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This question already has an answer here:

For $\mathcal L = -\frac14 F_{\mu\nu}F^{\mu\nu}$ I would appreciate some help evaluating

$$\frac{\partial \mathcal L}{\partial(\partial_{\mu}A_{\nu})}.$$

I've found

$$\frac{\partial \mathcal L}{\partial(\partial_{\mu}A_{\nu})} = -\frac14\frac{\partial}{\partial(\partial_{\mu}A_{\nu})}(\partial_{\lambda}A_{\sigma} - \partial_{\sigma} A_{\lambda})(\partial_{\lambda} A_{\sigma} - \partial_{\sigma} A_{\lambda}) $$

$$ = -\frac14 \frac{\partial}{\partial(\partial_{\mu}A_{\nu})} \left( 2\partial_{\lambda}A_{\sigma}-2\partial_{\lambda}A_{\sigma}\partial_{\sigma}A_{\lambda}\right)$$ $$ = -\frac14 4\left(\partial_{\nu} A_{\mu} - \partial_{\mu} A_{\nu}\right) \qquad(*) $$ $$ = F_{\mu\nu}$$

but I cannot understand the step taken to get to $(*)$. In my attempt, I take the chain rule to obtain

$$\frac{\partial \mathcal L}{\partial(\partial_{\mu}A_{\nu})} = -\frac12 F_{\mu\nu} \frac{\partial(\partial_{\alpha}A_{\beta} - \partial_{\beta} A_{\lambda})}{\partial(\partial_{\mu} A_{\nu})}$$

$$= -\frac12 F_{\mu\nu}(1) - \frac12 F_{\mu\nu} \frac{\partial(\partial_{\beta}A_{\lambda})}{\partial(\partial_{\mu}A_{\nu})}$$ $$ = -\frac12 F_{\mu\nu} - \frac12F_{\mu\nu}\delta_{\beta\mu} \delta_{\lambda\nu} = -\frac12 F_{\mu\nu} - \frac12 F_{\beta\lambda}$$

which seems to give me almost what I want, but I must have run into an error because this expression no longer makes sense, as each term should be indexed by $\mu,\nu$.

So my 2 questions are: how do we get to $(*)$ and where did my attempt go wrong?

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marked as duplicate by Qmechanic Jan 7 at 7:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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To begin, the antisymmetry of $F_{\mu\nu}$ implies $\mathcal{L}=-\frac{1}{2}\partial_\mu A_\nu F^{\mu\nu}$. We now differentiate with respect to $\partial_\rho A_\sigma$. By the product rule, the result is $$-\frac{1}{2}(\delta_\mu^\rho \delta_\nu^\sigma F^{\mu\nu} + \partial_\mu A_\nu (g^{\mu\rho}g^{\nu\sigma}-g^{\nu\rho}g^{\mu\sigma}))=-\frac{1}{2}(F^{\rho\sigma}+\partial^\rho A^\sigma - \partial^\sigma A^\rho)=-F^{\rho\sigma}.$$

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First write $F^2 = F^{\lambda \sigma} F_{\lambda \sigma}$ as $$ F^2 = g^{\lambda \alpha} g^{\sigma \beta} F_{\alpha \beta} F_{\lambda \sigma} $$ and show that $$ \frac{\partial }{\partial\left(\partial_\mu A_\nu\right)} F^2 = 2 g^{\lambda \alpha} g^{\sigma \beta} F_{\alpha \beta} \frac{\partial }{\partial\left(\partial_\mu A_\nu\right)} F_{\lambda \sigma} $$ Then use $$ \frac{\partial \left(\partial_\alpha A_\beta \right)}{\partial\left(\partial_\mu A_\nu\right)} = \delta^\mu_\alpha \delta^\nu_\beta $$ Alternatively, vary the action directly.

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