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<p>I am studying <a href="http://en.wikipedia.org/wiki/Mesoscopic_physics" rel="nofollow">mesoscopic physics</a>/quantum transport. Now I am wondering (out of interest): how did this field emerge and what made it such a huge field? I couldn't find this somewhere clear on the web and my book lacks an intro in which this is explained.</p>
| 2,907 |
<p>I'm solving an exercise about Hamiltonian equations. I have followed the proceeding below. The results given by the book are different to mine because its first result is the half of mine (and the second one linked to the first one is different to mine). I think that my proceeding is correct and so I can't understand...</p>
<blockquote>
<p><em>Given these two Hamiltonian equations:</em></p>
<p>$$\tag{1} \dot p ~=~ - \alpha pq,$$
$$\tag{2} \dot q ~=~\frac{1}{2} \alpha q^2.$$</p>
<p><em>Find $q(t)$ and p$(t)$, considering initial conditions $p_0$ and $q_0$.</em> </p>
</blockquote>
<p>I have integrated the second equation and obtained:</p>
<p>$$\tag{3} q(t)~=~\frac{2q_0}{2-q_0 \alpha (t-T_0)}$$</p>
<p>Then I have pugged this, in the second canonical eq, and I have obtained:</p>
<p>$$\tag{4} p(t)~=~p_0(2-q_0 \alpha (t-t_0))^2.$$</p>
<p>The solutions given by the book are:</p>
<p>$$\tag{5} q(t)~=~\frac{q_0}{1- \frac{1}{2} \alpha q_0 (t-t_0)},$$
$$\tag{6} p(t)~=~p_0[1-\frac{1}{2} \alpha q_0 (t-t_0)]^2.$$ </p>
<p>I can obtain the solutions of the book if I divide numerator and denominator of $q$ for 2.. but.. can I do it? </p>
<p>Is my proceeding correct? </p>
| 2,908 |
<p>If I take a sponge and place it in a shallow dish of water (i.e. water level is lower than height of sponge), it absorbs water until the sponge is wet, including a portion of the sponge above the water level. In other words, it seems the sponge pulls some water from the bath up into itself, doing work, and the water gains some gravitational potential energy.</p>
<p>Where does the energy required to do this work come from? My suspicion is that the answer involves physical and/or chemical bonds between the water and the sponge, or possibly the change in the surface area to volume ratio of the water.</p>
| 2,909 |
<p>On reading Feynman's lecture on physics, in the geometrical optics section he said that a curve which focuses all the rays coming from a point to another fixed point beyond the refracting surface perfectly is a complicated fourth degree curve which is actually the locus of all point with its distance $Op$ from one point, $O$ plus the distance $O_1p$, from another point $o_1$ times a number $n$ (refractive index of the material beyond the refractive surface) is a constant independent of point $p$ on the curve.</p>
<p>But I don't know how to derive this curve, or as a matter of fact, how to even begin to. Any derivation would be greatly appreciated (simple preferred).given below is the excerpt from feynman's lectures which is what i reffered to in my question.<img src="http://i.stack.imgur.com/QnY7n.png" alt="enter image description here"></p>
<p><img src="http://i.stack.imgur.com/1pFTy.png" alt="enter image description here"></p>
| 2,910 |
<p>I'm new at classical mechanics but the text book says there is the torque in the spinning top which generated only by gravitation. It is hard to explain the situation, I've add the link.</p>
<p><a href="http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/imgmech/toppre.gif" rel="nofollow">http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/imgmech/toppre.gif</a></p>
<p>There is no rotation axis between $r$(length between the point and center of mass) and $F$(gravitation), and I don't clearly understand the procession of the torque.</p>
<hr>
<p>I think the precssion of the spinning top is generated by the friction between the surface of the top(close with the point) and ground.</p>
<p><a href="http://www2.picturepush.com/photo/a/13200825/640/13200825.png" rel="nofollow">http://www2.picturepush.com/photo/a/13200825/640/13200825.png</a></p>
| 2,911 |
<p>In hydraulic analogy one compares electrical circuits with water circuits. For the electric case the formula $P = U \cdot I$ for the electric power holds. The analogous formula for water flow would be $P = \Delta p \cdot I_W$ where $\Delta p$ ist the pressure difference and $I_W$ the flow rate of the water through the pipe. I have some questions about this:</p>
<ul>
<li>under what circumstances/assumptions does this analogous formula hold</li>
<li>$P$ in the electric case can be interpreted as the energy per second which is <em>dissipated for example in a resistor</em>. Is there a similar interpretation in the water case and why does it hold?</li>
<li>with the assumptions from above, how can one derive the formula from first principles (e.g. from Bernoulli-equation or even from Navier-Stokes)?</li>
<li>with the assumptions from above, is there a nice conceptual argument, why the formula holds in the water case?</li>
</ul>
| 2,912 |
<p>If an atom is the smallest particle in the universe, does that mean that the spaces in between the atoms in water when being heated (expansion) are vacuum?</p>
| 2,913 |
<p>I am studying quantum physics and I have a question: what is the physical explanation for electrons having less energy than photons with the same wavelength?</p>
<p>Energy of a photon : $E = h c/\lambda$.<br>
Energy of an electron: $E = h^2/(2m\lambda^2)$</p>
| 2,914 |
<p>I am studying a bit granular dynamics and I have seen that two spheres of radius $R$ in contact with a contact area of radius $a$ would need an applied force $F$ on this two spheres that is nonlinear in the depth of deformation $\delta$ as it goes as:</p>
<p>$F \sim \delta^{3/2}$</p>
<p>To be honnest, I am not really interested in the full calculation as I am pretty sure I will forget it within two days and plus te full calculation probably would not give me a lot of insight on what is happening.</p>
<p>One way "a la de Gennes" that I have seen consist in relating the stess $\sigma$ to the vertical deformation $\epsilon$ via $\sigma = E \epsilon$.</p>
<p>Then people say that for spheres and if $a \ll R$ then </p>
<ul>
<li><p>1) $\epsilon \sim \delta /a$</p></li>
<li><p>2) $a \sim \sqrt{\delta R}$</p></li>
<li><p>$\Rightarrow \:\sigma \sim E\sqrt{\delta/R}$</p></li>
<li><p>3) $F = \pi a^2 \sigma \sim \pi R\delta \sqrt{\delta /R} \sim \pi E\sqrt{R}\delta^{3/2}$</p></li>
</ul>
<p>This final result is pretty close to the actual one. My point is that I don't understand the physics of the first equation as I am used to $\epsilon = \delta L/L$ which would give me $\epsilon \sim \delta/R$.</p>
<p><strong>Question:</strong></p>
<p>Is there any way to understand physically the fact that $\epsilon \sim \delta/a$?</p>
<p>Thanks very much for any answer.</p>
| 2,915 |
<p>Forgive me if this topic is too much in the realm of philosophy. John Baez has an interesting perspective on the relative importance of dimensionless constants, which he calls fundamental like alpha, versus dimensioned constants like G or c [ <a href="http://math.ucr.edu/home/baez/constants.html">http://math.ucr.edu/home/baez/constants.html</a> ]. What is the relative importance or significance of one class versus the other and is this an area that physicists have real concerns or expend significant research?</p>
<hr>
| 2,916 |
<p>Today, Korean media is reporting that a team of South Korean researchers solved <a href="http://en.wikipedia.org/wiki/Yang%E2%80%93Mills_existence_and_mass_gap" rel="nofollow">Yang-Mill existence and mass gap problem</a>. Did anyone outside Korea even notice this? I was not able to notice anything in US media. </p>
<p>The paper is </p>
<blockquote>
<p>Dimensional Transmutation by Monopole Condensation in QCD. Y. M. Cho, F. H. Cho, and J. H. Yoon. <a href="http://dx.doi.org/10.1103/PhysRevD.87.085025" rel="nofollow"><em>Phys. Rev. D.</em> <strong>87</strong> no. 8, 085025 (2013)</a>. <a href="http://arxiv.org/abs/1206.6936" rel="nofollow">arXiv:1206.6936</a> [hep-th].</p>
</blockquote>
<p>Is this just an ordinary "good" paper that is nothing close to solving Yang-Mills existence and mass gap problem? Or is it purported solution?</p>
<p>Note that the arxiv preprint is 2012's but the news of publication in Physical Review D is April, 2013.</p>
| 2,917 |
<p>I have read the paper</p>
<blockquote>
<p>Theoretical Study of Population Inversion in Graphene under Pulse Excitation. A. Satou, T. Otsuji and V. Ryzhii. <a href="http://dx.doi.org/10.1143/JJAP.50.070116" rel="nofollow"><em>Jpn. J. Appl. Phys.</em> <strong>50</strong> no. 7, pp. 070116-070116-4 (2011)</a>.</p>
</blockquote>
<p>Can a graphene laser really be made?</p>
| 2,918 |
<p>I would like to learn the basics about how interstellar matter contracts into stars under the influence of gravity.</p>
<p>Some of my questions:</p>
<p>Let's assume an ideal and infinite large cloud of equally distributed hydrogen atoms of zero K temperature. Will it collapse into a star if there is a small inhomogenity anywhere, for example a single dust grain, or a single cubic meter of space having more than the average number of hydrogen atoms?</p>
<p>What is the effect of the temperature? Obviously if the interstellar matter is hot, the individual hydrogen atoms have a high speed so it would need much more gravity to condense into a star.</p>
<p>What is the effect of ionization, or what happens if I consider molecules instead of single atoms, or even larger objects such as dust grains.</p>
<p>I would appreciate it if you can point me to relevant physics textbooks or online courses.</p>
| 2,919 |
<blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="http://physics.stackexchange.com/questions/2392/have-the-rowan-university-hydrino-findings-been-replicated-elsewhere">Have the Rowan University “hydrino” findings been replicated elsewhere?</a> </p>
</blockquote>
<p>I came across this controversial company: <a href="http://blacklightpower.com/" rel="nofollow">Black Light Power</a>
I want to know if there is any validity of the scientific claims made by this company and its founder Dr. Mills - especially concerning the existence of "hydrinos." </p>
| 60 |
<p>I can't remember what the equation is for the relationship between the force exerted by a coil and the stress in terms of the current density, the radius and the length.</p>
<p>I was thinking just standard F = AB^2 / 2*4*pi*10^-7(Permetivity Of Free Space), but I can't for the life of me figure out how to calculate this force in terms of current density...</p>
<p>Or lorentz law?</p>
<p>Any help would be greatly appreciated</p>
| 2,920 |
<p>Can we use entangled particles to transmit information or data such as TCP/UDP packets?</p>
<p>If so why hasn't this been done yet? Surely the costs of bringing this to market are much cheaper than laying submarine cables?</p>
| 61 |
<p>recently i wrote an assignment in one of the questions i got two different answers for the same question by considering a straight line motion</p>
<p><strong>The question was somewhat like this</strong></p>
<p><strong>THIS IS A HOMEWORK QUESTION AND I AM NOT ASKING THE EXACT ANSWER I JUST WANT GUIDANCE</strong></p>
<blockquote>
<p>A particle is thrown obliquely with a velocity of $15\sqrt{5}$ $metres$ $per$ $second$ and 45 degrees to the horizontal. $100$ meters from the point of projection starts stairs in which each step is $1$ meter in height and $1$ meter in width. Keeping this in consideration, which step should the particle hit?</p>
</blockquote>
<p>I calculated the horizontal range of this projectile from the formula
$$R=\frac{u^2\sin(2\theta)}g=112.5\,m$$</p>
<p>prescribed value of $g$ is $10$ metres per second square </p>
<p>That means the particle would have to hit any one of the steps.</p>
<p>Proceeding forwards with the formula of
$$x \tan\theta\left(1-\frac{x}R\right)=y$$
where $x$ is the horizontal displacement of the particle and $y$ is the vertical displacement.</p>
<p>Through which I can calculate its height, $y=11.111$, and as we know that the remaining horizontal distance to be covered by the particle before coming to the level of point of projection is $112.5-100=12.5$
by constructing a right angled triangle like this </p>
<p><strong>I know that particle won't travel a straight path in the influence of gravity, but as the distance is not too far from it's final point we can consider its path to be straight for some time.</strong></p>
<p><img src="http://i.stack.imgur.com/6Ktpw.png" alt="enter image description here"></p>
<p>as the point will be at the midpoint of rectangle and making the rectangle into four triangles</p>
<p>By Pythagorean theorem, we get the hypotenuse in every triangle as $8.35 meters$ then the point where they meet is approximately 8.35 meters on the line of stairs.</p>
<p><img src="http://i.stack.imgur.com/0AWp6.png" alt="enter image description here"></p>
<p>By converting this value into steps i.e. as we know one step measures $\sqrt{2}$ then for 8.32 it would be $5.883128419$ $meters$. Therefore, it would hit the wall of fifth step.</p>
<p><strong>But would this approximation be correct as we considered a straight line motion of the particle so would it hit 5th step or 6th step</strong></p>
<p><strong>how can I look at the problem and others like it so that I can imagine and formulate a solution on my own?</strong></p>
<p>i just have an idea of pythagorean theorem tosolve this and nothing else</p>
| 2,921 |
<p>A squeezed vacuum state is produced by applying a squeezing operator $S$ on the vacuum state $|0 \rangle$:</p>
<p>\begin{eqnarray}
S | 0 \rangle = \sum_n C_n |n \rangle
\end{eqnarray}</p>
<p>My question is, from $|0 \rangle$ (which means we have zero photons) we produce the state $| n \rangle$ which has $n$ number of photons. Does it mean we create photons from vacuum by squeezing?</p>
| 2,922 |
<p>EXAFS (Extended X-Ray Absorption Fine Structure), core electron is excited to the conduction band. The oscillations in the absorption coefficient persist for 100s of eV after the edge. </p>
<p>Taking fourier transform of the exafs function gives the R space peak positions that are related to the bond lengths.
