# Metric for a rotating star

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.

• 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. – Jepsilon Jun 23 at 21:12
• 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 – Jepsilon Jun 23 at 21:16
• @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. – StephenG Jun 24 at 7:03
• @Jepsilon yeah I would expect those difficulties to arise but I got the impression that other unexpected complications may arise in solving this problem – AoZora Jun 24 at 7:31
• 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. – Slereah Jun 24 at 8:59

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.