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<p>The decay mode of Carbon-8 is listed as 'XP' in this <a href="http://ie.lbl.gov/education/parent/C_iso.htm" rel="nofollow">table</a>. None of the references I looked at listed XP as a decay mode. What is it></p>
| 4,187 |
<p>What complications arise when examining the statistical mechanics of a system under the influence of gravity? Is it significantly different from the classical treatment of statistical mechanics?</p>
| 4,188 |
<p>As far I as I know, and from naturalness considerations, a 125 GeV Higgs mass is rather large for the MSSM. This is because in the MSSM $$m_h^2 \lesssim M_z^2 \cos^22\beta + \Delta$$ where $\Delta$ represents top/stop loop corrections to the Higgs. It takes the form $$\Delta \sim \ln(\frac{m_\text{stop}}{m_\text{top}}) + \text{mixing}$$ </p>
<p>Moreover, the Z mass is determined by the Higgs mass parameter, $m_{H_u},$ and $\mu$ (in the large $\tan\beta$ limit. i.e. $\tan\beta \geq 10$). And $m_{H_u} \sim - m_\text{stop}$, so the larger $m_\text{stop}$ the larger the fine-tuning in the MSSM. Now, only by large $m_\text{stop}$ (and mixing) one can reach a Higgs mass greater than LEP/LHC bounds. But this is already associated with large fine-tuning as I said. </p>
<p>So, in this view, a 125 GeV Higgs mass is large for the MSSM. (At least for the cMSSM.)</p>
<p>However, I realised that some people think 125 GeV is much less than what the MSSM predicts! Now this I don't understand at all. So could someone please explain what this view is based on?</p>
| 4,189 |
<p>I have a wet teabag in an empty cup. If I will hold the teabag and touch the wall of cup with it, it will stick to the cup, like there would be glue or some magnetic field, but there's just water.</p>
<p><img src="http://i.stack.imgur.com/HaBZ4.jpg" alt="enter image description here"></p>
<p>So, why does it happens? Why can just a little bit of water hold the teabag stronger than the gravity of an entire planet so it will not fall to the bottom of the cup, or even slide down a little bit?</p>
<p>Why is it so strong?</p>
| 4,190 |
<p>Say you have a chemical compound made up of one or more radioactive nuclei. If the nucleus decays, does the compound? </p>
<p>Possible outcomes I can think of:</p>
<ol>
<li><p>the compounds continues to exist if a bonding is still possible between the decay product and the rest of the original compound.</p></li>
<li><p>the compound just ceases to exist.</p></li>
<li><p>the decay product forms a new compound with some fraction of the remainder of the original one but discards some other part.</p></li>
</ol>
<p>Generally, it will do whatever results in the least energetic state, but is there some kind of regularity to it?</p>
| 739 |
<p>(little background: I'm trying to develop a small, quick 'n dirty static physics engine to determine whether a stacking of boxes is stable).</p>
<p>If I have a 3D rigid box (with the bottom in the horizontal plane), resting on n points (at [xn, yn]), and we apply a downward force F at [xF, yF], how can I calculate the resulting forces Fn at these n points?</p>
<p>If the system is in equilibrium then sum(Fn) should be equal to F.
However, I must also consider that the system might not be in equilibrium (for instance, if all xn < xF), so I probably can't use equilibrium equations. I'd still like to know the forces in that case, though, so that I can calculate the resulting torque.</p>
<p>Is there an easy formula for this? </p>
| 4,191 |
<p>I was reading today about the birth of the Universe and the conjectures about the matter that was supposed to exist at the moment of the Big Bang and what can be measured now.</p>
<p>There seems to be some sort of discrepancy between the calculated amount of matter in the Big Bang and the amount that can be measured in the visible Universe.</p>
<p>Why is that so?</p>
<p>And further more, why can't be this "missing matter" have been devoured by black holes through out the Universe?</p>
| 4,192 |
<p>Cepheids are used to evaluate distances. What is the math and physics behind their use?</p>
| 4,193 |
<p>I was wondering if it were possible to calculate the drag coefficient by allowing an object to reach terminal velocity. Can you rearrange the terminal velocity formula to give the drag coefficient?</p>
| 4,194 |
<p>Why nature made conjugates?
such as momentum and position, time and energy? Why not energy and position even both do not commute. .</p>
<p>thanks in advance</p>
| 128 |
<p>I understand that the larger the mass the greater gravity is and the slower time is, as well the faster an object is traveling the slower time passes. My question is that since the faster an object travels the more mass it has, is the increase in mass the reason for the change in time, or is it the velocity?</p>
| 4,195 |
<p>Why did nature decide to make conjugate of position to be momentum? Since energy and position do not commute, why not energy? What determines the pairing of time with energy and momentum with position?</p>
| 128 |
<p>A satellite is in a circular orbit when its engines turn on to exert a small force in the direction of the velocity for a short time interval. Is the new orbit further or closer to the Earth?</p>
<p>The solution is that the new orbit is further away (which is also intuitive) and is justified by stating that there is a positive increase in the total energy which is given by the formula:
$$\:E_{T}=-\frac{GMm}{2r}$$
And that states that an increase in the total energy would result in a larger radius. Thus, the new orbit is further away.</p>
<p>However, the problem arises when I look at this equation which relates the speed with the radius:</p>
<p>$$\:v^2=\frac{GM}{r}$$
Since the small force was in the direction of the velocity, an increase in veloctiy should result in a DECREASE in the radius meaning that the new orbit is closer in according to this equation. </p>
<p>Why is it that the first equation is correct to use, while the second one is wrong? Why is the second equation not working here?</p>
<p>Thanks!</p>
| 4,196 |
<p>can be a theory with an infinite number of divergent integrals of the form</p>
<p>$$ \int \frac{d^{p}k}{k^{m}} $$</p>
<p>for m=1 , 2 , 3 , 4 ,...... so the theory would be IR non renormalizable and you would need and infinite set of operations to cure all the IR divergences </p>
<p>thanks.</p>
| 4,197 |
<p>Forgetting Hooke's law for a minute why, from a microscopic perspective (preferably quantum) on up to a macroscopic one, does a spring under tension exert a force?</p>
<p>I was thinking that there might be an analogy between the high and low pressure states of an air wave and the density of the mass distribution of the compressed and stretched states of an iron spring, but I don't know nearly enough about solid state physics to even guess.</p>
| 4,198 |
<p>How much energy in form of heat does a human body emit at rest level?</p>
| 4,199 |
<p>As I understand it, black holes have an absolute event horizon and an apparent horizon specific an observer. In addition to black holes, an apparent horizon can come from any sustained acceleration. Firstly, there is the <a href="http://en.wikipedia.org/wiki/Event_horizon">"particle horizon" of the universe</a>, which is the furthest co-moving distance from which light can reach us before the acceleration of the expansion of the universe prohibits it, then there is the case of a constantly accelerating observer in just regular flat space.</p>
<p>My zeroth question is if the Hawking radiation from a black hole could be said to come from the apparent or absolute horizon - which is it? Obviously the former would be specific to each observer.</p>
<p>For the case of the universal acceleration, dark energy, cosmological constant (please help me with terminology as appropriate): One could ask "do we observe radiation from the horizon?", for which we already have an available answer in the form of the cosmic microwave background (CMB). But I am curious if this would match the temperature and intensity you would expect if you treated the edge of the universe as a black hole horizon. Are these two be equivocal in some sense?</p>
<p>Finally, I'm most interested to ask what would occur for the case of a constantly accelerating observer. An apparent horizon exists behind the observer beyond which light may never reach. Could it be possible the Hawking radiation is observed as coming from behind?</p>
<p><img src="http://i.stack.imgur.com/IJRj1.png" alt="Accelerating Observer"></p>
<p><strong>Could P possibly observe Hawking radiation coming from behind?</strong></p>
<p>Granted, if you took the simple version of the temperature of the Hawking radiation ($T = \frac{\hbar c^3}{8 \pi G M k_B}$) and plugged in some super large mass, you will correspondingly get some super small temperature which will not agree with the 2.7 K of the CMB. Of course, the two cases I'm discussing do not have any definable mass that can be used. I would imagine that in the case of a simply accelerating particle, a greater acceleration would correspond with greater Hawking radiation, which is why it's not an issue for normal stuff, but if we accelerated something small with great force, could it observe radiation from that (false) event horizon?</p>
| 4,200 |
<p>Recently, I read about dipole and multipole bound anions. Dipole bound anions are those, if I understood correctly, when an electron is attached electrostatically on a neutral molecule which is polar. If the molecule is having a quadrupole moment, the anion is said to be a quadrupole bound anion. These bound electrons will have less binding energy compared to the electrons in molecular orbital. My question is, these dipole, quadrupole and other multipole bound electrons behave like normal electrons in orbitals except that it has less binding energy or different? </p>
<p>To be more clear, do these electrons have specific orbitals with orbital angular momentum or not? </p>
<p>If so, are these orbitals like normal molecular orbitals (S P D F)? </p>
<p>Can we assign electronic states to these electrons?</p>
<p>Also, is it possible to excite any electron in the molecule (say valance electron) to this bound state (dipole or any multipole bound state)?</p>
| 4,201 |
<p>This is probably a really silly confusion I have about the definition of “coordinate differentials”, which I thought were things like $dx,dy,dz$
etc. The Minkowski line element $$ds^{2}=c^{2}dt^{2}-dx^{2}-dy^{2}-dz^{2}$$
defines the Minkowski metric $$\left[\eta_{\mu\nu}\right]=\left(\begin{array}{cccc}
c^2 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1
\end{array}\right).$$
Using index notation, the line element can be written as $ds^{2}=\eta_{\mu\nu}dx^{\mu}dx^{\nu}$.
In textbooks I have seen the terms $dx^{\mu},dx^{\nu}$
called “coordinate differentials”, which seems OK except $dx^{0}=cdt$.
I realise this is trivial, but is it correct to call $cdt$
a “coordinate differential”? To me it looks like a coordinate differential $dt$
multiplied by $c$. </p>
| 4,202 |
<p>The Poisson - Boltzmann equation is to find the potential whose charge density obey the Boltzmann distribution. If the potential is not large compare to $k_BT$, the PBE can lead to the Debye–Hückel equation. However if the potential is large and the system is high density, its result is <a href="http://pubs.acs.org/doi/abs/10.1021/ct3001156" rel="nofollow">unreal</a>.</p>
<p>I can see the problem is coming from Boltzmann part, not Poisson, but I can't find any document states that Boltzmann distribution is not suitable for high energy.</p>
| 4,203 |
<p>I'm looking for a treatment of the original basic <a href="http://en.wikipedia.org/wiki/Kaluza%E2%80%93Klein_theory" rel="nofollow">Kaluza-Klein theory</a>.
Can someone recommend a review article or something?</p>
| 4,204 |
<p>In aircraft literature, what does cross-track and along-track wind directions mean? Please explain in terms of the aircraft's motion relative to the wind direction. I can hazard a guess: along-track is wind in the direction of the aircraft's velocity, and cross-track would be perpendicular side-to-side (or is it top-bottom) wind direction wrt to aircraft velcoity direction? Not sure...some explanation and clarification would help. Thanks. </p>
| 4,205 |
<p>It has always been claimed decoherence solves the problem of the preferred-basis for observed probability distributions, but why should this be the case? If there is only one world, and there are probabilities for certain outcomes, why should the basis in which the probabilities are observed coincide with the pointer basis determined dynamically by decoherence?</p>
| 4,206 |
<p>I have seen <a href="http://www.physics.sfsu.edu/~lea/courses/ugrad/460notes5.PDF" rel="nofollow">this derivation</a>: </p>
<p><img src="http://i.stack.imgur.com/goWYX.png" alt="enter image description here"></p>
<ul>
<li><p>I want to estimate what is the intensity of the electrical field as
function of $r$ the distance from the radiated source ?</p></li>
<li><p>I think it is can modled as pointed source which implies a spherical
wave whose intensity goes like $1/r^2$. But I think I remember
from school it should be fall like $1/r^3$.</p></li>
</ul>
| 4,207 |
<p>Let's suppose a cellular automaton has a value $b(r,t)$ belongs to $Q$ at site $r$ and time $t$, where $Q$ is the set of possible states at each site. Let $N(r, t)$ be the values of the states of all the sites in some (e.g., Von Neumann or Moore) neighborhood of $r$ at time $t$, taken in some canonical order. </p>
<p>The cellular automaton rule is then
$$b(r, t + 1) = F(N(r, t))$$
where $F$ is the update function.</p>
<p>Let us suppose that $Q = \{0, 1\}$ so that we have a bit at each site. One way to construct reversible cellular automata (RCA) is to look at rules of the form $b(r, t + 1) = F(N(r, t)) \text{ xor } b(r, t − 1)$</p>
<p>To see why this is reversible, how to solve for $b(r, t −1)$ in terms of $b(r, t)$ and $b(r +1, t)$. Does it mean that this obeys a condition even stronger than reversibility?</p>
<p>One apparent disadvantage of this approach seems to be that computation of $b(r, t+1)$ requires knowledge of the state at time $t−1$ in addition to that at time $t$. Is there way to fix this problem by adding extra state at a site so that the state at time $t + 1$ depends only on that at time $t$?
