# Derivatives with Two Indices in Electromagnetic Lagrangian [duplicate]

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I was reading about the derivation of Maxwell's equations from an electromagnetic Lagrangian density from Sean Carroll's Spacetime and Geometry: An Introduction to General Relativity. The Lagrangian density $$\mathcal{L}$$ is given by

$$\mathcal{L} = -\frac{1}{4} F^{\mu\nu}F_{\mu\nu} + A_{\mu}J^{\mu},$$ where the symbols have their usual meanings.

In eq. (1.166) he writes the following equation:

$$\frac{\partial F_{\alpha\beta}}{\partial(\partial_{\mu}A_{\nu})} = \delta^{\mu}_{\alpha} \delta^{\nu}_{\beta} - \delta^{\mu}_{\beta} \delta^{\nu}_{\alpha}.$$

As $$F_{\alpha\beta} = \partial_{\alpha}A_{\beta} - \partial_{\beta}A_{\alpha}$$, the following results is used:

$$\frac{\partial (\partial_{\alpha} A_{\beta})}{\partial(\partial_{\mu}A_{\nu})} = \delta^{\mu}_{\alpha} \delta^{\nu}_{\beta}$$

How to prove this result?

## marked as duplicate by Qmechanic♦Jan 7 at 7:24

• This speaks to what you'll do next rather than your question here, but when you try to obtain the Euler-Lagrange equations, you may find it helpful to rewrite the Lagrangian density as $\mathcal{L}=-\frac{1}{2}\partial_\mu A_\nu F^{\mu\nu}+A_\mu J^\mu$. – J.G. Jan 7 at 6:42
• @J.G. Thanks, I have done it. (By any chance, if you have a copy of Schwartz's QFT, then have a look at eq. (3.43). I could not figure out how he wrote the Lagrangian as follows: $\mathcal{L} = -\frac{1}{2} (\partial_{\mu} A_{\nu})^2 + \frac{1}{2} (\partial_{\mu} A_{\mu})^2 - A_{\mu} J_{\mu}$ . ) – omehoque Jan 7 at 7:27
• @omehoque I suspect a total derivative has been added. Subtract one expression for $\mathcal{L}$ from the other to see. – J.G. Jan 7 at 7:36

I don't know exactly what kind of proof you're looking for (like what level/how involved). But it's the same reason that

$$\frac{\partial x^\mu}{\partial x^\nu} = \delta^\mu_\nu.$$

But now our "variable", instead of $$x^\mu$$ is a two indexed object. For instance, say we consider the function $$x^\mu x^\nu$$, and we wanted

$$\frac{\partial( x^\mu x^\nu)}{\partial (x^\lambda x^\sigma)}$$

this answer would be zero unless both indices matched since we are taking the partial with respect to that. As such we would get

$$\frac{\partial( x^\mu x^\nu)}{\partial (x^\lambda x^\sigma)} = \delta^\mu_\lambda \delta^\nu_\sigma$$

We multiply the delta functions because we need both the indices to be equal. I'm sure you could prove this with the chain rule and product rule if you'd like, though in your case the derivative would slightly complicate things. However the result generalizes.