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If we want to describe a static spherically symmetric star we can use a metric which matches the Schwarzschild solution with correct mass on the outside of the star but differs from Schwartzschild in the inside of the matter distribution.

Basically we solve the Einstein equations with a source $T_{\mu\nu}$, for instance $$T_{\mu\nu}=(\rho+p)u_{\mu}u_{\nu}+p\,g_{\mu\nu}$$ where $u_{\mu}$ has zero spatial components, meaning it is the velocity in a static fluid (this can also be seen as a consequence of Einstein equations).

Can we do something similar for a rotating star using the metric for a Kerr black hole?

I heard that it is a much more difficult problem and I would like to understand how difficult it is (Is it possible?) and what makes it so difficult.

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    $\begingroup$ Well the fact that you are solving $R_{\mu\nu}-\frac12 R g_{\mu\nu} = T_{\mu\nu}$ instead of $R_{\mu\nu}=0$ is already a step up in difficulty. Another issue is that even $T_{\mu\nu}$ depends on the metric tensor so the only thing that you usually hope on knowing is written in terms of your unknown. In terms of known solutions I don't know if it has ever been solved, however one thing that we can say is that in the limit $J\rightarrow 0$ it should approximate the Schwarzschild interior solution. $\endgroup$
    – Jepsilon
    Commented Jun 23, 2019 at 21:12
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    $\begingroup$ That was comparing the hypothetical Kerr interior to the known Kerr exterior. If we compare the two Kerr ones to the two Schwarzschild ones, added complexity arises due to the reduction in symmetry (Schwarzschild is spherically symmetric and Kerr is axially symmetric). This prevents you from reducing the "degrees of freedom" of the solution so to speak $\endgroup$
    – Jepsilon
    Commented Jun 23, 2019 at 21:16
  • $\begingroup$ @Jepsilon Please consider changing your comments to a proper answer. In general answering in comments is strongly discouraged on Physics SE as it can leave questions unanswered on the system (and prevents the OP from accepting an answer). You can also get more reputation points for an answer than a comment. $\endgroup$
    – StephenG - Help Ukraine
    Commented Jun 24, 2019 at 7:03
  • $\begingroup$ @Jepsilon yeah I would expect those difficulties to arise but I got the impression that other unexpected complications may arise in solving this problem $\endgroup$
    – AoZora
    Commented Jun 24, 2019 at 7:31
  • $\begingroup$ The Kerr metric isn't the unique axisymmetric stationary vacuum metric, unlike the Schwarzschild metric. The generic solution is the Tomimatsu-Sato metric, which is parametrized by the multipolar moments. $\endgroup$
    – Slereah
    Commented Jun 24, 2019 at 8:59

2 Answers 2

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Well the fact that you are solving $R_{\mu\nu}-\frac12 R g_{\mu\nu} = T_{\mu\nu}$ instead of $R_{\mu\nu}=0$ is already a step up in difficulty. Another issue is that even $T_{\mu\nu}$ depends on the metric tensor so the only thing that you usually hope on knowing is written in terms of your unknown. In terms of known solutions I don't know if it has ever been officially solved, however one thing that we can say is that in the limit $J\rightarrow 0$ it should approximate the Schwarzschild interior solution.

That was comparing the hypothetical Kerr interior to the known Kerr exterior. If we compare the two Kerr ones to the two Schwarzschild ones, added complexity arises due to the reduction in symmetry (Schwarzschild is spherically symmetric and Kerr is axially symmetric). This prevents you from reducing the "degrees of freedom" of the solution so to speak.

I don't really know if there are any actual difficulties other than just being more of a chore to work through (even Schwarzschild exterior is quite a challenge let alone anything less idealised). With a quick Google search "Kerr interior solution" there were at least 3 publications in the top three so a solution could be available. However I haven't personally read through them so I cannot really say anything about their validity.

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    $\begingroup$ Changed my comments to an answer by Mr. StephenG's request in the comments. I posted a comment rather than an answer because it was a while since I looked at GR so I was kinda nervous. But I guess since the reception was OK it's worth posting as a full answer. $\endgroup$
    – Jepsilon
    Commented Jun 24, 2019 at 21:26
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    $\begingroup$ Thanks for the effort! On the other hand, I found claims that a rotating interior matching Kerr is currently still not available. For instance on Wikipedia en.wikipedia.org/wiki/Asymptotically_flat_spacetime in the section "Criticism". Unfortunately until now I have not had the time to look into the subject further. $\endgroup$
    – AoZora
    Commented Jun 25, 2019 at 14:27
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It looks like there is already an interesting solution available.

https://arxiv.org/abs/1701.02098

This assumes an anisotropic fluid in the interior and claims that it satisfies the strong energy condition. The only other solution I have seen involves one that had some unphysical properties that I can't recall. Anyway, here is a link just in case:

https://arxiv.org/abs/1705.06496

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