Are there other functions besides $\text{xor}$ that could be used to create RCA in the above fashion for $Q = \{0, 1\}$?</p>
| 4,208 |
<p>Don't be a $\frac{d^3x}{dt^3}$</p>
<p><img src="http://i.stack.imgur.com/Ptblb.jpg" alt="enter image description here"></p>
<p>What does it all mean?</p>
| 4,209 |
<p>A gun has a recoil speed of 2 m/s when firing. If the gun has a mass of 2kg and the bullet has a mass of 10g (0.01 kg) what speed does the bullet come out at? </p>
<p>The gun has zero total momentum before firing and afterwards the gun has negative acceleration. </p>
<p>So far:</p>
<p><strong>Conservation of momentum</strong>: $m_1v_1 = m_2v_2.$</p>
<p>We have the recoil speed of $2\,$m/s.
The mass of the gun is equal to $2\,$kg.</p>
<p>Plus the bullet's total mass which is 0.01 kg.</p>
<p>$$2 \frac{m}{s}\cdot 2.01\,\text{kg} + (0.01\,\text{kg} \cdot v)$$
$$=4.02\,\text{kg}\frac{m}{s} + (0.01\,\text{kg} \cdot v)$$</p>
<p>That's as far as I can go.</p>
| 4,210 |
<p>If we have a planar and harmonic EM wave, with $B$ field:</p>
<p>$$B=A\left(\begin{array}{c}
1\\
i\\0
\end{array}
\right)e^{-i(\omega t-\vec k\cdot\vec r)}$$</p>
<p>and with it's corresponding $E$ field. This is a circularly polarised wave, but that field does not have 0 divergence, the three components of it, when taking real part, are:</p>
<p>$$x=\cos (\vec k\cdot\vec r-\omega t)$$
$$y=i\cdot i\sin(\vec k\cdot\vec r-\omega t)=-\sin(\vec k\cdot\vec r-\omega t)$$
$$z=0$$</p>
<p>Tha divergence won't be 0 unless $\vec k=(0,0,a)$, for some $a$. so what's the problem here? Isn't that a wave unless it spread on the $z$ axis? If the $E$ field is just the same but with different phase, I guess the same thing would have to hold if the wave had not matter around: $\nabla\cdot E =0$, right?</p>
| 4,211 |
<p>I want to show that the 2-d wave equation is invariant under a boost, so, the starting point is the wave equation</p>
<p>$$\frac{\partial^2\phi}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2} $$</p>
<p>and the Lorentz transformation:</p>
<p>$$t'=\gamma(t-\frac{v}{c^2}x) \\ x'=\gamma(x-vt)$$</p>
<p>My question is <strong>should I write $\displaystyle\frac{\partial}{\partial t}$ as a derivative with respect to $x'$ and $t'$ and then substitute?</strong></p>
<p>Work done so far</p>
<p>$$\frac{\partial}{\partial t}=\frac{\partial }{\partial x'}\frac{\partial x'}{\partial t}+\frac{\partial}{\partial t'}\frac{\partial t'}{\partial t}=-\gamma v\frac{\partial}{\partial x'}+\gamma\frac{\partial}{\partial t'} $$</p>
<p>$$\frac{\partial^2}{\partial t^2}=\frac{\partial }{\partial t}\left(\frac{\partial}{\partial t} \right)=\frac{\partial}{\partial t}\left( -\gamma v\frac{\partial}{\partial x'}+\gamma \frac{\partial}{\partial t'}\right)= \\
=-\gamma v\frac{\partial}{\partial x'}\left( -\gamma v\frac{\partial}{\partial x'}+\gamma \frac{\partial}{\partial t'}\right)+\gamma\frac{\partial}{\partial t'}\left( -\gamma v\frac{\partial}{\partial x'}+\gamma \frac{\partial}{\partial t'}\right)=\\ =
\gamma^2v^2\frac{\partial^2}{\partial x'^2}-2\gamma^2v\frac{\partial ^2}{\partial x'\partial t'}+\gamma^2\frac{\partial^2 }{\partial t'^2}$$</p>
<p>-Edit-</p>
<p>The same applies to $x$</p>
<p>$$\frac{\partial}{\partial x}=\frac{\partial }{\partial x'}\frac{\partial x'}{\partial x}+\frac{\partial }{\partial t'}\frac{\partial t'}{\partial x}=\gamma\frac{\partial}{\partial x'}-\frac{\gamma v}{c^2}\frac{\partial }{\partial t'} $$</p>
<p>$$\frac{\partial^2}{\partial x^2}=\frac{\partial}{\partial x}\left( \frac{\partial }{\partial x}\right)=\frac{\partial}{\partial x}\left(\gamma\frac{\partial}{\partial x'}-\frac{\gamma v}{c^2}\frac{\partial }{\partial t'} \right)= \\ =
\gamma\frac{\partial}{\partial x'}\left(\gamma\frac{\partial}{\partial x'}-\frac{\gamma v}{c^2}\frac{\partial }{\partial t'} \right)-\frac{\gamma v}{c^2}\frac{\partial }{\partial t'}\left(\gamma\frac{\partial}{\partial x'}-\frac{\gamma v}{c^2}\frac{\partial }{\partial t'} \right)=\\=
\gamma^2\frac{\partial^2 }{\partial x'^2}-2\frac{\gamma^2v}{c^2}\frac{\partial^2 }{\partial x'\partial t' }+\frac{\gamma^2v^2}{c^4}\frac{\partial^2}{\partial t'^2}$$</p>
<p>Edit 2 with the hints given by nervxxx</p>
<p>The wave equation becomes</p>
<p>$$\frac{\gamma^2v^2}{c^2}\frac{\partial^2 \phi}{\partial x'^2}-\frac{2\gamma^2v}{c^2}\frac{\partial ^2 \phi}{\partial x'\partial t'}+\frac{\gamma^2}{c^2}\frac{\partial^2 \phi}{\partial t'^2}=\gamma^2\frac{\partial^2 \phi}{\partial x'^2}-2\frac{\gamma^2v}{c^2}\frac{\partial^2 \phi}{\partial x'\partial t' }+\frac{\gamma^2v^2}{c^4}\frac{\partial^2\phi}{\partial t'^2}$$</p>
<p>$$ \frac{\gamma^2 v^2}{c^2}\frac{\partial ^2 \phi}{\partial x'^2}+\frac{\gamma^2}{c^2}\frac{\partial^2 \phi}{\partial t'^2}=\gamma^2\frac{\partial^2 \phi}{\partial x'^2}+\frac{\gamma^2v^2}{c^4}\frac{\partial ^2\phi}{\partial t'^2}$$</p>
<p>But I still don't get... since all $\gamma^2$ cancel</p>
<p>Final edit. done!</p>
<p>$$ \frac{\gamma^2 v^2}{c^2}\frac{\partial ^2 \phi}{\partial x'^2}-\gamma^2\frac{\partial^2 \phi}{\partial x'^2}=\frac{\gamma^2v^2}{c^4}\frac{\partial ^2\phi}{\partial t'^2}-\frac{\gamma^2}{c^2}\frac{\partial^2 \phi}{\partial t'^2}$$</p>
<p>$$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} $$</p>
<p>$$ \left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{v^2}{c^2}\frac{\partial ^2 \phi}{\partial x'^2}-\left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{\partial^2 \phi}{\partial x'^2}=\left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{v^2}{c^4}\frac{\partial ^2\phi}{\partial t'^2}-\left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{1}{c^2}\frac{\partial^2 \phi}{\partial t'^2}$$</p>
<p>$$ \left( \frac{v^2}{c^2-v^2}\right)\frac{\partial ^2 \phi}{\partial x'^2}-\left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{\partial^2 \phi}{\partial x'^2}=\left( \frac{v^2}{c^2-v^2}\right)\frac{\partial ^2\phi}{\partial t'^2}\frac{1}{c^2}-\left( \frac{1}{1-\frac{v^2}{c^2}}\right)\frac{1}{c^2}\frac{\partial^2 \phi}{\partial t'^2} $$</p>
<p>$$\frac{\partial \phi^2}{\partial x'^2}\left(\frac{v^2}{c^2-v^2}- \frac{1}{1-\frac{v^2}{c^2}}\right)=\frac{\partial^2 \phi}{\partial t'^2}\frac{1}{c^2}\left(\frac{v^2}{c^2-v^2}- \frac{1}{1-\frac{v^2}{c^2}}\right) $$</p>
<p>$$ \frac{\partial^2\phi}{\partial x'^2}=\frac{1}{c^2}\frac{\partial^2\phi}{\partial t'^2} $$</p>
| 4,212 |
<p>The Earth carries a negative electric charge of roughly 500 thousand Coulombs (according to different sources I've seen). If I touch the Earth I should therefore pick up some of this electric charge (through conduction) and become negative charged. Assuming the earth can modeled as a conducting sphere with radius $\small 6371 [\text{km}]$ and me as a conducting sphere with radius $\small 1 \small[\text m]$, around how much negative charge would I accumulate? The reason I ask is because I'm trying to prove to myself that grounding does indeed render a charged object neutral (i.e. transfers all the object's charge to the Earth). Using the well known equation for two connected conducting spheres with different radii
(see Example 3-13 on page 115 in David Cheng's "Field and Wave Electromagnetics, 2nd Ed."), I calculate $\small 0.0785 [\text C]$, which is way too big and must be wrong. </p>
<p>Here is my calculation:</p>
<p>$V_\text{sphere}=k\times Q_1/r_1 $ (potential of conducting sphere with radius $r_1$ and and net charge $Q_1$)
$V_\text{earth}=k\times Q_2/r_2$ (potential of conducting sphere with radius $r_2$ and and net charge $Q_2$)</p>
<p>where $k$ is a constant. If the sphere touches the earth then their potentials ($V_\text{sphere}$ and $V_\text{earth}$) must be equal,assuming that the charges on the spherical conductors may be considered as uniformly disturbed. Setting $V_\text{Sphere}=V_\text{earth}$, we get:</p>
<p>$Q_1/r_1=Q_2/r_2$</p>
<p>Setting $\quad Q_1+Q_2=Q_\text{total}$,</p>
<p>yields:$\quad Q_1=Q_\text{total}\times r_1/(r_1+r_2)$</p>
<p>Substituting </p>
<p>$Q_\text{total}=500,000[\text C],\quad r_1=1 [\text m],\quad r_2=6371000 [\text m]$ </p>
<p>I get:</p>
<p>$Q_1=0.0785[\text C]$. </p>
<p>I feel this number is way too large to be correct. If you take coupling into account (by modeling earth as PEC plate, the charge I calculate only gets larger!). What am I doing wrong here? There seems to be no way you accumulate $\small-0.0785 [\text C]$ of charge by touching the earth. </p>
| 4,213 |
<p>I'm wondering about the exact reason why anyons escape the spin-statistic theorem (SST), see <em>e.g.</em> <a href="http://en.wikipedia.org/wiki/Spin%E2%80%93statistics_theorem">http://en.wikipedia.org/wiki/Spin–statistics_theorem</a>. </p>
<p>I've read somewhere (the wikipedia page is sufficient I believe to understand this point, to be honest I don't remember where I read this) that the reason is just that anyons do not belong to the relativistic sector. Despite Lorentz invariance is still the cornerstone of the SST, anyons can have fractional exchange statistic because they belong to the Galilean invariance. See also <a href="http://physics.stackexchange.com/questions/23338/">this SE question</a> about the SST and its demonstration.</p>
<p>On the other hand, it seems to me that this argument might well be wrong. Indeed, I've the feeling that the exact reason is that the permutation group is not enough to understand the exchange of two particles. If one discusses the rotation group instead of the permutation group, then one ends up with the Leinaas and Mirheim [please see the bottom of the question for the reference] argument, saying that in 2D, the homotopy group of SO(2) is $\mathbb{Z}$, no more $\mathbb{Z}_{2}$ as for SO($n$) with $n\geq 3$.</p>
