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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?
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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.
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revised What happens to the magnetic field in this case?
added 34 characters in body
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answered What happens to the magnetic field in this case?
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