Electromagnetism – Perfect Conductor in a Time-Dependent Magnetic Field

Question

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?

Answer 1

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.