Proving modified Maxwell action is gauge invariant

[DaniV]

Homework Statement

Consider the quantum field modified Maxwell action in 2 + 1 dimensions:
$S=\int (-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\frac {\theta}{2}\epsilon^{\alpha\mu\nu}A_\alpha F_{\mu\nu})\,d^3x$
when $\theta$ is dimensionful constant and $\epsilon^{\alpha\mu\nu}$ Levi-Civita symbol
Show that this action is gauge invariant.

Relevant Equations

$F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu$
gauge transformation: $A^\mu \to A^\mu + \partial ^\mu \chi$ when $\chi$ is scalar function.

I want to show that the action staying the same action after taking $A^\mu \to A^\mu + \partial ^\mu \chi$, for the first term I suceeded in showing the invariance using the fact $[\partial ^ \mu , \partial ^\nu]=0$ but for the second term I'm getting: $\epsilon^{\alpha\mu\nu}A_\alpha F_{\mu\nu} \to \epsilon^{\alpha\mu\nu}A_\alpha F_{\mu\nu} +\epsilon^{\alpha\mu\nu}(\partial_\alpha \chi )F_{\mu\nu}$ so I don'`t understand how to show the last step: $\epsilon^{\alpha\mu\nu}(\partial_\alpha \chi) F_{\mu\nu}=0$

Thanks.

 

[TSny]

how to show the last step: $\epsilon^{\alpha\mu\nu}(\partial_\alpha \chi) F_{\mu\nu}=0$

This expression will not equal zero, in general.

In order for the action to be invariant under the transformation, it is not necessary that the integrand of the action integral be invariant. It is sufficient for the integrand to be invariant up to a divergence term. Using the divergence theorem, the integral of the divergence term will equal zero if the fields vanish at infinity sufficiently rapidly.

 

[DaniV]

This expression will not equal zero, in general.

In order for the action to be invariant under the transformation, it is not necessary that the integrand of the action integral be invariant. It is sufficient for the integrand to be invariant up to a divergence term. Using the divergence theorem, the integral of the divergence term will equal zero if the fields vanish at infinity sufficiently rapidly.

how to show that this expression is a total derivative so we can use divergence theorem, I tried to do integration by parts and I got:
$\int \epsilon^{\alpha \mu \nu}(\partial_\alpha \chi )F_{\mu\nu}\, d^3x = \oint \epsilon^{\alpha \mu \nu} \chi F_{\mu\nu} d\vec S$ $+\int \epsilon^{\alpha \mu \nu} \chi (\partial_\alpha F_{\mu\nu})\, d^3x$
the first term of integration is zero due to boundry condition but the second one is not total derivative because $\chi$ is outside the derivative..

 

[TSny]

Write out the second one in terms of the 4-vector potential $A$.