3
$\begingroup$

The Komar mass of some spacetime is defined as an integral (volume or surface, depending on its formulation): https://en.wikipedia.org/wiki/Komar_mass

The de-Sitter metric in static coordinates is ($\Lambda > 0$ here): \begin{equation}\tag{1} ds^2 = (1 - \frac{\Lambda}{3} \, r^2) \, dt^2 - \frac{1}{1 - \frac{\Lambda}{3} \, r^2} \, dr^2 - r^2 \, d\Omega^2. \end{equation} This metric has a timelike Killing vector for $r < \ell$, where $\ell = \sqrt{3/\Lambda}$: \begin{equation}\tag{2} \xi^{\mu} = (1, 0, 0, 0). \end{equation} Thus $\xi^{\mu} \, \xi_{\mu} = g_{00}$. It is spacelike for $r > \ell$.

AFAIK, the Komar mass is defined only for asymptotically flat spacetimes, which the de-Sitter spacetime isn't. So I believe that the Komar mass cannot be defined for the de-Sitter spacetime. Is this true?

If the Komar mass was defined for the de-Sitter spacetime, it should already be well known today, but I can't find it anywhere. So I guess that the answer to my question is affirmative, but I need a clear confirmation of it.

If an observer (say at $r = 0$) has only access to the region inside the horizon located at $r = \ell$, could we define a Komar mass (or an energy of some sort) for that observer?

$\endgroup$

1 Answer 1

4
$\begingroup$

One could define "Komar-like" quantities for spacetimes which admit Killing vector fields, in general. But, whether they make sense is another question.

It is known that there is no well-defined notion for the mass in asymptotically de Sitter spacetime since in such spacetimes the Killing vector of time translational symmetries is spacelike at future null infinity, and not timelike. All notions of the mass in GR that we know are defined for the (asymptotically) timelike translational Killing vector fields. However, the mass is well-defined in the anti-de Sitter spacetimes because there exists the asymptotically timelike Killing vector fields. In this case, the usual definition of the Komar mass needs a small correction. See for instance this and this, for more information.

$\endgroup$
2
  • $\begingroup$ You just saved my M.Sc. thesis with those refs. Thank you! :D $\endgroup$
    – Y2H
    Commented Feb 4, 2020 at 8:25
  • $\begingroup$ Glad to hear that! $\endgroup$
    – Astrolabe
    Commented Feb 7, 2020 at 14:57

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.