-A^2,B^2,\ldots)$. Furthermore, let's assume that we have some perfect fluid whose rest frame corresponds to zero coordinate velocity in these coordinates. The velocity vector of this rest frame is $v^\mu=(A^{-1},0,0,0)$, or, lowering an index, $v_\mu=(-A,0,0,0)$. Simply plugging in to the expression for $T$, we have $T_{00}=A^2\rho$, $T_{11}=B^2P$, $T^0_0=-\rho$, $T^1_1=P$. So this is the justification for Brown's use of the mixed-index form of $T$ -- it simply happens to be the one in which the factors of $A$ and $B$ don't appear. But that doesn't mean that the mixed-index form is the "real" one. What a static observer actually measures is the scalars $\rho$ and $P$. Similarly, it's not really $\Delta x^0$ or $\Delta x_0$ that an observer measures on a clock, it's the scalar $\Delta s$. When we say that coordinate differences correspond to upper-index vectors $\Delta x^i$, we're actually making a much more complicated statement that refers not to a simple measurement with a single device but rather to some much more extensive setup involving surveying, gyroscopes, and synchronization of clocks.</p>
<p>I think one of the pitfalls here is that I wasn't keeping in mind the fact that in an equation like $T_{\mu\nu}=\operatorname{diag}(\ldots)$, the right-hand side circumvents the rules of index gymnastics. This makes it preferable to work with equations like $T_{\mu\nu}=(\rho+P)v_\mu v_\nu+P g_{\mu\nu}$, where both sides are valid index-gymnastics notation, and we can raise and lower indices at will without worrying about invalidating the equation.</p>
<p>Finally, in the example above, suppose that we have an asymptotically flat spacetime, and suppose that at large distances we have $A^2\rightarrow 1$ and $B^2\rightarrow 1$. Then $|T^0_0/T_{00}|=A^{-2}>1$ corresponds to a gravitational redshift factor seen by an observer at infinity.</p>
<p>Related: <a href="http://physics.stackexchange.com/questions/76048/type-valence-of-the-stress-tensor">Type/Valence of the stress tensor</a></p>
<p><em>References</em></p>
<p>Brown, "Tensile Strength and the Mining of Black Holes," <a href="http://arxiv.org/abs/1207.3342" rel="nofollow">http://arxiv.org/abs/1207.3342</a></p>
<p>Burke, <em>Spacetime, Geometry, Cosmology</em>, 1980</p>
<p>Misner, Thorne, and Wheeler, <em>Gravitation</em>, 1973</p>
<p>Rindler, <em>Essential Relativity</em>, 1997</p>
| 2,100 |
<p>I am working on this lab that involved gathering data from two different sources. It involved gathering reaction times from a device and from a web application which was put into our data sets.</p>
<p>It is asking about anticipated events and if they agreed with the experimental uncertainty. After Googling these terms, I still found the explanations very unclear. Can someone explain how I could go about answering this?</p>
| 2,101 |
<p>In the movie <em>Transporter 3</em> a submerged car is floated to the surface by filling a large bag with air from the tyres.</p>
<p>I know that movies are about the worst places to get examples of physics in action, and my first thought was that if the air from the tyres was enough to inflate the bag and lift the car then wouldn't the air do that while still in the tyres? But then I wondered: is compressed air less buoyant than air at atmospheric pressure?</p>
| 2,102 |
<p>I am a high school student doing an IB Extended Essay investigation concerning the resonant frequencies of <a href="http://en.wikipedia.org/wiki/Ernst_Chladni" rel="nofollow">Chladni</a> plates of differing materials and sizes. Would someone please explain the definition of radial and diametric nodes ($n$ and $m$), and how to determine them? In addition, would someone be willing to provide a derivation of Chladni's equation $$f \sim (m + 2n)^2?$$</p>
<p>Here is a link with related mathematical resources:
<a href="http://www.nhn.ou.edu/~johnson/Education/Juniorlab/Chladni/2001-ChladniPlates-CurtisParry.pdf" rel="nofollow">http://www.nhn.ou.edu/~johnson/Education/Juniorlab/Chladni/2001-ChladniPlates-CurtisParry.pdf</a></p>
| 2,103 |
<p>Yesterday my brother asked me how orbits work. Suppose for the sake of the question that you are trying to put a rocket in orbit around the Earth. I explained that orbiting is essentially being in free fall while going very fast sideways, so that by the time you fall, there's no ground anymore and you keep going.</p>
<p>He said he understands that. His question was about why, after doing a full revolution around the Earth, you end up in the same place with the same velocity instead of doing, say, an inwards spiral. I admit I was stumped by this. Obviously, conservation of angular momentum prevents you from spiraling down to the center, but I see no reason why, at least in principle, you couldn't have a spiral orbit with an amplitude that increases and decreases periodically. </p>
<p>Is there an explanation for this that doesn't involve actually doing the math? Of course, the equations say this doesn't happen, but that doesn't help me understand any better, nor does it help me explain it to my brother.</p>
| 2,104 |
<p>What is theory behind free energy perturbation? Is it way too difficult to understand? Can someone explain it in simple terms.</p>
| 2,105 |
<blockquote>
<p>I want to calculate the expectation value of a Hamiltonian. I have a wave function that is $$\psi = \frac{1}{\sqrt{5}}(1\phi_1 + 2\phi_2).$$</p>
</blockquote>
<p>I want to know if I set this up properly. The Hamiltonian is $\hat H \left(x, \frac{\hbar \partial^2}{2m\partial x^2}\right)$. To get an expectation value I need to integrate this: </p>
<p>$$\int \psi^* \hat H \psi dx.$$ </p>
<p>Since the wavefunctions are normalized and real I can go with $\psi^* = \psi$. </p>
<p>OK, so I put together the integral. </p>
<p>$$\int \frac{1}{\sqrt{5}}(\phi_1 + 2\phi_2)\frac{\hbar}{2m} \frac{1}{\sqrt{5}}(\phi_1'' + \phi_2'') dx=\frac{\hbar}{2m}\frac{1}5\int(\phi_1 + 2\phi_2)(\phi_1'' + 2\phi_2'')dx,$$</p>
<p>and I know the wavefunction for $$\phi_n = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$$ so $$\phi_1 = \sqrt{\frac{2}{L}}\sin\left(\frac{\pi x}{L}\right)$$ and $$\phi_2 = \sqrt{\frac{2}{L}}\sin\left(\frac{2\pi x}{L}\right).$$ </p>
