# Electromagnetic tensor [closed]

How to prove the equality in this? $$F^{\mu\nu}F_{\mu\nu}=g^{\mu\alpha}g^{\nu\beta}\left(\partial_{\alpha}A_{\beta}-\partial_{\beta}A_{\alpha}\right)\left(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}\right)=2g^{\mu\alpha}g^{\nu\beta}\left(\partial_{\nu}A_{\mu}\right)\left(\partial_{\beta}A_{\alpha}-\partial_{\alpha}A_{\beta}\right)$$

• Have you tried expanding the second expression? – DomDoe Sep 3 '19 at 8:16
• I have but I still don't get it. – Kirby Sep 3 '19 at 8:18
• What is your field, Professor? Anyway, keep trying. There is no point to the exercise if you don't work it out yourself. – my2cts Sep 3 '19 at 11:16
• Do you understand that a contracted index can be relabeled using a different letter? – G. Smith Sep 3 '19 at 15:52
• @my2cts, I'm a professor in general relativity. – Kirby Sep 4 '19 at 10:53

Good news everybody! I solved the problem! The trick was that you had to write it open into four terms and then change the sum indices of two of those terms (one with plus and one with minus in front of the term). So you can change the indices for example in this way: $$\alpha \rightarrow \beta, \beta \rightarrow \alpha,\mu \rightarrow \nu, \nu \rightarrow \mu$$. Here is the solution: $$\begin{eqnarray*} F^{\mu\nu}F_{\mu\nu} & = & g^{\mu\alpha}g^{\nu\beta}\left(\partial_{\alpha}A_{\beta}-\partial_{\beta}A_{\alpha}\right)\left(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}\right)\\ & = & g^{\mu\alpha}g^{\nu\beta}\partial_{\alpha}A_{\beta}\partial_{\mu}A_{\nu}-g^{\mu\alpha}g^{\nu\beta}\partial_{\alpha}A_{\beta}\partial_{\nu}A_{\mu}-g^{\mu\alpha}g^{\nu\beta}\partial_{\beta}A_{\alpha}\partial_{\mu}A_{\nu}+g^{\mu\alpha}g^{\nu\beta}\partial_{\beta}A_{\alpha}\partial_{\nu}A_{\mu}\\ & = & g^{\mu\alpha}g^{\nu\beta}\partial_{\alpha}A_{\beta}\partial_{\mu}A_{\nu}-g^{\mu\alpha}g^{\nu\beta}\partial_{\alpha}A_{\beta}\partial_{\nu}A_{\mu}-g^{\mu\alpha}g^{\nu\beta}\partial_{\alpha}A_{\beta}\partial_{\nu}A_{\mu}+g^{\mu\alpha}g^{\nu\beta}\partial_{\alpha}A_{\beta}\partial_{\mu}A_{\nu}\\ & = & 2g^{\mu\alpha}g^{\nu\beta}\partial_{\alpha}A_{\beta}\left(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}\right). \end{eqnarray*}$$