# Where is the singularity located in a Kerr black hole?

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

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.

• 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? Jan 4, 2020 at 2:08
• @ReneKail: what are "spherical coordinates" in Kerr spacetime? The underlying geometry is not spherically symmetric. Jan 6, 2020 at 14:29
• 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? Jan 6, 2020 at 19:13
• @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. Jan 7, 2020 at 18:26
• 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 $\pi/2$ Jan 7, 2020 at 18:32