<p>I can plug these in and do the integral, and I wanted to check that was the right thing to do. I suspect there is an easier method, though. But if this will work then I can say "great, I at least understand this enough to do the problem." </p>
| 2,106 |
<p>Suppose that completely stopping a subatomic particle, such as an electron, could happen under certain conditions. What would be likely ways to get an electron to be perfectly still, or even just stop rotating the nucleus and collapse into it by electromagnetic forces? What would likely be required, below absolute-zero temperatures? Negative energy? Or could a 0 energy rest state not exist in any form, of any possible universe imaginable?</p>
<p>Let's say there was a magnetic field of a certain shape that we could postulate that is so intensely strong that if we put an electron in the center of it, it could not move at all in any direction. Would the energy requirement of the field be infinite? What would be the particle's recourse under this condition?</p>
<p>Further, suppose it were possible and one could trap an electron and stop all motion completely. What would this do to Heisenberg's Uncertainty Principle and/or Quantum Mechanics, because its position and momentum (0) would both be known? If it can be done, is Quantum Mechanics no longer an accurate model of reality under these conditions? Could we say QM is an accurate model under most conditions, except where it is possible to measure both position and momentum of a particle with zero uncertainty?</p>
<h1>Clarification:</h1>
<p>Please assume, confined in a thought experiment, that it IS possible to stop a particle so that is has 0 fixed energy. This may mean Quantum Mechanics is false, and it may also mean that under certain conditions the uncertainty commutation is 0. ASSUMING that it could physically be done, what would be likely to do it, and what would be the implications on the rest of physics?</p>
<h1>Bonus Points</h1>
<p>Now here's the step I'm really after - can anyone tell me why a model in which particles can be stopped is so obviously not the reality we live in? Consider the 'corrected' model is QM everywhere else (so all it's predictions hold in the 'normal' regions of the universe), but particles can be COMPLETELY stopped {{inside black holes, between supermagnetic fields, or insert other extremely difficult/rare conditions here}}. How do we know it's the case that because the uncertainty principle has lived up to testing on earth-accessable conditions, that it holds up under ALL conditions, everywhere, for all times?</p>
| 2,107 |
<p>if I have a not necessarily homogenous electric field of a charge distribution in an electrolyte and i want to find out what the electric field at some position in the electrolyte is. is there any equation that i could use to consider also electric screening by the ions, so that I will get a different electric field that is lowered by the screening?</p>
<p>So basically I am looking for an equation that takes an electric field or potential and gives me the screened one.</p>
| 2,108 |
<p>I am a class 9 student in India. I want to become a physicist. What should I study after class 10? What are my options for colleges and universities? What should I do after my education?
What are the fields to do research in post-graduation?</p>
| 40 |
<p>I know that if a star collapses into a volume with radius less or equal to the Schwarzschild radius $r_s=\frac{2GM}{c^2}$ then a black hole is created and it has the same mass of the star that gave it origin. But is there a way to calculate the mass of a black hole without knowing the volume of the star?</p>
| 2,109 |
<p>This is a question about an historical theory of gravitation, studied by Einstein quite a bit <em>before</em> he settled on General Relativity. At that time, Einstein did not know that gravity was a consequence of curved space-time. He identified the variations of gravity with the variations of light speed in a gravitational field. <br></p>
<p>In March 1912, Einstein postulated a first equation for static gravitational field, derived from the Poisson equation
$$\Delta c = kc\rho \tag{1}~,$$ where $c$ is light speed, $\rho$ is mass density and $\Delta$ is Laplacian.</p>
<p>Two weeks later, he modified this equation by adding a nonlinear term to satisfy energy-momentum conservation :
$$\Delta c = k\big(c\rho+\frac{1}{2kc} (\nabla c)^2\big)~. \tag{2}$$
Einstein's argument is the following:</p>
<p>The force per unit volume in terms of the mass density $\rho$ is $f_a$
$= \rho \nabla c$. Substituting for $\rho$ with $\frac{\Delta c}{kc}$ [equation (1)], we find $$f_a = \frac{\Delta c}{kc} \nabla c~.$$</p>
<p>This equation must be expressible as a total divergence (momentum conservation) otherwise the net force will not be zero (assuming $c$ is constant at infinity).
Einstein says: </p>
<blockquote>
<p>"In a straightforward calculation, the equation <strong>(1)</strong> must be
replaced by equation <strong>(2)</strong>."</p>
</blockquote>
<p>I never found the straightforward calculation. That's something that's actually hard for me!</p>
<p><b>addendum</b> <br>
The solution given and explained by @Gluoncito (see below) answers perfectly my question. However, it is likely that it is not the demonstration of Einstein for at least one reason : It is not a <em>straightforward calculation</em>.<br> Historically, Abraham, a german physicist, was the first to generalize the Poisson equation by adding a term for the energy density of the gravitational field (coming from $E=mc^2$). He published a paper in january 1912 containing a static field equation with the term :
$\frac{c^2}{\gamma}(\nabla c)^2 $ different but not far away from the Einstein term. After the publication of Einstein, Abraham claimed That Einstein copied his equation. I believe Einstein was at least inspired by Abraham. To what extent, I don't know. </p>
| 2,110 |
<p>Does gravity slow the speed that light travels? Can we actual measure the time it takes light from the sun to reach us? Is that light delayed as it climbs out of the sun's gravity well?</p>
| 2,111 |
<p>Is it more efficient to stack two Peltier modules or to set them side by side?