<a href="http://cars9.uchicago.edu/xafs_school/Material/Lectures/Intro%20Newville.pdf" rel="nofollow">http://cars9.uchicago.edu/xafs_school/Material/Lectures/Intro%20Newville.pdf</a></p>
<p>My questions are: </p>
<ol>
<li><p>In the case of pure metal are the R space peaks shifted from their expected positions.</p></li>
<li><p>Is there data base for pure metals bond lengths extracted from EXAFS experiments. </p></li>
</ol>
| 2,923 |
<p>I am attaching a section from a text book (Conical Intersections Electronic Structure, dynamics and spectroscopy: David R Yarkony & Horst Koppel).</p>
<p><img src="http://i.stack.imgur.com/rZway.jpg" alt="enter image description here"></p>
<p>Here I am not understanding the so called 'Non Adiabatic Coupling Term"... From eqn (7.b), this term depends only on nuclear kinetic energy and nuclear coordinates. It doesn't involve anything from electronic coordinates!!! Then how come it is a coupling term? How this term describes the dynamical interaction between the electronic and nuclear motion? Also, why it is called non adiabatic?</p>
| 2,924 |
<p>We have done an experiment heating water with a wire element and have determined it takes 21.5964 watts to heat 200ml of water to a specific temperature in 10 minutes. However the power of the electrical input is 29.9859 watts. It asks for an explaination of why they are different but I'm a bit lost</p>
| 2,925 |
<p>If you know that the radioactive source is, for example, Cesium-137, is it possible to extrapolate a relationship between the count rate and radiation intensity? If it is not possible, what is the minimum information you need to determine this, e.g. activity?</p>
<p>Furthermore, are the units for radiation intensity still $\text{W m}^{-2}$?</p>
| 2,926 |
<p>How do we know that virtual-particle's virtual-pressure differences caused the plates to collide in the Casimir Effect Experiments. </p>
<p>What about micro-gravitation-fields produced but the plates themselves pulling each other towards one another?</p>
<p>Or Van Der Waal's Forces or some other forces?</p>
| 2,927 |
<p>I know it's a simple and basic question but would someone show me how to evaluate $[\hat{p}_x,\hat{p}_y]$?</p>
| 2,928 |
<p>This question is a continuation of "<a href="http://physics.stackexchange.com/questions/132654/can-a-third-type-of-electrical-charge-exist">Can a third type of electrical charge exist?</a>" and specifically <a href="http://physics.stackexchange.com/questions/132654/can-a-third-type-of-electrical-charge-exist#comment270844_132659">this comment</a>.</p>
<p>I know the common knowledge that there is 1 kind of electric charge and thus 1 kind of photon; there are 2 kinds of weak isospin and thus 3 kinds of weak bosons; there are 3 kinds of color charge and thus 8 kinds of gluons. But I wonder if this is fixed.</p>
<p>How I understand this: if there are $n$ possible charges then there are $n$ identical fermion fields each carrying its own kind of charge. Then the gauge field acts on them changing their phases (only! to conserve the probability, probably). So its group should be some group acting on $\mathbb{C}^n$ and conserving the norm. Usually the maximal such group is considered, that is $\mathrm{U}(1)$, $\mathrm{SU}(2)$ and $\mathrm{SU}(3)$ respectively.</p>
<p>1 . May some subgroup of the maximal one be used instead? For example, $\mathrm{SO}(n)$ or $\mathrm{SU}(m)$ for $m<n$?</p>
<p>2 . May fermions make up not the fundamental representation of the gauge group but some other representation?</p>
<p>Positive answers would lead to possible less numbers of photons.</p>
<p>Next, I notice that the maximal group for several charges is $\mathrm{U}(n)$, but in weak and color interactions $\mathrm{SU}(n)$'s are used instead. $\mathrm{U}(n)=\mathrm{U}(1)\rtimes\mathrm{SU}(n)$, and as I understand, the common phase factor cannot belong to every interaction in the Standard Model, so it belongs only to the electromagnetic interaction, and other interactions lose it, and so their maximal groups are restricted to $\mathrm{SU}(n)$'s.</p>
<p>3 . Is this right?</p>
<p>If yes, then in the case we imagine more kinds of <em>electric</em> charge, then the electromagnetic gauge group can be $\mathrm{U}(n)$ instead of $\mathrm{SU}(n)$, and the number of photons $n^2$ instead of $n^2-1$ respectively.</p>
| 2,929 |
<p>I am simulating a disordered ising-like model in 2d whose phase transition is expected to be continuous, whose universality class is as yet unknown. By plotting the Specific heat scaling function, i.e., $C L^{-\alpha/\nu}$ vs $tL^{1/\nu}$, I find that the ratio ($\alpha/\nu$) is $\approx 2.44$. Is there a previously studied universality class which has this value for $\alpha/\nu$ ?</p>
| 2,930 |
<p>I am working on the Ginzburg-Landau model for Charge density waves, and I am carrying out the sum of Green's functions to calculate the terms in the GL model. I have the following question:</p>
<p>Is the sum's order over $ \vec{k} $ (or eventually $ \vec{r} $) and $\omega_n$ important? Mathematically the question is the following,</p>
<p>$$
\sum_{\vec{k}} \sum_{\omega_n} =^? \sum_{\omega_n} \sum_{\vec{k}}
$$</p>
<p>If it is not, when does it happen or under which conditions there is a difference?</p>
<p>Thanks</p>
<p>Al</p>
| 2,931 |
<p>In quantum field theory, one defines a particle as a unitary irreducible representations of the Poincaré group. The study of these representations allows to define the mass and the spin of the particle. However, the spin is not defined the same way for massive particles (where the eigenvalue of the Pauli-Lubanski vector squared are $-m^2 s(s+1)$ where $$s = -S, -S +1, \cdots, S,$$ and $S$ is the spin of the particle (and $m$ the mass)) and massless particles (where the helicity has eigenvalues $\pm \lambda$ and $S=\left|\lambda\right|$ - not to mention continuous spin representations). </p>
<p>With this definition, the "spin" $S$ that appears in both cases doesn't seem to be exactly the same thing. The eigenvalue that labels the irreducible representations are not from the same operator in the massive and the massless case ... however, it's tempting (for me) to see the maximum value of these eigenvalues as the same physical quantity if both cases. </p>
<p>So I wonder if there is a more general definition that would embrace both massive and massless particles (even less practicable)?</p>
| 2,932 |
<p>I'm no knowledgeable pool player, but I've noticed that sometimes when the cue ball hits another pool ball, they roll together; and sometimes the cue ball bounces back. And I have a very, very rough sense that a hard, sharp, and even strike of the cue ball tends to make it bounce back more while a slower or more angled strike will make it roll forward after collision. Can anyone give a more rigorous analysis of the phenomena, or point me to a resource for this? I've tried googling but haven't see anything that really seems to address this as far as I can tell.</p>
<p>[Edit: Upon more contemplation, I suppose a more general question is: In a collision, what determines how much of the combined momentum of the system is distributed to the parts? So in cars colliding, or pool balls, or a skater on ice throwing a baseball--what features of the system determine the amount of momentum imparted to each component?]</p>
| 2,933 |
<p>So I've heard of meteorites "originating from Mars" (e.g. <a href="http://en.wikipedia.org/wiki/Allan_Hills_84001" rel="nofollow">AH84001</a>), but the phrase confuses me. I'm interested in what this means - have these rocks somehow escaped Mars' gravity and ended up here; or were they part of the material that Mars formed from, but did not end up as part of the red planet? Or another explanation?</p>
| 2,934 |
<p><em>(Warning! Newbie question coming up!)</em></p>
<h2>Background</h2>
<p>As seen on this picture of the cosmic microwave background (take from the <a href="http://en.wikipedia.org/wiki/Cosmic_microwave_background_radiation">Wikipedia entry</a> on the very same topic) there exists irregularities in the distribution of matter: <img src="http://i.stack.imgur.com/KjAdE.png" alt="Cosmic microwave background. Boom baby!"></p>
<p>To my knowledge, this is because the Universe wasn't formed with all the matter (or whatever you should call whatever existed right after the big bang) in a completely regular pattern (or if it was the case that anti-matter had this big fight with ordinary matter after the Big Bang, it's really not that important I guess), and I guess this is the reason why we have a highly diverse universe today, inhabited by galaxies, black holes, and Justin Bieber.</p>
<h2>Question</h2>
<p>What would the Universe look like if there were no irregularities in it from the get-go? Would we have a big massive black hole in the middle (if one even can talk about a middle), would the Big Bang never happen, or have I just misunderstood the whole thing, making my question completely nonsensical?</p>
| 2,935 |
<p>On November 7, 1631 Pierre Gassendi saw the transit of Mercury across the face of the Sun. How did he see it? I mean what instrument was used to reduce the apparent brightness of the Sun?</p>
| 2,936 |
<p>I carefully read the Wikipedia article <em><a href="http://en.wikipedia.org/wiki/Discovery_of_Neptune" rel="nofollow">Discovery of Neptune</a></em>, and I don't get what the irregularity of Uranus orbit was that lead to the discovery of Neptune. Years ago, I watched some educational film that schematically showed the irregularity as follows. When the two planets were "near" each other Neptune's gravity would temporarily pull Uranus out of its orbit and then Uranus would be somehow pulled back and continue orbiting along an elliptical orbit as if nothing happened. So Uranus' orbit would be perfectly elliptical except some rather short arc where it would be bent outwards.</p>
<p>This doesn't make sense - if there's some external force that "pulls" (even slightly) Uranus out of its orbit how would it get back into orbit? I suppose its orbit must have permanently (yet slightly) changed. So as time passes, successive deviations must accumulate each time Neptune passes Uranus and Uranus' orbit must get bigger (further from the Sun) and since Neptune is also attracted to Uranus by the very same gravity Neptune orbit must get smaller (closer to Sun).</p>
<p>What exactly is the irregularity and why is Uranus' orbit not permanently changed each time Neptune passes by?</p>
| 2,937 |
<p>If the earth was flat, would the transition between day and night be sudden?</p>
| 2,938 |
<p>When I read descriptions of the many-worlds interpretation of quantum mechanics, they say things like "every possible outcome of every event defines or exists in its own <em>history</em> or <em>world</em>", but is this really accurate? This seems to imply that the universe only split at particular moments when "events" happen. This also seems to imply that the universe only splits into a finite number of "every possible outcome". </p>
<p>I imagine things differently. Rather than splitting into a finite number of universes at discrete times, I imagine that at <em>every moment</em> the universe splits into an <em>uncountably infinite number of universes</em>, perhaps as described by the Schrödinger equation.</p>
<p>Which interpretation is right? (Or otherwise, what is the right interpretation?) If I'm right, how does one describe such a vast space mathematically? Is this a Hilbert space? If so, is it a particular subset of Hilbert space?</p>
