Past exam of the optics and electrodynamics course of the University of Cambridge / 2015 / Paper 1 / 1 / b / Solution

To first order in the applied fields, the carrier acceleration gives $m\dot{\mathbf{v}}=q\mathbf{E}$. The current density $\mathbf{j}=nq\mathbf{v}$ consequently obeys the first London equationTaking its curl and using Faraday's law yieldsFor an initially field-free, current-free preparation, the integration constant is zero. Equivalently, impose the second London equation appropriate to the Meissner state, $\nabla \times \mathbf{j}=-{ϵ}_0{\omega }_p^2\mathbf{B}$. In the final static state, Ampère's law and $\nabla \cdot \mathbf{B}=0$ then giveFor a planar half-space $z>0$ with a tangential surface field, the bounded solution isThe plasma-frequency form of London penetration depth therefore shows that an applied field is exponentially small several penetration depths inside the material.

The magnetic-history constant in an undamped plasma explains the required qualification. Undamped carrier motion alone fixes a time-independent value of $\nabla \times \mathbf{j}+{ϵ}_0{\omega }_p^2\mathbf{B}$; it does not force that value to vanish. A perfect conductor can retain a pre-existing field. The additional London equation describes London screening independently of that arbitrary ideal-conductor history. The original London paper explicitly distinguishes its description from acceleration theory.^[ref]