And why?</p>
<p>I have a small box that I want to cool down about 20 K below ambient -- cold, but not below freezing.
(I want to keep my camera cool, so I'm putting in this cool box.
The camera looks through a flat glass window on one side of the box).</p>
<p>The heatsink I have on hand is about twice as wide as the widest Peltier module I originally planned on using.
So there's room to put 2 Peltier modules side-by-side under the heatsink.
Or I could center a stack of 2 Peltier modules under the heatsink.
Which arrangement is more efficient?</p>
<p>I have to cut a bigger hole in the insulation for the side-by-side arrangement, so the unwanted heat that "back-flows" through the side-by-side arrangement is worse.
On the other hand, other effects are worse for the stacked arrangement.</p>
<p>(Is <a href="http://electronics.stackexchange.com/">http://electronics.stackexchange.com/</a> a better place to post questions about Peltier coolers?)</p>
| 2,112 |
<p>If we are to take two Hydrogen atoms and subject them to the same potential, then wouldn't both Hydrogen atoms be in the same exact quantum state? This bother me because no two identical fermions can be in the same quantum state! This seems to contradict the principle. This applies to any two elements or molecules that are subjected to the same potential.</p>
<p>Say these two Hydrogen atoms are located 1m from each other, then would the only way to distinguish them would be their spatial location?</p>
<p>What is the technical term for two seemingly identical things to be distinguishable by their location?</p>
| 2,113 |
<p>Every so often,* we get a question about what would happen should there be a change in a physical constant that contains dimensional information, such as $\hbar$, $c$, $G$, or often "the scale of the universe", and we often wind up reinventing the wheel to explain why this is not a well-posed question. I would therefore like to set this up as a canonical question to refer to later and save some duplication of effort. So:</p>
<p><strong>Why is it meaningless to speak about changes to a physical constant that contains dimensional information?</strong></p>
<p><sub>*
<a href="http://physics.stackexchange.com/questions/34874/has-the-speed-of-light-changed-over-time">These</a>
<a href="http://physics.stackexchange.com/questions/78635/what-would-happen-if-the-scale-of-some-atoms-changed">are</a>
<a href="http://physics.stackexchange.com/questions/721/if-the-size-of-universe-doubled">some</a>
<a href="http://physics.stackexchange.com/questions/33378/was-plancks-constant-h-the-same-when-the-big-bang-happened-as-it-is-today/33408">examples</a>
<a href="http://physics.stackexchange.com/questions/47259/if-everything-in-the-universe-doubled-in-size-overnight-would-it-be-noticeable">but</a>
<a href="http://physics.stackexchange.com/questions/24481/what-would-be-the-effects-of-an-inflating-universe">there's</a>
<a href="http://physics.stackexchange.com/questions/7285/are-there-measurable-effects-to-scaling-the-action-by-a-constant">more</a>
.
</sub></p>
| 2,114 |
<p>We all use deodorant and they always feel cold, why is that?</p>
<p>Is it because it is liquid inside the bottle and a gas when it is released?</p>
| 2,115 |
<p>The firewall paradox is a very hot topic at the moment (1207.3123v4). Everyone who is anybody in theoretical physics seems to be jumping into the action (Maldacena, Polchinski, Susskind to name a few). </p>
<p>However, I am unable to see the paradox. To me Hawking's resolution of the information paradox (hep-th/0507171) also resolves the so called firewall paradox. Hawking never says that the information is not lost on a fixed black hole background. In fact, he says the opposite. He says (page 3):
"So in the end everyone was right in a way. Information is lost in topologically non-trivial metrics like black holes. This corresponds to dissipation in which one loses sight of the exact state. On the other hand, information about the exact state is preserved in topologically trivial metrics. The confusion and paradox arose because people thought classically in terms of a single topology for spacetime."</p>
<p>To surmise, in quantum gravity you don't know if you actually have a black hole or not, so you have to include the trivial topologies, including those when there isn't a black hole there, in the amplitude. Only then do recover unitarity. </p>
<p>It seems to me that the error of the AMPS guys is that they use a fixed black hole background and assume conservation of information (i.e., late time radiation is maximally entangled with early time radiation). It is no wonder they are lead to a contradiction. They are simply doing the information paradox yet again. </p>
<p>They give a menu of implications in the abstract:
(i) Hawking radiation is in a pure state,
(ii) the information carried by the radiation is emitted from the region near the horizon, with low energy effective field theory valid beyond some microscopic distance from the horizon, and
(iii) the infalling observer encounters nothing unusual at the horizon.</p>
<p>But the obvious solution is that (i) is wrong. The radiation, within the semi-classical calculation in which they calculate it (i.e., not quantum gravity), is non-unitary.</p>
<p>So my question is, why is this a paradox? Something so obvious surely can not be overlooked by the ``masters'' of physics. Therefore I'd like to hear your opinions. </p>
| 2,116 |
<p>My physics teacher talked about the meeting of 2 parallel lines, and he said that it may occur in the infinity or something. I know that 2 parallel lines can meet in spherical geometry, (thanks to math stackexchange), but can such a thing occur in physics? (I really want to know so I could tell him).</p>
| 2,117 |
<p>Do Earth's rotation/revolution is regulated by other planets (and vice versa) in any way? How?</p>
| 2,118 |
<p>Recently, I was doing my homework and I found out that Torque can be calculated using $\tau = rF$.
