It is well known that the limit on the angular momentum for a Kerr black hole is given by \begin{equation} 0 \leq a^* \leq 1 \quad \text{where} \quad a^* \equiv \frac{cJ}{GM^2} \end{equation} and $J$ is the angular momentum and $M$ is the black hole mass. Now, $a^*$ can exceed unity for other objects just not black holes.

Suppose however that I have an object with $a^* > 1$ and I confine this object to fit into its Schwarzschild radius without changing $J$ and $M$ and therefore $a^*$. What will happen since I cannot have a black hole with $a^*$ larger than 1?


1 Answer 1


This is an interesting question that could really be formulated as follows:

When collapsing an object that has the ratio $a^* = cJ/(GM^2)$ larger than one, can an over-extremal black hole form?

There is no rigorous proof, but the answer seems to be no, at least in reasonable physical situations. The simple reason for that is that when you have an object with angular momentum, it wants to keep rotating, and it is kept from falling to the center by this tendency (this is often summarized by the statement that the "centrifugal force" keeps it from falling to the center).

If you make an object rotate a lot, it will fall apart and you simply cannot bring it to collapse. If you do manage to collapse some of an object that has a total $a^*>1$, it turns out that you always collapse a sub-partition of matter with $a^*<1$ and a another part of the matter with $a^*>1$ flies away (think supernova remnants).

Take for instance a neutron star, the object closest to a black hole in terms of its radius as compared to the gravitational radius (Schwarzschild radius). If you make the neutron star rotate beyond $a^*\gtrsim 0.6$, it will start shedding mass, since it is unable to keep the rotating matter from flying away by its immensely strong gravitational force.

Similarly, you could find many similar examples where all kinds of physical processes act against having $a^*>1$ as the object contracts to collapse radii. The point really is that to have $a^*\sim 1$ at the point when the matter is just reaching inevitable collapse, the matter has to be reaching speeds close to the speed of light. This provides many opportunities for it to pass its angular momentum to other matter in its vicinity.

Another, perhaps most salient point is that until the horizon has formed, there is no reason for the matter to contract inevitably. There is always the option to fly out by some physical process. However, matter with $a^*>1$ never has a horizon formed around itself, so there is really no inevitable reason to why it should form a naked curvature singularity (the over-extremal Kerr space-time with $a^*>1$).

(This is not to say you can never form over-extremal Kerr singularities in some scenarios, but it seems that it requires an artificial idealized setup that has essentially been designed to create the curvature singularity by a pathology that was in the setup essentially from the beginning. If you consider physical setups, you do not get these results.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.