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: enter image description here

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?


1 Answer 1


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
$$ \int_{\Omega} \sqrt{g} \nabla_\mu (T^{\mu\nu}\xi_\nu)= \int_{\Omega} \partial_\mu (\sqrt{g}T^{\mu\nu}\xi_\nu)= \int_{\partial\Omega} (\sqrt{g}T^{\mu\nu}\xi_\nu)dS_\mu=0 $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.