This means the units of torque are Newton meters. Energy is also measured in Newton meters which are joules.</p>
<p>However, torque isn't a measure of energy. I am really confused as why it isn't measured in Joules.</p>
| 784 |
<p>Roughly speaking, we define a <a href="http://en.wikipedia.org/wiki/Manifold" rel="nofollow">manifold</a> $M$ to be covered by a set of charts $\{(U_i , \varphi_i)\}$ such that locally the $n$-dimensional manifolds looks like $\mathbb{R}^n$. One of the conditions is that all the $U_i$ are open sets of the topology of the manifold.</p>
<p>Why do we require the manifold to be a topological space? And why do we want $U_i$ to be open sets? What are the implications of these requirements on physics? (It appears to me that without these conditions the manifold still looks locally like $\mathbb{R}^n$.)</p>
<p>Edit to make my question more concrete: are there any physical theories that use manifolds that are not topological spaces? For instance, what would happen to general relativity if the spaces are diffeomorphic to each other, instead of homeomorphic (see answer below by Robin Ekman)?</p>
| 2,119 |
<p>When an electron absorbs a photon, does the photon become electron "stuff" (energy); or, is it contained within the electron as a discrete "something"?</p>
| 2,120 |
<p>I'm trying to understand the definition given on my electromagnetism course for the current density. More specifically, I want to know why, as defined below, the current density is given the name "current density."</p>
<p>On my course, the current density is $\vec{j}(t,\vec{x}):=\rho (t,\vec{x}) \vec{v} (t,\vec{x})$ where $\vec{v} (t,\vec{x})$ is the velocity field governing how the charged particles move. I'm trying to get some intuition for what this quantity is. </p>
<p>To give an example of what I'm talking about, in classical mechanics where you have momentum equal to mass multiplied by velocity, the definition makes sense intuitively because momentum is the oomph you will feel if an object hits you, and you feel that oomph more if either the mass or velocity of the object increases. So I have a really tangible idea of what momentum is.</p>
<p>Wikipedia describing current density: "In electromagnetism, and related fields in solid state physics, condensed matter physics etc. current density is the electric current per unit area of cross section." This justifies calling it a density (as it's an area density by defn.). I'm trying to understand what a current density could be, and in my head I've got an idea of a cross-sectional area with some fluid flowing through it (it's the same scenario in which I picutre Gauss' Law). I'm not sure what current at a point is, so I don't really understand what a current density could be! Then I need to relate this to the definition I've been given on my course somehow. Thanks for any help!</p>
| 2,121 |
<p>When resistances connected in series then why through each resistances the different potential difference occurs? </p>
| 2,122 |
<p>If space is not actually a void, then there's something shooting around in it, I'm guessing a lot like molecules in a gas. Could we apply a $(PV/T = k)$ -like gas law to it. Obviously we're not measuring a regular everyday gas but some theoretical "fluid" that we don't know enough about yet. Also regular radiation might also be part of the total pressure, and there's a <a href="http://en.wikipedia.org/wiki/Radiation_pressure" rel="nofollow">wikipedia article</a> on that.</p>
<p>What are your thoughts on this?</p>
<p>If there is such a fluid, and it obeys the gas law, then there's either an infinite amount of this fluid in our currently observable universe or the universe is finite in volume.</p>
| 2,123 |
<p>Does the resistance of an analog current meter increase or decrease when it is set to a more sensitive scale (lower range)?</p>
| 2,124 |
<p>Since the $\vec{E}$ field inside a "perfect" conductor is zero, do the electrons(the current) flow only on the outer surface? This has bothered me since I studied electromagnetism. </p>
<p>Thank you for your time.</p>
| 2,125 |
<p>Are there any QM effects that have been/could be measured from interactions involving non-charged particles? </p>
<p>Elementary QM is all about the electron energy levels in the atom, photon - atom interactions, etc. </p>
<p>When one looks at the nucleus, its all about quark interactions - which are also charged particles. </p>
<p>I can think of some theoretical ones - like neutrinos orbiting a mass, but they would be hard to impossible to measure. Another possibility is the strong / weak nuclear force - but that always happens with particles that are also charged. </p>
<p>In the end we always need matter built instruments to see a result - that's fine. You could for instance observe photons coming from some distant (metres to Mpc) away region where some interaction occurred.</p>
| 2,126 |
<p>So I was reading about <a href="http://www.nasa.gov/mission_pages/spitzer/news/spitzer20111221.html" rel="nofollow">GN-108036</a> this morning and for some reason I thought of something which I can't quite wrap my head around and make sense of. It's early morning so maybe coffee hasn't kicked in yet.</p>
<p>If it starts with a central big bang and we now look at something that is 12.9 billion light years away, how can we be here already to "observe" the light if the universe is approx. 13 billion years old?</p>
<p>In other words:</p>
<p>Big Bang, stuff starts expanding.<br>
What will form GN-108036 is part of it<br>
What will eventually form earth is part of it.<br>
Nothing can travel faster than light yet;</p>
<p>12.9 million years after we are at this point "ahead" of the light from GN-108036 to see the light reaching us?</p>
| 2,127 |
<p>In <a href="http://en.wikipedia.org/wiki/Catching_Fire" rel="nofollow">Catching Fire</a>, the second episode of the Hunger Games trilogy, one of the characters (Beetee) proposes a plan to kill some of the remaining tributes. He proposes wrapping a special wire that he has designed to withstand the vast energy of lightning around a tall tree and running it down to a salt water lake. Then when lightening strikes the tree (which it does predictably every twelve hours) anyone in the lake will be killed. (N.B. This is not Beetee's real plan, but it is the one he wants his allies to believe.)</p>
<p>I don't think his plan would work at all. Assuming that his wire can carry the enormous current delivered by a lightening strike (this is fiction after all) surely the current would short to ground through the salt water rather than taking a path through anyone standing or swimming in the water. The analogy I have in mind is that of birds standing untroubled on live electricity wires. They are unharmed because they do not provide a path to earth. But another analogy might be dropping a mains electric heater into a bath when you are in it. I think that kills you (though I'm not sure why.)</p>
| 650 |
<p>While reading with my son about how a Mars-like planet collided with the early Earth that resulted in our current moon, it said the initial debris also formed a ring, but that ring ended up getting absorbed by the Earth and the Moon.</p>
<p>I couldn't answer his question then why Saturn still has rings. Shouldn't Saturn's rings be clumping into Moons or getting absorbed by Saturn's gravity?</p>
| 2,128 |
<p>In my particle physics lecture, <a href="http://en.wikipedia.org/wiki/Faddeev%E2%80%93Popov_ghost" rel="nofollow">ghost fields</a> were briefly mentioned. </p>
<p>As far as I understand, these come up when computing cross sections by the path integral method, to compensate for equivalent contributions due to gauge freedom.</p>
<p>I'm still not sure how these equivalent contributions come up in the calculations.
