In the classroom my teacher stated that the Gauge-fixing term in the action $$\frac{1}{2\alpha}\int d^4x (\partial_\mu A^\mu(x))^2$$ transforms under $A_\mu(x) \rightarrow A_\mu(x)+\partial_\mu \theta(x)$ as: $$\frac{1}{\alpha}\int d^4x(\partial_\mu A^\mu(x))(\partial_\nu \partial^\nu \theta(x))$$ when inserting the transformation in the first equation I get the additional term: $$\int d^4x (\partial_\mu\partial^\mu \theta(x))^2.$$ I was wondering why this term is null; any hint is appreciated.

EDIT: It was an infinitesimal transformation: with $\theta$ small higher order of $\theta$ were discarded.

  • 2
    $\begingroup$ Maybe it is an infinitesimal transformation, i.e. $\theta$ is assumed to be small, such that only the first-order term in $\theta$ is kept? $\endgroup$
    – Leo L.
    May 12, 2020 at 16:50
  • $\begingroup$ It was an infinitesimal transformation, I modified the question and accepted the only answer $\endgroup$
    – Ringo_00
    May 13, 2020 at 19:57

1 Answer 1


Can you write it as a total divergence? If so, then you can use the divergence theorem to argue that it goes to 0 at the surface, which you can take to infinity.


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