TITLE: Divergenceless of energy momentum tensor for any metric $g_{\mu\nu}$
QUESTION [2 upvotes]: As suggested by @my2cts, from this post, I want to know if the divergenceless of energy-momentum energy tensor is valid for any metric $\eta_{\mu\nu}$ (i.e for example with $\eta_{\mu\nu}=g_{\mu\nu}$)? 
Here the formula with $\eta_{\mu\nu}$ (I think the author has taken a Minkowski pseudo-metric in this formula but I am not sure)
$$\nabla_{\mu} T^{\mu \nu} = 0\tag{1}$$
$$T^{\mu \nu} = \frac{1}{\mu_{0}}\Big[F^{\alpha \mu} F^{\nu}_{\alpha} - \frac{1}{4}\eta^{\mu \nu}F^{\alpha \beta}F_{\alpha \beta}\Big]\tag{2}$$
Is this also the case in not-vacuum space?

REPLY [2 votes]: Diffeomorphism invariance of the matter theory$^1$ implies$^2$ 
$$ \nabla_{\mu} T^{\mu\nu}~\stackrel{m}{\approx}~0 \tag{*}$$
for an arbitrary metric $g_{\mu\nu}$. Note that the derivative in eq. (*) is a covariant Levi-Civita derivative rather than a partial derivative. Also note that $T^{\mu\nu}$ is here the symmetric metric/Hilbert SEM tensor. For details, see e.g. my Phys.SE answer here.
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$^1$ Concretely OP's matter theory is Maxwell theory in curved spacetime.
$^2$ The $\stackrel{m}{\approx}$ symbol means equality modulo matter eom. Matter fields means in this context anything other than gravitational fields, e.g. EM fields.

REPLY [1 votes]: Yes. The Einstein tensor has zero divergence for any metric. The Einstein field equations equate the Einstein tensor to the stress-energy, so if the field equations hold, the stress-energy must be divergenceless.
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