Angular momentum for the Kerr solution of a rotating blackhole

I am reading 't Hooft's noted on Black holes, where he quotes the Kerr metric for a black hole rotating about the z-axis as follows:

He later says: "The parameter a can be identified with the angular momentum i.e. $$J=aM$$"

How can I prove this statement? How is the angular momentum tensor defined for an arbitrary metric satisfying Einstein's equations? Do, I have to find all the components of $T_{\mu \nu}$ by first calculating the curvature tensor or is there an easier way?

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The Kerr solution is a vacuum solution to Einstein's equation. If you calculate all of the values of $T_{\mu \nu}$, you will find that they are all zero. To calculate the angular momentum of the spacetime, you will need to use the ADM formalism on a constant-time slice of the metric, and then calculate what the generator of rotational translations on the ''sphere at infinity'' is. Since we classically equate this with angular momentum, we can therefore conclude that this generator is the angular momentum contained within the spacetime.
It should be noted that as we've fixed the coordinates in a sensible way, we only need to deal with the connection between the scalar quantity $aM$ and rotations in the $\varphi$-direction. –  Chris White Jun 25 '13 at 18:07