# Divergenceless of energy momentum tensor for any metric $g_{\mu\nu}$

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?

• 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. – G. Smith Nov 9 '18 at 2:43

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.

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.

• -@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 ? – youpilat13 Nov 8 '18 at 20:34
• 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. – Ben Crowell Nov 8 '18 at 21:38