| 908 |
<p>When I push a wall with my hand, the wall does the same and because <strike>both forces cancel out</strike> the sum of the forces applied by me on the wall and the wall on me is zero, plus the friction on the ground to keep me from sliding, neither me or the wall is displaced from its position.</p>
<p>Is there a preference over action or reaction? My doubt is: is the interaction between the bodies instantaneous, there is no time delay between action and reaction? And if there is no delay, does it matter if I'm assuming that I pressed the wall first or the wall did it first?</p>
| 2,939 |
<p>I am perplexed by recent papers by 't Hooft giving an explicit construction for an underlying deterministic theory based on integers that is indistinguishable from quantum mechanics at experimentally accessible scales. Does it mean that it is deterministic complexity masquerading as quantum randomness?</p>
<p><a href="http://arxiv.org/abs/1204.4926">http://arxiv.org/abs/1204.4926</a></p>
| 2,940 |
<p>I like some 'science' in my 'science fiction', so I started crunching out the kinematic equations for some of the scenarios my characters are getting involved in, and ran smack dab into an issue. (Please excuse my formatting, I can't figure out sub/superscript notations)</p>
<p>$v_f = \sqrt{2ad}$, with $v_i$ assumed to be 'close enough' to zero. That's easy enough to crunch the math on.</p>
<p>Except... With sufficiently high values for $a$ or $d$, you start crowding, or even violating, the light speed limit.</p>
<p>And I can't find the equations to help handle relativistic distortion! I know they exist because I remember working with relativity when I took my physics class, years and years ago.</p>
| 2,941 |
<p>I'm reading a book about string theory, and it tells me in the future it could be possible to detect existence of other bubble universes through cosmic background radiation. Is this true? What could we potentially see?</p>
| 2,942 |
<p>Are timelike diffeomorphisms really redundancies in description in quantum gravity? Certainly Yang-Mills gauge transformations can be considered redundancies in description. Ditto for p-form electrodynamics. Even spatial diffeomorphisms too. What about timelike diffeomorphisms? If they are, do the only real degrees of freedom in quantum gravity lie on the conformal boundaries, which are invariant under diffeomorphisms? This would be an extreme version of holography.</p>
| 2,943 |
<p>My second grader thought making a <a href="http://en.wikipedia.org/wiki/Homopolar_motor" rel="nofollow">homopolar motor</a> for her science experiment would be fun. And, it was. Now I am trying to explain how it works and the Lorentz force. Please help me by giving me a very simple explanation to what is turning out to a very complex theory (or law). </p>
| 2,944 |
<p>Imagine a boxcar with a lamp on the left end and a mirror on the right end, so that a light signal can be sent down and back. The boxcar is moving to the right with velocity v. They find the length of the boxcar with respect to the car itself and then with respect to an observer on the ground using time dilation. They use the round trip time, which is fine and makes sense. However, what if you don't use the round trip time? What if you just measure the distance and time it takes light to travel to the end of the boxcar and thats it? For an observer on the car, it would be $t'=d'/c$. For an observer on the ground, it would be the distance he measures the boxcar to be plus how much the boxcar has progressed in time t. So for him it would be $t=(d+vt)/c$. Now, if you use the fact that Lorentz contraction says that d' and d should be related by $d'=\gamma d$, and plug d into the time it takes the light to reach the end according to the observer and relate $t$ to $t'$, you do not get the time dilation relationship. You actually just get $t=t'\sqrt{\frac{c+v}{c-v}}$. This is a problem because $t$ and $t'$ should be related by gamma, not this other factor. I thought the time it takes for light to travel to the end in the frame of the car should be related by gamma to the time it takes to reach the end in the frame of the ground. Can you please explain why this is not so?</p>
| 2,945 |
<p>I would like to ask how the result of interference changes with the change of polarisation angle difference? Obviously we get the best results for 2 parallel polarised beams, and no intensity interference effects for orthogonally polarised, however what happens in the middle? I heard it's changing with $cos\Delta\phi$, so the result would be $I=I_1+I_2+\sqrt (I_1 I_2)cos \Delta \phi$ . Is it true and why? I tried to calculate it on my own, but I got something like $\sqrt(90°-\Delta\phi)/90°$ which gives the same results for 0, 45 and 90 degrees, but that's it :P</p>
| 2,946 |
<p>I have a particle near two Schwarzschild black holes. Let the black holes remain at rest so that only the particle is moving for the observer. We are in a plane. I calculate the distance travelled by the particle in one frame for each black hole, using Schwarzschild solution. Now the problem is if it is possible to sum the velocities. We can name the variables $\Delta\tau_1$, $\Delta\tau_2$, $\Delta r_1$, $\Delta r_2$, $\Delta\phi_1$, $\Delta\phi_2$, and $\Delta t_1=\Delta t_2$, which are converted in Cartesian coordinates. Their values are known, but that I need is to find the values the "resultant" velocity if it is possible. How should I proceed?</p>
<p>If it is not possible, however is there a method to find the velocity of a particle near two Schwarzschild black holes if sufficient data are known?</p>
| 2,947 |
<p>recently, I stumbled accross a concept which might be very helpful understanding quasiparticles and effective theories (and might shed light on an the question <a href="http://physics.stackexchange.com/questions/2506/how-to-calculate-the-properties-of-photon-quasiparticles">How to calculate the properties of Photon-Quasiparticles</a>): the <strong>spectral function</strong> $$A\left(\mathbf{k},\omega \right) \equiv -2\Im G\left(\mathbf{k},\omega \right)$$</p>
<p>as given e.g. in <a href="http://prb.aps.org/abstract/PRB/v77/i8/e081412" rel="nofollow">Quasiparticle spectral function in doped graphene</a> (<a href="http://arxiv.org/PS_cache/arxiv/pdf/0708/0708.1133v2.pdf" rel="nofollow">on arXiv</a>). </p>
<p>It is widely used in <strong>many-body physics</strong> of interacting systems and contains the information equivalent to the <a href="http://en.wikipedia.org/wiki/Green%27s_function" rel="nofollow">Greens function</a> $G$. For free particles, $A$ has a $\delta$-peaked form and gets broader in the case of interactions.</p>
<p>The physical interesting thing is, as I read, quasiparticles of interacting systems can be found if $A$ is also somehow peaked in this case. I don't understand this relationship, hence my question:</p>
<blockquote>
<h3>What is the relation of the spectral function's peak to the existence of quasiparticles in interacting systems?</h3>
</blockquote>
<p>Thank you in advance<br>
Sincerely</p>
<p>Robert</p>
| 2,948 |
<p>The question relates to <a href="http://physics.stackexchange.com/questions/23083/why-is-the-higgs-boson-spin-0">this post</a>.</p>
<p>Spontaneous symmetry breaking is somehow a volume effect, that in-principle only happens at infinity large system. Weinberg in the second volume of his QFT used a chair demostrated the lost of rotational symmetry due to large scale of the chair.</p>
<p>p. 163</p>
<blockquote>
<p>We do not have to look far for examples of spontaneous symmetry
breaking. Consider a chair. The equations governing the atoms of the
chair are rotationally symmetric, but a solution of these equations, the
actual chair, has a definite orientation in space. Here we will be concerned
not so much with the breaking of symmetries by objects like chairs, but
rather with the symmetry breaking in the ground state of any realistic
quantum field theory, the vacuum.</p>
</blockquote>
<p>p.164-165</p>
<blockquote>
<p>Spontaneous symmetry breaking actually occurs only for idealized systems
that are infinitely large. The appearance of broken symmetry for a
chair arises because it has a macroscopic moment of inertia $I$, so that
its ground state is part of a tower of rotationally excited states whose energies are separated by only tiny amounts, of order $\frac{{\hbar}^2}{I}$. This gives the state vector of the chair an exquisite sensitivity to external perturbations; even very weak external fields will shift the energy by much more
than the energy difference of these rotational levels. In consequence, any
rotationally asymmetric external field will cause the ground state or any
other state of the chair with definite angular momentum numbers rapidly
to develop components with other angular momentum quantum numbers.
The states of the chair that are relatively stable with respect to small
external perturbations are not those with definite angular momentum
quantum numbers, but rather those with a definite orientation, in which
the rotational symmetry of the underlying theory is broken.
For the vacuum also,</p>
</blockquote>
<p>To a very good approximation, the chair is described by quantum electrodynamics
$$L= -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} + i \bar{\psi}(\gamma^{\mu}D_{\mu}-m) \psi $$
The rotation of chair corresponds to $SO(3)$ group as a subgroup of the Lorentz group $SO(1,3)$. If vacuum is similar with the chair, the rotational symmetry could also break together with global rigid $O(N)$ symmetry. The standard argument that Higgs boson being a scalar, is to keep Lorentz invariance. If the $SO(3)$ rotational symmetry is lost, then this argument is not necessarily valid. Is there any other rationalization for the Higgs boson being scalar?</p>
<p>If vacuum is <em>not</em> similar with the chair, why only global rigid $O(N)$ symmetry is broken but not $SO(3)$? Is that a pure guess then confirmed experimentally?</p>
| 2,949 |
<p>Excuse me for long prehistory. Maybe it can be useful for someone.</p>
<p>I was little confused with spinor indices when getting an expression relating the spinor and antisymmetric tensors. An antisymmetric tensor $M_{\mu \nu}$ have an expression (in spinor formalism)</p>
<p>$$
h_{ab\dot {a}\dot {b}} = \left((\sigma^{\mu})_{\alpha \dot {\alpha}}(\sigma^{\nu})_{\beta \dot {\beta }} - (\sigma^{\mu})_{\beta \dot {\beta}}(\sigma^{\nu})_{\alpha \dot {\alpha }}\right)M_{\mu \nu} = \varepsilon_{\dot {\alpha} \dot {\beta}}h_{(\alpha \beta )} + \varepsilon_{\alpha \beta}h_{(\dot {\alpha} \dot {\beta })}, \qquad (.1)
$$
where second identity is the decomposition into irreducible spin coefficients, and
$$
\varepsilon^{\alpha \beta} = \varepsilon^{\dot {\alpha }\dot {\beta }} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}^{\alpha \beta}, \quad \varepsilon^{\alpha \beta} = -\varepsilon_{\alpha \beta},
$$
$$
h_{(\alpha \beta)} = -\frac{1}{2}\varepsilon^{\dot {\alpha} \dot {\beta} }h_{(\alpha \beta)\dot {\alpha }\dot {\beta }}, \quad h_{(\dot {\alpha }\dot {\beta })} = -\frac{1}{2}\varepsilon^{\alpha \beta }h_{\alpha \beta (\dot {\alpha }\dot {\beta })}, \qquad (fixed)
$$
$$
h_{[\alpha \beta]\dot {\alpha }\dot {\beta }} = \frac{1}{2}\left( h_{\alpha \beta \dot {\alpha }\dot {\beta }} - h_{\beta \alpha \dot {\alpha }\dot {\beta }} \right), \quad h_{(\alpha \beta)\dot {\alpha }\dot {\beta }} = \frac{1}{2}\left( h_{\alpha \beta \dot {\alpha }\dot {\beta }} + h_{\beta \alpha \dot {\alpha }\dot {\beta }} \right).
