Under $A_\mu\mapsto A_\mu+\partial_\mu\alpha$, the field strength is unchanged, but the mass term in the <Proca action> changes by
$$
\frac{m^2}{2}\{2A^\mu\partial_\mu\alpha+(\partial\alpha)^2\},
$$
which is not generally a total derivative. The mass therefore breaks <gauge invariance>.

The <Euler-Lagrange field equation> is
$$
\partial_\mu F^{\mu\nu}+m^2A^\nu=0.
$$
Taking its divergence and using the antisymmetry of $F^{\mu\nu}$ gives $m^2\partial_\nu A^\nu=0$. Since $m>0$, the <Lorenz constraint in Proca theory> follows. Substitution back into the field equation then gives
$$
(\Box+m^2)A^\nu=0,
\qquad
\partial_\nu A^\nu=0.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 301 / 1 / a / Solution

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The Proca action adds $m^2A_\mu A^\mu/2$ to the Maxwell Lagrangian. It describes a massive spin-one field and explicitly breaks the Abelian gauge symmetry of the massless theory.


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 Incoming links: Proca action