Question
When calculating the hamiltonian for the free Electromagnetic Field with Lagrangian density
$$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$
Using Noether's theorem I found the answer to be
$$T^{\mu\nu}=-F^{\mu\sigma}\partial^{\nu}A_{\sigma} – \eta^{\mu\nu}\mathcal{L}$$
However, this cannot be true because the energy-momentum tensor is symmetric but the expression that I have calculated is not symmetric. I've looked online and found that it should be
$$T^{\mu\nu}=\eta_{\sigma\lambda}F^{\mu\sigma}F^{\lambda\nu}- \eta^{\mu\nu}\mathcal{L}$$
My Working
I began with Noether's theorem for the energy-momentum tensor
$$T^{\mu\nu}=\frac{\partial \mathcal{L}}{\partial (\partial_\mu A_\sigma)}\partial^\nu A_\sigma – \eta^{\mu\nu}\mathcal{L}$$
Now I substituted in the lagrangian for the free field which gives us
$$T^{\mu\nu}=\frac{\partial}{\partial (\partial_\mu A_\sigma)}[-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}]\partial^\nu A_\sigma – \eta^{\mu\nu}\mathcal{L}$$
Factoring the constant outside of the derivitive leaves the expression as
$$T^{\mu\nu}=-\frac{1}{4}\frac{\partial}{\partial (\partial_\mu A_\sigma)}[F_{\alpha\beta}F^{\alpha\beta}]\partial^\nu A_\sigma – \eta^{\mu\nu}\mathcal{L}$$
We can now use the chain rule to seperate the multiplied terms
$$T^{\mu\nu}=-\frac{1}{4}F^{\alpha\beta}\frac{\partial F_{\alpha\beta}}{\partial (\partial_\mu A_\sigma)}\partial^\nu A_\sigma -\frac{1}{4}F_{\alpha\beta}\frac{\partial F^{\alpha\beta}}{\partial (\partial_\mu A_\sigma)}\partial^\nu A_\sigma – \eta^{\mu\nu}\mathcal{L}$$
Now if we substitute the definition for the $F_{\alpha\beta}$ tensor
$$T^{\mu\nu}=-\frac{1}{4}F^{\alpha\beta}\frac{\partial}{\partial (\partial_\mu A_\sigma)}(\partial_{\alpha}A_{\beta}-\partial_{\beta}A_{\alpha})\partial^\nu A_\sigma -\frac{1}{4}F_{\alpha\beta}\frac{\partial}{\partial (\partial_\mu A_\sigma)}(\partial^{\alpha}A^{\beta}-\partial^{\beta}A^{\alpha})\partial^\nu A_\sigma – \eta^{\mu\nu}\mathcal{L}$$
The second term in the expression can be lowered by the minkowski metric
$$T^{\mu\nu}=-\frac{1}{4}F^{\alpha\beta}\frac{\partial}{\partial (\partial_\mu A_\sigma)}(\partial_{\alpha}A_{\beta}-\partial_{\beta}A_{\alpha})\partial^\nu A_\sigma -\frac{1}{4}\eta^{\alpha\alpha'}\eta^{\beta\beta'}F_{\alpha\beta}\frac{\partial}{\partial (\partial_\mu A_\sigma)}(\partial_{\alpha'}A_{\beta'}-\partial_{\beta'}A_{\alpha'})\partial^\nu A_\sigma – \eta^{\mu\nu}\mathcal{L}$$
To evaluate the derivitives we can look at what happens if we look at the expression
$$\frac{\partial}{\partial x^\alpha} x^\beta$$
The derivitive should be $0$ for all indexes exept for when $\alpha=\beta$ which is the definition of $\delta_{\alpha}^{\beta}$. So this means that
$$\frac{\partial}{\partial x^\alpha} x^\beta=\delta_{\alpha}^{\beta}$$
If we look at taking the derivitive with respect to a rank two tensor
$$\frac{\partial}{\partial G^{\alpha\beta}} G^{\sigma\gamma}$$
This means that $\alpha = \sigma$ and $\beta = \gamma$ which can be written as
$$\frac{\partial}{\partial G^{\alpha\beta}} G^{\sigma\gamma} = \delta_{\alpha}^{\sigma}\delta_{\beta}^{\gamma}$$
However we know that $\frac{\partial}{\partial (\partial_\mu A_\sigma)}$ is simpily the derivitive with respect to a rank 2 tensor so applying the results yeilds
