Why is the mass of a Kerr black hole proportional to its angular momentum? I'm a third year mathematics undergrad, and have just started the module General Relativity and spacetime geometry, I also have a keen interest in black holes.
However I would like to know why and how the mass of the Kerr black hole is proportional to it's angular momentum, and also inversely proportional to it's Schwarzchild radius?
 A: Your statement is not true.  First, note Sofia's point that $J = Ma$, where $a$ is the angular momentum parameter inserted into the standard Kerr solution.  Then to see that the claims in the OP are wrong, simply note that as $a\rightarrow 0$, the angular momentum goes to zero, but the mass does not.  Meanwhile, the radius of the black hole horizon (I hesitate to say schwarzschild radius for a non-schwarzschild black hole) is given by:
$$r = M \pm \sqrt{M^{2}-a^{2}}$$
Which does not have a simple proportionality/inverse proportionality relationship with the angular momentum or the angular momentum per unit mass parameter $a$.
A: The proportionality between angular momentum and mass of the Kerr black hole can be shown directly by performing the Komar integral for angular momentum. In fact, you find that $J=Ma$. The parameter $a$ in the Kerr metric is thus the angular momentum per unit mass.
I am not familiar with the relation between $M$ and $R_S$
A: arXiv:gr-qc/9501002. From this essay you could find that angular momentum is proved to be proportional to its mass.
