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I don't know if the following is correct, i want to compute the following derivative

$$\frac{\partial }{\partial (\partial_{\mu}A_{\nu})}\left(\partial^{\alpha}A^{\beta}\partial_{\alpha}A_{\beta} \right)$$

So I first introduce the metric in order to lower the first two indices

$$\partial^{\alpha}A^{\beta}\partial_{\alpha}A_{\beta}=\eta^{\alpha\gamma}\eta^{\beta\delta}\partial_{\gamma}A_{\delta}\partial_{\alpha}A_{\beta} $$

and then

$$\begin{align}\frac{\partial }{\partial (\partial_{\mu}A_{\nu})}\left(\eta^{\alpha\gamma}\eta^{\beta\delta}\partial_{\gamma}A_{\delta}\partial_{\alpha}A_{\beta}\right) &= \eta^{\alpha\gamma}\eta^{\beta\delta}\frac{\partial }{\partial (\partial_{\mu}A_{\nu})}\left(\partial_{\gamma}A_{\delta}\partial_{\alpha}A_{\beta}\right) \\ &=\eta^{\alpha\gamma}\eta^{\beta\delta}\left(\partial_{\alpha}A_{\beta}\delta_{\mu \gamma}\delta_{\delta \nu}+\partial_{\gamma}A_{\delta}\delta_{\alpha\mu}\delta_{\beta\nu}\right) \\ &=\eta^{\alpha\mu}\eta^{\beta\nu}\partial_{\alpha}A_{\beta}+\eta^{\mu\gamma}\eta^{\nu\delta}\partial_{\gamma}A_{\delta} \\ &=2\partial^{\mu}A^{\nu}\end{align}$$

I was troubled because in the metric sometimes the first index is the summed one and other times the second. Is it correct that $A^{\alpha}=\eta^{\alpha \beta}A_{\beta}=\eta^{\beta \alpha}A_{\beta}$?

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6  
Yup you got it! –  joshphysics Mar 8 '13 at 20:49
1  
Thanks, I was troubled because in the metric sometimes the first index is the summed one and other times the second. Is this right? $A^{\alpha}=\eta^{\alpha \beta}A_{\beta}=\eta^{\beta \alpha}A_{\beta}$ –  Jorge Mar 8 '13 at 21:12
1  
Yes; the Minkowski metric is symmetric, so you can switch the two indices. –  joshphysics Mar 8 '13 at 21:22
2  
Hi Nivalth - questions that just ask if your work is correct or not aren't really the kind of thing we do here, but luckily you mentioned what was confusing you in a comment, so I edited it into the question. @joshphysics you might want to turn your comments into an answer. –  David Z Mar 10 '13 at 5:33

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