The electromagnetic field tensor is given by $F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$, and it appears in the Lagrangian as $$L = -\frac{1}{4}F_{\mu\nu}^2 - A_{\mu}J_{\mu}.$$ Schwartz's QFT textbook says that (chapter 8, page 116) $$F_{\mu\nu}^2 = 2(\partial_{\mu}A_{\nu})^2 - 2(\partial_{\mu}A_{\mu})^2,$$ but I do not see how that is correct.

The first step of the expansion gives $F_{\mu\nu}^2 = (\partial_{\mu}A_{\nu})^2 - \partial_{\mu}A_{\nu}\partial_{\nu}A_{\mu} - \partial_{\nu}A_{\mu}\partial_{\mu}A_{\nu} + (\partial_{\nu}A_{\mu})^2$, and I see that the first and last term add up to produce $2(\partial_{\mu}A_{\nu})^2$. However, the second term $\partial_{\mu}A_{\nu}\partial_{\nu}A_{\mu}$ when expanded has a term like $\partial_{0}A_{1}\partial_{1}A_{0}$, and I don't see how this is present in $(\partial_{\mu}A_{\mu})^2 = (\partial_{0}A_{0} + \partial_{1}A_{1} + \partial_{2}A_{2} + \partial_{3}A_{3})^2$.


1 Answer 1


All of these quantities are underneath a $d^4x$ integral, since $S = \int d^4x\, \mathcal{L}$, so we can integrate by parts twice: $$(\partial_\mu A^\nu)(\partial_\nu A^\mu) = -A^\nu (\partial_\mu \partial_\nu A^\mu) = (\partial_\nu A^\nu)(\partial_\mu A^\mu) = (\partial_\mu A^\mu)^2.$$ So you can always swap the locations of two derivatives under an integral. For some discussion of how this process changes the field theory, see my question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.