London equations

Remarks

In a static situation, Ampère's law $\nabla \times \mathbf{B} = \mu_0 \mathbf{j}_s$ from Maxwell's equations and $\nabla \cdot \mathbf{B} = 0$ turn the second equation into $\nabla^2 \mathbf{B} = \mathbf{B} / \lambda_L^2$. Example: let the superconductor fill $x > 0$ and let $\mathbf{B} = B(x) \mathbf{e}_z$ with $B(0) = B_0$ and $B$ bounded. The equation becomes $B'' = B / \lambda_L^2$, whose bounded solution is $B(x) = B_0 e^{-x/\lambda_L}$. Then $\mathbf{j}_s = \mu_0^{-1} \nabla \times \mathbf{B} = (B_0 / (\mu_0 \lambda_L)) e^{-x/\lambda_L} \mathbf{e}_y$, and one checks $\nabla \times \mathbf{j}_s = -(n_s q^2/m) \mathbf{B}$. Non-example: a perfect conductor obeys the first equation but not the second. For it, $\partial \mathbf{B} / \partial t$ decays inside, so the field is frozen in, but a field present before cooling is not expelled. With the vector potential $\mathbf{A}$ in the London gauge, the two equations combine into $\mathbf{j}_s = -(n_s q^2 / m) \mathbf{A}$.