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Jul
22
answered Does something like Pascal's Law apply in a Swimming Pool? If not, how does pressure spread?
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.
Jul
18
revised What happens to the magnetic field in this case?
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Jul
18
answered What happens to the magnetic field in this case?
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20
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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
4
revised Where's the energy in a boosted capacitor?
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Mar
2
answered Estimate number of hairs on human head
Mar
2
revised Practical method to weigh human limbs with common household items?
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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.
Mar
2
revised Practical method to weigh human limbs with common household items?
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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