# Derivative of the electromagnetic tensor invariant $F_{\mu\nu}F^{\mu\nu}$

The electromagnetic field tensor is $F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu$. I am trying to calculate the quantity $$\frac{\partial(F_{\alpha\beta}F^{\alpha\beta})}{\partial(\partial_{\mu}A_{\nu})}.$$ This calculation arises when trying to derive the electromagnetic equations of motion (i.e. Maxwell's equations) from the Lagrangian $\mathcal{L}=C F_{\mu\nu}F^{\mu\nu}$. According to P. 14 of these online notes, this derivative is $$\frac{\partial(F_{\alpha\beta}F^{\alpha\beta})}{\partial(\partial_{\mu}A_{\nu})}=2F^{\alpha\beta}\frac{\partial F_{\alpha\beta}}{\partial(\partial_\mu A_{\nu})} .$$

This result surprises me. I can use the product rule to find

$$\frac{\partial(F_{\alpha\beta}F^{\alpha\beta})}{\partial(\partial_{\mu}A_{\nu})}=F_{\alpha\beta}\frac{\partial(F^{\alpha\beta})}{\partial(\partial_{\mu}A_{\nu})}+F^{\alpha\beta}\frac{\partial(F_{\alpha\beta})}{\partial(\partial_{\mu}A_{\nu})}$$

and it is clear that if $F_{\alpha\beta}\frac{\partial(F^{\alpha\beta})}{\partial(\partial_{\mu}A_{\nu})}=F^{\alpha\beta}\frac{\partial(F_{\alpha\beta})}{\partial(\partial_{\mu}A_{\nu})}$ then you get the desired result. However, I can't see why this is true. In particular, I don't understand how to take the derivative $$\frac{\partial(F^{\alpha\beta})}{\partial(\partial_{\mu}A_{\nu})}=\frac{\partial(\partial^{\alpha}A^{\beta}-\partial^{\beta}A^{\alpha})}{\partial(\partial_{\mu}A_{\nu})}=\frac{\partial(\partial^{\alpha}A^{\beta})}{\partial(\partial_{\mu}A_{\nu})}-\frac{\partial(\partial^{\beta}A^{\alpha})}{\partial(\partial_{\mu}A_{\nu})}$$ where the downstairs part has lower indices and the upstairs part has upper indices.

• Can't you just use the metric to lower the indices in the numerator? May 22 '18 at 20:39

For your first question: the components of the metric don't depend on $\partial_\mu A_\nu$, or for that matter, anything at all. So we have, e.g. $$J^\mu \partial K_\mu = J^\mu \partial (\eta_{\mu\nu} K^\nu) = J^\mu \eta_{\mu\nu} \partial K^\nu = J_\nu \partial K^\nu = J_\mu \partial K^\mu$$ where $\partial$ stands for any kind of derivative whatsoever and $J$ and $K$ are arbitrary. The proof for your case is identical.
For your second question: simply do the same thing. Note that $$\frac{\partial J^\nu}{\partial J_\mu} = \frac{\partial (\eta^{\rho\nu} J_\rho)}{\partial J_\mu} = \eta^{\rho\nu}\frac{\partial J_\rho}{\partial J_\mu} = \eta^{\rho \nu} \delta^\mu_\rho = \eta^{\mu\nu}$$ where you can adapt this reasoning to your own example. After you do this a couple times, it becomes completely second nature, and you won't have to write out the steps. Everything works out exactly how you would expect, just "lining up the indices", $$\frac{\partial J^\nu}{\partial J_\mu} = \eta^{\mu\nu}, \quad \frac{\partial J^\nu}{\partial J^\mu} = \eta^{\nu}_\mu, \quad \frac{\partial J_\nu}{\partial J_\mu} = \eta_\nu^\mu, \quad \frac{\partial J_\nu}{\partial J^\mu} = \eta_{\mu\nu}$$ where, in order to write all four results the same way, I defined $\eta^\mu_\nu = \delta^\mu_\nu$.
$$F^{ab} = g^{ac}g^{bd}F_{cd}$$