How can Kerr black holes have a 'speed limit' to how fast they can spin?

Question

Obviously, the speed of light is a presumed limit, at least for 'physical' objects moving 'within' ('through'?) spacetime, but...

In recent news some scientists believe they have figured out the approximate angular velocity of our own supermassive black hole, Sagittarius A*, and stated that it is about 0.9 on a scale from 0 to 1....

However, they said that a black hole spinning at a speed of 1 on the scale is not necessarily rotating at the speed of light, let alone faster, because they believe there is an intrinsic limit to how fast a black hole of a particular size (mass? or volume?) can actually spin...

How can there be an absolute upper limit to a black hole's spin?

Also, how could it be limited (or even slowed) by interactions with the outside universe? The surface(s) of a black hole are not entirely 'physical' like normal objects, as these same articles make clear.... So how can a black hole become tidally locked to an outside object or exchange angular momentum with it?

Answer 1

Yes, general relativity does not contain anything preventing spacetime from expanding or moving superluminally (e.g. "warp drive spacetimes"): normally other constraints (e.g. the lack of exotic matter) are needed to keep things from going crazy.

However, the Kerr black hole spacetime has the spin limitation (1) simply because it will not be a black hole for higher spins, and (2) there is no way of spinning up a Kerr black hole beyond it by dumping in spinning particles or fields.

The metric in Boyer-Lindquist coordinates is $$ds^2 = -\left(1-\frac{r_s r}{\Sigma}\right)c^2 dt^2 + \frac{\Sigma}{\Delta}dr^2 + \Sigma d\theta^2 + \left(r^2+a^2+\frac{r_s r a^2}{\Sigma}\sin^2 \theta\right)\sin^2\theta d\phi^2 - \frac{2r_s r a \sin^2\theta}{\Sigma} c dt d\phi$$ where $\Sigma = r^2+a^2\cos^2\theta$ and $\Delta=r^2-r_sr+a^2$, and $r_s$ is the Schwarzschild radius for the mass. Looking for event horizons ($g_{rr}\rightarrow \infty$) gives $r_H =(1/2)(r_s \pm \sqrt{r_s^2-4a^2})$, or, if we rescale into natural units $G=M=c=1$, $r_H=1\pm\sqrt{1-a^2}$. This corresponds to two event horizons, of which the outer is the one we care about.

This shows that if $a>1$ there is no event horizon. As the hole increases spin from a low $a$ the event horizon moves inward until it is at half of the non-spinning Schwarzschild radius. Then there is no longer any event horizon.

The standard hand-wave is that this produces a naked singularity and that is bad, so it must... somehow... not happen. Or at least it is an unphysical solution we should not pay attention to just as we rightly ignore other unphysical solutions (but then again, black holes were once regarded as unphysical, so this heuristic has a bad track record).

So the trivial answer is that $a>1$ solutions are simply not black holes.

Note that this does not happen because something was moving at the speed of light, although it is the case that for $a=1$ the innermost stable circular orbit of a massive object corresponds to the innermost circular photon orbit (kind of! see (Bardeen, Press & Teukolsky 1978) for some of the nontrivial details).

So, formally, there is nothing stopping Kerr solutions from being weird if not "black hole"-ish. However, if we try to create a superextremal black hole by dumping in particles or fields, it turns out that it is impossible. The reasons are a bit subtle. Intuitively, the faster the hole spins the harder it is to get a particle to fall in, and most particles add more mass (lowering $a$) than spin. There are tricks to get it to absorb a test particle and overspin, but backreaction prevents this. This result can be strengthened to more general cases.

Here the speed limit is not due to anything moving faster than light, just the "conspiracy" that it becomes impossible to dump more spin into the hole as it becomes extremal. And the reason for that is that the spacetime just gets the wrong shape for letting anything in (there are no geodesics we can throw paricles along to be absorbed).

Answer 2

As an addendum to @Anders Sandberg's excellent answer: The maximum limit on the spin is a consequence of two (not fully proven) conjectures that each imply the existence of this limit individually.

The first is the weak cosmic censorship conjecture. In short what this conjecture says is that all singularities in general relativity should be hidden ("censored") by event horizons. Consequently, observers that do not themselves venture into these regimes are never confronted with "naked" singularities. As stated in Sandberg's answer, when $a>1$ the horizon in the Kerr solution disappears. That answer also referred to some results in the literature showing how this censorship is enforced by nature for Kerr black holes. Notably, proving these results is extremely complicated and technical rather than following from some rather straightforward argument.

The second principle is the third law of black hole mechanics. This conjecture states that it is impossible to construct a black hole with surface gravity (and therefore Hawking temperature) equal to zero. The surface gravity of Kerr black hole $\kappa$ is given by $$ \kappa = \frac{\sqrt{1-a^2}}{2 r_+}.$$ So, the surface gravity goes to zero exactly when $a\to 1$. So the third law forbids $a$ from ever reaching 1 (let alone getting even bigger). The results cited by Sandberg, again show how this is enforced in practice (although they do not actually exclude the possibility of reaching surface gravity exactly zero). Again, that Kerr exactly obeys this conjecture appears highly non-trivial.

Answer 3

Spin too fast and you have a "naked" singularity

Basically, yeah, that's exactly what's limiting the rotational speed of a black hole.

Rotating black holes, also known as Kerr black holes, have a spin parameter value that is between either 0 and 1. A black hole with a spin parameter value "0" is a Schwarzschild black hole, i.e. a "static" black hole, that does not spin$^1$.

A spin parameter value of "1" is the absolute limit of rotating black hole. That is because $SPV$ "1" is equal to $1c$, or about 100% the speed of light, which is impossible to attain by nearly every simple object in the universe, forget black holes.

For example, the Milky Way's supermassive black hole, Sagittarius A* has a SPV of 0.1, which means it's spinning at about 10% the speed of light. Pretty impressive, but dull compared to some other black holes.

M87 has a SPV of about 0.90 ± 0.05, which means it's rotating at about 90% of the speed of light, extremely rapid, but just a short way from the maximum speed limit.

However, here is where really weird things start to happen.

Normally, when a black hole rotates, it drags the spacetime along with it, forming a donut-like region of warped-spacetime, known as the ergosphere:

However, if you make the black hole spin real fast, then the ergosphere starts flattening out. This effect causes the event horizon of the black hole to shrink. If it speeds up, it shrinks even more.

Eventually, at SPV 1 and above, the event horizon converges at the singularity.

There you have it, a singularity in the nude (No NSFW reference, what are you talking about?). This naked singularity is highly erratic, as it can randomly spew out completely random (yes I am repeating that TWICE) pieces of information. To give you an idea of how erratic a naked singularity is, here is this graph/comic/idk? thingy that illustrates it.

But, as explained, that cannot happen. You need a black hole to spin at the maximum speed limit, which is impossible, as it requires it to rotate at the speed of light or even higher, which cannot occur in nature.

So there you have it.

Moral: If you're a black hole, don't spin wayyy fast, or you might become "naked".

Some videos which can provide more information on this topic, if you are interested: 1 and 2.

_{1. It can also be a static Reissner-Nordstrom black hole, one that is charged. Charged black holes can also exist, one which possess either a positive or negative charge. Also, charged black holes can rotate as well.}