Deriving Maxwell's equations from the Lagrangian

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Homework Statement

Given $\mathcal{L} = -\frac{1}{2}(\partial_\mu \mathcal{A}_\nu)(\partial^\mu \mathcal{A}^\nu)+\frac{1}{2}(\partial_\mu \mathcal{A}^\mu)^2$, compute $\frac{\partial{\mathcal{L}}}{\partial(\partial_\mu \mathcal{A}_\nu)}$.

Relevant Equations

Euler-Lagrange equations of motion.

This isn't a homework problem (it's an example from David Tong's QFT notes where I didn't understand the steps he took), but I am confused as to how exactly to take the partial derivative of the Lagrangian with respect to $\partial(\partial_\mu \mathcal{A}_\nu)$. (Note the answer is: $-\partial^\mu \mathcal{A}^\nu+(\partial_\rho \mathcal{A}^\rho)\eta^{\mu \nu}$)

 

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Write $(\partial_\mu \mathcal{A}_\nu)(\partial^\mu \mathcal{A}^\nu) = (\partial_\lambda \mathcal{A}_\rho)(\partial_\sigma \mathcal{A}_\tau) \eta^{\sigma \lambda} \eta^{\tau \rho}$ and use the relations $\frac{\partial}{\partial(\partial_\mu \mathcal{A}_\nu)}(\partial_i \mathcal{A}_j) = \delta^i_\mu \delta^j_\nu$ to show that the partial of the first term of $\mathcal{L}$ with respect to $\partial_\mu \mathcal{A}_\nu$ is $-\partial^\mu \mathcal{A}^\nu$.

Similarly write $(\partial_\mu \mathcal{A}^\mu)^2 = (\partial_\rho A_\sigma \cdot \eta^{\rho \sigma})^2$ and use the chain rule to get that the partial of the second term of $\mathcal{L}$ with respect to $\partial_\mu \mathcal{A}_\nu$ equals $\partial_\rho \mathcal{A}^\rho \cdot \eta^{\mu\nu}$.

 

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Write $(\partial_\mu \mathcal{A}_\nu)(\partial^\mu \mathcal{A}^\nu) = (\partial_\lambda \mathcal{A}_\rho)(\partial_\sigma \mathcal{A}_\tau) \eta^{\sigma \lambda} \eta^{\tau \rho}$ and use the relations $\frac{\partial}{\partial(\partial_\mu \mathcal{A}_\nu)}(\partial_i \mathcal{A}_j) = \delta^i_\mu \delta^j_\nu$ to show that the partial of the first term of $\mathcal{L}$ with respect to $\partial_\mu \mathcal{A}_\nu$ is $-\partial^\mu \mathcal{A}^\nu$.

Similarly write $(\partial_\mu \mathcal{A}^\mu)^2 = (\partial_\rho A_\sigma \cdot \eta^{\rho \sigma})^2$ and use the chain rule to get that the partial of the second term of $\mathcal{L}$ with respect to $\partial_\mu \mathcal{A}_\nu$ equals $\partial_\rho \mathcal{A}^\rho \cdot \eta^{\mu\nu}$.

Thank you so much for your help, I have a question after using the chain rule on the second term. After expanding as you suggested and using the chain rule, I get: =$(\partial_\rho \mathcal{A}_\sigma \eta^{\rho \sigma}) \cdot \eta^{\rho \sigma} \delta_\mu^\rho \delta_\nu^\sigma$ but this means we must substitute $\rho=\mu$ and $\sigma=\nu$ everywhere to satisfy the delta, however this gives $(\partial_\mu \mathcal{A}^\mu) \cdot \eta^{\mu \nu}$ which isn't the same as the answer right? I'm struggling to find where I'm going wrong here.

 

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Note that $\eta^{\mu\nu}$ are constants, so they have zero derivatives. The partial derivative of $(\partial_\rho \mathcal{A}_\sigma \cdot \eta^{\rho\sigma})^2$ with respect to $\partial_\mu\mathcal{A}_\nu$ is $$2(\partial_\rho \mathcal{A}_\sigma \cdot \eta^{\rho \sigma})\cdot \partial_\mu\mathcal{A}_\nu(\partial_i \mathcal{A}_j \cdot \eta^{i j}) = 2(\partial_\rho \mathcal{A}_\sigma \cdot \eta^{\rho \sigma})\delta_{\mu i} \delta_{\nu j} \eta^{ij}$$ Can you reduce the latter expression further?