I have to show that the Maxwell action $$S=-\frac{1}{4}\int d^4x F^{\mu\nu}F_{\mu\nu}\,$$ is invariant under translation: $\delta_aA_\mu=a^\nu \partial_\nu A^\mu$ with $a^\mu$ as arbitrary and constant 4-vector.

I just pluged in the defintion of $F^{\mu\nu}=(\partial^\mu A^\nu-\partial^\nu A^\mu)$ and tried to show $\delta S=0$. I end up with something like $\delta S=\int \partial^\mu a_\rho\partial^\rho A^\mu(\partial_\mu a^\rho \partial_\rho A_\nu-\partial_\nu a^\rho\partial_\rho A_\mu)$ but now I can't just rename the indices, can I?

Maybe there is even a more direct way?

EDIT: Ok my equation for $\delta S$ seems to be wrong. Following DavidHs suggestion: $$\delta \mathcal{L}=\mathcal{L}'-\mathcal{L}=-\frac{1}{4}F'^{\mu\nu}F'_{\mu\nu}+\frac{1}{4}F^{\mu\nu}F_{\mu\nu}\\=-\frac{1}{2}\partial_\mu(A_\nu+a^\rho \partial_\rho A_\nu)(\partial^\mu A^\nu+ \partial^\mu a_\rho \partial^\rho A^\nu- \partial^\nu A^\mu -\partial^\nu a_\rho \partial^\rho A^\mu)+\frac{1}{4}F^{\mu\nu}F_{\mu\nu}\\ =-\frac{1}{2}(a_\rho \partial_\mu A_\nu \partial^\mu \partial^\rho A^\nu - a_\rho\partial_\mu A_\nu \partial^\nu \partial^\rho A^\mu+ a^\rho \partial_\rho \partial_\mu A_\nu \partial^\mu A^\nu - a^\rho\partial_\rho \partial_\mu A_\nu \partial^\nu A^\mu)\\=-\frac{1}{2}\partial^\rho(a_\rho\partial_\mu A_\nu F^{\mu\nu})$$

  • $\begingroup$ if $a^\mu$ is constant, then $ \partial^\mu a_\rho$ is zero. $\endgroup$ – mike stone May 5 '19 at 14:19

I don't think the expression for $\delta S$ you provide is correct --- for a start it contains the index $\rho$ four times.

Remember that you only need to show that $\delta S = 0$ for an infinitesimal transformation, so you can drop terms that are $O(a^2)$. Try computing the first order correction to the Lagrangian $\mathcal{L}$ and showing that it may be written as a total derivative (i.e. $\delta \mathcal{L} = \partial_\rho[\text{something}]$).

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  • $\begingroup$ Yes, my first equation seems wrong. I tried your suggestion, now it seems correct. $\endgroup$ – clearseplex May 6 '19 at 11:17

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