Say you want to compute the cross section of an electron anti-electron annihilation into a photon near a massive nucleus. I assume you would then take the sum of the path integrals over all possible Feynman diagrams corresponding to that interaction.</p>
<p>QED has charge symmetry (i.e. $U(1)$), so the physics stay the same if the two charges are swapped.</p>
<p>Where concretely do we have to introduce these ghost fields, and why can we not simply combine the Feynman diagrams that are graph-isomorphisms of each other into a single equivalence class and count the contribution of each equivalence class only once into the final probability amplitude.</p>
<p>Please keep the answers simple, as I'm not a physics major.</p>
| 2,129 |
<p>In the YouTube video <a href="https://www.youtube.com/watch?v=_yEu2R1gYSs" rel="nofollow">Monster magnet meets computer</a>, the south pole of a 1 T (roughly) neodymium magnet is held in front of a CRT. Assuming the CRT produces electrons of 30 keV, and that the screen is 0.2 m from the end of the electron gun, how much energy does each electron have when it hits the screen when the magnet is present? To simplify the question, assume the screen has no thickness, and electrons from the electron gun are headed straight for the middle of the screen. Simple high school physics isn't enough to answer this due to the velocities involved, so I can't answer this myself.</p>
| 2,130 |
<ol>
<li><p>What is the meaning <a href="http://en.wikipedia.org/wiki/Generalized_coordinates" rel="nofollow">generalised coordinates</a> in Classical Mechanics?</p></li>
<li><p>How is <a href="http://en.wikipedia.org/wiki/Lagrangian_mechanics" rel="nofollow">Lagrangian formalism</a> different from <a href="http://en.wikipedia.org/wiki/Hamiltonian_mechanics" rel="nofollow">Hamiltonian formalism</a>?</p></li>
<li><p>How are they related to <a href="http://en.wikipedia.org/wiki/Hamilton%27s_principle" rel="nofollow">Hamilton's Principle</a>?</p></li>
<li><p>How are they related to <a href="http://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation" rel="nofollow">Euler-Lagrange equation</a>?</p></li>
</ol>
| 2,131 |
<p>Recently I read <a href="http://blogs.discovermagazine.com/badastronomy/2011/11/17/huge-lakes-of-water-may-exist-under-europas-ice/">this entry</a> by Dr. Phil Plait (a.k.a. The Bad Astronomer). He is talking about a lake of water contained within the ice layer of Europa.</p>
<blockquote>
<p>the lake is completely embedded in the ice shell. In general, the ice is very thick, explaining the usual look of Europa’s surface. But in some spots, just below the ice, the ice has melted. The ice above this underground lake is much thinner, perhaps only 3 km (about 2 miles) thick, explaining the chaotic surface in those localized spots. </p>
</blockquote>
<p>Like in this artist's conception:</p>
<p><img src="http://i.stack.imgur.com/Wei7H.jpg" alt="Europa Lake"></p>
<p>Dr. Plait talks about the importance of this discovery, but doesn't delve much into HOW this happens. Have any papers, hypotheses, or guesses been made at how this lake (and presumably others like it) were made?</p>
| 2,132 |
<p>I understand now how I can derive the lowest energy state $W_0 = \tfrac{1}{2}\hbar \omega$ of the <a href="http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator" rel="nofollow">quantum harmonic oscillator</a> (HO) using the ladder operators. What is the easiest way to now derive possible wavefunctions - the ones with <a href="http://en.wikipedia.org/wiki/Hermite_polynomials" rel="nofollow">Hermite polynomials</a>?</p>
<p>I need some guidance first and then I will come up with a bit more detailed questions.</p>
| 2,133 |
<p>There is a section of my notes which I do not understand, hopefully someone here will be able to explain this to me. The notes read (after introducing the uncertainty operator):</p>
<blockquote>
<blockquote>
<p>If the state $\chi_A$ is an eigenstate of $\hat O_A$ then the uncertainty is zero and we measure it with probability 1. However, if $\hat O_B$ is another observable which does not commute with $\hat O_A$, then the uncertainty in any simultaneous measurement of the two observables will be infinite.</p>
</blockquote>
</blockquote>
<p>I understand the first sentence, but I can't see how to justify/prove the second one. Can someone tell me how the second sentence is justified, please?</p>
| 2,134 |
<p>I thought electronics is mainly discussed in engineering majors. But in my university Electronics and Digital Electronics is also compulsory for physics majors.I searched other's syllabus, but I didn't found any electronics course. In MIT opencourseware also there is no electronics course in physics.</p>
<p>Is there any phenomenon in electronics which is fundamental in our understanding of universe?</p>
| 2,135 |
<p>In <a href="http://arxiv.org/pdf/1003.1366.pdf">this</a> paper on p42, it is explained that when starting with a bare action that contains a standard kinetic term, this kinetic term attains a correction in the course of the RG flow which can be denoted by $1/Z_{\Lambda}$, such that the effective kinetic term at a scale $\Lambda$ can be written as</p>
<p>$$
\frac{1}{2Z_{\Lambda}}\int\frac{d^dp}{(2\pi)^d}\phi(-p,\Lambda)p^2\phi(p,\Lambda)
$$</p>
<p>the effective four point interaction would be</p>
<p>$$
\frac{1}{4!Z_{\Lambda}^2} \,\, \int\limits_{p_1,\ldots,p_4}\phi(p_1,\Lambda)\ldots\phi(p_4,\Lambda)\hat{\delta}(p_1+ ... p_4)