$$
So, for $h_{(\alpha \beta)}$ I get, using $(.1)$
$$
h_{(\alpha \beta )} = -\frac{1}{8}\varepsilon^{\dot {\alpha } \dot {\beta } }\left( (\sigma^{\mu})_{\alpha \dot {\alpha}}(\sigma^{\nu})_{\beta \dot {\beta}} + (\sigma^{\mu})_{\beta \dot {\alpha}}(\sigma^{\nu})_{\alpha \dot {\beta}} - (\sigma^{\mu})_{\beta \dot {\beta}}(\sigma^{\nu})_{\alpha \dot {\alpha }} - (\sigma^{\mu})_{\alpha \dot {\beta}}(\sigma^{\nu})_{\beta \dot {\alpha }}\right)M_{\mu \nu}. \qquad (.2)
$$
Then one can show, that
$$
\varepsilon^{\dot {\alpha } \dot {\beta } } (\sigma^{\mu})_{\alpha \dot {\alpha}}(\sigma^{\nu})_{\beta \dot {\beta}} = (\sigma^{\mu}\tilde {\sigma }^{\nu })_{\alpha \beta },
$$
where
$$
(\tilde {\sigma}^{\mu})^{\dot {\beta } \beta } = \varepsilon^{\beta \gamma}\varepsilon^{\dot {\beta} \dot {\gamma}}(\sigma^{\mu})_{\gamma \dot {\gamma }} = \left( \hat {\mathbf E }, -\hat {\mathbf \sigma } \right). \qquad (fixed)
$$
But I don't understand, how to show it. Can you help me?</p>
<p><strong>Addition 1</strong>. </p>
<p>The correct defs of $h_{(\alpha \beta)}, h_{(\dot {\alpha} \dot {\beta})}$ are
$$
h_{(\alpha \beta)} = -\frac{1}{2}\varepsilon^{\dot {\alpha} \dot {\beta }}h_{(\alpha \beta)\dot {\alpha }\dot {\beta }}, \quad h_{(\dot {\alpha }\dot {\beta })} = -\frac{1}{2}\varepsilon^{\alpha \beta }h_{\alpha \beta(\dot {\alpha }\dot {\beta })},
$$
"thanks" to my inattention. Also I changed the def of $\tilde {\hat {\sigma}}^{\mu}$ on correct def.</p>
<p><strong>Addition 2</strong>.</p>
<p>I got an answer, thanks to Trimok.</p>
<p>First, to simplify the work with each term I need to multiply $(.2)$ by $\varepsilon^{\delta \beta}$:
$$
\varepsilon^{\delta \beta }h_{(\alpha \beta )} =
$$
$$
= -\frac{1}{8}\varepsilon^{\delta \beta }\varepsilon^{\dot {\alpha } \dot {\beta } }\left( (\sigma^{\mu})_{\alpha \dot {\alpha}}(\sigma^{\nu})_{\beta \dot {\beta}} + (\sigma^{\mu})_{\beta \dot {\alpha}}(\sigma^{\nu})_{\alpha \dot {\beta}} - (\sigma^{\mu})_{\beta \dot {\beta}}(\sigma^{\nu})_{\alpha \dot {\alpha }} - (\sigma^{\mu})_{\alpha \dot {\beta}}(\sigma^{\nu})_{\beta \dot {\alpha }}\right)M_{\mu \nu}.
$$
For first and second terms, for example,
$$
\varepsilon^{\delta \beta }\varepsilon^{\dot {\alpha } \dot {\beta } }(\sigma^{\mu})_{\alpha \dot {\alpha}}(\sigma^{\nu})_{\beta \dot {\beta}} = (\sigma^{\mu} )_{\alpha \dot {\beta }}(\tilde {\sigma}^{\nu})^{\dot {\beta }\delta } = (\sigma^{\mu}\tilde {\sigma}^{\nu})_{\alpha}^{\quad {\delta}},
$$
$$
\varepsilon^{\delta \beta }\varepsilon^{\dot {\alpha } \dot {\beta } }(\sigma^{\mu})_{\beta \dot {\alpha}}(\sigma^{\nu})_{\alpha \dot {\beta}} = -\varepsilon^{\delta \beta }\varepsilon^{\dot {\beta } \dot {\alpha } }(\sigma^{\nu})_{\alpha \dot {\beta}}(\sigma^{\nu})_{\alpha \dot {\beta}} = -(\tilde {\sigma}^{\mu})^{\dot {\beta} \delta}(\sigma^{\nu})_{\alpha \dot {\beta}} = -(\sigma^{\nu}\tilde {\sigma}^{\mu})_{\alpha}^{\quad \delta}.
$$
So
$$
\varepsilon^{\delta \beta }h_{(\alpha \beta )} = -\frac{1}{8}\left( 2 (\sigma^{\mu}\tilde {\sigma}^{\nu})_{\alpha}^{\quad {\delta}} - 2(\sigma^{\nu}\tilde {\sigma}^{\mu})_{\alpha}^{\quad \delta} \right) M_{\mu \nu}.
$$
Second, I can multiply all expression by $\varepsilon_{\gamma \delta}$ and use identity $\varepsilon_{\alpha \beta}\varepsilon^{\gamma \beta} = -\delta^{\quad \gamma}_{\alpha}$:
$$
\varepsilon_{\gamma \delta}\varepsilon^{\delta \beta }h_{(\alpha \beta )} = h_{(\alpha \gamma )} = -\frac{1}{4}\left( (\sigma^{\mu}\tilde {\sigma}^{\nu})_{\alpha \gamma} - (\sigma^{\nu}\tilde {\sigma}^{\mu})_{\alpha \gamma }\right)M_{\mu \nu}.
$$</p>
| 2,950 |
<p>We say the universe is expanding, and by expanding we mean the distance between objects gets larger over time. We call that "Metric Expansion of the Universe". So far so good. I kind of get the idea about of distances getting larger. </p>
<p>Now, I think of a balloon's surface and the distance between two arbitrary points on the surface gets larger as the metric expansion happens. But, in order for metric expansion to happen, doesn't the universe really expand INTO something. Balloon's surface expands into air so there's no problem imagining it, but how about the universe itself? </p>
<p>Also, do we mean the whole universe or observable universe when we say the universe is expanding? Both maybe?</p>
<p>Edit: Also, I know some multiverse theories that try to explain it, but the idea of universe is expanding has been there before multiverse was even considered, so I guess it can be explained without multiverse theories. </p>
| 62 |
<p>I am trying to investigate the relationship between velocity and drag force of falling object in air. I used video analysis from logger pro to determine the displacement of the falling object at different time, and then differentiation to find velocity and acceleration. Drag force at each time interval is found by subtracting the resultant force from the weight. I found a smooth relationship between velocity and drag force, but it is similar to a hyperbolic tangent function which is totally different from the drag equation.<br>
How come??</p>
| 2,951 |
<p>Wave equations take the form:</p>
<p>$$\frac{ \partial^2 f} {\partial t^2} = c^2 \nabla ^2f$$</p>
<p>But the Schroedinger equation takes the form:</p>
<p>$$i \hbar \frac{ \partial f} {\partial t} = - \frac{\hbar ^2}{2m}\nabla ^2f + U(x) f$$</p>
<p>The partials with respect to time are not the same order. How can Schroedinger's equation be regarded as a wave equation?</p>
| 188 |
<p>In this <a href="http://www.cs.princeton.edu/courses/archive/fall05/frs119/papers/feynman85_optics_letters.pdf" rel="nofollow">paper</a>, Feynman gave the idea of creating a time-independent Hamiltonian from a quantum circuit. Is there anyway to say that these Hamiltonians will always be Hermitian? Moreover, will these Hamiltonians be always unitary?</p>
| 2,952 |
<p>I am a beginner to study QFT and confused about path integral for boson or fermion.</p>
<p>I have read about the path integral for single particle, and finished some problems. But I cannot understand the next chapter which is about path integral for boson and fermion. </p>
<p>Here, I am confused for a long time. Could you please tell me the difference between the two kind of path integrals? What is the point of path integral for boson and fermion?</p>
<p>I find that there are big differences in forms between single particle, boson and fermion. I am not understand why the book uses different forms to discuss them. Even the path integral for the spin system has a new form to discuss.</p>
<p>In my opinion, I think the biggest difference between single particle and boson (fermion) is statistics, but how to consider about the statistical properties in path integral? Just using one symbol for differential, $D$, to take the place of the original symbol, $d$, is enough?</p>
| 2,953 |
<p>Is there a way to view the Airy Pattern as an infinite sum of Fraunhoffer diffraction patterns? I don't know where the 1.22 would come from then. Is there something inherently wrong with collapsing diffraction slits to infinitely small slits?</p>
| 2,954 |
<p>There is this thing I got confused:</p>
<p>Microcausality is the statement that spacelike separated local field variables commute so that we can specify field variables on a spatial slice as a complete base. It is usually referred to as a statement about locality---if microcausality is broken then the "local" operators are not that "local".</p>
<p>There is another statement about the notion of "locality" in S-matrix language---an S-matrix have poles corresponding to particle exchange, and the residue factorizes into S-matrices of sub scattering processes in the limit that these processes happen far from each other. It is in the line of cluster decomposition principle.</p>
<p>So my question is: do these two statements somehow have connections to each other, or even are equivalent? Or they are simply two very different statement and not connected at all?</p>
| 2,955 |
<blockquote>
<blockquote>
<p>Suppose an observer $\mathcal{O}$ uses the coordinates $t$, $x$, and that another observer $\mathcal{O}'$, with coordinates $t'$, $x'$, is moving with velocity $\mathbb{v}$ in the $x$ direction relative to $\mathcal{O}$. Where do the coordinate axes for $t'$ and $x'$ go in the spacetime diagram of $\mathcal{O}$?</p>
<p>$t'$-axis: This is the locus of events at constant $x'=0$ (and $y'=z'=0$, too, but we shall ignore them here), which is the locus of the origin of $O'$'s spatial coordinates. This is $O's$'s world line, and it looks like that shown in the figure below.
