Let's say we try to remove the event horizon of a Kerr black hole by throwing in matter with some large angular momentum. If it starts with GM > a, could we increase a at all? Would such a particle be able to enter the black hole?
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1$\begingroup$ There's no known process by which this can happen using non-exotic matter without having the additional angular momentum end up being shed off in the form of gravitational radiation. $\endgroup$– Jerry SchirmerMay 2, 2014 at 15:47
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2 Answers
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To expand on Jerry's comment a bit, there is a review article that describes the underlying physics here (caution PDF). The article is based on R. M. Wald's paper Gedanken experiments to destroy a black hole, Ann. Physics 82 548-556, though the paper is behind a paywall so that's not much use unless your college has access. The article actually describes a Kerr-Newman black hole, where the infalling particle can be charged and the attempt to create a naked singularity is to both increase the angular momentum and the charge.
As I'm sure you'd have guessed, the calculation is involved, but basically Wald concludes that you can't add enough charge because the field of the Kerr-Newman black hole ends up repelling your charged particle. You can't add enough angular momentum because the extra angular momentum stops the particle falling through the horizon.
answered May 2, 2014 at 16:17
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Regarding violations of cosmic censorship, I suggest the paper Black Strings and P-Branes are Unstable by Gregory and Laflamme in 1997 available here which explores an instability of solutions in type II string theory by employing standard perturbation theory to linear order. They note in the conclusions:
It's a more exotic approach to obtaining a naked singularity, but if these solutions do exist in Nature, e.g. black strings (first proposed in Kaluza-Klein theory),
$$\mathrm{d}s^2 = \underbrace{\mathrm{d}S^2}_{\text{Schwarzschild}}-\underbrace{\mathrm{d}\sigma^2}_{\text{extra dimension}}$$
they may lead to naked singularities through their instability. The thermodynamic argument for the instability itself is that other configurations are entropically favorable.
