The Komar mass is a means of measuring gravitational mass in spacetime. Via Wikipedia (https://en.wikipedia.org/wiki/Komar_mass) it is stated as (for a stationary metric):

$$m=\int\rho d\mathrm{vol}=\intop\sqrt{g_{00}}(2T_{\mu\nu}-Tg_{\mu\nu})e^{\mu}e^{\nu}d\mathrm{vol}$$

where $T_{\mu\nu}$ is the stress energy tensor, $g_{\mu\nu}$ is the stress energy tensor, $e^{\mu}$ is a unit time-like vector and $d\mathrm{vol}$ is the three-volume form. Considering only orientable metrics, we can write:

$$d\mathrm{vol}=\sqrt{g_{11}g_{22}g_{33}}e^{1}\wedge e^{2}\wedge e^{3}$$

Where, for simplicity we've considered a diagonalized metric. When we apply this to the above:

$$m=\intop\sqrt{g_{00}}(2T_{\mu\nu}-Tg_{\mu\nu})e^{\mu}e^{\nu}\sqrt{g_{11}g_{22}g_{33}}e^{1}\wedge e^{2}\wedge e^{3}$$


Which now appears as a full spacetime integral. Is this correct? I may have messed something simple up (such as a contraction somewhere), even then a variation of this appears valid. Tips would be greatly appreciated!


1 Answer 1


Komar mass is only well defined, i.e. Invariant, in a stationary spacetime, i.e., one admitting a timelike Killing vector. Your derivation seems to be good only for a static spacetime, i.e., no rotations, so a Kerr metric for instance would not be admitted. Not only that, you seemed to assume a diagonalized metric also in the space coordinates, not sure if that constrains it further (it may not, not sure).

The Wikipedia reference does show an invariant version of the mass formula for only stationary (and no other assumption) spacetime.

I'd not not know if your derivation, once you got to an invariant formation by assuming lots of zeros, is still right for a general stationary spacetime. you could assume more zeros, like T = 0, and that would certainly not prove the equation was right then In general

  • $\begingroup$ Thank you. I'm pursuing a line of thought that: $$\rho^{2}=T^{\mu\nu}T_{\mu\nu} $$ Functions as an effective gravitational mass density, and it's naturally invariant. For that expression, there should be no static limitations. I was seeing how similar it is to the Komar mass. $\endgroup$
    – R. Rankin
    Jul 15, 2016 at 6:30
  • $\begingroup$ Sorry, the above is factored such that $$\rho=e^{\nu}T_{\nu}^{\mu}e_{\mu}$$ which is not the same as the trace BTW: Due to it's symmetric nature, the metric is always diagonalizable $\endgroup$
    – R. Rankin
    Jul 15, 2016 at 6:37
  • $\begingroup$ Having a timelike Killing vector I don't think means that symmetric. If rotation it has to have some cross term I think $\endgroup$
    – Bob Bee
    Jul 15, 2016 at 15:35
  • $\begingroup$ "Mathematically, spacetime is represented by a 4-dimensional differentiable manifold M and the metric is given as a covariant, second-order, symmetric tensor on M, conventionally denoted by g" en.m.wikipedia.org/wiki/Metric_tensor_(general_relativity) That's Wikipedia, but every text book I've ever read said the metric is symmetric, unless you leave relativity and start playing with nonzero torsions. $\endgroup$
    – R. Rankin
    Jul 15, 2016 at 19:21
  • $\begingroup$ Sure, the metric is symmetric in that sense. But that does NOT mean you can diagonize it. Symmetries in general relativity space times (yes, without torsion) will give you certain conditions, like a timelike symmetry implies you can find a coordinate system where the metric is independent of time. A rotating otherwise symmetric body will have, as the vacuum metric outside, the Kerr metric. It has components like g sub t sub phi. That is the same as g sub phi sub t because there is no torsion. But it is not diagonal. That is what having cross terms means, not that those are not symmetric. $\endgroup$
    – Bob Bee
    Jul 15, 2016 at 19:30

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