I Maxwell Tensor Identity Explained: Deriving Formula 8.23 in Schawrtz's Book

[dm4b]

Hello,

In Schawrtz, Page 116, formula 8.23, he seems to suggest that the square of the Maxwell tensor can be expanded out as follows:

$$-\frac{1}{4}F_{\mu \nu}^{2}=\frac{1}{2}A_{\mu}\square A_{\mu}-\frac{1}{2}A_{\mu}\partial_{\mu}\partial_{\nu}A_{\nu}$$

where:

$$F_{\mu\nu}=\partial_{\mu} A_{\nu} - \partial_{\nu}A_{\mu}$$

For the life of me, I can't seem to derive this. I get close, but always with an extra unwanted term, or two.

Anyone have a hint on the best way to proceed?

Thanks!

 

[dm4b]

... keeping it under the integral (of S) and differentiating by parts works out here. However, is there a way to achieve this with just tensor manipulation? I thought so, but I may not be remembering correctly.

 

[king vitamin]

However, is there a way to achieve this with just tensor manipulation?

No, you need to integrate by parts. The equality sign there is a bit misleading.

 

[haushofer]

The equality is modulo a total derivative.