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The definition of Komar mass in GR is associated with one asymptotically flat end. However, a hypersurface may contain more than one end, such as the spacelike Einstein-Rosen bridge in Kruskal Spacetime which has two ends.

Therefore, the total mass is conserved by the Stokes theorem and current conservation, but how is Komar mass associated with each end conserved?

In the example of Kruskal spacetime with parameter $M$ in the Schwarzschild metric, the total mass integral is certainly zero for the vacuum solution, with the Komar mass associated with two ends are $M$, $ -M$ respectively. So we know that $ M+(-M)=0$ is conserved, but why $M$ or $-M$ is conserved?

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    $\begingroup$ If "Kruzkal spacetime" is just Schwarzschild, why would the Komar mass be $0$? $\endgroup$
    – Ryan Unger
    May 5, 2016 at 0:41
  • $\begingroup$ @ocelo7 the komar mass is not 0, but why is it conserved? $\endgroup$
    – Shadumu
    May 5, 2016 at 8:52
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    $\begingroup$ So what is $M+(-M)=0$ supposed to mean? It's conserved because the spacetime is static. $\endgroup$
    – Ryan Unger
    May 5, 2016 at 13:19

2 Answers 2

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Komar mass is associated to one asymptotically flat end. So a wormhole has two Komar masses, one for each side, and in principle they can even be different! Indeed the simplest wormhole you can imagine is simply two copies Schwarzschild spacetimes glued together at the would-be horizon.

As an aside, even time can run differently on the two sides (often termed universes) of the wormhole. That's why it's very easy to build time machines with wormholes!

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  • $\begingroup$ What do you mean by saying that time runs differently on both sides? $\endgroup$
    – Khushal
    Apr 29, 2018 at 10:23
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    $\begingroup$ @Khushal , an observer in one universe will see the clock in the other universe running faster/slower. $\endgroup$
    – Rexcirus
    Jul 4, 2018 at 19:50
  • $\begingroup$ Having different Komar masses on each end is only possible for non vacuum solutions. $\endgroup$
    – TimRias
    Feb 12 at 21:11
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The conservation of the Komar mass on each end is still the consequence of the existence of a conserved current. Suppose you have two different surfaces $\Sigma_1$ and $\Sigma_2$ enclosing one asymptotically flat end. You can then find a hypersurface $V$ that has these surfaces as its edges. The volume Komar integral is prortional to the energy-momentum tensor, and hence must be zero in a vacuum spacetime. Consequently, $\Sigma_1$ and $\Sigma_2$ must have the same surface Komar integrals (once you account for the relative orientations).

The same logic tells you that the opposite asymptotically flat ends of the Krystal spacetimes must have opposite and equal in magnitude Komar charges, since you can connect closed surfaces on each end with a (vacuum) hypersurface.

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