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

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?

• You can try using the divergence theorem: $\int_{M} d^4x\, \partial_\mu(\Gamma j^\mu)=\int_{\partial M} dS\,\Gamma n^\mu j_\mu$, where $M$ is all of space-time and $\partial M$ its boundary. What is your current on the boundary (at spatio-temporal infinity, if you like)? – Zachary Nov 21 '19 at 5:16
• That had crossed my mind. But I'm unsure what the boundary condition at infinity would even look like? – user502382 Nov 21 '19 at 5:18
• Physically, the current must vanish at infinity, so that the fields also go to zero. – Zachary Nov 21 '19 at 5:22
• As a side remark, it is suspect that to prove a local symmetry one has to integrate over all space and assume boundary conditions at infinity.. – my2cts Nov 21 '19 at 5:25
• By the fields, do you mean the electromagnetic field tensor $F_{\mu\nu}$? I am still not completely confident why this needs to be zero at infinity. – user502382 Nov 21 '19 at 5:43

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.