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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?

Frobenius
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Dwagg
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  • There are a couple missteps here, and I think you can debug by just checking the indices explicitly at each and every step. For example, your first step has mismatched indices in the numerator. If you fix these more simple errors first it'll be easier to address your deeper confusion. – knzhou Sep 14 '17 at 12:29
  • Possible duplicates: https://physics.stackexchange.com/q/3005/2451 , https://physics.stackexchange.com/q/34241/2451 , https://physics.stackexchange.com/q/51169/2451 , https://physics.stackexchange.com/q/64272/2451 and links therein. – Qmechanic Sep 14 '17 at 12:47
  • If $\mu$ and $\nu$ are the indices in your derivative, they cannot appear in the lagrangian itself. – Omry Sep 14 '17 at 12:49

2 Answers2

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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 $$-\frac{1}{2}(\delta_\mu^\rho \delta_\nu^\sigma F^{\mu\nu} + \partial_\mu A_\nu (g^{\mu\rho}g^{\nu\sigma}-g^{\nu\rho}g^{\mu\sigma}))=-\frac{1}{2}(F^{\rho\sigma}+\partial^\rho A^\sigma - \partial^\sigma A^\rho)=-F^{\rho\sigma}.$$

J.G.
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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.

Eric Angle
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