The variation of the Lagrangian density for the canonical energy-momentum tensor

Question

I expanded the Lagrangian to this form $$ \mathcal{L} = -{1 \over 4} F^{\mu \nu} F_{\mu \nu} = ... = - {1 \over 2} (\partial^{\mu} A^{\nu} \partial_{\mu} A_{\nu} - \partial^{\mu} A^{\nu} \partial_{\nu} A_{\mu}). $$

Now, I know that the result of ${{ \delta \mathcal{L}} \over {\delta (\partial_{\mu} A_{\lambda})}} $ is $ - F^{\mu \lambda}$, but I dunno exactly why. I tried to put the Lagrangian density there $$ {{\partial \mathcal{L}} \over {\partial (\partial_{\mu} A_{\lambda})}} = - {1 \over 2} {{\partial} \over {\partial (\partial_{\mu} A_{\lambda})}} (\partial^{\mu} A^{\nu} \partial_{\mu} A_{\nu} - \partial^{\mu} A^{\nu} \partial_{\nu} A_{\mu}) = - {1 \over 2} \cdot 2 (\partial^{\mu} A^{\lambda} - \partial^{\lambda} A^{\mu}) $$ but I can't see whether is sufficient or I probably got the indices wrong.

Answer 1

The answer is correct but you shouldn't have got it from the method you used (yours would be a factor of $1/2$ out). To do it correctly you need to lower all the indices of $\partial^{\mu} A^{\mu}$ in order to properly do the partial derivative: i.e. $$ \partial^{\mu} A^{\nu} = g^{\mu \rho} g^{\nu \sigma} \partial_{\rho} A_{\sigma} \ .$$ Now if you use this for both terms in $F^{\mu \nu}$, and then do the partial derivative of this with respect to $\partial_{\mu} A_{\nu}$ using the chain rule etc, you will get the correct result.