<p><strong>I'm wondering about the quantum field community perspective ?</strong> Do they continue to discuss the Lorentz invariance as the key point to obtain SST, or do they generally accept the Leinaas and Mirheim point of view [see below].</p>
<p>The two arguments might be almost equivalent, except that I believe the permutation group does not care about the space dimension. Moreover, I think it has no one-dimensional-non-trivial-projective representation, as required for anyons to exist, am I correct ? </p>
<p>NB: There are some alternative proof of the SST linked from the Wikipedia page about it. I gave the link at the beginning of this question. </p>
<ul>
<li>Leinaas, J. M., & Myrheim, J. <em>On the theory of identical particles.</em> Il Nuovo Cimento B, 37(1), 1–23 (1977). <a href="http://dx.doi.org/10.1007/BF02727953">http://dx.doi.org/10.1007/BF02727953</a> </li>
</ul>
| 4,214 |
<p>As I understand it, when we say that the $SU(2)_{L} \times U(1)_{Y}$ is broken via the Higgs mechanism, this is because the symmetry acts on the Higgs mass in a way that would change it's value. If we want to pick a particular model we need to pick a fixed value of the Higgs mass, and this is only possible if we say $SU(2)_{L} \times U(1)_{Y}$ is broken to $U(1)_{em}$. The Lagrangian is always invariant to $SU(2)_{L} \times U(1)_{Y}$ (even if the Higgs mass changes under $SU(2)_{L} \times U(1)_{Y}$ the Lagrangian is such that it remains invariant). The symmetry breaking is just neccesary in choosing a theory with one particular value of the Higgs vev. </p>
<p>Real life does appear to have a fixed Higgs mass so we require that $SU(2)_{L} \times U(1)_{Y}$ breaks to $U(1)_{em}$. But then aren't we saying that observable physics is described by $SU(3)_{C} \times U(1)_{em}$, why does the weak interaction still work in our universe that contains a fixed Higgs mass?</p>
| 4,215 |
<p>If you happen to have the Third Edition of Classical Electrodynamics by John David Jackson, turn to section 11.8, as that's where I'm getting all this from. If not, you should still be able to follow along.</p>
<p>In said section, Jackson gives us this equation that relates any physical vector <strong>G</strong> in a rotating vs. non-rotating reference frame:</p>
<p>$\left(\frac{d\mathbf{G}}{dt}\right)_{nonrot} = \left(\frac{d\mathbf{G}}{dt}\right)_{rest frame} + \boldsymbol{\omega}_T \times \mathbf{G}$</p>
<p>where</p>
<p>$\boldsymbol{\omega}_T = \frac{\gamma^2}{\gamma+1}\frac{\mathbf{a}\times\mathbf{v}}{c^2}$</p>
<p>"where <strong>a</strong> is the acceleration in the laboratory frame," according to the textbook. Also, gamma is defined using <strong>v</strong>, the velocity of the particle as measured in the lab frame.</p>
<p>Ok. So I decided to check this by setting <strong>G</strong> = <strong>x</strong>, the position vector, for a particle that is undergoing circular motion in the laboratory frame. So we have</p>
<p>$\left(\frac{d\mathbf{x}}{dt}\right)_{nonrot} = \mathbf{v}$</p>
<p>and</p>
<p>$\left(\frac{d\mathbf{x}}{dt}\right)_{rest frame} = 0$ because the particle doesn't have any velocity in its own frame (right?).</p>
<p>So far so good (I think). Now, this implies that $\boldsymbol{\omega}_T \times \mathbf{x} = \mathbf{v}$. So if we can verify this using the definition of $\boldsymbol{\omega}_T$, we're golden. However, if you use the fact that $|a| = \frac{v^2}{|x|}$ for circular motion as well as the fact that <strong>a</strong> is perpendicular to <strong>v</strong>, and that <strong>a</strong> is (anti)parallel to <strong>x</strong>, and carefully apply the right hand rule, you'll find, after the algebraic dust settles, that</p>
<p>$\boldsymbol{\omega}_T \times \mathbf{G} = (1-\gamma)\mathbf{v}$</p>
<p>So this is definitely a contradiction. Because it implies that $\mathbf{v} = (1-\gamma)\mathbf{v}$. Can anyone tell me where this went horribly horribly wrong? I worked on this with my professor for two hours yesterday and we couldn't figure it out.</p>
| 4,216 |
<p>Let's say I have a free jet of air leaving a pipe into the atmosphere. I know that the static gauge pressure at the pipe exit is equal to the atmospheric. But what about the static gauge pressure 10 meters away if the air is still traveling as a free jet? Is it still atmospheric?</p>
| 4,217 |
<p>Why is gravity such a weak force? </p>
<p>It becomes strong for particles only at the Planck scale, around $10^{19}$ $\text{GeV}$, much above the electroweak scale ($100$ $\text{GeV}$, the energy scale dominating physics at low energies). </p>
<p>Why are these scales so different from each other? What prevents quantities at the electroweak scale, such as the Higgs boson mass, from getting quantum corrections on the order of the Planck scale? </p>
<p>Is the solution super symmetry, extra dimensions, or just anthropic fine-tuning?</p>
<p>Can we relate few problems of quantum mechanics with gravity ?</p>
<p>Despite the fact that there is no experimental evidence that conflicts with the predictions of general relativity, physicists have found compelling reasons to suspect that general relativity may be only a good approximation to a more fundamental theory of gravity. The central issue is reconciling general relativity with the demands of quantum mechanics. Well tested by experiment, quantum mechanics is the theory that describes the microscopic behavior of particles. In the quantum world, particles are also waves, the results of measurements are probabilistic in nature, and an uncertainty principle forbids knowing certain pairs of measurable quantities, such as position and momentum, to arbitrary precision. The Standard Model is the unified picture of the strong, weak, and electromagnetic forces within the framework of quantum mechanics. Nonetheless, theoretical physicists have found it to be extremely difficult to construct a theory of quantum gravity that incorporates both general relativity and quantum mechanics.</p>
<p>At the atomic scale, gravity is some $40$ orders of magnitude weaker than the other forces in nature. In both general relativity and Newtonian gravity, the strength of gravity grows at shorter and shorter distances, while quantum effects prevent the other forces from similarly increasing in strength. At a distance of approximately $10^{-35}$ $\text{m}$, called the Planck length, gravity becomes as strong as the other forces. At the Planck length, gravity is so strong and spacetime is so highly distorted that our common notions of space and time lose meaning. Quantum fluctuations at this length scale produce energies so large that microscopic black holes would pop into and out of existence. A theory of quantum gravity is needed to provide a description of nature at the Planck length. Yet, attempts by researchers to construct such a theory, analogous to the Standard Model of particle physics, have lead to serious inconsistencies.</p>
| 4,218 |
<blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="http://physics.stackexchange.com/questions/10670/what-nonlinear-deformations-will-a-fast-rotating-planet-exhibit">What nonlinear deformations will a fast rotating planet exhibit?</a> </p>
</blockquote>
<p>This is really a basic physics question that I wanted to check myself on because I'm not entirely sure of the correct answer. I asked a puzzler question on the last episode of <a href="http://podcast.sjrdesign.net/shownotes_008.php" rel="nofollow">my podcast about hollow Earth claims</a>. I asked people to imagine a sphere with a solid shell that was filled with a compressible material, and this was spun very rapidly. (1) Would a hollow form, and (2) what shape would that hollow take on if it did form?</p>
<p>My thinking is that, yes, a hollow would form, but it would not be spherical as the hollow Earthers claim. It would be a biaxial ellipsoid with the short axis oriented along the poles and the long axes to the equator. This is because the centrifugal forces at the equator due to the spin would be <strong>much</strong> greater than at the poles, so there would be no "outward force" at the poles.</p>
<p>Is this correct? If so, is that a good explanation for a lay audience? If not correct, then what is the solution (it's been a long time since I took classical mechanics)? Thanks!</p>
| 129 |
<p>I am studying scattering theory right now in my QM class, and I'm attempting the Griffiths problem 11.4 as an exercise (it's not for homework). The problem is: Consider the case of low-energy scattering from a spherical delta-function shell: $V(r) = \alpha \delta(r-a)$. Calculate the scattering amplitude $f(\theta)$, the differential cross-section, and the total cross-section. </p>
<p>He outlines the following method in the section: in the exterior region where $V(r) = 0$, you get an expression for the wavefunction in terms of some complicated Hankel function expression
$$
\psi(r,\theta) = A \sum i^l (2l+1)[j_l(kr) + ika_lh_l^{(1)}(kr) ] P_l(cos \theta).
$$
You then match boundary conditions with the explicit solution to the wavefunction inside the region where $V(r) \neq 0$. </p>
<p>I'm confused how to carry this out. It seems that the solution of the wavefunction inside the region $r<a$ is your typical plane wave $f(r) = \sin(kr)/r$. Then I'm thinking that you should match $f(a) = \psi(a)$ and $\psi'(a)-f'(a) = - \frac{2m\alpha}{\hbar}^2 f(a)$. However, this doesn't appear to give me the right answer.. </p>
<p>So is this the correct approach (if not, how do you go about these kinds of problems)? </p>
| 4,219 |
<p>Why galaxies move away from each other in general on the astronomical scale?
