# What is the action for a rotating blackhole with angular momentum $J$?

The action for an electromagnetic theory is given by $$S_{EM}[g^{\mu\nu},A^{\mu}]=-\frac{1}{4\mu_0}\int F_{\alpha\beta}F^{\alpha\beta}\sqrt{-g}d^4x$$ and the action for charged relativistic dust is $$S_q[x^{\mu},A^{\mu}]=-\int\rho_{EM}v^{\mu}A_{\mu}\sqrt{-g}d^4x$$ where $A^{\mu}$ is the electromagnetic field (potential), $g^{\mu\nu}$ is the gravitational field, $\rho_{EM}$ is the charge density of dust, $v^{\mu}=\frac{dx^{\mu}}{dt}$ is the $4$-velocity of dust, and $F_{\alpha\beta}=\nabla_{\alpha}A_{\beta}-\nabla_{\beta}A_{\alpha}$ is the electromagnetic field tensor. Using these as general references for the form of an action, is it then possible to write down an action for a rotating blackhole with angular momentum $J$?

Thanks in advance!

• FWIW, the Kerr metric extremizes the Einstein-Hilbert action. – Qmechanic Mar 26 '18 at 17:08
• Thanks. Do you have any references which show this? @Qmechanic – Sergio Charles Mar 26 '18 at 17:29
• Pretty much any textbook in GR? The Kerr-Metric solves the vacuum Einstein equations, which can be obtained by varying the Einstein-Hilbert action. Hence the Kerr metric extremizes the Einstein-Hilbert action. – mmeent Mar 27 '18 at 7:56
• Yes, I know how to derive the vacuum EFEs from the Einstein-Hilbert action. However, I am asking if one may find gμν due to Kerr exactly from the EH Formulation. But, in retrospect, it is a silly question. @mmeent – Sergio Charles Mar 29 '18 at 0:04