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It is known that - When a star collapses during the formation of the black hole, the black hole obtains the spin of the star which it collapsed from...

What I'd like to know is, If this spin accelerates as a result of angular momentum (if any), What effects could this rapid rotation have on the black hole, its gravity or anything else around it?

Waffle's Crazy Peanut
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Alex Voinescu
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The black hole gains the angular momentum of the matter that falls into it, minus the anglar momentum of the outgoing gravitational radiation. There are a lot of questions currently on the site regarding the effects of a spinning black hole. The primary one is that the black hole will cause the space around it to co-rotate with the hole.

Zo the Relativist
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This is question is both open theoretically and experimentally, although there is some astrophysical data.

After a black hole forms, the matter is squeezed as it falls into an accretion disk, and the matter that is absorbed is rotating very fast by the time it falls in. The dynamics of the accretion disk are complicated, because there are horrendous fields caused by ionization and charge separation. This makes it difficult to achieve consensus about whether the matter falling into a small black hole causes its spin to slow down or speeds up.

The experimental consensus seems to currently be that the black holes that power active galactic nuclei are spinning at close to extremality, meaning they have close to the maximum amount of angular momentum for their mass. One reason to expect this is that, absent quantum mechanics, the black hole can only generate energy to the extent that it is spinning. The AGN's are generating a huge amount of energy, so something must be maintaining the spin, and this should be the infalling matter.

It is my opinion that there is still uncertainty regarding the emission of highly spinning black holes, because there is no certainty about what happens to infalling matter. One has reasons to suspect that this matter is emitted from the black hole nonthermally, more or less as it came in, after doing a traversal of the interior regions. If this is so, you must take into account the spin-up/spin-down effects of the in-out matter, and this requires the classical limit of quantum black hole, something which is nearly, but not quite, available, thanks to string theory.

