From Andrew Hamilton's black hole image collection
There are, for most practical purposed, two qualitatively distinct types of black holes: Schwarzschild - which are spherical and not spinning and Kerr - axisymmetric and spinning
So... I know how to spin up a non-spinning black hole, you drop matter in with some finite angular momentum.
I also know how to extract angular momentum from a black hole, eg through the Penrose process
What I do not know is how, or if, you can go exactly back to a non-spinning black hole.
Why is this? (J. Pedersen pointed this issue out to me and figured out many of the concerns, and possible solution approaches).
Classically this is trivial, the spin-up and spin-down processes are symmetric - or, trivially, you can just add matter with exactly the opposite angular momentum to despin the black hole.
But, relativistically there is a non-trivial issue: the Schwarzschild solution is not a smooth limit of the Kerr solution as the spin parameter goes to zero.
This is true for several reasons: for one thing, the Schwarzschild singularity is spacelike and the Kerr singularity is timelike!
Another issue is that the Schwarzschild singularity is a point, the Kerr singularity is a ring: I can take a point and wiggle it to make it into a ring, so spin-up is ok; but how to do you "open" a ring singularity to shrink it back to a point? You have to change the topology of space-time to do so and there does not seem to be a smooth way to do it and a discontinuous change seems physically unsatisfactory.
One possibility is that Schwarzschild solutions don't actually exist, they are always minimal Kerr solutions, but that makes a strong quantum mechanical statement, roughly equivalent to it being impossible to reach zero spin stationary states. (Since the problem is true independent of mass...)
Alternatively we can build a zero spin black hole but for some subtle reason the despin can never be exactly complete (since classically the sign of the angular momentum of mass added can be either sign this is not satisfactory).
I don't see a nice solution.
But, this really gets interesting when you think about a Kerr black hole evaporating: how do you reach the final Planck state without either violating the maximal spin limit or introducing correlated emission so as to conspire not to avoid spin bounds? I can see some conceivable solutions, but none are very satisfactory and serious considerations seem to imply either violations of thermodynamic principles, or holography.
One glimmer of a solution is that the Kerr solution is striclty valid only for static metrics (and uniquely so), whereas the final stage of evaporation is not static - this does not work for the broader issue of despin, since that can be done as classically and quasi-statically as we please.
I'm puzzled, this is one of these niggling issues that has started to really bother me.
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Interesting, one possible way out is that the inner horizon of the Kerr Black hole (and consequently the region behind it) is known to be unstable to perturbations. So a small time dependence on top of the Kerr black hole (which you'll need in order to spin it down) will turn the inner horizon into spacelike singularity, and then the timelike ring singularity is unphysical. In that sense the slightly perturbed Kerr black hole is not that different from Schwarzschild, and the spin-down is a continuous process.
Makes sense?
(see, you got me worried...)
Remind me, is the inner horizon unstable to infinitesimal perturbations?
Because the angular momentum transfer can be done as slow as you like; I guess it reduces to the "why you can't thread the middle" problem, so yes?
Even if you can turn the inner horizon spacelike, I still don't understand the topological change at the end - how do you "open the ring" in the Kerr singularity and map it to a point?
This is worrying me too, since I can see how to start wiggling the point into a ring as you spin it up, and the operation classically must be reversible.
I'm inclining to the "there is no zero spin state" solution, ie that the exact Schwarzschild is unphysical. But that has worrying implications for quantum mechanics.
Well, not if it is stringy I suppose. Then we're just left with the problem of the extremal black hole, I'll get to that later. Requires thought.
Yep, the inner horizon is surface of infinite blue shift, so even a single particle will destabilize it. There is some numerical evidence that generically the inner horizon will be hidden behind a spacelike singularity. I would just say then that the spacetime structure of the unperturbed Kerr BH is unphysical, and with infinitesemal perturbation it is more or less the same as the unrotationg case.