Kerr black holes (and Kerr-Newman black holes), instead of the "point" singularity theorized in spherically symmetric black holes, instead have a "ring" singularity, spread along the equatorial plane of rotation. any particle inside a spherically symmetric black hole will tend to go towards the center. But what if a particle is in the exact center of a Kerr black hole? Will it go towards any point in the ring singularity, or will it stay where it is, unless it is perturbed in one direction?

  • $\begingroup$ With a particle in the center, the setup has the same symmetry as the Kerr solution. So it would be an equilibrium, although likely an unstable one (I am not sure how to prove it is unstable, so also look forward to the answers). $\endgroup$ – CuriousKev Sep 28 '14 at 2:31
  • $\begingroup$ Here: en.wikipedia.org/wiki/Kerr_metric#Kerr_black_holes_as_wormholes it sounds like the interior solution is not well understood. So I guess someone would have to figure out what a reasonable stable solution matching to the Kerr exterior would be, before anyone could discuss particle motions in this region. $\endgroup$ – CuriousKev Sep 28 '14 at 2:47
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    $\begingroup$ Since the interior of a stellar sized black hole can not be probed, we may never know directly. We may be able to infer it theoretically, but not with any theory that is available today. $\endgroup$ – CuriousOne Sep 28 '14 at 5:36
  • $\begingroup$ I am not sure if you can even analytically describe the center....so the only possibility is a numerical calculation, which should imply that the particle will move only when perturbed, as you cannot have a well defined Lagrangian to track the particle's motion through the singularity. My guess is it will keep falling through concentric rings so that ultimately, it is at a definite point equivalent to the center of a spherically symmetric black hole.....but it is purely a guess. $\endgroup$ – GRrocks Oct 3 '14 at 9:35

A maximally extended Kerr solution has multiple horizon. The outer horizon is an event horizon and the inner horizon is a Cauchy horizon.

Outside both is a normal type region of spacetime. In between the two the r coordinate is a timelike coordinate so you must continue decreasing r if you entered with a decreasing r.

But inside the Cauchy horizon means the singularity is on your past light cone. So its not really sensible to say you know what would happen.

Another indicator of a problem is that on the other side of the ring singularity is a region with closed timelike curves. So you can see a singularity (what does that look like) and you can see a region beyond it with time travel (what does that look like). But you can still have a manifold and a metric, this is after all a solution for an external black hole so it just was always there, spinning around.

If you look at the center of the ring you can have a geodesic that stays at the center. But is that really meaningful?


The particle would be in an unstable equilibrium (there would be no prefered direction if you are at rest), but if it were slightly off center it accelerates outwards (gravity is repulsive there, see the d²r/dτ² in the numeric display). If I start at r=0, θ=0 the particle stays where it is, but if I displace it slightly (r=1/1000GM/c², θ=1/1000rad, v=0) it accelerates outwards.

The escape velocity (to escape the time reversed white hole, see the last paragraph) in the center goes to 0, so a very small displacement is already enough to shoot the particle away. The coordinates used are Kerr Schild (click on the image to enlarge):

black hole off center trajectory

The frame dragging rotation in the center also vanishes, so if the test particle is small enough it may stay in the exact center without spinning, see ω fdrag and v fdrag (angular and true frame dragging velocity) in the numeric display (the local velocity components on the bottom right are relative to local ZAMOs, which are possible inside the inner ergosphere).

The acceleration is always in the positive r direction, since when you pass through the ring into negative space the sign of gravity is reversed. So if the displacement is in direction of the negative space you still get accelerated in the direction of positive space.

Then the particle goes back up again until it reaches the inner side of the Cauchy horizon, where its proper time goes to zero while the Finkelstein time continues, which means that the particle gets hit by all the light that falls into the black hole at once, so he gets fried due to the infinite blueshift.

If you somehow survive the infinite blueshift on the Cauchy horizon your proper time gets reversed relative to the universe after you saw all its future in one moment, and you can escape the black hole again, which is now a white hole in your frame of reference. But that happens far away from the center, where everything should be fine for the observer.


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