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 Jun 7 comment Capturing (perturbatively) non-equilibrium field theory effects using “elementary” methods @Bubble As I understand, the perturbative prodecures in standard intro to qft textbooks only apply if we are interested in the fate of a system long after it is perturbed, i.e. the initial and final states are both free states. Here, op is interested in calculating the transient time evolution of a system(the $\psi$ field), where the perturbation(due to $\chi$) is never turned off. So in this sense the problem is a non-equilibrium one, am I right? Jul 19 comment What happens to the magnetic field in this case? @PeterShor yes, but there's no problem with it.So $E=Constant-U$ and $dE=-dU$, if we remove a segment $dE$ is positive so $dU$ is negative.Thus the potential energy decreases as it should,and there's no problem.It's all because of the (-) sign. But of course we've to include the field energy in the "meat" of the magnet to get the total energy. I was trying to say that the extractable potential energy is non-zero, so when the pieces blow apart they do not get kinetic out of nothing. Btw I made a mistake when I wrote "Any other terms in $E$ are infinitely large", that is only for point charges. Jul 19 comment What happens to the magnetic field in this case? So we cannot say that a configuration with zero total magnetic field everywhere has lower potential energy than configuration with non-zero magnetic field. And when we integrate $-\frac{μ_0}{2} \int (Σ_{i≠j} B_i B_j) dV$, we have to cover the region of magnetization(the "meat"). The magnetic fields in this region are very likely to change with different configurations and thus is the potential energy contribution. So we have two layers of shield that can protect the conservation of energy from being violated. Jul 19 comment What happens to the magnetic field in this case? @PeterShor the total magnetic field does not say anything about the potential energy of the system. The potential energy is not given by $E=\frac{\mu_0}{2}\int B_{tot}^2 dV$, but it is given by the cross terms $-\frac{μ_0}{2} \int (Σ_{i≠j} B_i B_j) dV$. Any other terms in $E$ are infinitely large and independent of configurations. When the magnetic field is zero, we only have $B_{tot}=Σ B_i$. But it does not mean that every single $B_i$ is zero, thus there is no reason to say that $-\frac{μ_0}{2} \int (Σ_{i≠j} B_i B_j) dV$ is zero. Mar 6 comment Where's the energy in a boosted capacitor? I'm sorry I still don't understand what were you trying to say in your second last comment, after re-reading it several times. Thanks for the paper, actually I already know about stress tensor. I'm just trying different approach here. Mar 4 comment Where's the energy in a boosted capacitor? @LarryHarson yes I think it's somewhat clearer that way. I've modified my answer to include that, thanks for the suggestion. I have rechecked it again, I still think that event 1 happens before event 2. The rule of thumb is "rear clock ahead." Mar 2 comment Practical method to weigh human limbs with common household items? @0xC0000022L Hey I made stupid mistakes in my previous answer, sorry for that. I've updated my answer with an "uglier" one. Feb 27 comment Whistle Physics Yes it is a kind of vortex shedding. Perhaps this is what you are looking for? www2.ibp.fraunhofer.de/akustik/ma/pipesound/animEdgeTone.mpeg Feb 27 comment Where's the energy in a boosted capacitor? I'm not sure where my argument is flawed. The same conclusion can be reached by calculating the momentum of the rod and using the fact that the rest mass of the rod is zero to calculate the energy of the rod in the boosted frame. Feb 26 comment Where's the energy in a boosted capacitor? There won't be any physical change happening on the capacitor during this changing process. The only thing is that the energy/mass that was hidden in rod 1 is simply transferred to rod 2. The purpose of changing rod is just to determine the amount of energy hidden inside the rod. The rod is part of the system and we need to include it in the energy calculation. Feb 26 comment Where's the energy in a boosted capacitor? I think it is not quite a problem, we can always recreate the starting condition. Suppose that initially there is already a rod between the plates, call this rod 1. Then at $t=0$, this rod suddenly disappears and rod 2 suddenly appears to replace it. From the point of view of the rest frame of capacitor, doing this won't change the energy of the system. And thus it must be so in any other frame. In the moving frame, rod 2 does not receive work from the left plate and the right plate simultaneously. Thus some energy is transferred into rod 2. And this energy must be taken into account. Feb 26 comment Where's the energy in a boosted capacitor? I was't saying that the capacitor is accelerated. I mean the plates are attracting each other through electrostatic forces, so if there is nothing that counters these forces(i.e. a rigid rod) the plates will crash into each other. Now if there is a rod between the plates everything looks the same as in the problem description, I know that we are just viewing it in a moving frame. But the rod plays a role here. In the frame where the capacitor appears to be moving, the rod does work on the capacitor. And the total work turned out to negative, as a result the rod's mass is increased. Feb 22 comment What happens to the magnetic field in this case? Yes all the field lines coming out from the north poles will end up in the south poles by squeezing through the small opening. However the magnetic flux itself is weakened as a whole when we put the pieces together. The outer field exists in order to preserve $\int \vec{B}.\vec{dl}=0$. By putting the pieces together we decrease the $\int \vec{B}.\vec{dl}$ inside the "meat" of the magnet, therefore the field outside the magnet is reduced. So even though the field lines get squeezed more, the number of field lines coming out of the north poles is also decreased. Feb 21 comment What happens to the magnetic field in this case? Also we don't need so much energy to insert the last piece of magnet, the field is already quite weak before the insertion. Can the downvoter please explain? Feb 21 comment What happens to the magnetic field in this case? I don't see why it will create a non-physical magnetic configuration. Can you specify what rule might be violated ? I think it's okay for the magnetic field outside to disappear entirely, it doesn't mean that the interaction energy will also go away with it. Higher potential energy(depends on negative of the cross term of magnetic energy density) doesn't always mean higher total field energy, because only part of the field energy is responsible for interactions(force). The final configuration will have a very high potential energy and tends to blow up, but there's no singularity involved here. Feb 18 comment Can't seem to reconcile geometric optics and wave optics @Thomas I've updated my answer to talk about this Sep 9 comment What are these rays that appear in photograph of sun? Maybe they are talking about different stuffs, they are many other phenomena similar to this. But I think in most daily life cases, using ordinary camera, this is the right explanation. I have done some experiments to prove it using my ipod. Sep 9 comment What are these rays that appear in photograph of sun? you can see similar effects using bare eyes even without squinting, and if you tilt your head, the rays will also rotate with it. You can only see this using bare eyes at night because usually this reflected light's intensity is much weaker than that of other sources of light. If you squint your eyes, different effect takes place, I think it is related to eye liquid's movement. Jul 15 comment Frequency of Vortex Shedding Thank you, things are clearer to me now. But I need to think about it for a while.. Jul 14 comment Glass - paper: Stevin's Law They should work together. The surface tension force itself is extremely weak, it loses against water's total weight even for small radius cases. You can see how little amount of water an open straw can hold (even if it is made of glass). The air pressure difference holds most of the water's weight, and the surface tension takes care of the rest which is very small. So in terms of supporting the water's weight, the surface tension is negligible. But it does an important job to prevents water leaking from sideways and it also prevents water to penetrate through little holes in the paper