Perfect conductor in a time-dependent magnetic field I have read about how a perfect conductor responses to a time-dependent magnetic field. It was stated that the "equation of motion" of such a perfect conductor is given by
$$\frac{dj(r, t)}{dt}=ne^2E(r,t),$$
as well as this equation follows from $j(r,t)=nev(r,t)$, where $v$ is the drift velocity.
I don't understand why the first equation follows from the second one. Can somebody explain this to me?
 A: This is one of the London equations.
This comes from modeling the electrons as free electrons.
As such the motion of a single electron is
$\vec{F} = e\vec{E}$
$m\frac{d \vec{v}}{dt} = e\vec{E}$
$m\frac{d ne\vec{v}}{dt} = ne^2\vec{E}$
$m\frac{d \vec{J}}{dt} = ne^2\vec{E}$
$ \frac{d\vec{J}}{dt} = \frac{ne^2}{m}\vec{E}$
Behaviour of magnetic fields in perfect conductors
$ \frac{d \nabla × \vec{J}}{dt} = \frac{ne^2}{m}\nabla × \vec{E}$
$ \frac{d \nabla × \vec{J}}{dt} = - \frac{ne^2}{m} \frac{\partial \vec{B}}{\partial t}$
$\nabla × \vec{J} = -\frac{ne^2}{m}\vec{B}$
Finding $\nabla × \vec{J}$:
$\nabla × (\nabla × \vec{B} = \mu_0 \vec{J})$
$\nabla (\nabla \cdot \vec{B}) - \nabla^2 \vec{B} = \mu_0 \nabla × \vec{J}$
$- \nabla^2 \vec{B} = \mu_0 \nabla × \vec{J}$
Sub in:
$\nabla^2 \vec{B} = \frac{\mu_0 n e^2}{m} \vec{B}$
Whose solution is an exponential decay of the magnetic field. Which is why "there is no magnetic  field inside a superconductor". This makes sense, as any changing external magnetic field induces an electric field, which generates a current to oppose the flux
There are a few assumptions in this derivation however.
