# Integral form of energy-momentum tensor conservation (Stokes' theorem)

Is there a way in which the conservation law of the energy momentum tensor $$\nabla _\nu T^{\mu\nu}=0$$ can be written in integral form using Stokes' theorem, namely as something roughly similar to:

$$0 = \int_\Omega d^4x \sqrt{-g}\nabla _\nu T^{\mu\nu} = \int_{\partial \Omega} ?$$

In particular i'm imagining a case where we choose the region $$\Omega$$ such that $$T^{\mu\nu}$$ vanishes on the boundary except for some "initial" and "final" spacelike slices as in this image:

so what we get is a relation between the initial and final spatial integrals. In flat spacetime I guess this just gives the expected conservation of the total 4-momentum, if we choose the slices to be at constant time coordinate $$t$$ :

$$\int_{V_1} d^3x T^{0\mu} - \int_{V_0} d^3x T^{0\mu} = 0$$

but what is the general version of this ? i.e. what is the equivalent relation when (a) spacetime is not flat or (b) the integration region is arbitrary?

To get from a "covariantly conserved" energy-momentum tensor to an actual conserved quantity, you need to have a spactime symmetry. This requires a Killing vector field $$\xi_\mu$$ such that the Lie derivative $${\mathcal L}_\xi g_{\mu\nu}\equiv {\nabla_\mu} \xi_\nu+\nabla_\nu \xi_\mu=0$$. Then $$0=\nabla_\mu T^{\mu\nu}$$ implies that $$0=\nabla_\mu (T^{\mu\nu}\xi_\nu) = \frac 1{\sqrt{g}} \partial_\mu (\sqrt{g}T^{\mu\nu}\xi_\nu)$$ and hence
$$\int_{\Omega} \sqrt{g} \nabla_\mu (T^{\mu\nu}\xi_\nu)= \int_{\Omega} \partial_\mu (\sqrt{g}T^{\mu\nu}\xi_\nu)= \int_{\partial\Omega} (\sqrt{g}T^{\mu\nu}\xi_\nu)dS_\mu=0$$