In a rotating Kerr black hole is the ringlike singularity situated between the inner and the outer event horizon of the black hole?


1 Answer 1


No, it is inside of the inner horizon, located at Boyer-Linquist coordinate $r=0$ (note that these coordinates do not have the same coordinate singularity at $r=0$ that standard spherical coordinates have)

One might ask, how is this a ring, then? The easiest way to show this is to realize that if you set $M=0$, the Boyer-Linquist metric has no curvature singularity, and if you take $ r=0, M=0$, then

$$ds^{2} = -\frac{\Delta}{\rho^{2}}\left(dt - a^2\sin^{2}\theta d\phi\right)^2 + \frac{sin^{2}\theta}{\rho^{2}}\left(\left(r^{2} + a^{2}\right)d\phi - a dt\right)^{2} + \frac{\rho^{2}}{\Delta}dr^{2} + \rho^{2}d\theta^{2}$$

where $\rho^{2} = r^{2} + a^{2}\cos^{2}\theta$ and $\Delta = r^{2} + a^{2} - 2Mr$

simply becomes

$$ds_{\rm ind}^{2} = -dt^{2} + a^{2}\sin^{2}\theta d\phi^{2} + a^{2} \cos^{2}\theta d\theta^{2}$$

Finally, realizing that the coordinate singularity only happens at $\rho=0$, which requires that you have $\cos\theta = 0$, and setting $t=constant$, you have:

$$ds^{2}_{\rm ind} = a^{2}d\phi^{2}$$

which is pretty obviously the metric for a ring of radius $a$.

Oh, to finish this and show that this is definitely inside the inner horizon, remember that the horizon is located at the location $\Delta = r^{2} + a^{2} - 2Mr = 0$, which the quadratic equation gives as located at:

$$r = M \pm \sqrt{M^{2} - a^{2}}$$

which has an inner r value greater than zero unless $a=0$, in which case, we have a Schwarzshild black hole, which is known to not have an inner horizon.

  • $\begingroup$ I posted the above question because I learned that r(ring) = aM, that is always larger (or equal) than the inner horizon distance measured from the center of the Kerr BH in spherical coordinates. Can this be considered correct or not? $\endgroup$
    – Rene Kail
    Jan 4, 2020 at 2:08
  • $\begingroup$ @ReneKail: what are "spherical coordinates" in Kerr spacetime? The underlying geometry is not spherically symmetric. $\endgroup$ Jan 6, 2020 at 14:29
  • $\begingroup$ From the mathematics above it follows that the ring singularity in the equatorial plane of the Kerr BH has a radius R = a . But calculation shows that this radius is always larger (or equal) than the radius r- of the inner event horizon. How then is this compatible with the statement that the ring singularity lies within the inner event horizon? $\endgroup$
    – Rene Kail
    Jan 6, 2020 at 19:13
  • $\begingroup$ @ReneKail: the inner horizon is not a sphere (in fact, for sufficiently large $a$, the Kerr horizon doesn't even embed into $R^3$). and the $r$ coordinate of the inner horizon is always larger than the value $r=0$ for the ring singularity. These are not flat coordinates in flat spacetime we're talking about here. $\endgroup$ Jan 7, 2020 at 18:26
  • $\begingroup$ in particular, near the ring, the $\theta$ coordinate ceases to behave like an angle. One "passes through the ring" by going through $r=0$ while taking a $\theta$ value other than $\pi/2$ $\endgroup$ Jan 7, 2020 at 18:32

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