Which answer is correct of the following?</p>
<p>It is because of the big bang theory, everything is just moving away from everything else on the astronomical scale, because of the explosion that happened in the past, just by inertia.</p>
<p>It is because space is being created all the time between galaxies, that is why they look receding from each other. Like drawing dots on a rubber sheet, then start stretching that rubber sheet. Dots are not moving, just more and more space is created between them.</p>
<p>Which answer is the right one? and if the 2nd answer is the right one, is it only space that is being created or spacetime?</p>
| 4,220 |
<p>In Steven Weinberg's Lecture on Quantum Mechanics (p. 342), he writes:</p>
<blockquote>
<p>The correlation between the spins of the two particles can be
expressed as the average value of the product of the $\hat{a}$
component of the spin of particle 1 and the $\hat{b}$ component of the
spin of particle 2:</p>
<p>$\langle (\textbf{s1} \cdot \hat{a}) (\textbf{s2} \cdot \hat{b})
\rangle = -\frac{\hbar^2}{4} \int{d\lambda\ \rho(\lambda) S(\hat{a},
\lambda) S(\hat{b}, \lambda)}$</p>
<p>where $\hat{a}$ and $\hat{b}$ are any two unit vectors.</p>
<p>In quantum mechanics, we have:</p>
<p>$\langle (\textbf{s1} \cdot \hat{a}) (\textbf{s2} \cdot \hat{b})
\rangle = -\frac{\hbar^2}{4} \hat{a} \cdot \hat{b}$</p>
<p>There is no obstacle to constructing a function $S$ and a probability
density $\rho$ for which are equal for any single pair of directions
$\hat{a}$ and $\hat{b}$.</p>
</blockquote>
<p>Can someone explain to me precisely how this is done and what $S(\hat{a}, \lambda)$ and $\rho(\lambda)$, along with the hidden variables $\lambda$ would be?</p>
<p>My guess is that since we want something that is rotationally invariant since it depends only on the dot product, we want some $S$ and hidden variables $\lambda$ that act nicely under rotations.</p>
<p>On the other hand, since the first equation is a local hidden variable theory, I thought this was not possible by Bell's inequalities...</p>
| 4,221 |
<p>In equation of <a href="http://en.wikipedia.org/wiki/Gibbs_free_energy#Graphical_interpretation" rel="nofollow">Gibbs free energy</a> change $(\Delta G) = (\Delta H)-T(\Delta S)$, is $T$ the system's or surrounding's temperature? </p>
<p>Edit:Oh sorry I am not clear earlier now I get clear question.I know we have to calculate gibbs free energy of system but criterion for spontaneity of ΔStotal should be greater than zero.When we relate it to gibbs free energy to show that gibbs free energy change should be always negative we keep both system and surrounding temperature same with pressure also constant.Then how process can occur?Wikipedia says that it is chemical potential that undergo changes there but what about gibbs free energy change.</p>
<p>I think I made too hurry to ask question when I am studying but I am running out of time so I ask here.</p>
<p><strong>NEW EDIT TO QUESTION</strong> :I found that above equation can only be applied to open systems but the quoted statement at top of <strong><em>Free energy of reactions</em></strong> section taken from
<a href="http://en.wikipedia.org/wiki/Gibbs_free_energy/" rel="nofollow">Wikipedia Article</a></p>
<blockquote>
<p>To derive the Gibbs free energy equation for an <strong><em>isolated system</em></strong>, let Stot be the total entropy of the isolated system, that is, a system that cannot exchange heat or mass with its surroundings. According to the second law of thermodynamics:</p>
</blockquote>
<p>the word isolated system means that the formula $\Delta G=\Delta H-T\Delta S$ is derived for isolated system.But I think exactly in opposite way that it is true for open systems.Read Above question before this New EDIT to understand my question.Please Help me.</p>
| 4,222 |
<p>I am a graduate student in pure mathematics, during my study on <a href="http://en.wikipedia.org/wiki/Ricci_flow" rel="nofollow">Ricci Flow</a> I faced some functional known as <a href="http://en.wikipedia.org/wiki/Energy_functional" rel="nofollow">energy functional</a>. For example Einstein-Hilbert functional is called an energy functional, also in Perelman's works $\mathcal{F}(g,f)=\int_M(R+|\nabla f|^2)e^{-f}d\mu$ is introduced as an energy functuional, where $M$ is a closed manifold, $g$ is Riemannian metric, $R$ is Ricci scalar, and $f$ is any function that in the physics literature is called <a href="http://en.wikipedia.org/wiki/Dilaton" rel="nofollow">dilaton</a>. </p>
<p>I do not know why these functionals are attributed to the energy concept and why does $f$ show dilaton concept? </p>
<p>Can anyone help me?</p>
| 4,223 |
<p>The Source of Water is Pond where the water level do change randomly. A pipeline of 200mm pipe diameter run for around 2500 m. The Height difference between the level of the inlet pipe and the level of the exit pipe is around 9m. In other word: Water need to run by gravity for 2500 m. Difference in height is 9m. </p>
<p>My question: How I could calculate the water volume/flow received at the exit pipe as function of the height of the water above the inlet pipe. </p>
<p>Ex: If the water in the Pond is covering the inlet pipe by say 10 cm, what is the expected flow at the exit pipe. </p>
| 4,224 |
<p>Lets say I enter a closed room with the walls and everything in it (including me and my eyes) at thermal equilibrium. Its a very hot room, but my super-eyes still work at 5000 degrees Kelvin.</p>
<p>I have two propositions which are in conflict.</p>
<p>The first is that I should be effectively blind when I am in the room. For me to see anything at all, the pigments in my retina would have to absorb some of the thermal radiation and convert it into electrical impulses for my brain. (The same applies for anything which detects radiation). But we can't extract energy from the heat without a cooler object, and there is none, so we can't use the energy in the black-body radiation for vision. There is no available free heat energy; everything is at thermal equilibrium. What is happening is that the pigments in my retina are generating black body radiation at the same rate as they are absorbing black body radiation from the object being looked at, so no net chemical change can occur and hence no vision.</p>
<p>However, I also know that objects have different emissivity, which means some objects in the room will be brighter than others. If I am looking at an object of lower emissivity than the pigments in my eyes, my eyes should radiate more power out than they absorb from the object, and vice versa for an object of higher emissivity. This means that if I look at different things, the pigments in my eyes will react differently to different objects, meaning vision is possible. But this contradicts the first proposition.</p>
<p>Can somebody explain this "paradox" ?</p>
| 4,225 |
<p>Lets say we have a tunelling problem in the picture, where $W_p$ is a finite potential step: </p>
<p><img src="http://i.stack.imgur.com/aFgQ4.png" alt="enter image description here"></p>
<p>If particle is comming from the left a general solutions to the Schrödinger equations for sepparate intervals I, II and II are: </p>
<p>\begin{align}
\text{I:}& & \psi_1 &= \overbrace{A e^{i\mathcal L x}}^{\psi_{in}} + \overbrace{Be^{-i \mathcal L x}}^{\psi_{re}}& \mathcal L &= \sqrt{\tfrac{2mW}{\hbar^2}}\\
\text{II:}& & \psi_2 &= C e^{\mathcal K x} + De^{-\mathcal K x}& \mathcal K &= \sqrt{-\tfrac{2m(W-W_p)}{\hbar^2}}\\
\text{III:}& & \psi_3 &= \underbrace{E e^{i \mathcal L x}}_{\psi_{tr}}& &\\
\end{align}</p>
<p>Where $\psi_{in}$ is an incomming wave, $\psi_{re}$ is a reflected wave and $\psi_{tr}$ is transmitted wave. I used the boundary conditions and got a system of 4 equations: </p>
<p>\begin{align}
{\tiny\text{boundary}}&{\tiny\text{conditions at x=0:}} & {\tiny\text{boundary conditions}}&{\tiny\text{at x=d:}}\\
A + B &= C + D & Ce^{\mathcal K d} + De^{-\mathcal K d} &= E e^{i \mathcal L d}\\
i \mathcal L A - i \mathcal L B &= \mathcal KC - \mathcal K D & \mathcal K C e^{\mathcal K d} - \mathcal K D e^{-\mathcal K d}&= i \mathcal L E e^{i \mathcal L d}
\end{align}</p>
<p>So now i decided to calculate coefficient of transmission $T$: </p>
<p>\begin{align}
T &= \dfrac{|j_{tr}|}{|j_{in}|} \!=\! \Bigg|\dfrac{\dfrac{\hbar }{2mi}\! \left( \dfrac{d\overline{\psi}_{tr}}{dx}\, \psi_{tr} - \dfrac{d \psi_{tr}}{dx}\, \overline{\psi}_{tr} \right)}{\dfrac{\hbar}{2mi} \!\left( \dfrac{d\overline{\psi}_{in}}{dx}\, \psi_{in} - \dfrac{d\psi_{in}}{dx}\, \overline{\psi}_{in} \right) }\Bigg| \!=\! \Bigg|\dfrac{\frac{d}{dx}\big(\overbrace{Ee^{-i\mathcal L x}}^{\text{konjug.}}\big) Ee^{i\mathcal L x} - \frac{d}{dx} \left( Ee^{i\mathcal L x}\right)\! \overbrace{Ee^{-i\mathcal L x}}^{\text{konjug.}}}{ \frac{d}{dx}\big(\underbrace{Ae^{-i\mathcal L x}}_{\text{konjug.}}\big) Ae^{i\mathcal L x} - \frac{d}{dx} \left( Ae^{i\mathcal L x}\right)\! \underbrace{Ae^{-i\mathcal L x}}_{\text{konjug.}}}\Bigg|\! = \nonumber\\
&=\Bigg|\dfrac{-i\mathcal L Ee^{-i\mathcal L x} E e^{i \mathcal L x} - i\mathcal L E e^{i \mathcal L x} Ee^{-i \mathcal L x}}{-i \mathcal L A e^{-i\mathcal L x} Ae^{i \mathcal L x} - i \mathcal L A e^{i \mathcal L x}Ae^{-i \mathcal L x} }\Bigg|=\Bigg|\dfrac{-i\mathcal L E^2 - i\mathcal L E^2}{-i \mathcal L A^2 - i \mathcal L A^2}\Bigg|=\Bigg|\dfrac{-2 i \mathcal L E^2}{-2i\mathcal L A^2}\Bigg| = \frac{|E|^2}{|A|^2}
\end{align}</p>
<p>It accured to me that if out of 4 system equations i can get amplitude ratio $E/A$, i can calculate $T$ quite easy. Could anyone show me how do i get this ratio?</p>
| 4,226 |
<p><a href="http://www.youtube.com/watch?v=UJJuq_pcyIQ" rel="nofollow">http://www.youtube.com/watch?v=UJJuq_pcyIQ</a></p>
<p>What exactly is going on in the video example? I understand the phenomena occurs because of magnetism but I am trying to figure out the mechanics behind this sculpture. There obviously is a magnet underneath but what is it doing? Is it moving? Is it getting some type of charge? Any insight is great!</p>
| 4,227 |
<p>Lets say we have a particle of mass $m_1$ which has a kinetic energy $W_{k1}$. This particle collides with another same particle. How can i calculate mass $m_2$ and the speed $v_2$ of the particle which is formed out of the colliding two particles? </p>
<p>I know that relativistic preservation of energy $W$ and momentum $p$ must hold and i know these equations: </p>
<p>\begin{align}
W^2 &= p^2c^2 + {W_0}^2\\
W &= W_k + W_0\\
W &= mc^2 \gamma(v)\\
p &= mv\gamma(v)\\
\end{align}</p>
<hr>
<p>My attempt was something like this: </p>
<p>\begin{align}
p_2 &= p_1\\
p_2 c&= p_1 c\\
{p_2}^2 c^2&= {p_1}^2 c^2\\
{W_1}^2-{W_{01}}^2 &= {W_2}^2-{W_{02}}^2 \longleftarrow \smash{\substack{\text{Doesnt it hold that $W_1 = W_2$?}\\ \text{Can i cross them out to get}\\\text{the line below?}}}\\
{W_{01}}^2 &= {W_{02}}^2\\
\left(2m_1c^2\right)^2 &= \left(m_2 c^2\right)^2\\
4{m_1}^2c^4 &= {m_2}^2 c^4\\
m_2 &= 2m_1
\end{align}</p>
<p>Is this even correct? My professor said it is not... Why?</p>
| 4,228 |
<p>In 2009, Rowan University released a paper claiming to replicate Blacklight Power's results on energy generation using hydrino states of the hydrogen atom. <a href="http://www.blacklightpower.com/pdf/BLPIndependentReport.pdf" rel="nofollow">The paper</a> appears to describe the procedure in every detail as far as my untrained eye can tell.</p>
<p>Has anyone else attempted to replicate these results and did they succeed?</p>
| 60 |
<p>So when you're looking at B-H curves for ferromagnetic substances, you often see these magnetic hysteresis curves, which occur, I gather, largely because of domain formation which has some reversible and some irreversible components:</p>
<p><img src="http://i.stack.imgur.com/DNjCd.png" alt="A magnetic hysteresis curve."></p>
<p>I've been reading through many papers, web sites and <a href="http://rads.stackoverflow.com/amzn/click/0120932709">this book (Hysteresis in Magnetism)</a>, but I haven't seen any equations for how to generate these curves. I recognize that there may be no easy way to express the entire curve in a single equation, but clearly people are generating these plots, and I think they're doing it from some equations, presumably based on some characteristic parameters like the saturation remanance, the coercive force, etc. Is there anyone out there who can help me with this?</p>
<p>Personally, my first choice would be an equation that actually describes the system given some set of material conditions (subject to some constraints is fine, too) but if that's quite complicated and for the most part, people just stitch together two <a href="http://en.wikipedia.org/wiki/Sigmoid_function">sigmoid functions</a> until it looks about right, then I'm fine with that answer as well, so long as there's some justification for why this is done in there somewhere.<sup>1</sup></p>
<p><sup>1</sup><sub>Full disclosure: I am in thesis deadline mode and I'm currently probably a 5 on the <a href="http://www.stanford.edu/~dement/sss.html">Stanford Sleepiness Scale</a>, so I apologize if some of the words in this post don't make sense - I'll try and edit it if people find that I haven't adequately conveyed my question.</sub></p>
| 4,229 |
<p>I want to calculate the work done by friction if the length $L$ of uniform rope on the table slides off. There is friction between the cord and the table with coefficient of kinetic friction $\mu_k$.</p>
<p>$$ W = \int F \cdot d \vec{s}$$
I think it would be:
$$ W_{fr} = \frac M L g \int_{0}^{L} dx$$</p>
<p>But the solutions (which could be mistaken) say:
$$ dW_{fr} = \mu_k \frac M L g \, x \, dx$$
which is then integrated.</p>
<p>Should there be an $x$ in the integral? I don't think there should be because you are summing up over an infinitesimal displacement $dx$ and the force of friction is not proportional to the displacement at any instant (I think).</p>
<p><img src="http://i.stack.imgur.com/lKzuO.png" alt="enter image description here"></p>
| 4,230 |
<p>For an experiment I will hopefully be soon conducting at Johns Hopkins I need two different lenses.</p>
<p>The first needs to allow all wavelengths above 500 nm to pass (thus a high pass filter) and cut off everything else.<br>
The second needs to allow all wavelengths below 370 nm to pass (thus a low pass filter) and cut off everything else.</p>