$$T^{\mu\nu}=-\frac{1}{4}F^{\alpha\beta}(\delta_{\alpha}^{\mu}\delta_{\beta}^\sigma-\delta_{\beta}^{\mu}\delta_{\alpha}^\sigma)\partial^\nu A_\sigma -\frac{1}{4}\eta^{\alpha\alpha'}\eta^{\beta\beta'}F_{\alpha\beta}(\delta_{\alpha'}^{\mu}\delta_{\beta'}^\sigma-\delta_{\beta'}^{\mu}\delta_{\alpha'}^\sigma)\partial^\nu A_\sigma – \eta^{\mu\nu}\mathcal{L}$$
We can raise the indexes of the Electromagnetic Field Strengrh tensor with the minkoiski metric
$$T^{\mu\nu}=-\frac{1}{4}F^{\alpha\beta}(\delta_{\alpha}^{\mu}\delta_{\beta}^\sigma-\delta_{\beta}^{\mu}\delta_{\alpha}^\sigma)\partial^\nu A_\sigma -\frac{1}{4}F^{\alpha'\beta'}(\delta_{\alpha'}^{\mu}\delta_{\beta'}^\sigma-\delta_{\beta'}^{\mu}\delta_{\alpha'}^\sigma)\partial^\nu A_\sigma – \eta^{\mu\nu}\mathcal{L}$$
Since we are summing over $\alpha'$ and $\beta'$ we can replace them with $\alpha$ and $\beta$
$$T^{\mu\nu}=-\frac{1}{4}F^{\alpha\beta}(\delta_{\alpha}^{\mu}\delta_{\beta}^\sigma-\delta_{\beta}^{\mu}\delta_{\alpha}^\sigma)\partial^\nu A_\sigma -\frac{1}{4}F^{\alpha\beta}(\delta_{\alpha}^{\mu}\delta_{\beta}^\sigma-\delta_{\beta}^{\mu}\delta_{\alpha}^\sigma)\partial^\nu A_\sigma – \eta^{\mu\nu}\mathcal{L}$$
Since we now have like terms we can combine them which gives
$$T^{\mu\nu}=-\frac{1}{2}F^{\alpha\beta}(\delta_{\alpha}^{\mu}\delta_{\beta}^\sigma-\delta_{\beta}^{\mu}\delta_{\alpha}^\sigma)\partial^\nu A_\sigma – \eta^{\mu\nu}\mathcal{L}$$
If we apply the kroneker delta we get that
$$T^{\mu\nu}=-\frac{1}{2}(F^{\mu\sigma}-F^{\sigma\mu})\partial^\nu A_\sigma – \eta^{\mu\nu}\mathcal{L}$$
Finaly since $F^{\mu\sigma}$ is anti symetric
$$T^{\mu\nu}=-F^{\mu\sigma}\partial^\nu A_\sigma – \eta^{\mu\nu}\mathcal{L}$$
Best Answer
This is the Belinfante-Rosenfeld procedure (note that it is not necessary to invoke spin currents for the free electromagnetic theory). Just to spell things out, this involves modifying the canonical energy-momentum tensor by adding a divergenceless term (an alternative method is using the Hilbert definition, see this question for the relation between the two):
$$T^{\mu\nu}=-F^{\mu\sigma}\partial^{\nu}A_{\sigma} - \eta^{\mu\nu}\mathcal{L} +C^{\mu\nu},$$
where $\partial_\mu C^{\mu\nu}$ is a zero extremal variation of the Lagrangian, and thus preserves the conservation of $T_{\mu\nu}$. This does modify the $T^{0\nu}$ densities, but as G. Smith says:
Anyway, if we make the ad hoc choice of $C^{\mu\nu}$ as $F^{\mu\sigma}\partial_\sigma A^\nu$, you can see that $$ \partial_\mu C^{\mu\nu}= (\partial_\mu F^{\mu\sigma})\partial_\sigma A^\nu+F^{\mu\sigma}\partial_\mu\partial_\sigma A^\nu=0, $$ since the first term is zero by the equations of motion, while the second term involves a contraction between a symmetric and antisymmetric indices, which can be shown to be zero. Hence the inclusion of this term modifies the energy-momentum tensor to be:
$$T^{\mu\nu}=F^{\mu\sigma}\partial_\sigma A^\nu-F^{\mu\sigma}\partial^{\nu}A_{\sigma} - \eta^{\mu\nu}\mathcal{L} \\ = \eta_{\sigma\lambda} (F^{\mu\sigma}\partial^{\lambda}A^{\nu}-F^{\mu\sigma}\partial^\nu A^\lambda) - \eta^{\mu\nu}\mathcal{L} \\ = \eta_{\sigma\lambda}F^{\mu\sigma}F^{\lambda\nu}- \eta^{\mu\nu}\mathcal{L} $$
which is exactly what you set out to show.