# Rotating black hole: fall along symmetry axis

I understand rotating black holes are very complex to describe. Yet, I have question based on few statements I believe true:

-Singularity is circular
-Symmetry should be conserved
-Energy should be conserved

If a particle in vacuum falls along the rotational axis into a rotating black hole, it cannot hit singularity, because of symmetry. Symmetry axis passes through the center of the circular singularity, it does not pass through singularity.

If the fall is in vacuum the particle cannot dissipate energy, it does not hit the singularity.

As a necessary consequence It seems to me that a particle should go outside the black hole on the other side.

I know a non-zero event horizon exists even for rotating black holes, so there is something wrong with my arguments.

Can anybody clarify the situation?

• Singularity is circular. Axis goes through the center of the circle, which is not singularity, singularity is the circumference of the circle. A particle on the axis has no preferred direction to go off the axis to meet singularity. Commented May 17, 2017 at 7:36
• +1. Yes, it makes sense. I assumed that by circular you meant spherical, I had never heard of this before!
– user126422
Commented May 17, 2017 at 10:52

You are quite correct. An object following an axial trajectory into a Kerr black hole passes through the centre of the singularity and out the other side. It re-emerges from the black hole into a region of spacetime commonly called the Kerr negative universe. In fact the trajectory doesn't have to be exactly axial as any trajectory close enough to the axis will also do this.

This is possible in all charged or rotating or both black holes. While we're used to the idea that nothing can escape from a black hole this is an oversimplification. For example see my answer to Entering a black hole, jumping into another universe---with questions.

But ...

You need to be clear that the Kerr metric, like the Schwarzschild and Reissner-Nordström metrics are mathematical ideals that do not exist in the real world. These metrics describe time independent objects i.e. objects that have existed for an infinite time in the past and will continue to exist for an infinite time into the future. Since our universe is of finite age there are no Kerr black holes - only objects closely approximated by the Kerr metric.

Plus it's far from obvious that even in the ideal case the trajectory in and out of the black hole could exist. We generally take the black hole metric as a fixed background i.e. we assume the infalling matter doesn't perturb it. But for the trajectories that pass in and out of the black hole it isn't clear this is a safe assumption.

And finally the Kerr negative universe has some rather odd properties and it isn't clear that it's physically meaningful. I have to confess I don't know much about this so I can't comment further, though I imagine Googling would turn up more information if you're interested.

In any case we will never observe an object passing through a black hole because for us the object takes an infinite time to even reach the horizon let along pass through it. While the maths tells that the falling object reaches and passes the horizon in a finite (and short!) time as measured by any clocks they carry what happens to them after that is irrelevant to us.