(<em>A First Course In General Relativity</em>, Bernard Schutz, Second Edition, p. 6)</p>
</blockquote>
</blockquote>
<p>Since an observer is at the origin of his coordinate system, and $\mathcal{O'}$ is moving with a relative velocity of $v$ in the $x$ direction relative to $O$, I am perfectly fine with the fact that the line which we call $t'$-axis in the diagram is the worldline of $\mathcal{O'}$ in the spacetime diagram drawn by $\mathcal{O}$. However, I do not at all get the point of why we say that this worldline of $\mathcal{O'}$ is the "$t'$-axis" of $\mathcal{O'}$ in the spacetime diagram drawn by $\mathcal{O}$, and why it has to be that way; "tilted". Can someone please explain the reasons to me? </p>
<p><img src="http://i.stack.imgur.com/liU96.jpg" alt="enter image description here"> </p>
| 2,956 |
<p>A guy who has a career in medical physics named Pierre-Marie Robitaille argues in two recently published papers in “Progress in Physics”, that the CMB is not from the big bang but from the oceans. </p>
<p>The first paper is entitled WMAP: A Radiological Analysis. This work analyzes the WMAP images based on accepted standards for image acquisition and processing. It demonstrates that it is not appropriate to evaluate cosmological parameters based on measurements from either WMAP or COBE.</p>
<p>In the second paper, evidence is presented for the reassigment of the CMB to the oceans of the Earth. This work demonstrates that the Earth cannot be modeled as a 285K source as the COBE team assumes. This is postulated by a third paper authored by Dmitri Rabounski, a Russian theoretical physicist in "The Relativistic Effect of the Deviation between the CMB Temperatures Obtained by the COBE Satellite". </p>
<p>Robitaille <a href="http://youtu.be/3Hstum3U2zw" rel="nofollow">also claims that Kirchhoff's law of thermal emission is invalid</a> and this <a href="http://youtu.be/i8ijbu3bSqI" rel="nofollow">recent video</a> is of him claiming that Penzias and Wilson measured water on Earth.</p>
<p>These people also base their case on 2 questions:
(1) When you put a glass of water inside a microwave oven and turn it on does the water in the glass reflect the microwaves or does it absorb them?
(2) Is a powerful absorber of microwaves also a powerful emitter thereof?</p>
<p>Can anyone debunk and address these claims?</p>
| 355 |
<p>I have heard that the presence of an extremely strong gravitational field possesses the capacity to warp or tear spacetime and to potentially create a wormhole. Is any energy lost when spacetime is ripped, and if so is there a constant energy per volume required to 'rip' a given region of spacetime? Otherwise, is there some minimum gravitational field strength required to rip spacetime?</p>
| 2,957 |
<p>I would like to compare my result in an order of magnitude. So, How can I estimate the strength of the electric field in a typical Si PN-junction?</p>
| 2,958 |
<p>Let's assume we have a capacitor of capacitance $C$ and potential difference $U$. After charging it we disconnect it. Then we put a dielectric between the plates. I know that capacitance will increase by $C * k$, however what happens with the charge and potential difference on it.</p>
<p>Let's say $k = 2$. Will $q$ double or will $ U $decrease to a half ?</p>
| 2,959 |
<p>I had very well read that when charge is stored on a rough surface, the leakage is very high from the pointed tips of such surfaces, by a phenomenon called action of points. But now, I've come to know that if a surface is over-polished (or say ideally polished), it becomes very difficult to store charge on it. I'm confused with this. What might be the reason to say so? I do think that by over polishing, you destroy the dielectric layer on the surface of such object (where we tend to store charge) and charge can't be stored easily. But I'm not satisfied with my own logic. So, can someone please clarify it?</p>
| 2,960 |
<p>I'm reading these <a href="http://www.maths.dur.ac.uk/~dma0saa/lecture_notes.pdf" rel="nofollow">notes</a> - page 8 and 9 - and I'm a bit confused.</p>
<p>If we consider a field $\phi$ (which can be either bosonic or fermionic) transforming as:
\begin{equation}
\phi(x) \rightarrow \phi(x) + \delta \phi (x)
\end{equation}
with:
\begin{equation}
\delta \phi^a = t^a \phi(x)
\end{equation}
where $t^a$ is the generator of the transformation. The generators satisfy the Lie algebra:
\begin{equation}
[t^a,t^b] = if^{abc} t^c \tag{$*$}
\end{equation}
Let us suppose that the above transformation is a symmetry transformation such that the Noether charge corresponding to this symmetry is given by:
\begin{equation}
Q^a = \int \mathrm{d}^3 \mathbf{x} \; \pi \delta\phi^a = \int \mathrm{d}^3 \mathbf{x} \; \pi t^a \phi
\end{equation}
where $\pi$ is the canonical momentum density. It is then possible (but tedious) to show that the charges satisfy the so-called charge algebra:
\begin{equation}
[Q^a,Q^b] = i f^{abc} Q^c \tag{1}
\end{equation}
Until this point I understand it. But then the notes say on page 8:</p>
<blockquote>
<p>[...] the charges generally have to satisfy the same algebras as the generators – in fact
it is only because of this that the symmetry has any useful physical meaning. In particular
it is the charges which are the physical observables that participate in interactions rather
than gauge fields for example.</p>
</blockquote>
<p>I don't really understand what is meant with the above statement. What does the quote have to do with the fact that Noether charges obey equation $(1)$?</p>
<p>Edit: I understand that the charges satisfy the same Lie algebra as the generators. But according to the quote above, if understand it correctly, we should also expect this based on logical/physical reasons. Apparently, according to the notes, "it is only because of this that the symmetry has any useful physical meaning." I don't understand why this is the case.</p>
| 2,961 |
<p>Given a general metric, $g_{ab}$ I can select an orthonormal basis $\omega^{a}$ such that,</p>
<p>$$g_{ab} = \eta_{ab}\omega^a \otimes \omega^b$$</p>
<p>where $\eta_{ab}$ = $\mathrm{diag}(1,-1,-1,-1).$ We may conveniently compute the spin connection and curvature form by employing Cartan's equations. The problem I have lies in getting back to the coordinate basis. I know the general formula, </p>
<p>$$R^{\mu}_{\nu \lambda \tau} = (\omega^{-1})^{\mu}_a \, \omega^b_\nu \, \omega^c_\lambda \, \omega^d_\tau \, R^{a}_{bcd}$$</p>
<p>where the l.h.s. $R^{\mu}_{\nu \lambda \tau}$ is in the coordinate basis. The objects $\omega^b$ are familiar, they're just the orthonormal basis, so what is the object $\omega^b_\nu$ (with the extra index)? The $(\omega^{-1})^{\mu}_a$ are the inverse vielbeins? How are these obtained?</p>
<p>I've visited several sources, including Wikipedia, but it's still not 100% clear. I'd appreciate any clarification, especially a small explicit example if possible.</p>
| 2,962 |
<p>I have read that <strong>Gaussian surface</strong> cannot pass through discrete charges. Why is it so?
I have even seen in application of Gauss' Law when we imagine a Gaussian Surface passing through a charge distribution, e.g. in case of infinite plane charge carrying sheet .</p>
<p>If it cannot pass through discrete charges how do we use it in continuous charge distributions
as same '<em>objection</em>' must be there for it also.</p>
<p>Please explain the reason.
<img src="http://i.stack.imgur.com/t7Yjt.png" alt=""></p>
<p>Here $E \rightarrow \infty$ as, $r\rightarrow 0$</p>
<p>If this is ambiguity then this must be same in continuous charge distribution , otherwise please state it more clearly because we can define charge to be a spherical ball and half charge can be considered inside surface (as in pic and even agreed by @JoshuaBarr).</p>
| 2,963 |
<p>Consider the following game show: two friends Tom and Jerry (X and Y) are selected from an audience to compete for a grand prize, a brand new Ferrari. </p>
<p>The game description: </p>
<ol>
<li><p>The two contestants are space-like separated.</p></li>
<li><p>Each contestant will be asked one of three questions {A, B, C} and the questions which are asked of the two contestants need not be the same. Each of these questions has two possible answers, which we will quantify as {+1, -1}. </p></li>
<li><p>Repeat 2 a large number of times.</p></li>
<li><p>When the two contestants are asked the same question, they must always give the same answer. </p></li>
<li><p>When the two contestants are asked different questions, they must agree $ \frac 14 $ of the time. </p></li>
</ol>
<p>Now, how can Tom and Jerry win this game?</p>
<p>It's simple: They create a strategy, whereby they pre decide on a set of answers to give to the three questions {A, B, C}. This will guarantee that they always give the same answer when they are asked the same question. For example, they may decide on {+1, +1, -1} to {A, B, C}. Let us denote these answers as $v_i(\alpha) = \pm1$ for $\mathcal i$ = X, Y, $\alpha$ = A, B, C. </p>
<p>This will not allow Tom and Jerry to satisfy 4. </p>
<p>$\mathscr Theorem:$ </p>
<p>There do not exist random variables $v_i(\alpha), \mathcal i$ = X, Y, $\alpha = A, B, C$ such that: </p>
<p>$$ 1. v_i(\alpha) = \pm1 $$
$$ 2. v_X(\alpha) = v_Y(\alpha)\forall \alpha $$
$$ 3. Pr(v_X(\alpha) = v_Y(\beta)) = \frac 14 \forall \alpha, \beta, \alpha \neq \beta $$</p>
<p>$\mathscr Proof:$</p>
<p>Assume for contradiction that there do exist random variables $v_i(\alpha), \mathcal i$ = X, Y, $\alpha = A, B, C$ such that (1-3 hold).</p>
<p>Since $ v_i(\alpha)$ can only take on the two values $\pm1$, we must have $Pr(v_X(A) = v_X(B)) + Pr(v_X(A) = v_X(C)) + Pr(v_X(B) = v_X(C)) \geq 1 $</p>
<p>By condition 2, we then have $Pr(v_X(A) = v_Y(B)) + Pr(v_X(A) = v_Y(C)) + Pr(v_X(B) = v_Y(C)) \geq 1$</p>
<p>Now, by condition 3, we have $ \frac 14 + \frac 14 + \frac 14 \geq 1$ a contradiction.</p>
<p>But, if you look at the predictions of quantum mechanics, it is possible to satisfy (1-3). Experiments have validated
quantum mechanics, thus the correlations achieved cannot be due to a pre existing strategy. Then we must wonder how could Tom and Jerry
always assure that property (2) holds, if they can not simply pre decide on a set of answers to give to the three questions {A, B, C}.
It must be that when Tom and Jerry are asked a question, they communicate with each other what question is being asked, and agree to
an answer, for if not, one would have $Pr(v_X(\alpha) = v_Y(\alpha)) = \frac 12$</p>
<p>$\mathscr Bell's\; Theorem\; Implication:$
Quantum mechanical correlations are not due to pre existing properties $\Rightarrow$ there must exist an exchange of information
between entangled subsystems about what properties are being measured on them, and what values of these properties are to be taken on.
Combining this with rule (1) of the game implies that this must take place faster than light, recent experiments overwhelmingly suggest
instantaneously.</p>
<p>My question is why is this salient point so muddled in the literature? Bell's theorem is often stated as follows. </p>
<p>$\mathscr No\; theory\; of\; local\; hidden\; variables\; can\; produce\; the\; predictions\; of\; quantum\; mechanics$</p>
<p>That's missing the point. The hidden variables (pre existing properties) are a derived consequence of no exchanging of information
about what properties are being measured and what values to take on. Bell simply showed that pre existing properties fail as an explanation.
