2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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)? $\endgroup$ – Zachary Nov 21 '19 at 5:16
  • $\begingroup$ That had crossed my mind. But I'm unsure what the boundary condition at infinity would even look like? $\endgroup$ – user502382 Nov 21 '19 at 5:18
  • $\begingroup$ Physically, the current must vanish at infinity, so that the fields also go to zero. $\endgroup$ – Zachary Nov 21 '19 at 5:22
  • 1
    $\begingroup$ 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.. $\endgroup$ – my2cts Nov 21 '19 at 5:25
  • $\begingroup$ 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. $\endgroup$ – user502382 Nov 21 '19 at 5:43
1
$\begingroup$
  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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.