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?

• The mass and the angular momentum are independent properties. But there is a lower bound on the ratio (M^2)/J, below which it's not a black hole any more, because something that falls in isn't irreversibly trapped (no event horizon). It would have something to do with the twisting of space (frame dragging) due to the spinning, creating escape trajectories in this "overextreme" case en.wikipedia.org/wiki/Kerr_metric#Overextreme_Kerr_solutions ... The question deserves an exact answer. – Mitchell Porter Jan 24 '14 at 21:47
• As for the relation between mass and radius, the "radius" is the distance from the center to the event horizon, i.e. to the point of no return for objects that fall in, the place where the escape velocity equals the speed of light. If a black hole gains mass by absorbing something, its gravitational field will be stronger (because there is more mass) and it will get bigger, the point of no return will now be further out. So the radius goes up with the mass; it's directly, not inversely, proportional... – Mitchell Porter Jan 24 '14 at 22:09
• As for why the Schwarzschild radius is proportional to the mass (and not e.g. proportional to some power of the mass), well, gravitational force is proportional to mass even in Newton's theory. It would have something to do with that, but expressed in the language of general relativity. Again, this deserves an exact answer... – Mitchell Porter Jan 24 '14 at 22:23
• But explaining the exact constant of proportionality may be more difficult. See the discussions at physics.stackexchange.com/q/33473 ... the explanations are a bit stiff and unilluminating, and the exact calculations only apply in the "weak-field" case. So there are mathematical depths here that are not yet understood... – Mitchell Porter Jan 24 '14 at 22:26
• ... and I would guess that the real answer may come from the holographic principle and the "Kerr/CFT duality", which is still a subject of research... I may try to turn all this into a proper answer. – Mitchell Porter Jan 24 '14 at 22:27

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 $a$.

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 $R_S$

arXiv:gr-qc/9501002. From this essay you could find that angular momentum is proved to be proportional to its mass.

• arxiv.org/abs/gr-qc/9501002 – beats angel Jun 5 '18 at 11:40
• Hi, welcome to physics SE! Please post single-link answers as comments instead. If you don't have the reputation to do that, try answering other questions first. – user191954 Jun 5 '18 at 12:40