<p>My knowledge of optics is middling. I know that good old glass cuts of UV light, but I was hoping for something more specific. Does anyone know of the theory necessary to "tune" materials to make such filters?</p>
<p>Truth be told, I'm an experimentalist, so simply giving me a retail source that has such lenses would get me to where I need to go! But learning the theory would be nice as well.</p>
<p>Thanks,
Sam</p>
| 4,231 |
<p>You build a <a href="http://en.wikipedia.org/wiki/Dyson_sphere">Dyson sphere</a> around a star to capture all its energy. The outer surface of the Dyson sphere still radiates heat at much higher temperature than the cold space background, so you're easy to detect.</p>
<p>But you'd like to stay hidden. So you cool the outer surface of the Dyson sphere to near cold space background. Of course you still need to radiate your excess heat somewhere, so you plan to radiate it off in directed beams, away from the directions of the nearby solar systems, to stay hidden from your neighbors at least.</p>
<p><strong>Questions</strong>:</p>
<ol>
<li>Is such directed radiation of excess heat allowed by known laws of physics?</li>
<li>Would the energy of the star be sufficient for running the cooling system?</li>
</ol>
| 4,232 |
<p>An ideal spring is attached to a wall, and the other end is attached to a mass $m$. The spring is initially compressed a distance $x$. After it is released, the mass collides with another mass $2m$ at a distance $x/2$ to the right of the spring equilibrium. The collision is inelastic and they slide together. How far will they slide before coming to a momentary stop?</p>
<p><strong>My work</strong></p>
<p>Since the collision is inelastic then mechanical energy is not conserved. And since the two-mass system has an external force, momentum is not conserved. So one has to use only mechanical principles to solve this.</p>
<p>We can compute the velocity of the mass at the moment of impact using simple harmonic motion, $r(t) = -x\cos\left( t\sqrt{k/m}\right)$ so that $x/2 = -x\cos \left(t^{*}\sqrt{k/m}\right)$. With this I could do an awkward solution for time and maybe continue from there.</p>
<p>At this point, though, I'm not perfectly clear on how to move forward. Do I just assume that momentum is conserved for a very, very short time period during the collision and use that to find the velocity of the combined mass system, then put that into another simple harmonic motion problem?</p>
| 4,233 |
<p>Suppose I have an atom at rest which is at energy level $E_i$. Would it be possible to raise it to the next higher level $E_{i+1}$ by shooting a photon of energy $E_{i+1}-E_i$ at it?</p>
<p>I ask because the photon will impart some momentum to the atom, so the atom will end up with some kinetic energy after the absorption, and clearly this kinetic energy must come out of the photon's energy budget. So if the energy levels $E_i$ and $E_{i+1}$ do not already account for this kinetic energy [and I don't see how they could], I would need to use a photon of energy slightly higher than $E_{i+1}-E_i$. </p>
<p>Conversely, if the atom is at rest at energy level $E_{i+1}$, and it emits a photon to fall back to energy level $E_i$, I assume that the emitted photon will have energy slightly lower than $E_{i+1}-E_i$, to account for the kinetic energy of the atom after the emission. Is that correct?</p>
<p>I was led to this question while thinking about the energy balance in laser cooling.</p>
| 130 |
<p>The <a href="https://en.wikipedia.org/wiki/Bekenstein_bound">Bekenstein bound</a> sets the maximum amount of information that can be contained in a region of space/energy, and is usually referred to in the same way as computer storage density:</p>
<blockquote>
<p>For example, a single hydrogen atom, if it were to code as much information as permitted by the Bekenstein Bound, would code about 4×10<sup>6</sup> bits of information, since the hydrogen atom is about one Ångström in radius, and has a mass of about 1.67×10<sup>−27</sup> kilograms. (<a href="http://arxiv.org/pdf/astro-ph/0111520.pdf">source</a>)</p>
</blockquote>
<p> </p>
<blockquote>
<p>Nature permits a surprising amount of information to be encoded before the Bekenstein bound is reached. For example, a hydrogen atom can encode about 1 Mb of information — most of a floppy disk. (<a href="http://books.google.com/books?id=Kp6g79LuKWEC&lpg=PA54&ots=axR1DK3S8l&dq=Bekenstein%20bound%20hydrogen%20atom&pg=PA54#v=onepage&q=Bekenstein%20bound%20hydrogen%20atom&f=false">source</a>)</p>
</blockquote>
<p> </p>
<blockquote>
<p>The "Bekenstein bound" leaves room for a million bits in a hydrogen atom (<a href="http://books.google.com/books?id=fduW6KHhWtQC&lpg=PA166&ots=SvrByr246k&dq=Bekenstein%20bound%20hydrogen%20atom&pg=PA166#v=onepage&q=Bekenstein%20bound%20hydrogen%20atom&f=false">source</a>)</p>
</blockquote>
<p>But what does this really mean? How could <em>any</em> information be stored in a hydrogen atom?</p>
| 4,234 |
<p>For an outside observer the time seems to stop at the event horizon. My intuition suggests, that if it stops there, then it must go backwards inside.
Is this the case?</p>
<p>This question is a followup for the comment I made for this question: <a href="http://physics.stackexchange.com/questions/23118/are-we-inside-a-black-hole">Are we inside a black hole?</a></p>
<blockquote>
<p>Food for thought: if time stops at the event horizon (for an outside observer), for inside, my intuition suggests, time should go backwards. So for matter, that's already inside when the black hole forms, it won't fall towards a singularity but would fall outwards towards the event horizon due to this time reversal. So inside there would be an outward gravitational force. It would be fascinating if it turns out that all this cosmological redshift, and expansion we observe, is just the effect of an enormous event horizon outside pulling the stuff outwards.</p>
</blockquote>
<p>So from outside: we see nothing fall in, and see nothing come out.</p>
<p>And from inside: we see nothing fall out, and see nothing come in.</p>
<p>Hopefully the answers make this clear, and I learn a bit more about the GR. :)</p>
| 4,235 |
<p>Some sources say that when a photon hits the PV cell, it breaks apart electron-hole pairs. Other sources say that photons <em>create</em> electron-hole pairs. Can anyone explain which one is right? I've read several explanations of what goes on in the solar cell, but they don't seem very clear. To me, who has little prior knowledge, a few aspects of the many different explanations seem to contradict each other. I don't know if it's because some of the explanations said things a certain way for simplicity's sake, but this part seems too significant to ignore.</p>
<p>For example,
<a href="http://www.solarenergyexperts.co.uk/buyersguides/photovoltaic-glass-how-does-it-work/" rel="nofollow">http://www.solarenergyexperts.co.uk/buyersguides/photovoltaic-glass-how-does-it-work/</a>: "When photons (light particles) from the sun hit the cell, the energy breaks up the paired particles. The freed electrons go into the n-type layer, while the holes go down into the p-type layer."</p>
<p>science.howstuffworks.com/environmental/energy/solar-cell3.htm : "When light, in the form of photons, hits our solar cell, its energy breaks apart electron-hole pairs."</p>
<p>solarjourneyusa.com/bandgaps.php has a hole section called "<em>Generation</em> of electron-hole pairs"</p>
<p>This one basically seems to be talking about the same thing, but doesn't really mention electron-hole pairs, and the explanation seems more confusing to me, but anyway, <a href="http://www2.pv.unsw.edu.au/nsite-files/pdfs/UNSW_Understanding_the_p-n_Junction.pdf" rel="nofollow">http://www2.pv.unsw.edu.au/nsite-files/pdfs/UNSW_Understanding_the_p-n_Junction.pdf</a>:
"...broken bonds created by the light act as holes ... and these holes are also free to move throughout the material. Electrons and holes created in this way are physically near each other: for every electron excited by the light there is a corresponding hole generated. These electrons and holes can remain excited only for a short period of time. In a process called recombination, excited electrons stray too close to holes and the two fall back into bonded positions."
This one seems to be saying that photons break apart the bonds between the electron and its atom, which creates a free electron and a hole, i.e. another one saying that an electron-hole pair is created.</p>
<p>solarcellcentral.com/junction_page.html : "When photons hit the solar cell, free electrons attempt to unite with holes in the p-type layer." </p>
<p>www.solarenergy.net/Articles/how-photovoltaic-cells-work.aspx : "When enough photons are absorbed by the negative layer of the photovoltaic cell, electrons are freed from the negative semiconductor material."</p>
<p>So what actually happens when light hits a solar cell?</p>
| 4,236 |
<p><strong>Question:</strong></p>
<blockquote>
<p>consider a long, straight wire of cross-sectional area $A$ carrying a
current $i$. Let there be $n$ free electrons per unit volume. An observer
places himself on a trolley moving in the direction opposite to the
current with a speed $v = \frac{i}{nAe}$ and separated from the wire by a
distance $r$. The magnetic field seen by the observer is very nearly</p>
</blockquote>
<p><strong>My Answer:</strong></p>
<blockquote>
<p>Zero. Because current is $neAv$ where $v$ is drift velocity of electrons. Relative velocity between him and electrons is zero. So, no flow of charge through any cross-section according to him. So no current. So no magnetic field.</p>
</blockquote>
<p><strong>Actual answer:</strong></p>
<blockquote>
<p>$\frac{\mu\ i}{2\cdot\pi\cdot r}$ where $\mu$ is the permeabilty of free space.</p>
</blockquote>
| 4,237 |
<p>I'm doing an experiment where I bring a probe very close to a well full of fluid and then very slowly lower it to obtain some force deformation values. The material behaves very much like a fluid and when I bring the probe close to the fluid, it sort of jumps up and grabs the probe and forms like a "column?" that actually pulls on the probe rather than pushing on it. </p>
<p>Anyway, I'm trying to coat the probe with an oil or something that will prevent this, but I don't know how to look into it because I don't know what its called or how its described in the literature. Any ideas on what this phenomena is called?</p>
| 4,238 |
<p>The last decades there has been a lot of research going on in the the area of three dimensional gravity. The motivation, I understand, is threefold:</p>
<ol>
<li><p>Whereas gravity is not perturbatively renormalizable in four spacetime dimensions, in three dimensions it is. To make it even more interesting it has black hole solutions and it is exactly solvable. This opens the way to to study quantum black holes. This make three dimensional gravity a very interesting system on itself.</p></li>
<li><p>Through the AdS/CFT correspondence there is a connection between conformal field theories (CFT) in two dimensions and gravity in three dimensions. CFT's are important in condensed matter physics and one can use 3D gravity to learn more about them.</p></li>
<li><p>Gravity in three dimensions is simpler to deal with then gravity in four dimensions. Therefor it can be used as a toy model for gravity in four dimensions. </p></li>
</ol>
<p>I am wondering what are the most important insights that 3d gravity brought in these respects? In particular I am interested in point three: did 3d gravity provide any new view on 4d gravity so far?</p>
| 4,239 |
<p>I'm trying to get motivated in learning the Atiyah-Singer index theorem. In most places I read about it, e.g. wikipedia, it is mentioned that the theorem is important in theoretical physics. So my question is, what are some examples of these applications?</p>
| 4,240 |
<p>Why is the critical point for the phase diagram of pure water degrees of freedom equal to 0? Maybe, you know what is the mathematical explanation for the fact that the number of degrees of freedom at the critical point is 0? What else is affected by the lack of degrees of freedom in addition to the Gibbs phase rule? </p>
| 4,241 |
<p>Consider two objects presented in the figure below. Objects have equal masses and are separated by a distance of 60 light seconds. </p>
<p>Assume that we move left object by 3 light seconds to the left in 30 seconds. This requires energy input, lets say it's equal to X. Change in potential energy is also equal to X.</p>
<p>Now, the right object does not 'know' yet that the left object was moved. Left object was moved 30 seconds ago and this information requires 60 seconds to reach the right object. </p>
<p>We have 2 options:</p>
<p><strong>A:</strong> We can move the right object by 3 light seconds to the right with the speed of 0.1c now. </p>
<p><strong>B:</strong> We can move it with the same speed and by the same distance later - let's say after 60 seconds.</p>
<p>Clearly case A requires the same energy input as when moving left object = X. This is because right object still 'thinks' that the left mass in the same place.</p>
<p>In the case B the right object 'knows' that the left mass was move so we will need less energy to move it. Energy needed < X.</p>
<p>The end state in A and B is the same. Why it is possible to achieve it with 2 different energy inputs? If we choose method A the increase in potential energy is smaller than energy input X. Where is the missing energy?</p>
<p>Maybe the 'speed of gravity' is infinite?</p>
<p><img src="http://i.stack.imgur.com/dgCfA.png" alt="2 objects"></p>
| 4,242 |
<p>There have been a few <a href="http://www.crunchgear.com/2010/12/10/navys-mach-7-railgun-fires-a-round-at-33000-megajoules/">articles about the Navy's new Mach 7 33 Megajoule railgun.</a> As a physics teacher, I have a couple of questions about this, and was hoping for some help. </p>
<ol>
<li><p>Is the kinetic energy of the projectile 33 MJ, or is that the energy that was delivered from the capacitor bank that was used to fire the projectile, so that some of it would have been lost in resistive heating, etc. </p></li>
<li><p>The photos of the projectile show a cloud of hot gas behind the projectile? What is this? superheated air? Plasma from the electrical breakdown inside the railgun? What causes this? </p></li>
</ol>
| 4,243 |
