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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?

Qmechanic
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rainman
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1 Answers1

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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.

InertialObserver
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