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

Physics Asked by user341440 on July 22, 2021

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

$$ 0 = int_Omega d^4x sqrt{-g}nabla _nu T^{munu} = int_{partial Omega} ? $$

In particular i’m imagining a case where we choose the region $Omega$ such that $T^{munu}$ 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 $t$ :

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

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?

One Answer

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 ${mathcal L}_xi g_{munu}equiv {nabla_mu} xi_nu+nabla_nu xi_mu=0$. Then $0=nabla_mu T^{munu}$ implies that $$ 0=nabla_mu (T^{munu}xi_nu) = frac 1{sqrt{g}} partial_mu (sqrt{g}T^{munu}xi_nu) $$ and hence
$$ int_{Omega} sqrt{g} nabla_mu (T^{munu}xi_nu)= int_{Omega} partial_mu (sqrt{g}T^{munu}xi_nu)= int_{partialOmega} (sqrt{g}T^{munu}xi_nu)dS_mu=0 $$

Correct answer by mike stone on July 22, 2021

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