<p>One of the most disturbing aspects of General Relativity is the 'Marble versus Wood' duality of the theory: Matter creates curvature, and curvature doesn't create curvature (at least not directly)</p>
<p>The holographic principle suggests a reason for this division: matter fields and gravity are paired to structures that exist in different dimensions: so 4D gravity (if taken over a finite region $\Omega$ of spacetime) should be paired with some (abstract or actual) 3D (2 spatial+1 time?) boundary $\partial \Omega$ where a QFT of some sort exists, but the curvature in this 4D spacetime is dictated by 4D distributed matter fields, which themselves might be related to some 5D gravity (possibly actual, but maybe unobservable) that has the 4D spacetime as part of its boundary. </p>
<p>Arguably one could extend this idea to lower and higher dimensional mappings. Besides the pragmatic but inherently hard question about what could be gained by such insight, I'm asking if</p>
<blockquote>
<p>Has anyone tried to formulate physical theory in terms of a hierarchy
of such mappings?</p>
</blockquote>
<p>It seems that such formulation would have to deal somehow with the fact that the boundary operator is nilpotent ($\partial \partial \Omega = 0$) when defining the hierarchy</p>
<p>So, if the set $HM(D)$ gives all the possible meaningful holographic mappings between compact manifolds of dimension $D$ and their boundaries, then one can imagine that a full description of physics could be achieved by experimentally finding the elements $hm_0, hm_1...hm_X$ with $hm_i \in HM(i)$ and $X$ some maximum dimension above which all different mappings become completely unobservable at 4D</p>
| 4,244 |
<p>A wire placed in a magnetic field and has current flowing, it will experience the Lorentz force, work is done by the electric field and energy is conserved. Being converted from the power source to KE, and the wire moves.</p>
<p>How can we apply Poynting's theorem to state that this system's energy will always be conserved?</p>
<p>To fully observe the conservation of energy from the power source is equal to the work done by the wire when it gains KE.</p>
| 4,245 |
<p>I found the following explanation for <a href="http://en.wikipedia.org/wiki/Chirality_%28physics%29">chirality</a> for spin 1/2 particles <a href="http://www.quantumdiaries.org/2011/06/19/helicity-chirality-mass-and-the-higgs/">here</a></p>
<blockquote>
<p><img src="http://i.stack.imgur.com/mUD0C.png" alt="enter image description here"></p>
<p>What happens when you rotate a left- vs right-chiral fermion 360
degree about its direction of motion. Both particles pick up a -1, but
the left-chiral fermion goes one way around the complex plane, while
the right-chiral fermion goes the other way. The circle on the right
represents the complex phase of the particle's quantum state; as we
rotate a particle, the value of the phase moves along the circle.
Rotating the particle 360 degrees only brings you halfway around the
circle in a direction that depends on the chirality of the fermion.</p>
</blockquote>
<p>My question is: Isn't the way the phase of the wavefunction changes related to the direction of rotation of the particle, which is an external parameter depending on how you rotate it? Is there a better explanation for chirality?</p>
| 4,246 |
<p>I was reading <a href="http://www.lightreading.in/document.asp?doc_id=226833&site=lrindia&" rel="nofollow">this article</a>. There is a statement "It is a well-known fact that the telecom towers mounted with antennas in the lower frequency bands can cover far greater areas than those using the 1800 frequency bands." Is this accurate/wrong/a simplified version. One point which made me think that it is wrong is the fact that Satellite TV uses high frequency bands for communication</p>
| 4,247 |
<p>I've recently been reading (at a basic level) about the double slit experiment and how the mere act of observing can cause the "wave function to collapse", as they say. I find this mind-blowingly fascinating, and it naturally sparks some questions.</p>
<p>In particular, consider the following scenario: I perform the double slit experiment and detect which slit was traversed, sending the data to some computer. This would cause the wave function to collapse, as I understand. Now, imagine that I delete the data as it is being read by the computer. If I am not mistaken, the wave function would not collapse in this case, since there would be no way for me to determine which slit was traversed. <em>Now,</em> imagine that I am under the <em>belief</em> that the data is being deleted when in reality it is being saved to some disk. What would happen in this case? Assuming I never come to know about the existence of the saved data, would the wave function collapse?</p>
| 4,248 |
<p>Is photon interaction , electrostatic interaction outside the nucleus and gravitational interaction is all due to electromagnetic waves ? and CAN be identified as with the de Broglie waves ?</p>
<p>I thought of a theory in which is assuming that photon interaction , electrostatic interaction outside the nucleus and gravitational interaction is all due to electromagnetic waves and can be identified as with the de Broglie waves in order to explain how the force of gravitation act between particles . So can this assumption stand a chance of being theoretically and experimentally correct .
Please add your comments regarding this in your answer and please explain why or why not .</p>
| 4,249 |
<p>Let $\mathbf{E}(r,t),\mathbf{B}(r,t)$ be two vector fields (in $\mathbb{R}^3$), s.t. they satisfy fot $t=0$ the equations:</p>
<ol>
<li><p>$\nabla \cdot \mathbf{B}(r,0)=0.$ </p></li>
<li><p>$\nabla \cdot \mathbf{E}(r,0)=\frac{\rho(r,0)}{\epsilon_0}.$</p></li>
</ol>
<p>The question now is:</p>
<p>What properties have $\mathbf{E}$ and $\mathbf{B}$ got in order to satisfy equation 1 and 2 for all $t>0$?</p>
<p>I think $\mathbf{B}$ must be independent of $t$, i.e. $\mathbf{B}(r,t)=\mathbf{B}(r,0)$ for all $t>0$. But what about $\mathbf{E}$?</p>
<p>Doesn't the answer depend on the function $\rho(r,t)$?</p>
| 4,250 |
<p>Conventional wires become very lossy at high frequencies. </p>
<p><strong>Is this because of the skin effect?</strong></p>
<p>Plasmon based computer chips, since plasmons can support much higher frequencies (~100 THz range). </p>
<p>Aren't free electron oscillations in a metal plasmons. What is the difference and why can they support such high optical frequencies?</p>
| 4,251 |
<p>Two equal masses of mass M are glued to a massless hoop of radius R is free to rotate about its center in a vertical plane. The angle between the masses is 2$\theta$. Find the frequency of oscillations.</p>
<p>My attempt at a solution is ($\phi$ is the displaced angle):
$$x_{m1}=-R\sin (\theta -\phi)$$
$$y_{m1}=-R\cos(\theta-\phi)$$
$$x_{m2}=R\sin(\theta+\phi)$$
$$y_{m2}=-R\cos(\theta +\phi)$$</p>
<p>I got my lagrangian and then found the Euler Lagrange equations in terms of $\phi$. I found that
$$\ddot {\phi}=\frac{-g\cos(\theta)}{R}\sin\phi$$
So for small angles, $\sin(\phi)\approx\phi$ which means $\omega^2=\frac{g\cos(\theta)}{R}$</p>
<p>Can anyone let me know if this makes sense. I think I did it correctly, but I am not completely sure. Thanks!</p>
<p>EDIT:: Lagrangian:
$$L=T-U=mR^2\dot{\phi}^2+2mg\cos(\theta)\cos(\phi)$$</p>
| 4,252 |
<p>Everybody has been taught at one point, "oh the universe expands, but that doesn't mean that everything is expanding uniformly, since that means we can't detect the expansion, but only that huge galaxies are moving away from each other".</p>
<p>But I'm rather confused. Can't the expanding universe simply be thought as a coordinate axis that expands? The definition of "1" on the coordinate axis constantly expands? After all, spacetime itself is expanding, and not only "average distance of galaxies".</p>
<p>So our Planck length would expand, light wavelengths expand, we expand, etc. But clearly this isn't happening. Of course, the handwavy explanation is "electromagnetism/gravity overcomes expansion at small scales", but the electromagnetic and gravity forces are defined in terms of constants relative to our "coordinate intervals", so if spacetime itself is expanding "under the feet" of gravity, shouldn't gravity not be able to do anything about it?</p>
<p>We would also be unable to measure the expansion of spacetime since all of our measurement benchmarks and tools are also expanding.</p>
<p>Clearly, this isn't happening. What is the <em>precise</em> reason?</p>
| 56 |
<p>This is related to this question:<a href="http://physics.stackexchange.com/questions/88826/does-the-weight-of-an-hourglass-change-when-sands-are-falling-inside">Does the weight of an hourglass change when sands are falling inside?</a></p>
<p>At Brigham Young University, there is a display consisting of a sealed off liter bottle with a sunken sealed off hourglass. When one turns over the bottle, the hourglass floats as the sands flow down. Eventually, the hourglass reaches some transition, and precipitately changes from floating to sinking.</p>
<p>It has a sign next to it saying that it's for a fishing lure,and that you should not reveal the answer if you get it.</p>
<p>What causes it to change from floating to sinking so quickly? </p>
| 4,253 |
<p>The <a href="http://en.wikipedia.org/wiki/Uncertainty_principle" rel="nofollow">uncertainty principle</a> states that,</p>
<p>$$\sigma _{{x}}\sigma _{{p}}\geq {\frac {\hbar }{2}}.$$</p>
<p>It is mentioned from many sources that the probability distribution of the particle position and momentum would follow a Gaussian distribution.</p>
<p>Why is it a Gaussian distribution? is this the distribution that minimizes uncertainty? Is this distribution definitely the case for the uncertainty principle or can it be different under different conditions? Has this been proven?</p>
<p>What are the formulas for position and momentum probability distributions of a free particle? How is this derived from the wave function? What would be the formulas of the probability distributions for the position and momentum for a system of 2 identical bosons separated by a distance $R$?</p>
| 4,254 |
<p>I read somewhere that the tidal forces between the earth and the moon causes the equality between the 2 periods of the moon and that every planet-satellite system will evolve to this condition (like pluto-charon of you consider pluto a planet). Can somebody give me some references (books, articles) discussing this idea and why doesn't this happen in a star-planet system?</p>
| 4,255 |
<p>The global conformal group in 2D is $SL(2,\mathbb{C})$. It consists of the fractional linear transforms that map the Riemann sphere into itself bijectively and is finite dimensional. </p>
<p>However, when studying $CFT_2$ people always use the full Virasoro algebra, not just the $L_{0,\pm1}$ which actually exponentiate to invertible transformations. I would like to know why people consider the other $L_n$'s to be symmetries of the theory. </p>
<p>I am aware that Ward identities are local statements, and that I can consider a coordinate patch where the additional conformal transformations are bijective in order to derive relations between correlation functions in this patch. I am also familiar with the representations of the Virasoro algebra and how constraining the symmetry is.</p>
<p>However, we are doing quantum mechanics, and a symmetry of the theory should take me from one physical state to another. In addition, the symmetry should have an inverse which undoes this transformation. This means that the physical Hilbert space should organize itself into representations of the symmetries of the theory. However, the local conformal transformations cannot have inverses, and so they do not form a group as far as I know. So why is it assumed that the states of a $CFT_2$ should organize themselves into representations of the Virasoro algebra? (I am aware $L_{n \leq -2}|0\rangle\neq 0$, $\langle 0|L_{n\geq 2} \neq 0$ so all but the $L_{0,\pm 1}$ are "spontaneously broken" on the in/out vacuum, but this is not relevant since the states of the theory are still assumed to assemble into Verma modules since it is assumed that the Virasoro was a symmetry of the theory which is just violated by the vacuum).</p>
<p>My question basically boils down to: How can I have symmetries of a theory which are not invertible? I'd appreciate any comment's that clear up my confusion.</p>
| 4,256 |
<p>I understand that a general interpretation of the $1/r^2$ interactions is that virtual particles are exchanged, and to conserve their flux through spheres of different radii, one must assume the inverse-square law. This fundamentally relies on the 3D nature of space.</p>
<p>General relativity does not suppose that zero-mass particles exchanged. What is the interpretation, in GR, of the $1/r^2$ law for gravity? Is it come sort of flux that is conserved as well? Is it a postulate?</p>
<p>Note that I am not really interested in a complete derivation (I don't know GR enough). A physical interpretation would be better.</p>
<p>Related question: <a href="http://physics.stackexchange.com/q/89/24774">Is Newton's Law of Gravity consistent with General Relativity?</a></p>
| 4,257 |
<p>This in the context of the AdS/CFT correspondence. I am reading <a href="http://arxiv.org/abs/hep--th/9905111.pdf" rel="nofollow">this</a> review on AdS/CFT Aharony et. al. (The MAGOO review) The abstract can be found <a href="http://arxiv.org/abs/hep-th/9905111" rel="nofollow">here</a></p>
<p>Equation (2.50) of the above paper lists the $R$-symmetry groups of extended supersymmetries in $AdS_{p+2}$. He uses the following notation for the group: </p>
<blockquote>
<p>$AdS_5: SU(2,2 | {\cal N}/2),~{\cal N}=2,4,6,8$</p>
</blockquote>
<p>Can anyone explain this notation to me?</p>
| 4,258 |
<p>Is it possible to quick freeze half of an unopened bottle of water or soda by putting half of the bottle in a subfreezing solution liquid nitrogen or other solution without freezing the whole bottle?</p>
| 4,259 |
<p>A hollow cylinder, with no top or bottom, of radius $R$ and length $L$ is uniformly charged with density $\sigma>0$. I have to find the point on space where a point charge $q>0$ has to be drawn from $\infty$ with speed $v\rightarrow0$ so that the work done is maximum, so I have to find the point with maximum potential.