Consequently, we must have the implication above. </p>
<p>Credit to Tim Maudlin for the game description. </p>
| 2,964 |
<p>The contraction in most images looks like a gravity well. Is it that you use negative energy to BOTH contract and expand space or positive and negative to contract (positive energy) and then expand with then negative energy?</p>
| 2,965 |
<p>In a different thread, a user stated the following in respect of events preceding or following other events:</p>
<blockquote>
<p>However, if the two events are causally connected ("event A causes event B"), the causal order is preserved (i.e. "event A precedes event B") in all frames of reference.</p>
</blockquote>
<p>My question is what does "causally connected" really mean? What does "causes" mean? Further, given that we know that we can have instantaneous effects in typical quantum processes (e.g. flip the polarizer, effect on another a reading light years away, even though we cannot transmit useful information with this), does that not constitute "causing" for the purpose of this statement?</p>
| 2,966 |
<p>I was wondering the other day about teletransportation (human). And I had the idea that as far as I know, matter is energy. So I was wondering if it's possible to excite matter so it turns into energy, energy which may be could be moved to another physical location and then it would be allowed to return to it's original form. Is this just plaint stupid or is it theoretically possible?</p>
| 2,967 |
<blockquote>
<p>A 1 meter long rod on the ice with mass $m_2=1$ kg is perpendicularly hit on one end by a point particle with mass $m_1=0.1$ kg. The collision is elastic and the point particle is bounced back in the same direction. After the collision the rod's frequency is $\nu =2$ Hz. What was the initial velocity of the point particle?</p>
</blockquote>
<p>My attempt:</p>
<p>Since the collision is elastic, the kinetic energy of the system is the same before and after the collision:
$$0.5m_1v_1^2=0.5J_2 \omega_2^2+0.5m_2v_2^2+0.5m_1v_3^2$$
Where $v_3$ is the velocity of the point particle after the collision.</p>
<p>Now, in the case of a rod:
$$J=\frac{1}{12}L^2m$$
And, we also know:
$$\omega_2=2 \pi \nu$$
And there are also no external forces, therefor the momentum of the system is the same before and after the collision:
$$m_1\vec{v_1}=m_1 \vec{v_3}+m_2\vec{v_2}$$
Here $v_1$ is the quantity we're looking for, $v_3$ is the point particle's velocity after the collision and $v_2$ is the velocity of the rod's center of mass. It follows:
$$\vec{v_2}=\frac{m_1 \vec{v_1}-m_1 \vec{v_3}}{m_2}$$
From this it follows:
$$0.5m_1v_1^2=\frac{1}{24}L^2m_2 4 \pi^2 \nu^2+0.5m_2 |\frac{m_1 \vec{v_1}-m_1 \vec{v_3}}{m_2}|^2+m_1v_3^2$$
This is 1 equation with 2 unknowns, and this is where I get stuck.
Any help is appreciated.</p>
| 2,968 |
<p>An air stream is created by a propeller at velocity $\nu = 93.6\frac{m}{s}$</p>
<p>To calculate force applied by the propeller I used this formula:</p>
<p>$$F=\frac{1}{2}\rho\nu^2A$$</p>
<p>where:</p>
<ul>
<li><p>$\rho = 1.29 kg/m^3$ is the air density</p></li>
<li><p>$\nu = 93.6\frac{m}{s}$ is the air velocity</p></li>
<li><p>$A = 0.54m^2$ is the area swept by propeller's blades (This is for simplicity reasons)</p></li>
</ul>
<p>Is this a correct way? I think it's correct because the answer results in $kg.m/s^2 = N$. By plugging the numbers I got a force of $F = 3051$ N</p>
<p><strong>How do I calculate momentum?</strong></p>
<p>I tried this:</p>
<p>$$p = m\cdot\nu$$</p>
<p>where $m$ is air mass. I calculated the mass this way:</p>
<p>$$m = \rho.A.\nu$$</p>
<p>Although this would give $kg/s$ because of velocity, but I'm working on 1 second, we simply eliminate the $s$ by multiplying by $t = 1$ sec, thus the mass is $m = 65.2kg$</p>
<p>$$p = 6102 \frac{kg.m}{s}$$</p>
<p>This is <strong>twice</strong> the force. Where did it go wrong?</p>
| 2,969 |
<p>Is there any way in which air can be made to reflect light? By making air denser or by any other way? The task is to project images in air...</p>
| 2,970 |
<p>How does a body on the surface of the earth apply a force on the ground? Since neither the body nor the surface of the earth is in motion, there is no acceleration and, hence, there can be no force because force equals mass times acceleration. </p>
| 63 |
<p>If <strong>light</strong> is switched ON, only for a second, and the distance between the observer and the light source is 10 million kilometers, can I still see the light spark?</p>
<p>For example, let's assume that the source is the Sun and the observer is a human eye.</p>
| 2,971 |
<p>If I understand entropy correctly, then for example two objects orbiting a centre of mass have lower entropy than when said objects eventually crash into each other and form a new one.</p>
<p>So let's say that a typical galaxy spirals around its centre of mass and eventually objects within it will fall into the center thus increasing its entropy.</p>
<p>But if the entropy of the Universe was somehow to be constant, then maybe that's why space is expanding? As each galaxy becomes more chaotic while objects are going closer and closer together these galaxies are at the same time becoming more and more spread apart thanks to expansion of space. </p>
<p>I don't know the exact calculations of entropy, but is it possible that there is a mechanism triggering space expansion as a reaction to gravity increasing local entropy?</p>
<p>Also gravity is increasing in intesity when the distance between objects is shorter, so the longer two masses are gravitating, the shorter the distance between them and then the more intensive gravity becomes over time. This might correspond then to the increasing speed of expansion of Universe, as it has to compensate faster to keep entropy constant.</p>
<p>Forgive me if what I said above is completely incorrect; I've been thinking about entropy and came with this idea, but I'm no professional physicist and would like to know if any of the above make sense.</p>
| 2,972 |
<p>Assume that a coin is placed on circular disk and now a disk is rotated with constant angular velocity.</p>
<p>If there is no friction between the surfaces of a disk and coin, according to theory the coin will move away from centre of disk. But I have confusion here that the centripetal and centrifugal forces are of equal magnitude so why latter comes to play effectively?</p>
| 64 |
<p>Does anyone know, is a model with lagrangian
$\mathcal{L} = \frac{(\partial_{\mu}\phi_a)^2}{2}-\frac{m^2 \phi_a^2}{2}-\frac{\lambda}{8N}(\phi_a \, \phi_a)^2$ renormalizable? I'm using BPHZ scheme and everything is OK in one loop. But it seems to me (may be I'm mistaken) that the scheme breaks down even in two loops. I will be grateful for links on books or lectures where such a model is considered.</p>
| 2,973 |
<blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="http://physics.stackexchange.com/questions/48543/do-some-half-lives-change-over-time">Do some half-lives change over time?</a> </p>
</blockquote>
<p>Would it be possible to considerably speed up the decay rate of an isotope?<br>
Considerably meaning more then a 1 or 2% increase in decay rate.</p>
| 65 |
<p>At the page 336 of Hawking, Ellis: <em>The Large Scale Structure of Space-Time</em>, the Gauss-Bonnet theorem is stated as</p>
<p>$$\int_H \hat{R}\ d\hat{S} = 2\pi \chi(H) \qquad (1)$$</p>
<p>with</p>
<p>$$\hat{R} = R_{abcd} \hat{h}^{ac} \hat{h}^{bd}$$</p>
<p>and induced metric on the horizon $\hat{h}_{ab}$,</p>
<p>$$\hat{h}_{ab} = g_{ab} + \ell_a n_b + n_a \ell_b \ ,$$</p>
<p>where $\ell^a$ and $n^a$ is a pair of future-directed null vectors on the horizon.</p>
<p>Is there a missing factor of <strong>2</strong> on the RHS of equation (1)? </p>
<p>The (2-dimensional) <a href="http://en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem" rel="nofollow">Gauss-Bonnet theorem</a> in the literature is usually stated using "Gaussian curvature" $K = R/2$, so I'm suspecting in this "hidden factor" (compare it, for example, with Heusler: <em>Black Hole Uniqueness Theorems</em>, equations (6.23)--(6.26)).</p>
| 2,974 |
<p>Admittedly, Nuclear Physics is not my strength. I'm writing a simulation to model alpha-decay. So far, I have looked up the values of the kinetic energy of the alpha particles that are emitted in a certain decay. Now I have seen this a lot of times, for instance for 212-Polonium:</p>
<p><strong>$^{212}$Po -10.3649 MeV</strong></p>
<p>the the value given is the <a href="http://en.wikipedia.org/wiki/Mass_excess" rel="nofollow">mass excess</a>. Then I looked into a Nuclear Physics book (Krane) and found:</p>
<p>$T_\alpha = \frac{Q}{1 + m_\alpha / m_{x'}}$</p>
<p>where $T_\alpha$ is the kinetic energy of the alpha particle, Q is the <a href="http://en.wikipedia.org/wiki/Q_value_%28nuclear_science%29" rel="nofollow">Q-value</a>, $m_\alpha$ is the mass of the alpha particle and $m_{x'}$ is the mass of the daughter nucleus.</p>
<p>First I thought, Q and $\Delta M$ are simply related by a factor of $c^2$. But that does not seem to be the case, since my calculations are wrong.</p>
<p>How can I calculate the kinetic energy of the alpha-particle <strong>only given the information above</strong>?</p>
| 2,975 |
<p>I am reading Bailin and Love's review on Kaluza Klein theories. On section 4.1 they start talking about infinitesimal isometries generated "with a particular generator $t_a$ of the isometry group".</p>
<p>$I+id\sigma{}t_a$ $y^n\to{}y'^n=y^n+d\sigma{}\xi_a^n(y)$</p>
<p>where the $\xi_a^n(y)$ are just killing vectors.</p>
<p>Immediately thereafter it says</p>
<blockquote>
<p>For any representation of the non-Abelian isometry group, the eigenvalues of a diagonal generator $t_a$ will be integral multiples of some lowest (positive) eigenvalue $g_{min}$</p>
</blockquote>
<p>Why is that so?</p>
<p>And also, I have a suspicion that where it says representation of the non-Abelian (Lie) group it really means representation of the Lie algebra associated. Is my suspicion right?</p>
| 2,976 |
<p>I'm not sure the Physics StackExchange is the perfect place for this environmental/applied physics question, but as I found no forum more fitting I ask my question here. Otherwise please move my question.</p>
<p><strong>Main Question</strong></p>
<p><strong>In which direction will the fumes move when leaving the exhaust of a car?</strong> No aerial movement or wind is assumed. On cold winter days it can be seen that the initial direction is horizontal or downwards, forming small fume clouds behind the car. Where to will the fumes move from there on?</p>
<p>The fumes assumingly have a higher temperature than the surrounding air, indicating they will rise. Nevertheless cooling effects and density may eliminate or surpass this effect.</p>
<p><strong>Sub-question</strong></p>
<p>My office is in 3rd floor next to a main traffic road. Will parts of the fumes enter the room when I open the window? (This is where my initial inspiration to this question comes from)</p>
| 2,977 |
<p>This is a Homework problem so please feel free to not answer and just give pointers.</p>
<p>A localized wavepacket is given as:</p>
<p>$$\Phi(r,t=0 ) = \frac { e^{-\large\frac {r^2}{2s^2}} e^{\large\frac{i\pi x}{4a}} } {(2\pi^{\frac{3}{4}}s^\frac{3}{2})}$$</p>
<p>Find the group velocity (given by: $v_g = \frac{d\omega}{dk}$ (I can't see how this will help))</p>
<p>Also, given a time-dependent homogeneous electric field in z-direction $E_z(t) = E_0sin(\omega t)$, give the center of mass movement $r_0(t)$ of the wavepacket for $t>0$, neglecting all the scattering events.</p>
| 2,978 |
<p>I once worked as a kitchen porter over a winter season.<br>
We had fun with thermal temperature guns (like <a href="http://www.amazon.co.uk/tag/infrared%20thermometer/products" rel="nofollow">these</a>) which I learned can be used for measuring the temperature of something a reasonable distance away (aside from the obvious use of <a href="http://en.wikipedia.org/wiki/Laser_tag" rel="nofollow">laser tag</a>), which to my mind is pretty impressive.</p>
<p>How do they work?</p>
| 2,979 |
<p>I usually wonder which thermodynamics first law is better to use ? </p>
<p>The one given by physics : $\Delta U=Q-\Delta W$</p>
<p>or the one by chemistry : $\Delta U=Q+\Delta W$</p>
<hr>
<p>In other words, should I take the gas as my system and take every parameter in its terms?</p>
| 2,980 |
<p>Suppose that an asteroid) that moves at a velocity $v$ passes near to a planet: now I know that the gravity force of the planet will curve the trajectory of the asteroid so I tried to make a draw of the asteroid while passing near the planet using vectors but as result the planet <strong>ALWAYS</strong> falls into the planet that is obviously wrong because the asteroid in real life can also pass the asteroid. Can someone please help me and maybe show some math to calculate the orbits?</p>
| 2,981 |
<p>How does one know whether, in treating a certain problem, one should consider particles as waves or as point-like objects? Are there certain guidelines regarding this?</p>
| 2,982 |
<p>It maybe a stupid question, but from the Ehrenfest's theorem, we have
\begin{eqnarray*}
\frac{d\langle A\rangle}{dt} &=& \left\langle\frac{\partial A}{\partial t}\right\rangle + \frac{1}{i\hbar}\left\langle[A,H]\right\rangle
\end{eqnarray*}
The if we apply it to the Hamiltonian,
\begin{eqnarray*}
\frac{d\langle H\rangle}{dt} &=& \left\langle\frac{\partial H}{\partial t}\right\rangle + \frac{1}{i\hbar}\left\langle[H,H]\right\rangle
\end{eqnarray*}
But since the last term vanishes
\begin{eqnarray*}
\frac{d\langle H\rangle}{dt} &=& \left\langle\frac{\partial H}{\partial t}\right\rangle
\end{eqnarray*}
But in general cases, the expectation value of the time derivative of the Hamiltonian is not zero, i.e. in the infinite potential well.
