Showing that the $A$-$j$ coupling in classical electromagnetism is gauge invariant

Question

I am attempting an early exercise from Altland's Condensed Matter Field Theory. The electromagnetic field's action is given as: $$S[A]=\int d^4x(c_1F_{\mu\nu}F^{\mu\nu}+c_2A_\mu j^\mu),$$ and I wish to show that the second term is invariant under a gauge transformation $A_\mu\to A_\mu + \partial_\mu \Gamma$. It is hinted that I should use integration by parts and the continuity equation $\partial_\mu j^\mu$. Doing so, I can show that: $$c_2\int d^4x(A_\mu+\partial_\mu\Gamma)j^\mu=c_2\int d^4x A_\mu j^\mu+c_2\int d^4x\partial_\mu(\Gamma j^\mu),$$ where I now must show that the second term on the right-hand side is zero to prove gauge invariance. I am not sure how to show this - how should I proceed? Is there some boundary condition that I'm missing?

Answer 1

1. Assuming the continuity equation $d_{\mu}J^{\mu}=0$, the gauge symmetry is more precisely a gauge quasi-symmetry, meaning that the action is only invariant up to boundary terms.

2. It seems relevant to stress that no boundary conditions are imposed in Noether's theorems. In contrast, boundary conditions are necessary for the principle of stationary action. This point is also made in my Phys.SE answer here.