What I did was find the electric field, using Gauss's law, and then set $V(\infty)=0$:</p>
<p>$0<\rho < R: E=0 \\R<\rho < \infty: E=\frac{\sigma R}\rho$</p>
<p>so $V(\rho)=-\int\limits_\infty^\rho{E.dl}$ .</p>
<p>For $R<\rho<\infty$ I get $V(\rho)=\sigma R \ln(\infty/\rho) $, so here I don't know what to do.
Maybe I shouldn't have used Gauss's law, and calculate $E$ the 'classic' way, integrating over the cylinder, but the integral looks so complicated. This problem was given on an exam and trust me, if the integral is actually that complicated, I wasn't supposed to do it that way.
So, what should I do?</p>
| 4,260 |
<p>This is about the same paper as this thread: <a href="http://physics.stackexchange.com/questions/11645/some-questions-about-chapter-i-1-by-minahan-of-the-review-of-ads-cft-integrab">Some questions about chapter I.1 (by Minahan) of the "Review of AdS/CFT Integrability"</a> but it was never answered. </p>
<p>I have some different questions about it and I'll separate them into a couple posts if need be. I'd also be grateful if anyone can recommend other introductions or reviews for understanding N=4 generally and the Minahan review <a href="http://arxiv.org/pdf/1012.3983v2" rel="nofollow">http://arxiv.org/pdf/1012.3983v2</a> in particular. Some of the algebra/group theory was particularly hard for me to follow (highest-weight reps, Cartan subalgebras...).</p>
<p>Some questions I'm particularly intrigued/troubled by are:</p>
<ol>
<li><p>After (3.1) he says an operator $O(x)$ having dimension $\Delta$ means that when $x\rightarrow \lambda x$, then "$O(x)$ scales as $O(x) \rightarrow \lambda^{-\Delta} O(\lambda x) $." Should this be $O(x) \rightarrow \lambda^{-\Delta} O(x) $? If we say that $O(x)$ is some polynomial of degree $n$ in $x$, then after the rescaling $O(x)$ will be a polynomial of degree $n$ in $\lambda x$. So we'd have $O(x) \rightarrow O(\lambda x) \sim \lambda^n O(x)$. Then if we identify $-\Delta = n$ we have $O(x) \rightarrow O(\lambda x) \sim \lambda^{-\Delta} O(x)$. Am I missing something?</p></li>
<li><p>How does he get eq. (3.2)? It apparently follows from $D$ being the generator of scalings, by which he says he means that $O(x) \rightarrow \lambda^{-iD} O(x) \lambda^{iD}$. I'm confused by this, too, as I expect to see the generator exponentiated by $e$, not $\lambda$. I'd expect something like $e^{-i\lambda D} O(X) e^{i\lambda D}$, with $D$ as the generator and $\lambda$ as the parameter. </p></li>
<li><p>Later, in eq. (3.9), he introduces the $R_{IJ}$ as the $SO(6)$ R-symmetry generators, as well as some matrices $\sigma^{IJ}$ that he only addresses later. Here $I, J = 1...6$. I don't understand the notation. Why are there two indices on these guys? And if it's an $SO(6) \sim SU(4)$ symmetry group, then there should only be 15 generators. So are some of these $R$ and $\sigma$ redundant? Because naively it would appear that we have $6\times 6=36$ of each. I suspect that I'm missing something about how to understand these indices. </p></li>
<li><p>Kind of the same thing as 3. In (3.14) he gives some of those $\sigma^{IJ}$ and states that they are the generators in the fundamental $SU(4)$ representation. Why? Where did these come from?</p></li>
</ol>
<p>I'll stop for now and post any other questions I have in another thread so as not to go overboard.</p>
| 4,261 |
<p>Lets say we have a potential step with regions 1 with zero potential $W_p\!=\!0$ (this is a free particle) and region 2 with potential $W_p$. Wave functions in this case are: </p>
<p>\begin{align}
\psi_1&=Ae^{i\mathcal L x} + B e^{-i\mathcal L x} & \mathcal L &\equiv \sqrt{\frac{2mW}{\hbar^2}}\\
\psi_2&=De^{-\mathcal K x} & \mathcal K &\equiv \sqrt{\frac{2m(W_p-W)}{\hbar^2}}
\end{align}</p>
<p>Where $A$ is an amplitude of an incomming wave, $B$ is an amplitude of an reflected wave and $D$ is an amplitude of an transmitted wave. I have sucessfuly derived a relations between amplitudes in potential step: </p>
<p>\begin{align}
\frac{A}{D} &= \frac{i\mathcal L-\mathcal K}{2i\mathcal L} & \frac{A}{B}&=-\frac{i \mathcal L - \mathcal K}{i \mathcal L + \mathcal K}
\end{align}</p>
<p>I know that if i want to calculate transmittivity coefficient $T$ or reflexifity coefficient $R$ i will have to use these two relations that i know from wave physics:</p>
<p>\begin{align}
T &= \frac{j_{trans.}}{j_{incom.}} & R &= \frac{j_{trans.}}{j_{incom.}}
\end{align}</p>
<hr>
<p><strong>Question 1:</strong> I know that $j = \frac{dm}{dt} = \frac{d}{dt}\rho V \propto \rho v \propto \rho k$ But what is a density $\rho$ equal to?</p>
<p><strong>Question 2:</strong> I noticed that $\mathcal L$ and $\mathcal K$ are somehow (i dont know how) connected to the wavevector $k$ from the equation in 1st question but how? How can i make it obvious?</p>
| 4,262 |
<p>I would like to find the equation of motion for the scalar field $\phi$ by varying the following action in General Relativity. </p>
<p>Special Relativity:
$$
S = -\tfrac{1}{2}\int d^4\xi\, \eta^{ab} \partial_a \phi\partial_b\phi
$$</p>
<p>General Relativity:
$$
S = -\tfrac{1}{2}\int d^4x \sqrt{g}\, g^{\mu\nu} \partial_{\mu} \phi\partial_\nu \phi
$$</p>
<p>I was able to get the correct equation using the covariant derivatives. Since they are constant with respect to the metric partial integration works and one obtains $ \Box\phi = 0$. But since for the scalar field the covariant and the partial derivates are the same I wanted to vary the action with the partial derivatives. </p>
<p>Note: $\phi \rightarrow \phi + \delta\phi$
$$
\begin{align}
\delta S &= -\tfrac{1}{2}\int d^4x \sqrt{g}\, (g^{\mu\nu} \partial_{\mu} \delta\phi\partial_\nu\phi + g^{\mu\nu} \partial_{\mu} \phi\partial_{\nu} \delta\phi ) \\
%
&=-\int d^4x \sqrt{g}\, g^{\mu\nu} \partial_{\mu} \delta\phi\partial_\nu\delta\phi \\
%
&=\int d^4x \, \partial_\nu(\sqrt{g}\,g^{\mu\nu}\partial_\mu\phi)\delta\phi
\end{align}
$$
The third line one obtains after partial integration. But at this point I'm stuck.
I know the following definition:
$$
\Box\phi = \tfrac{1}{\sqrt{g}}\partial_\mu(\sqrt{g}g^{\mu\nu}\partial_\nu\phi)
$$
But I'm not able to convert my obtained solution into this.
Thanks for your help!</p>
| 4,263 |
<p>What would happen if you were to accelerate one end of a material, say a steel rod for instance, at faster per second than the speed of sound in that material? </p>
<p>For example, if the speed of sound in steel is 6100m/s, what would happen if you pushed the end of that piece of steel at more than 6100m/s/s?</p>
<p>This only occurred to me because I heard that the speed of sound is the speed an impulse travels through a material, and I was wondering if there could be some sort of limit to acceleration which would arise from this. </p>
<p>I appologize in advance if I'm comparing apples with irrelevant oranges. Thanks!</p>
| 4,264 |
<p>I heard that this statement is correct. However, it seems odd to me. The number of possible microstates is still the same, so isn't the entropy constant?</p>
| 4,265 |
<p>For a spin 1/2 particle you have two spin states, either up or down which are orthogonal. But what about a spin 1 particle which has 3 spin states, either up, down, not up/not down?</p>
| 4,266 |
<p>Red colour bares 430-480THz and Violet bares 670-750THz according to Wikipedia.</p>
<p>What frequencies are the colours between red and violet? Will they be higher than red? or lower than violet?</p>
<p>What physically properties do opposite colours carry? Cyan is the opposite colour of red; yellow is the opposite colour of blue; green is the opposite colour of magenta. Can it be expressed mathematically?</p>
| 4,267 |
<p>Basically, What is the difference <a href="http://en.wikipedia.org/wiki/Diamagnetism" rel="nofollow">diamagnetism</a> and <a href="http://en.wikipedia.org/wiki/Superconductivity" rel="nofollow">superconductivity</a>?</p>
<p>As far as I understand, diamagnetism comes from the fact, that all electrons in a solid, when exerted by an external magnetic field, create so-called Eddy Currents which oppose the external magnetic field, right ?</p>
<p>And in a superconductor, the supercurrents, are actually are "normal" current in the material, which opposes the magnetic field.</p>
<p>But is that the whole story, or am I simplifying it too much ?</p>
| 4,268 |
<p>I recently studied that NASA has planned to tow and place it in the orbit of the moon.
My doubt is when asteroid is placed in the orbit near moon.since the gravitational field of earth is very high.what will it revolve around the moon or the earth.