$$</p>
<p>and analogously for higher order interactions. To get rid of the change done to the action described by the $Z_{\Lambda}$ dependent corrections, on redifine the field $\phi$ such as</p>
<p>$$
\phi \rightarrow \phi\left( 1- \frac{\eta}{2}\frac{\delta\Lambda}{\Lambda}\right)
$$</p>
<p>with the anomalous dimension </p>
<p>$$
\eta = \Lambda\frac{d\ln Z_{\Lambda}}{d\Lambda}
$$</p>
<p>I dont see why the anomalous dimension has to be defined like this in this case in order for the rescaling of the field leading to a cancellation of the changes to the action due to a renormalization step and like to see an explanation. How generally valid the above expression for the anomalous dimension anyway, is it "only" valid whan starting with bare actions that have a (standard) kinetic term? How can the anomalous dimension be calculated generally? I always thought that roughly speaking, when rescaling after the course graining step, the anomalous dimension "parameterizes" just the deviations from the canonical (some call them engineering) scaling dimensions that have to be applied.</p>
| 2,136 |
<p>When solving the heat equation,
$$
\partial_t u -\Delta u = f \text{ on } \Omega
$$
what physical situations are represented by the following boundary conditions (on $\partial \Omega$)?</p>
<ul>
<li>$u=g$ (Dirichlet condition),</li>
<li>$n\cdot\nabla u = h$ (Neumann condition),</li>
<li>$n\cdot\nabla u = \alpha u$ (Robin condition),</li>
<li>$n\cdot\nabla u = u^4-u_0^4$ (Stefan-Boltzmann condition).</li>
</ul>
<p>Are there other common physical situations where another boundary condition is appropriate?</p>
| 2,137 |
<p>In the context of this <a href="http://physics.stackexchange.com/questions/1428/a-question-on-evolution-of-mass-and-velocity-distributions-rhor-t-and-v">question</a> should mass distribution $\rho(r,t)$ and momentum distribution $p(r,t)$ be well behaved ? By 'well behaved' it is meant that derivatives of all orders exist everywhere.</p>
<p>I don't see any reason why they should be well behaved. For the equations (given in the answer (by Mark Eichenlaub) to the above mentioned question) to be consistent the conditions required are</p>
<ol>
<li><p>$\vec{p}$ should be continuously differentiable in order that $\nabla\cdot\vec{p}$ exists.</p></li>
<li><p>$\rho(r,t)$ should be partially differentiable wrt $t$ i.e., $\frac{\partial\rho(\vec{r},t)}{\partial t}$ exists.</p>
<p>Please give strong reasons as to why it is required that $\rho(r,t)$ and $\vec{p}(r,t)$ to be always well behaved ?</p></li>
</ol>
| 2,138 |
<p>Since any source of light will have a finite duration, the light emited won't have a particular frecuency. It will be a sum of different frequencies (infinite, I think) if we apply Fourier's series (integral).</p>
<p>Would this mean that any photon's frequency will have some uncertainty or something like the photon emitted would be a combination of different photons? </p>
| 2,139 |
<p>Assume the time standard clocks and any memories about the time standard are destroyed. Can we recover the time standard again exactly?</p>
<p>Recovering the time standard again means we can determine the date and time
that are exactly the same as the destroyed time standard clocks will show (if they are not destroyed). </p>
| 704 |
<p>I work on a drilling rig as a roughneck and we had a lecture today (at the office) about mechanical advantage in pulley systems. Now, I know that my boss is well educated in oil drilling, but my instincts tell me that he may have this one wrong.</p>
<p>A drilling rig works sort of like a crane in that it has a tall structure supporting a pulley system. There is a large winch permanently installed on the base platform and then it goes over the top of the structure (the crown of the derrick) and down through a floating sheave--this has a few wraps to give us more mechanical advantage. I am including pictures to help describe the situation. </p>
<p><img src="http://i.stack.imgur.com/naok2.jpg" alt="Floating Sheave (called the "blocks")"></p>
<p>Here the picture shows the floating sheave (the blocks) which we use to do most all of our operations. Most importantly, we use it to pick up our string of pipe that is in the ground.</p>
<p><img src="http://i.stack.imgur.com/nW0mS.gif" alt="The set up of a drilling rig"></p>
<p>As seen in this picture, the blocks hold the weight of the string of pipe. Now he told us that if the pipe get stuck in the hole (maybe it snags something or the hole caves in), that we lose all of our mechanical advantage. He said that is why the weight indicator will shoot up and go back down after it is freed. He said that because when the pipe is snagged in the hole then we are not dealing with a free floating sheave anymore and that is what is required to have a mechanical advantage.</p>
<p>I disagree with this because even if it is not free, there is still a mechanical advantage such that (say the normal mechanical advantage is 6 to 1) our pulling force is multiplied by 6. I would like somebody to confirm this for me.</p>
<p>First picture taken from www.worldoils.com on June 21, 2013 Second picture taken from www.PaysonPetro.com on June 21, 2013</p>
| 2,140 |
<p>Since I don't know the proper physical terms for this, I describe it in everyday English. The following has kept me wondering for quite some time and so far I haven't found a reasonable explanation.</p>