$$ \left\langle\frac{\partial H}{\partial t}\right\rangle=\int\Psi^*\frac{\partial H}{\partial t}\Psi dx=\int\sum_n c_n^* \psi_n^*e^{iE_n t/\hbar}\frac{\partial H}{\partial t}\sum_m c_m \psi_m e^{-iE_m t/\hbar}dx$$
$$ =\int\sum_n c_n^* \psi_n^*e^{iE_n t/\hbar}\sum_m c_m (H \psi_m) \frac{\partial }{\partial t}e^{-iE_m t/\hbar}dx$$
$$=\int\sum_n c_n^* \psi_n^*e^{iE_n t/\hbar}\sum_m c_m{1\over{i\hbar}}E_m^2\psi_m e^{-iE_m t/\hbar}dx$$
$$={1\over{i\hbar}}\sum_n\sum_m e^{i(E_n -E_m) t/\hbar}c_n^*c_m\int E_m^2 \psi_n^*\psi_m dx$$
$$={1\over{i\hbar}}\sum_n |c_n|^2E_n^2 $$
But since the expectation value of the Hamiltonian in the infinite well is a constant, it is obviously a contradiction. Is it impossible to apply the Ehrenfest's theorem to the Hamiltonian, or is there any mistake in my calculation?</p>
| 2,983 |
<blockquote>
<p>This question is part of <a href="http://meta.physics.stackexchange.com/a/5171/8563">this week's Journal Club session</a>.</p>
</blockquote>
<hr>
<p>These systems look ridiculously fun to construct. Could someone explain the particulars? What are the various types of solutions, and what are their dynamics? How are they different from previously discovered solutions?</p>
<p><img src="http://i.stack.imgur.com/2DPJi.jpg" alt="enter image description here"></p>
<p>It's been a while since I first saw the article, which is described nicely in</p>
<blockquote>
<p>Physicists Discover a Whopping 13 New Solutions to Three-Body Problem. Jon Cartwright, <a href="http://news.sciencemag.org/physics/2013/03/physicists-discover-whopping-13-new-solutions-three-body-problem" rel="nofollow">Science Now news, 8 March 2013</a>.</p>
</blockquote>
<p>The original paper is at</p>
<blockquote>
<p>Three Classes of Newtonian Three-Body Planar Periodic Orbits. Milovan Šuvakov and V. Dmitrašinović. <a href="http://dx.doi.org/10.1103/PhysRevLett.110.114301" rel="nofollow"><em>Phys. Rev. Lett.</em> <strong>110</strong> no. 11, 114301 (2013)</a>. <a href="http://arxiv.org/abs/1303.0181" rel="nofollow">arXiv:1303.0181</a>.</p>
</blockquote>
| 2,984 |
<p>Thanks if you take the time to read this. Here is the problem statement:
<img src="http://i.stack.imgur.com/Xne8Y.png" alt="problem statement"></p>
<p>The problem I'm getting is that I'm not getting the kinetic energy diagonal when I convert to the coordinates that diagonalize the potential energy. If you scroll all the way down you should see there's a cross derivative term involving the partial derivatives w.r.t $q_1$ and $q_2$. I haven't been able to spot any significant mistakes. So is there some other way to simultaneously diagonalize potential and the conjugate momenta in the new coordinates? Is there something I'm missing? How does the $\frac{1}{2(2M + m_2)}$ factor come into play?</p>
<p>First we make a transformation in the x coordinates so that the $a$'s go away. Namely
$$\begin{bmatrix} x_1^\prime \\x_2^\prime \\ x_3^\prime\end{bmatrix} = \begin{bmatrix} x_1 \\x_2 \\ x_3 \end{bmatrix} - \begin{bmatrix} a/2 \\-a/2 \\ -3a/2\end{bmatrix} $$
And I will refer to the $x_i^\prime$ coordinates without the prime hereafter.
$$(x_2 - x_1)^2 + (x_3 - x_2)^2 = x_1^2 +x_2^2 - 2x_1x_2 + x_2^2+ x_3^2 - 2x_2x_3 $$
There are multiple choices for the matrix which below which would produce the above algebraic expression. We want to choose the one which Hermitian, so that the corresponding transformation is unitary.
$$= \begin{bmatrix} x_1 &x_2 & x_3\end{bmatrix}\begin{bmatrix} 1 &-1 &0 \\ -1 &2 &-1 \\ 0 & -1 & 1 \end{bmatrix}\begin{bmatrix} x_1 \\x_2 \\ x_3\end{bmatrix} $$
$$det\begin{bmatrix} 1 - \lambda &-1 &0 \\ -1 &2- \lambda &-1 \\ 0 & -1 & 1- \lambda \end{bmatrix} = \left(1 - \lambda \right)^2\left(2 - \lambda \right) - \left(1 - \lambda \right) - \left(1 - \lambda \right) $$
\begin{align*} = \left(1 - \lambda \right)\left[\left(1 - \lambda \right)\left(2 - \lambda \right) -2 \right] &= \left(1 - \lambda \right)\left[2 - 3\lambda + \lambda^2 -2 \right] \\
& =\lambda\left(1 - \lambda \right)\left(\lambda - 3 \right) \end{align*}
So the set of eigenvalues is $\left\{0, 1, 3 \right\}$.
$$U = \begin{bmatrix} 1/\sqrt{3} &1/\sqrt{3} &1/\sqrt{3} \\ 1/\sqrt{2} &0 &-1/\sqrt{2} \\ 1/\sqrt{6} & -2/\sqrt{6} & 1/\sqrt{6} \end{bmatrix}$$
$$ U^{T}\begin{bmatrix} 0 &0 &0 \\ 0 &1 &0 \\ 0 & 0 & 3 \end{bmatrix}U = \begin{bmatrix} 1 &-1 &0 \\ -1 &2 &-1 \\ 0 & -1 & 1 \end{bmatrix}$$
$$\begin{bmatrix} q_1 \\q_2 \\ q_3\end{bmatrix} = U \begin{bmatrix} x_1 \\x_2 \\ x_3\end{bmatrix}$$
Next begins the calculation of the partial derivatives of $x_i$ in the new coordinate frame, as the first step to finding the KE in terms of the momenta conjugate to the new coordinates $q_i$.
$$\begin{bmatrix}\frac{\partial }{\partial x_1} & \frac{\partial }{\partial x_2} & \frac{\partial }{\partial x_3}\end{bmatrix} = \begin{bmatrix}\frac{\partial }{\partial q_1} & \frac{\partial }{\partial q_2} & \frac{\partial }{\partial q_3}\end{bmatrix}U$$
$$KE = -\hbar^2\begin{bmatrix}\frac{\partial }{\partial x_1} & \frac{\partial }{\partial x_2} & \frac{\partial }{\partial x_3}\end{bmatrix} \begin{bmatrix} \frac{1}{2M} &0 &0 \\ 0 &\frac{1}{2m_2} &0 \\ 0 & 0 & \frac{1}{2M} \end{bmatrix} \begin{bmatrix}\frac{\partial }{\partial x_1} \\ \frac{\partial }{\partial x_2} \\ \frac{\partial }{\partial x_3}\end{bmatrix} $$
$$ = -\hbar^2\begin{bmatrix}\frac{\partial }{\partial q_1} & \frac{\partial }{\partial q_2} & \frac{\partial }{\partial q_3}\end{bmatrix}U \begin{bmatrix} \frac{1}{2M} &0 &0 \\ 0 &\frac{1}{2m_2} &0 \\ 0 & 0 & \frac{1}{2M} \end{bmatrix}U^{T}\begin{bmatrix}\frac{\partial }{\partial q_1} \\ \frac{\partial }{\partial q_2} \\ \frac{\partial }{\partial q_3}\end{bmatrix} $$
$$ U \begin{bmatrix} \frac{1}{2M} &0 &0 \\ 0 &\frac{1}{2m_2} &0 \\ 0 & 0 & \frac{1}{2M} \end{bmatrix}U^{T}$$
$$= \begin{bmatrix} 1/\sqrt{3} &1/\sqrt{3} &1/\sqrt{3} \\ 1/\sqrt{2} &0 &-1/\sqrt{2} \\ 1/\sqrt{6} & -2/\sqrt{6} & 1/\sqrt{6} \end{bmatrix}\begin{bmatrix}
\frac{1}{2M\sqrt{3}} & \frac{1}{2M\sqrt{2}} &\frac{1}{2M\sqrt{6}} \\
\frac{1}{2m_2\sqrt{3}} & 0 &-\frac{2}{2m_2\sqrt{6}} \\
\frac{1}{2M\sqrt{3}} & -\frac{1}{2M\sqrt{2}} & \frac{1}{2M\sqrt{6}} \end{bmatrix}$$
$$ = \begin{bmatrix}
\frac{1}{3M} + \frac{1}{6m_2} & 0 & \frac{1}{3\sqrt{2}}\left(\frac{1}{M} - \frac{1}{m_2} \right) \\ 0 & \frac{1}{2M} & 0 \\
\frac{1}{3\sqrt{2}}\left(\frac{1}{M} - \frac{1}{m_2} \right) & 0 & \frac{1}{6M} + \frac{1}{3m_2} \end{bmatrix} $$
$$\frac{KE}{-\hbar^2} = \begin{bmatrix}\frac{\partial }{\partial q_1} & \frac{\partial }{\partial q_2} & \frac{\partial }{\partial q_3}\end{bmatrix}
\begin{bmatrix}\left(\frac{1}{3M} + \frac{1}{6m_2} \right)\frac{\partial }{\partial q_1} + \frac{1}{3\sqrt{2}}\left(\frac{1}{M} - \frac{1}{m_2} \right)\frac{\partial }{\partial q_3}\\ \frac{1}{2M}\frac{\partial }{\partial q_2} \\ \frac{1}{3\sqrt{2}}\left(\frac{1}{M} - \frac{1}{m_2} \right)\frac{\partial }{\partial q_1}+ \left(\frac{1}{6M} + \frac{1}{3m_2} \right)\frac{\partial }{\partial q_3}\end{bmatrix} $$
$$= \left(\frac{1}{3M} + \frac{1}{6m_2} \right)\frac{\partial^2 }{\partial q_1^2} + \frac{1}{2M}\frac{\partial^2 }{\partial q_2^2} + \left(\frac{1}{6M} + \frac{1}{3m_2} \right)\frac{\partial^2 }{\partial q_3^2}$$
$$+ \frac{\sqrt{2}}{3}\left(\frac{1}{M} - \frac{1}{m_2} \right)\frac{\partial^2 }{\partial q_1 \partial q_3} $$
So to me the fact that there is a cross term means that I won't able to express the Hamiltonian in terms of three independent oscillators.</p>
| 2,985 |
<p>Can the firewall be viewed as the holographic boundary?
