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

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 $a_{\ast }\equiv a/M$, it turns out that general relativity predicts the condition that $a_{\ast }\le 1$, or, equivalently, $J\le G{M}^2/c$. 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 $$r_{\pm }=M\pm {(M^2-a^2)}^{1/2}$$ 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 $a_{\ast }>1$, 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 $a_{\ast }$ 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 $a_{\ast }<0.98$ (McClintock et al. 2006). Different models by the group returned different precise values for $a_{\ast }$ (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.

posted about 6 years ago

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