<p>When you fill a ceramic cup with coffee and you click with the spoon at the bottom (from the top, through the coffee), each following tick, even when you pause for some seconds, will have a higher pitch. The following I've observed so far:</p>
<ul>
<li>works better with coffee than with tea (works hardly at all with tea)</li>
<li>works better with cappuccino than with normal coffee</li>
<li>doesn't work with just cold water</li>
<li>works best with ceramic cups, but some plastic cups seem to have the same, yet weaker, behavior</li>
<li>doesn't work on all types of cups, taller cups seem to work better</li>
<li>must have a substantive amount of liquid (just a drop doesn't make it sing).</li>
</ul>
<p>It must be something with the type of fluid, or the milk. I just poured water in a cup that had only a little bit fluffy left from a previous cappuccino, and it still worked. Then I cleaned it and filled it again with tap water and now it didn't work anymore.</p>
<p>Can someone explain this behavior?</p>
| 844 |
<p>Assuming a frictionless / "perfect" environment, and given a ball held in an elastic sling (like a hand-held catapult) <em>where the pocket is lighter than the projectile itself</em>, <strong>what is the point at which the ball separates from the sling pocket? At the start of the shot, or as the sling-pocket passes through its at-rest position?</strong></p>
<p>My thoughts -- please confirm or refute:</p>
<ul>
<li><p>Hooke's Law tells us that the force applied to the sling-pocket is directly proportional to the distance from the rest position.</p></li>
<li><p>Given the pocket's lower weight, I suppose then that it's velocity (resulting from the applied elastic force) would be greater than that of the projectile itself; thus it should be pushing the projectile (and thus be in contact with it) until such point as we pass the rest position for the first time (assuming sinusoid motion of the band as it vibrates)</p></li>
<li><p>In passing the rest position, the force now begins to be applied in the opposite direction and thus reduces the forward velocity of the pocket to be less than that of the projectile, which is travelling at a constant, maximal velocity.</p></li>
</ul>
<p>I am a programmer, not a physicist, so your patience is appreciated. Also note that I am trying to keep certain factors, eg. momentum, out of the picture here, as it is not necessary for what I am building (due to certain shortcuts taken). It is really just force, acceleration, mass and velocity I am concerned with.</p>
| 2,141 |
<p>Given a stationary 1-D wave function $\psi(x)$, how is the derivative in the momentum operator interpreted?</p>
<p>$$
\int_{-\infty}^\infty \psi^*(x) \hat{p} \psi(x) dx
= \int_{-\infty}^\infty \psi^*(x) (-i\hbar\nabla) \psi(x) dx
$$</p>
<p>Should the integral be interpreted as</p>
<p>$$-i\hbar\int_{-\infty}^\infty \psi^*(x) \psi'(x) dx$$</p>
<p>where $\psi'(x)$ is the derivative with respect to $x$?</p>
| 2,142 |
<p>Let’s consider this equation for a scalar quantity $f$ as a function of a 3D vector $a$ as:</p>
<p>$$ f(\vec a) = S_{ijkk} a_i a_j $$</p>
<p>where $S$ is a tensor of rank 4. Now, I’m not sure what to make of the index $k$ in the expression, as it doesn’t appear on the left-hand side. Is it a typo, meaning there is a $k$ missing somewhere (like $f_k$), or does it mean that it should be summed over $k$ like so:</p>
<p>$$f(\vec a) = \sum_i \sum_j \sum_k S_{ijkk} a_i a_j $$</p>
| 2,143 |
<p>Let's consider a situation: we have distant point source of unpolarized light in certain non-zero range of wavelengths (it's polychromatic). Let's divide this light into 2 beams depending on polarization direction with e.g. Wollaston prism. Then let's rotate the polarization plane of one of these beams by angle 90 degrees. Are resulting beams coherent (able to produce interference pattern)?</p>
<p>Thank you!</p>
| 2,144 |
<p>When solving problems in physics, one often finds, and ignores, "unphysical" solutions. For example, when solving for the velocity and time taken to fall a distance h (from rest) under earth gravity:</p>
<p>$\Delta t = \pm \sqrt{2h/g}$</p>
<p>$\Delta v = \pm \sqrt{2gh}$</p>
<p>One ignores the "unphysical" negative-time and positive-velocity solutions (taking x-axis as directed upwards normal to the earth's surface). However, this solution is not actually unphysical; it is a reflection of the fact that the equation being solved is invariant with respect to time-translation and time-reversal. The same equation describes dropping an object with boundary conditions ($t_i$ = 0, $x_i$ = h, $v_i$ = 0) and ($t_f$ = $|\Delta t|$, $x_f$ = 0, $v_f$ = $-\sqrt{2gh}$), or throwing an object backward in time with boundary conditions ($t_i$ = $-|\Delta t|$, $x_i$ = 0, $v_i$ = $+\sqrt{2gh}$) and ($t_f$ = 0, $x_f$ = h, $v_f$ = 0). In other words, both solutions are physical, but they are solutions to superficially different problems (though one implies the other), and this fact is an expression of the underlying physical time-translation and time-reversal invariance. </p>
<p>My question is: is there a more general expression of this concept? Is there a rule for knowing when or if an "unphysical" solution is or is not truly unphysical, in the sense that it may be a valid physical solution corresponding to alternate boundary conditions? </p>
| 2,145 |
<p>I heard once in a TED talk how Fizeau measured the speed of light in the 19th century. Here is the link </p>
