Deriving Electromagnetism energy-stress tensor in GR Please find the mistake in the following calculations.
We have $L=-F^{\mu\nu}F_{\mu\nu}$, and try to derive the energy-stress tensor using
$\delta(-g)^{1/2}=\frac{1}{2}(-g)^{1/2}g^{\mu\nu}\delta g_{\mu\nu}$
We will have 
$T^{\mu\nu}=\frac{\delta (L(-g)^{1/2})}{\delta g_{\mu\nu}}=(-g)^{1/2}[-\frac{1}{2}g^{\mu\nu}F^{\rho\sigma}F_{\rho\sigma}-2F^{\mu\sigma}F^{\nu\rho}g_{\rho\sigma} ]$
But if we use
$\delta(-g)^{1/2}=-\frac{1}{2}(-g)^{1/2}g_{\mu\nu}\delta g^{\mu\nu}$
We will have 
$T_{\mu\nu}=\frac{\delta (L(-g)^{1/2})}{\delta g^{\mu\nu}}=(-g)^{1/2}[+\frac{1}{2}g_{\mu\nu}F^{\rho\sigma}F_{\rho\sigma}-2F_{\mu\sigma}F_{\nu\rho}g^{\rho\sigma} ]$
There is a sign difference for the first term in the two results derived from different ways. The first result seems to be wrong. I cannot figure out where is the mistake.
 A: There seem to be two mistakes. 


*

*Your definitions of the energy-stress tensors are inconsistent. Let's take your second definition to be the correct one, which I write as $\delta\left(L(-g)^{1/2}\right)=T_{\mu\nu}\delta g^{\mu\nu}$. From this definition we can also find out what $T^{\mu\nu}$ is, using $T_{\mu\nu}=T^{\rho\sigma}g_{\mu\rho}g_{\nu\sigma}$. Plugging this in your second definition of the energy-stress tensor yields $$\delta\left(L(-g)^{1/2}\right)=T_{\mu\nu}\delta g^{\mu\nu}=T^{\rho\sigma}g_{\mu\rho}g_{\nu\sigma}\delta g^{\mu\nu}.$$ Now you can use the identity $g_{\mu\rho}g_{\nu\sigma}\delta g^{\mu\nu}=-\delta g_{\rho\sigma}$. You are probably aware of this, since you also wrote that $\delta(-g)^{1/2}=\frac{1}{2}(-g)^{1/2}g^{\mu\nu}\delta g_{\mu\nu}=-\frac{1}{2}(-g)^{1/2}g_{\mu\nu}\delta g^{\mu\nu}$ and the minus sign difference comes from this identity. Anyway, using this results in
$$\delta\left(L(-g)^{1/2}\right)=T_{\mu\nu}\delta g^{\mu\nu}=-T^{\rho\sigma}\delta g_{\rho\sigma}$$ and therefore 
$$T^{\mu\nu}=-\frac{\delta (L(-g)^{1/2})}{\delta g_{\mu\nu}}.$$ So assuming your second definition, your first definition of the stress-energy tensor is off by a minus sign. 

*Because of the point above, you get an additional overall minus sign in your first derivation. Comparing your two results then reveals there is a sign inconsistency in the second term, which comes from the variation of $F_{\mu\nu}F^{\mu\nu}$. Probably, in your second derivation you wrote 
$$\delta(F_{\mu\nu}F^{\mu\nu})=\delta\left(F_{\mu\nu}F_{\rho\sigma}g^{\rho\mu}g^{\sigma\nu}\right)=F_{\mu\nu}F_{\rho\sigma}(\delta g^{\rho\mu}g^{\sigma\nu}+g^{\rho\mu}\delta g^{\sigma\nu}),\;\;\;\;\;\;\;\;\;\;\;\;\; (1)$$ 
and renamed indices afterwards to get your result. Notice that you assume in the last step that $\delta F_{\mu\nu}=0$ when varying w.r.t. the metric. This is correct. In your first derivation you probably used 
$$\delta(F_{\mu\nu}F^{\mu\nu})=\delta\left(F^{\mu\nu}F^{\rho\sigma}g_{\rho\mu}g_{\sigma\nu}\right)=F^{\mu\nu}F^{\rho\sigma}(\delta g_{\rho\mu}g_{\sigma\nu}+g_{\rho\mu}\delta g_{\sigma\nu}).\;\;\;\;\;\;\;\;\;\;\;\;\; (2)$$ This means that in the last step you used that $\delta F^{\mu\nu}=0$. This is however not correct. Remember that the electromagnetic field tensor $F$ is a 2-form (it is defined as the exterior derivative of the 1-form, i.e., $F=dA$). This means that given some coordinates $x^\mu$, the components of $F$ are given by $F=F_{\mu\nu}dx^\mu dx^\nu$. If you didn't get that, the important thing is that the components of $F$ are defined with the indices down ($F_{\mu\nu}$). I don't need a metric tensor for this definition, so $\delta F_{\mu\nu}=0$ when varying w.r.t. the metric. However, to define the tensor $F^{\mu\nu}$, with the indices up, then I do need the metric tensor: $F^{\mu\nu}\equiv g^{\mu\alpha}g^{\nu\beta}F_{\alpha\beta}$. Therefore $\delta F^{\mu\nu}\neq 0$ when varying w.r.t. the metric! This is probably your mistake in your first derivation.
Thus, to fix your mistake in your first derivation, you should just not use Eq. $(2)$ above (since it is wrong), but rather just use Eq. $(1)$ above, and then use the identity $g_{\mu\rho}g_{\nu\sigma}\delta g^{\mu\nu}=-\delta g_{\rho\sigma}$ once again to obtain the other minus sign you need.
