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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?

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  • $\begingroup$ You mean "valid for any metric $g_{\mu\nu}$ (i.e. for example, with $g_{\mu\nu}=\eta_{\mu\nu})$". It's $g_{\mu\nu}$ that is a general metric tensor, and $\eta_{\mu\nu}$ that is a specific metric, namely the Minkowski metric. $\endgroup$ – G. Smith Nov 9 '18 at 2:43
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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.

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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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  • $\begingroup$ -@Ben Crowell thanks for this quick answer. And are there specific conditions on Faraday tensor $F_{\alpha\beta}$ to get this divergenceless with any metric ? $\endgroup$ – youpilat13 Nov 8 '18 at 20:34
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    $\begingroup$ And are there specific conditions on Faraday tensor Fαβ to get this divergenceless with any metric ? No, it holds for the reasons given in the answer. $\endgroup$ – Ben Crowell Nov 8 '18 at 21:38

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