# Can the event horizon of a black hole be distorted or destabilized by an extreme spin rate

Assuming that there is a means of increasing the spin rate of an existing black hole, what would happen to the shape of the event horizon as the spin rate was increased? In extremis what would eventually happen at very high spin rates? Could an extreme relativistic spin rate destabilise a black hole if the rotational energy imparted exceeded the energy content of the mass?

Assume that an arbitrarily large power source is available to spin up the black hole.

• Fascinating question that has nothing to do with worldbuilding. Care to clarify? – Chickens are not cows Mar 25 at 23:00
• Distorted yes, destabilized no. – Alexander Mar 25 at 23:01
• "An arbitrarily large power source" to destabilize a black hole like this might translate to "an infinitely large power source..." If a black hole is anything like a planet, then to spin a planet apart you need to get the rotational velocity to exceed the planet mass' gravitational pull. That puts it rotating at escape velocity. Escape velocity for a black hole is, well, at least c. However, you might find better responses on Stack Exchange Physics, this probably doesn't fit the Worldbuilding criteria. At least, not as it stands right now. – BMF Mar 25 at 23:01
• @Agrajag. Nothing directly to do with world building, however I assume it might be possible to generate some very unusual environments that might have world building potential. There again perhaps physics would be a better fit? – Slarty Mar 25 at 23:07
• @JustinThyme Frame-dragging exists, as well as inertial observers. – BMF Mar 26 at 0:13

## 1 Answer

The situation you're considering involves a rotating black hole characterized by the parameters $$M$$ and $$J$$, the mass and angular momentum of the black hole. The two are encapsulated in something called the Kerr parameter $$a$$, given by $$a\equiv cJ/GM$$ where $$c$$ and $$G$$ are the speed of light and the gravitational constant. If we define the parameter $$a_*\equiv a/M$$, it turns out that general relativity predicts the condition that $$a_*\leq1$$, or, equivalently, $$J\leq GM^2/c$$. This is known as the Kerr bound.

It turns out that the geometry of a rotating black hole is complicated. There are two event horizons, $$r_+$$ (the outer horizon) and $$r_-$$ (the inner horizon), given by $$r_{\pm}=M\pm\left(M^2-a^2\right)^{1/2}$$ Here, I'm working with geometrized units where $$G=c=1$$ and we can ignore the constants. The equation tells us that $$r_-. Now, if we let $$a_*>1$$, it turns out that $$r_{\pm}$$ is a complex number, which is unphysical. This is interpreted as implying a naked singularity, which is widely believed to be impossible. This should give you some physical intuition for the Kerr bound.

Now, we've observed systems where $$a_*$$ is very close to the Kerr bound, but doesn't pass it. GRS 1915+05 is perhaps the most commonly-cited example, with one group deriving a lower limit of $$a_*<0.98$$ (McClintock et al. 2006). Different models by the group returned different precise values for $$a_*$$ (although all larger than $$0.98$$ and less than $$1$$). I don't think anyone believes that GRS 1915+05 violates the Kerr bound.

The point of this is that black holes with Kerr parameters close to $$1$$ do exist and are indeed stable. Their event horizons are indeed different from those of non-rotating black holes (as is the case for any rotating black hole, not just those near the limit), so there is distortion but not instability per se.

• Naked singularities are the cosmic equivalent of a hangover piped on top of a week long migraine with a side order of brain freeze. Theoretically possible, but incredibly painful to contemplate and should never, ever happen. – Joe Bloggs Apr 2 at 8:58