Can anyone clarify my doubt ??</p>
| 4,269 |
<p>I am trying to prove the following equivalent form for long-range-order in superconductivity (Annett's book states something like this) : </p>
<p>\begin{equation}\lim_{|R-R'|\to \infty}<\psi^{\dagger}(R')\psi(R)> = <\psi^{\dagger}(R)\psi(R)> \end{equation}</p>
<p>where the brackets are evaluated on the BCS ground state : $|\psi_{BCS}>= ∏_{k}(u_k^{*}+v_k^* P_k^{\dagger})|0> $ ($|0> $ being the zero-temperature Fermi sea), $P_k^{\dagger}=c_{k\uparrow}^{\dagger}c_{-k\downarrow}^{\dagger}$ when : </p>
<p>$\psi^{\dagger}(R) = \int d^3r~\psi(r) \phi_{\uparrow}^{\dagger}
(R+r/2)\phi_{\downarrow}^{\dagger}(R-r/2)$</p>
<p>($\psi(r)$ being some wavefunction) and : </p>
<p>$\phi_{\sigma}(r) = \sum_{k}e^{ikr} c_{k\sigma}$ ($c$ being the commun fermion annihilator, and the $P$s commuting like bosons for this particular state).(Sorry for the lengthy definitions)</p>
<p>The gap function being defined as : $\Delta \propto \sum_k <P_k>$, we can see relatively easily that the right side of the first equation will be proportional to $|\Delta|^2$. But what I've found quite puzzling is what to do with the unhomogeneous terms (especially in the left term) where the $c$ and $c^{\dagger}$ do not form a pairing operator and the Fermi sea not being ideal to deal with them...</p>
<p>Could somebody help me on this calculation ? Or am I taking the problem on the wrong side ?</p>
| 4,270 |
<p>I am trying to solve this problem:</p>
<blockquote>
<p><em>Consider a rocket moving relative to an inertial frame $\mathcal{F}$ , such that its worldline is given by $$x^{\mu}=c^2/g(\sinh(g\tau/c),\cosh(g\tau/c)-1,0,0).$$ What are the components of four acceleration relative to the instantaneous rest frame of the rocket, $\mathcal{F}'$?</em></p>
</blockquote>
<p>I (think I) understand how to do this using Lorentz transformation: $dt/d\tau=(1/c) \cdot dx^0/d\tau=\cosh(g\tau/c)$. This is equal to $\gamma$ and it is then straightforward to compute 3-velocity and then use the Lorentz matrix to get $$dx'^\mu(\tau)/d\tau=(c,0,0,0) \quad
; \quad d^2 x'^\mu(\tau)/d\tau^2=(0,g,0,0)$$
However, when I first saw this I immediately thought that by definition the 4-velocity in the instantaneous rest frame of the rocket would be (c,0,0,0) because in $\mathcal{F}'$ the 3-velocity is zero and $\gamma$ would be 1 and I was wondering if this is a valid reasoning. Even if it is, <strong>why is the following wrong</strong>?</p>
<p>If $v'^\mu=dx'^\mu(\tau)/d\tau=(c,0,0,0)$, then $a'^\mu=dv'^\mu/d\tau=(0,0,0,0)$ since $c$ is a constant. But this contradicts the calculations given by the Lorentz transformation and I don't understand why and given this I have no idea of <strong>how to interpret 4-acceleration</strong>.</p>
| 4,271 |
<p>Apologies in advance for the long question.</p>
<p>My understanding is that in GR, massive observers move along timelike curves $x^\mu(\lambda)$, and if an observer moves from point $x^\mu(\lambda_a)$ to $x^\mu(\lambda_b)$, then his clock will measure that an amount of time $t_{ba}$ given by the curve's arc length;
$$
t_{ba} = \int_{\lambda_a}^{\lambda_b}d\lambda \sqrt{-g_{\mu\nu}(x(\lambda))\dot x^\mu(\lambda)\dot x^\nu(\lambda)}
$$
will have elapsed where $g_{\mu\nu}$ is a metric on spacetime with signature $(-,+,+,+)$. </p>
<p><strong>Why is this so?</strong> </p>
<hr>
<p>Here is how I would attempt to justify this fact in special relativity with $g_{\mu\nu} = \eta_{\mu\nu}$. Consider an inertial observer $O$ in $\mathbb R^{3,1}$, and suppose that this observer sees a clock, which I'll call observer $O'$ moving around on a curve $x^\mu(\lambda)$. If $O'$ were also an inertial observer, then given any event with coordinates $x^\mu$ as measured by $O$, observer $O'$ would measure the coordinates of the event to be $x'^\mu = \Lambda^\mu_{\phantom\mu\nu} x^\nu + x_0^\mu$ for some Lorentz transformation $\Lambda$. If $O'$ is not inertial, then this is no longer true, and there is some more complicated family of transformations, say $T_\lambda$ between events as seen by both observers.</p>
<p>I would argue, however, that if we were to partition the interval $[\lambda_a, \lambda_b]$ into a large number $N$ of intervals $I_1=[\lambda_a, \lambda_i], \dots, I_N=[\lambda_{N-1}, \lambda_b]$ with $\lambda_n = \lambda_a+n\epsilon_N$ and $\epsilon_N=(\lambda_b-\lambda_a)/N$, then on each interval $I_n$, $O'$ is approximately an inertial observer in the sense that
$$
T_{\lambda_n} = P_n + \mathcal O(\epsilon_N), \qquad (\star)
$$
for some Poincare transformation $P_n$. Then we would note that since $O'$ is stationary in his own reference frame, he measures his worldline to have the property $\dot x'^\mu(\lambda) = (\dot t(\lambda), \mathbf 0)$ so that
$$
I_{ba}=\int_{\lambda_a}^{\lambda_b}d\lambda \,\sqrt{-\eta_{\mu\nu}\dot x'^\mu\dot x'^\mu}
= \int_{\lambda_a}^{\lambda_b} d\lambda \, \sqrt{\dot t^2} = t(\lambda_b) - t(\lambda_a) = t_{ba}
$$
On the other hand the integral on the left can be written as a Riemann sum using the partition above, and we can invoke ($\star$) above to get
\begin{align}
I_{ba}
&= \lim_{N\to\infty}\left[\sum_{n=1}^N \epsilon_N\sqrt{-\eta_{\mu\nu}\dot x'^\mu(\lambda_n)\dot x'^\nu(\lambda_n)}\right] \notag\\
&= \lim_{N\to\infty}\left[\sum_{n=1}^N \epsilon_N\sqrt{-\eta_{\mu\nu}\dot x^\mu(\lambda_n)\dot x^\nu(\lambda_n)} + \mathcal{O}(\epsilon_N^2)\right] \notag\\
&= \int_{\lambda_a}^{\lambda_b}d\lambda \,\sqrt{-\eta_{\mu\nu}\dot x^\mu\dot x^\mu}
\end{align}
Combining these two computations gives the desired result.</p>
<p><strong>How do others feel about this argument?</strong></p>
<p>I'm not completely comfortable with it because of the assumption $(\star)$ I made on $T_\lambda$.</p>
<p>I imagine that in GR a similar argument could be made by invoking local flatness of the metric.</p>
| 4,272 |
<p>An object with a density of $x$ inside a liquid with a density greater than $x$ would float.
If we assume that both of these are positively charged and the object is in the middle of the liquid:</p>
<ol>
<li><p>Would the liquid touch the object?</p></li>
<li><p>Would the object still float up because of buoyancy?</p></li>
</ol>
| 4,273 |
<p>Usually what helps me understand a concept better in physics is to write a simulation of it. I've got to the point where I'm competent in the basics of special relativity, but, I can't figure out how to write a relativistic simulation! My normal approach in Newtonian mechanics is just to attach two objects together with a hookean spring force, maybe add a damping effect, and then just to use the calculated force to get acceleration, and numerically integrate over that.</p>
<p>The first issue is length contraction. The issue isn't as easy as a contraction of $1/\gamma$. If I have two point events simultaneous in some reference frame, corresponding to locations of masses connected by a spring, I do know their spacelike separation, which is an invariant. However, even though the proper length is invariant, if I change reference frames, the events are no longer simultaneous, so I can't really get a consistent definition from this... I figured that I could apply the spring force instantaneously, "faster than light", and show how when I do that that leads to violation of causality, but I can't even define force consistently!</p>
<p>I understand that if I use a wave equation or some sort of electromagnetic force, then I can have a force field that transforms correctly, but I really don't want to do this, because I'm not great at electromagnetism, and this is really just for me to better my understanding of special relativity. Plus it would be difficult computationally.</p>
<p>I haven't been able to make any headway on assuming that the masses are connected by a chain of springs, whose velocities relative to each other are $<<c$, but I think the solution may lie there.</p>
| 4,274 |
<p>In one of his lecture,Prof. Susskind mentioned that the event horizon "bulges" forward to meet any incoming radiation or matter; and it is a property of Einstein field equations. I have not come across any such property, and if it exists, shouldn't it belong to the Schwarzchild metric rather?
Please explain me that property in detail.</p>
| 4,275 |
<p>I notice a $2\pi$ term in the $\delta$-function when trying to construct an amplitude using the Feynman Rules. The $2\pi$ also appears as an integration measure to enforce normalisation in the phase space, what is the origin of this $2\pi$ term? Where does it come from? </p>
| 4,276 |
<p>I've heard two different descriptions of gravity, and I'm wondering how they work together.</p>
<p>The first is Gravitons:</p>
<blockquote>
<p>"The three other known forces of nature are mediated by elementary
particles: electromagnetism by the photon, the strong interaction by
the gluons, and the weak interaction by the W and Z bosons. The
hypothesis is that the gravitational interaction is likewise mediated
by an – as yet undiscovered – elementary particle, dubbed the
graviton. In the classical limit, the theory would reduce to general
relativity and conform to Newton's law of gravitation in the
weak-field limit." -- <a href="https://en.wikipedia.org/wiki/Graviton" rel="nofollow">Source</a></p>
</blockquote>
<p>I'll admit I don't know much about them, but I assume that they would work similarly how photons do in EM.</p>
<p>The second, which I understand more, is given by GR and is that space time is curves by mass-energy, sort of how putting a heavy object on a blanket curves it.</p>
<p>So, how would these work together? Would the curves space time be analogous to a "graviton field", where the more massive/energetic objects produce a stronger field which other objects are attracted to, and excitations in the field produce gravitons?</p>
| 131 |
<p>What sort of mode-locked laser systems allow for the production of single isolated pulse of light (as opposed to a train of more than one pulse), where the individual pulse duration is on order picoseconds ($\approx 10^{-12}$ seconds) to femtoseconds ($\approx 10^{-15}$ seconds)? Is there any reason I couldn't use a mode-locked Ti-sapphire laser to produce a single pulse of some desired / achievable duration in the femtosecond regime? Or is it generally challenging to drop the pulse repetition rate to a sufficiently small frequency to allow for shutters / etc. to filter out the rest of a pulse train?</p>
<p>My guess is that if we can get the pulse repetition rate down to the MHz regime, we can start to think about diverting or blocking additional pulses using voltage or mechanically switchable filters? What do people actually use?</p>
<p>I should also mention that I'm particularly interested in emission wavelengths in the range of something like $100 \space nm$ to $1.5 \space \mu m$ or so.</p>
<p>...</p>
<p>In terms of the different types of femtosecond lasers, there are fiber lasers, bulk lasers, dye lasers, semiconductor lasers, even free-electron lasers that one can use. Supposing I need something like a few hundred milliwatts to a watt of output power, and I'm interested in the visible to near-visible wavelength range, are there any families of femtosecond lasers particularly suited to the above task?</p>
| 4,277 |
<p>Malcolm Gladwell made a claim in a <a href="https://www.youtube.com/watch?feature=player_embedded&v=5L0GGfQblrc" rel="nofollow">recent talk</a> that a sling with a stone going at 30m/s has the same stopping power as a .45 calibre handgun.</p>
<p>How would I verify whether or not this claim is true - even given some assumptions, like the stone has a weight of 1lb or 2 lbs or whatever.</p>
<p>Not quite sure how to work this out.</p>
<p>Assume that the bullet is a 185grain 0.45 calibre bullet.</p>
| 4,278 |
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