Are two Maxwell Lagrangian forms equivalent without gauge fixing?

[jdstokes]

[SOLVED] Mandl and Shaw 5.1

To show

$-\frac{1}{2}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2}(\partial_\mu A^\mu)^2 \quad \mathrm{and} \quad -\frac{1}{2}\partial_\nu A_\mu \partial^\nu A^\mu$

represent the same Lagrangian it suffices to show that

$\partial_\nu A_\mu\partial^\mu A^\nu - \partial_\nu A^\nu \partial_\mu A^\mu$ is at most a 4-divergence.

The trouble is, I have no idea why this would be the case. Is this a matter of utilizing the product rule in some clever way?

Edit: Yes it is: factor out $\partial_\nu$.

 

[shadi_s10]

hi
does anybody have any suggestion to solve 2.4 too?!

 

[jmlaniel]

To show

$-\frac{1}{2}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2}(\partial_\mu A^\mu)^2 \quad \mathrm{and} \quad -\frac{1}{2}\partial_\nu A_\mu \partial^\nu A^\mu$

represent the same Lagrangian it suffices to show that

$\partial_\nu A_\mu\partial^\mu A^\nu - \partial_\nu A^\nu \partial_\mu A^\mu$ is at most a 4-divergence.

The trouble is, I have no idea why this would be the case. Is this a matter of utilizing the product rule in some clever way?

Edit: Yes it is: factor out $\partial_\nu$.

Can this be done without using the Lorentz gauge ($\partial_\mu A^\mu = 0$) or is it necessary imposed ?