# Why does kerr black holes have a ring shaped singularity rather than point?

I read about ring singularity on Wikipedia and didn't quite understand the flow of liquids inside a rotating body causing a ring shaped singularity with zero thickness. Moreover is it specifically ring shaped and not torus shaped in any case?

• There is an answer explaining (or outlining) the reason for the ring singularity as a result of the form of the Kerr metric and it's in answer to this question on Physics SE – StephenG Mar 30 '18 at 8:45

Firstly do not confuse the event horizon with the singularity. Wikipedia gives the formulae for the Kerr metric. There are a number of places where this formula appears to break down because you appear to be dividing by zero, essentially whenever $\Sigma = 0$ or $\Delta = 0$, in the notation of that page, corresponding to the ergosphere and the event horizon respectively. There are two event horizons and two ergospheres but they all have the topology of spheres, one inside another (with some touching at the poles). However those are not singularities in general because you can choose other coordinates for space and time where the division by zero goes away. However you can't do this at points on the equator of the inner ergosphere, which makes those points a ring singularity. It is a ring with zero thickness (at least until quantum effects come in) .There is a helpful picture