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From the Lagrangian in Maxwell theory

$$L = -\frac{1}{2}(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu}) + \frac{1}{2}(\partial_{\mu}A^{\mu})^2 - A_{\mu}J^{\mu}$$

I have to calculate $\frac{\partial L}{\partial (\partial_{\mu}A_{\nu})}$.

I understood that

$$\frac{\partial }{\partial (\partial_{\mu}A_{\nu})}\left(-\frac{1}{2}(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu}) \right) = -\partial^{\nu}A^{\mu}.$$

I don't get the second part though:

$$\frac{\partial}{\partial (\partial_{\mu}A_{\nu})}\left(\frac{1}{2}(\partial_{\mu}A^{\mu})^2\right) = (\partial_{\rho}A^{\rho})\eta^{\mu\nu}.$$

I am still surfing thought the indexes and tensor manipulations in derivatives and so on, but here I got stuck. Where does the metric (flat spacetime) tensor comes out from?

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    $\begingroup$ 1. This for sure has been asked here before, so use the search function 2. You should make sure to not use the same indices for the derivative, i.e. the indices in the Lagrangian are contracted, and to net get into trouble, you should use different indices. $\endgroup$
    – Tobias Fünke
    Commented Jan 20 at 10:15
  • $\begingroup$ @TobiasFünke I searched for it, I found lots of things but that one $\endgroup$
    – Heidegger
    Commented Jan 20 at 10:16
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    $\begingroup$ Well, I don't mean (although it is of course possible) the exact same question, but a very similar one. Anyway, hint: the metric raises and lowers indices. And in the worst case: write out the sum convention and plug in some values for the indices in the derivative to see how everything works out. $\endgroup$
    – Tobias Fünke
    Commented Jan 20 at 10:22
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/3005/2451 , physics.stackexchange.com/q/512402/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Jan 20 at 10:47

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