<p><a href="https://www.youtube.com/watch?v=F8UFGu2M2gM" rel="nofollow">https://www.youtube.com/watch?v=F8UFGu2M2gM</a></p>
<p>You can read about it here in Wikipedia:</p>
<p><a href="http://en.wikipedia.org/wiki/Fizeau%E2%80%93Foucault_apparatus" rel="nofollow">http://en.wikipedia.org/wiki/Fizeau%E2%80%93Foucault_apparatus</a></p>
<p>A short summary: He placed a kind of rotating wheel in front of a beam of light, and a mirror far away from these two things. The beam of light passes between two teeth of the rotating wheel, reaches the mirror and goes back from the original source. As the wheel is spinning very fast, during the time that the light has been travelling, the wheel has rotated a tiny bit, but enough to impede the passage of time through the point where it entered. Knowing the distance from the mirror as well as the speed at which the wheel is rotating, the speed of light can be easily calculated. The experiment is better explained in wikipedia, here I wrote a simplified version of it.</p>
<p>I loved the experiment, because it seemed fairly easy to reproduce, so I ordered a green laser pointer on Amazon, which can reach up to 10 km. As a proof of principle, I went with a friend in the night to a place where there is a good visibility. We began setting a mirror somewhere 500 metres away from the laser, but, even from that far, the light had scattered so much that it was impossible to collect the light with the mirror.</p>
<p>The laser is a very powerful one, of those that you can see the whole beam (usually used in astronomy). If I can't repeat the experiment using a laser like this, how on earth could Fizeau do that in the 18th century employing a much more rudimentary source of light and placing the mirror much farther away? It says in Wikipedia that the distance between them was like 8 km.</p>
| 2,146 |
<p>I have a very simple question.</p>
<p>Everyone must have seen the rainbow after rain. According to the theory the <a href="http://en.wikipedia.org/wiki/Rainbow#Explanation" rel="nofollow">rainbow</a> is created due to the passing of sunlight from small drops of water in the atmosphere(means by dispersion of light).</p>
<p>Now my what I want to know is that after rain the rain drops are present in the entire atmosphere. So the whole atmosphere should look colorful.
Why only a semicircular shape is formed (or is colorful).</p>
| 2,147 |
<p>As I understand, the kinetic energy of the proton beam in a hadron collider is quite large. Can you build a space propulsion system that is based on accelerating a proton bean to relativistic speeds and then using the resulting kinetic energy to propel a space vehicle? </p>
<p>Edit: In the large hadron collider, the kinetic energy of a proton reaches 7 Tev, which translate to about 6.71E+20 joules of kinetic energy per kg of accelerated protons.</p>
<p>If the Collider can accelerate 1 kg of proton every hour and point it downwards, then the resulting thrust could launch into orbit the entire collider structure.</p>
<p>Of course, the LHC cant accelerate 1 kg of protons in an hour, but maybe a derivative of it could and would be the basis of space propulsion system.</p>
<p>This is basically the idea. I know that it is enormously hard to build, but can it be done, in theory?</p>
| 2,148 |
<p>This is Problem 2.6 (b) in Griffiths, <em>Intro to QM:</em></p>
<blockquote>
<p>A particle in an infinite square well has its initial wave function an even mixture of the first two stationary states:</p>
<p>$\Psi(x,0) = A[\psi_1(x) + \psi_2(x)]$.</p>
</blockquote>
<p>Here is the part of the problem that I am having a little trouble with:</p>
<blockquote>
<p>(b) Find $\Psi(x,t)$ and $|\Psi(x,t)|^2$. Express the latter as a sinusoidal function of time, as in Example 2.1. To simplify the result, let $\omega \equiv \frac{\pi^2 \hbar}{2ma^2}$</p>
</blockquote>
<p>According to the answer key, even after $t=0$, the wave function continues to be a mixture of the first two stationary states. Why is that? I am having a little difficulty understanding this. Why can't it be a new 'mixture?'</p>
<p>Are my questions sufficiently clear?</p>
| 2,149 |
<p>Heat Flux= 10 MW
Inlet water temperature = 28 degree centigrade for cooling purpose through a tube passing through the centre of retangular block of length 50 mm,height 30 mm and width 30 mm
inner diameter of tube=10mm and outer diameter=12 mm
tube is made of copper.</p>
<p>Find out the heat transfer coefficient of water at 42 bar and given temperature </p>
<p>Find out Nusselt number and Reynold's number.</p>
| 2,150 |
<p>Assuming the following</p>
<ol>
<li><p>A universe is the surface from a bubble in hyperspace.
Inside a bubble there is nothing, only the surface represents a
universe. The size of the bubble is time.</p></li>
<li><p>Dark matter is from bubbles different then
our own bubble that only shares gravity.</p></li>
<li><p>The big bang are bubbles colliding, creating a other bubble.</p></li>
</ol>
<p>My problem is if bubbles do only share gravity why would they have a collision, I expect that they would just pass trough each other without creating a big bang?</p>
| 2,151 |
<p>Is it possible to deduce the Archimedes' law of the lever using only the laws of conservation of the classical mechanics? I never saw (which is strange), but I think that it's possible. </p>
| 2,152 |
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