Naively a hologram 3d image can not cross the hologram 2d surface that produces that image. According to the metaphor the boundary - 2d field quantum theory without gravity - could act as a firewall for the enclosed space - 3d string theory with gravity and black holes. Yet information won't be lost by the presence of the firewall, because the digital bits would be just the pixels of that firewall/boundary. Doesn't this settle the paradox of entanglements non-locality with quantum gravity?
" we must carefully impose the laws of thermodynamics and/or statistical physics to Maxwell's Demon himself and when we do so, his miraculous abilities to create a paradox evaporate." In the same way I'm asking to impose the boundary S matrix unitarity to the firewall ;-) The entangled partner behind the firewall is then a kind of fictitious hologram according to the Wheeler DeWitt diffeomorphism of quantum gravity, isn't it?</p>
| 2,986 |
<p>When talking about gamma ray or x-ray propagation in media, we usually talk about the mass attenuation coefficient, and we desire high density materials for shielding. This seems probabilistic: we increase the shielding to maximize the chance of EM radiation interacting with the media.</p>
<p>However, conductivity is a part of the attenuation coefficient. Why not use highly conductive materials? What justifies using high density lead over high conductivity?</p>
| 2,987 |
<p>In about 7 Billion years our planed will be consumed by the ever-growing sun, life would have become extinct long before that. That means that in several hundred thousand years we have a deadline to either:</p>
<p>1) Move earth to a higher orbit to keep it from heating up<br>
2) Move our civilization to Mars and beyond</p>
<p>Of course we are only beginning to understand some of the technologies which would be at work, but what do you think is the more efficient/likely solution to this issue?</p>
| 2,988 |
<p>I am solving a CLASSICAL an-harmonic oscillator problem with Hamiltonian given by
$H= (1/2)\dot{x}^2+(1/2)x^2-(1/2)x^4$ with all the constants (k's) and mass being taken as 1 (one).</p>
<p>I find that $x= \tanh(t/\sqrt{2})$ is satisfying the equation of motion. </p>
<p>But my question is how to incorporate the Hamiltonian, $H$ in to this solution so that by providing $H$ we can control the initial conditions of this problem. Or any other solution function that can have $H$ in it.</p>
<p>thanks in advance.</p>
<p>PS= In SHO (m=k=1) lets say $x=A\sin(t)$ then $A= \sqrt{2H}$, where $H$ is the total energy or the Hamiltonian. So $x=\sqrt{H}\sin(t)$.</p>
<p>I need a solution function like this.</p>
| 2,989 |
<p>Please consider the circuit diagram below, in particular, look at the capacitors enclosed by the green loop.</p>
<p><img src="http://i.stack.imgur.com/dybm1.jpg" alt="enter image description here"></p>
<p>*Note that the green loop and the (+) and (-) charges on the plates were drawn by me, the original circuit has none of these.</p>
<p>Is the charge distribution on the plates correct? The two capacitors in the green loop have a + and - plate connected by a wire, which by definition should be in series. The three -s that joins up at point <strong>d</strong> and evenly distribute itself (split) into two paths and adds charge to the two plate near point <strong>b</strong> (sorry for poor terminologies here)</p>
<p>But the book treats the green loop capacitors in parallel, which must imply the configuration. </p>
<p><img src="http://i.stack.imgur.com/3fzyw.jpg" alt="enter image description here"></p>
<p>I am awfully confused how $\ C_1$'s right (+) plate managed to induce $\ C_2$'s right plate to (-) and how the left plate of $\ C_2 $ has (+) plate when it's connected to the negative terminal of the battery. </p>
<p>Also, I've noticed that a lot of people often just attack these problems based on the geometry of the circuits and I think they would make the same mistake as my book.</p>
| 2,990 |
<p>In classical mechanics, if I want to view the Earth as the fixed center of the solar system, I must accelerate my reference frame to keep it centered on the Earth. That accelerated reference frame causes all sorts of messy fictitious forces that push the stars and planets around(<a href="http://physics.stackexchange.com/questions/10933">Why do we say that the earth moves around the sun?</a>). But in GR, since both the Sun and the Earth are in free-fall, I would think that the Earth's reference frame would be just as "natural" as the Sun's reference frame. In GR, would a geocentric model of the solar system would be "messy"*? If so, why?</p>
<p>*Let's avoid rotating reference frames by pretending that the Earth doesn't rotate sidreally.</p>
| 205 |
<p>I'm trying to solve this elementary problem. I'm studying mathematics, but there is a compulsory course in physics that has to be passed. I'm having an exam in 5 days and I have some doubts on problems like this:</p>
<blockquote>
<p><img src="http://i.stack.imgur.com/Y72Td.jpg" alt="enter image description here"></p>
<p>In the above system, both 1 and 2 have the same mass. 2 has an initial velocity of $2 \frac m s$ (meters over second)</p>
<p>$(a)$ Graph the free body diagram of each block. Indicate it's reaction-action pair.</p>
<p>$(b)$ Calculate the acceleration of 1 and describe analytically and graphically it's movement as a function of time.</p>
</blockquote>
<p>$(a)$ isn't too difficult:</p>
<p><img src="http://i.stack.imgur.com/68gEN.jpg" alt="enter image description here"></p>
<p>What is getting me a little troubled is $(b)$. My professor says we should take a "curved" $x$-axis in following the rope. I suppose then I take the $y$-axis as perpendicular to the $x$-axis.</p>
<p>I decompose the problematic $\vec w$ vectors via the angles given. I use $\sin 37º \approx 0,6$ and $\cos 53º \approx 0,8$.</p>
<p><img src="http://i.stack.imgur.com/jbZcn.jpg" alt="enter image description here"></p>
<p>EDIT: The normal vectors are wrong. They should be.</p>
<p>$$\eqalign{
& \vec N = mg\sin 37\hat j \cr
& {{\vec N}_0} = - mg\sin 37\hat j \cr
& \vec N' = - mg\sin 37\hat j \cr
& \vec N{'_0} = mg\sin 37\hat j \cr} $$
(there is a $-$ missing in $\vec w$, before $\cos 37$ which should be $i$, not $j$.)</p>
<p>Now I apply Newton's law, $\sum \vec F = m \cdot \vec a $, to my system. I'm interested only in $x$, so I have to use $\sum \vec F_x = m \cdot \vec a_x $. Since the conditions are ideal, $a$ is constant in all the system, $t$ is also constant. The equations are:</p>
<p>$$\eqalign{
& t - \left| w \right|\cos 37 = {m_1}a \cr
& \left| w \right|\sin 37 - t = {m_2}a \cr} $$</p>
<p>Now $|w| = m_2 g=m_1g$, so </p>
<p>$$\eqalign{
& t - m \cdot g \cdot \cos 37 = m \cdot a \cr
& m \cdot g \cdot \sin 37 - t = m \cdot a \cr} $$</p>
<p>So I find $a=-0.1 g\approx 1 \frac{m} {s^2}$</p>
<p>This means the acceleration of $1$ is in the positive direction of the $x$ axis. Is this correct? How can I move on with the problem?</p>
| 2,991 |
<p>As per Newton objects with mass attract each other, and per Einstein this is further explained by saying that mass warps space-time. So a massive object makes a "dent" into space-time,
a gravity well. I have taken to visualizing this as placing a object on a rubber sheet and the
resulting dent, being the gravity field. So obviously placing two objects
on the sheet not to far from each other will make the dents overlap, and the object will roll towards each other.
BUT for a BLACKHOLE, this is not a dent. It's a cut or rupture in the rubber sheet.
Furthermore, space-time is constantly falling INTO the blackhole, and everything else
that exists in space-time, including light.
So a blackhole is not just a super-massive object, it's really a hole, and how can a hole move?
How does it react to the gravity pull of a nearby object, when everything just falls thru it?
Thanks!</p>
| 2,992 |
<p>I'm using CERN library's libmathlib.a and libkernlib.a, downloaded from <a href="http://cernlib.web.cern.ch/cernlib/version.html" rel="nofollow">this site</a>, and running on Ubuntu linux to do <a href="http://mbtools.hepforge.org/" rel="nofollow">Mellin-Barnes integrals</a> with f77 fortran code in i386 mode (hence the <code>-m32</code>):</p>
<p>gfortran -m32 -O -o output MBpart1ep0.f -L. -lmathlib -lkernlib -Lcuba</p>
<p>But I'm getting errors like:</p>
<p>//usr/local/lib/libmathlib.a(cgamma64.o): In function <code>wgamma_':
cgamma64.F:(.text+0xd1): undefined reference to</code>s_wsfi'</p>
<p>It seems to be caused by an incompatibility to gfortran. Does anyone know where I can download a gfortran compatible version of <code>libmathlib.a</code> and <code>libkernlib.a</code>?</p>
| 2,993 |
<p>Is the concept of inertia still used? When is it useful as a <em>fictitious force</em>?</p>
<p>Can you list a few situations in which, if we didn't use this tool we might be in difficulty?</p>
| 2,994 |
<p>Reading a <a href="http://physics.stackexchange.com/questions/2460/time-travel-and-random-events-closed">previous closed question</a> an interesting variation has come to my mind.</p>
<p>Suppose that time travel to the past was possible:</p>
<ul>
<li>I wait for an atom to decay and measure the time, $t_{1a}$ </li>
<li>I travel back in time at $t_0<t_{1a}$</li>
<li>I wait for the same atom to decay and measure the time, $t_{1b}$</li>
</ul>
<p>Let's think about the two values, $t_{1a}$ and $t_{1b}$. </p>
<p>If they coincide, then from my point of view the measured $t_1$ time would not be governed by chance (the second time I would know $t_1$ <em>a priori</em>). It would therefore prove some form of existence of hidden variables. Would this violate any known laws of quantum mechanics? Would this prove the existence of hidden variables (e.g. Bohm's interpretation)?</p>
<p>If they are different, then what to make of $t_{1a}$ and $t_{1b}$? Which would be the <em>correct</em> value? To me this doesn't make any sense, but maybe it could be compatible with the multiverse interpretation (I don't know how though).</p>
<p>In other words, does this gedankenexperiment:</p>
<ol>
<li>...help us select viable QM interpretations and exclude others? Which?</li>
<li>...lead us to conclude that backwards information time travel is incompatible with QM?</li>
</ol>
| 2,995 |
<p>Some friends and I were at Disneyland this past week. We spent an hour arguing back and forth whether it is possible for someone to throw an NFL football over the Matterhorn? I think it is, but no one else really thought it was possible.</p>
<p>What do you guys think? Is this humanly possible?</p>
<p>The Matterhorn is 146 feet (44.5 metres) tall. Not sure how wide the base of the mountain is. It's ok if the football hits the other side of the mountain (as long as it clears the top).</p>
| 2,996 |
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