Doubt about Lagrangian Density for the Electromagnetic Field

Question

I have a struggle with the derivation of a term of the Electromagnetic Lagrangian.

It's known that $$\mathcal{L} = -(1/4)F^{\mu \nu} F_{\mu \nu}$$ for the free Electromagnetic field. There also appears in textbooks that $$\mathcal{L} = -(1/4)F^{\mu \nu} F_{\mu \nu} = -(1/2)(\partial _{\mu} A_{\nu})(\partial ^{\mu}A^{\nu}) + (1/2)(\partial _{\mu} A^{\mu})^2.$$ I know the definition of the strength tensor in terms of $A_{\mu}$, but I obtain the following:

$$\mathcal{L} = -(1/2)(\partial _{\mu} A_{\nu})(\partial ^{\mu}A^{\nu}) + (1/2)(\partial _{\mu} A_{\nu})(\partial ^{\nu}A^{\mu}).$$

Could someone derive that last term $(1/2)(\partial _{\mu} A^{\mu})^2$?

Answer 1

The Lagrangian density can be changed by a total derivative $\partial_\mu V^\mu$, for an arbitrary vector field $V^\mu$, without changing the resulting equations of motion. You need$$\partial_\mu V^\mu=\frac12\partial_\mu A^\mu\partial_\nu A^\nu-\frac12\partial_\mu A_\nu\partial^\nu A^\mu.$$You can check$$V^\mu=\frac12(A^\mu\partial_\nu A^\nu-A_\nu\partial^\nu A^\mu)$$works. This obtains four terms in $\partial_\mu V^\mu$, but two cancel in pairs. (If it's not obvious they do, relabel the indices in one of them.)

Answer 2

The second equation is incorrect therefore it cannot be derived. However when integrated over Minkowski space $$\int d^4x \, (\partial _{\mu} A^{\mu})^2 = \int d^4x \, (\partial _{\mu} A_{\nu})(\partial ^{\nu}A^{\mu}) \,. $$ This requires that the stock terms of the two necessary consecutive partial integrations vanish at the integration boundary, usually infinity. As the action integrals of the two forms in your second equation are equal, these forms lead to the same equation of motion.