Varying $A_\nu$ gives the <Maxwell equations>
$$
\partial_\mu F^{\mu\nu}=J^\nu,
\qquad
J^\nu=ie\bigl[(D^\nu\phi)^\dagger\phi-\phi^\dagger D^\nu\phi\bigr].
$$
On the vacuum manifold, choose unitary gauge so that $\phi=v$. Then $\mathbf J=-2e^2v^2\mathbf A$ and the <London equation> is
$$
\boldsymbol\nabla\times\mathbf J=-2e^2v^2\mathbf B.
$$
For a static configuration, $\boldsymbol\nabla\times\mathbf B=\mathbf J$ and $\boldsymbol\nabla\cdot\mathbf B=0$, so
$$
\nabla^2\mathbf B=2e^2v^2\mathbf B=\frac1{\xi^2}\mathbf B,
\qquad
\boxed{\xi=\frac1{\sqrt2ev}=\frac1{m_A}}.
$$
The magnetic field therefore decays exponentially inside the condensate. This is the <Meissner effect>, with penetration depth equal to the inverse gauge-boson mass.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 305 / 1 / ii / Solution

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The Meissner effect is the expulsion of magnetic field from the bulk of a superconductor. A static field decays from its surface over the London penetration depth.


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 Incoming links: Meissner effect