Ron Maimon
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  • I have never heard a single person talk about quantum gravitational effects being important for AGN. SMBH should be very, very, nearly classical at the horizon. – Zo the Relativist Sep 03 '12 at 03:37
  • @JerrySchirmer: You haven't heard it because I am the only one who says it, so stackexchange has the scoop. The quantum gravity is for coming out of a black hole. You can't calculate coming out classically. It doesn't come out semiclassically in this universe, but there is no other universe for it to come out in, so if it comes out, it must be in this universe. This process cannot be comprehended classically. Most people will tell you it doesn't come out quantum mechanically either, but bollocks, I say. – Ron Maimon Sep 03 '12 at 05:49
  • Numeric simulations plus the analytic solutions to the Kerr geodesic equation can produce the jets. And how do you get around the fact that the surface gravity, and thus the temperature of a $10^{6}M_{\sun}$ BH is essentially zero? – Zo the Relativist Sep 03 '12 at 14:22
  • @JerrySchirmer: "Surface gravity is zero" == "extremal black hole". Analytic solutions don't produce the jets, they produce the Penrose process which allows the jets in principle. I don't believe there are any simulations of jets, but whether there are or aren't, I think stuff falling into a cold black just comes out again. – Ron Maimon Sep 03 '12 at 17:39
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    Nope. The surface gravity is given at http://en.wikipedia.org/wiki/Surface_gravity#The_Kerr-Newman_solution. The larger the mass, the smaller the surface gravity, irrespective of the spin of the BH. And stuff will come out again, at very, very late times, and thermalized to the horizon temperature, which, once again, is very very low. The Kerr geodesic equations are very complex, and you can get jets from them. – Zo the Relativist Sep 03 '12 at 17:51
  • @JerrySchirmer: I know the standard story, and "nope" on your "nope". I am saying something different from what you heard at school. I am saying that the stuff comes out nonthermalized, as whatever crap it came in. I know you can get jets from Kerr geodesics, but these slow down the rotation, as Penrose energy extraction always does. In order to keep it near extremal, so that the energy production is always possible, you must assume that the accretion mechanism magically pumps up the black hole spin, even when it is near extremal. This is not supported by detailed simulations that I know of. – Ron Maimon Sep 04 '12 at 03:39
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    you don't know that. It's highly speculative, and not tied to any calculation I know of, or that you've provided. Extreme conclusions require extreme evidence. This answer is not science. – Zo the Relativist Sep 04 '12 at 03:48
  • @JerrySchirmer: It's my own calculations, and, in my opinion, it's not speculative. I think it is 4 sigma likely. I gave a sketchy argument here: http://physics.stackexchange.com/questions/35506/reasons-to-suspect-that-matter-is-emitted-from-black-holes-nonthermally/35518#35518 , and, if you want, I can show you how the geodesic go in and out. This solves mysterious things in AdS/CFT, but what I don't know (I wish I did) is the precise gluing map that tells you when objects come out exactly. This makes it vague, but I can't hide it, because I am pretty sure it's right. – Ron Maimon Sep 04 '12 at 03:56
  • Spurred by this question, I tried it today, and I just reproduced some things I did a few years ago (geodesics wandering around in and out of an extremal/near-extremal BH). The problem is that one needs a pure-exterior dual, which I am not sure is classical, in order to say what is the relation between the in-out frequency in proper time and the in-out frequency in asymptotic t-time. I think that this can be analyzed from classical normal modes, but these I did not compute for extremal RN. The higher dimensional branes are exactly analogous, but with dilaton and forms, and that's complicated. – Ron Maimon Sep 04 '12 at 03:59
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    Unpublished results that contradict what the entire astrophysics community thinks about realistic, physical black holes with no observational evidence? That's going to require more for me than vague heuristics on stack exchange. Also, the two horizons converge for extremal BH. – Zo the Relativist Sep 04 '12 at 15:24
  • @JerrySchirmer: It may require more for you, but it does not require more from nature. I figured out how to calculate the gluing yesterday, you can do it classically, and this is why I bumped up the certainty to 4-sigma. When I have the gluing, I'll be completely certain. The heuristics are not vague--- I can tell you the oscillation frequency of a geodesic as a function of proper time as a function of it's charge and angular momentum around the black hole. To find the gluing, you need the same oscillation frequency in t for two black hole merger, with one smaller than the other... – Ron Maimon Sep 04 '12 at 16:04
  • ... this is analytically solvable (shockingly) for extremal black holes, since the two-extremal-static black hole case has an exact solution which is simple. The oscillations in this case do not involve horizons joining, but the small horizon gets smeared on the larger, then bounces out. The frequency as a function of L and q can be calculated in t, and matching this to the proper time calculation gives you the gluing map (for extremal BHs). I calculated enough of this to be sure the gluing is consistent, but not enough to actually do it. The geodesic calculation is extremely simple. – Ron Maimon Sep 04 '12 at 16:07
  • @JerrySchirmer: You should also know that the two horizons do not converge at extremality--- they have the same "r" but the distance between them remains finite. This is because the metric in dr diverges. The geometry becomes cylinder-like at the extremal value. The downvotes are expected, but if you are downvoting this, you don't know what you're talking about. – Ron Maimon Sep 06 '12 at 01:14
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    @RonMainmon: not my downvotes. But I don't exactly feel sorry for you here. – Zo the Relativist Sep 06 '12 at 03:32
  • @JerrySchirmer: Don't feel sorry for me, figuring this stuff out has made me the as euphorically happy as I have ever been. I feel a little sorry for the downvoters, because they don't understand it, and a little worried about this site, which truly has the potential of being something extraordinary, but it requires enlightened voting. – Ron Maimon Sep 06 '12 at 05:36
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    write a technical level paper and put it on the arxiv. Stack Exchange explicitly prohibits posting of your own original unpublished research, particularly without stating that you are doing this. Also, if you have things going into the BH and leaving it, you have a CTC problem in the spacetime. – Zo the Relativist Sep 06 '12 at 15:37
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    @JerrySchirmer: I am wary of using Arxiv, as I don't have Arxiv permission and it is not open to anyone anymore. The first thing Cornell did when they got Arxiv is to shut down access by requiring an authority endorsement, and this stranglehold on submissions can be exploited for censoring people. I don't trust these people, they already forbid cold fusion papers. This stackexchange expressly encourages original research, that's the whole point. As far as "technical level", I don't know what that means--- I explained the method, the reason, but not the gluing. Maybe I'll go Vixra. – Ron Maimon Sep 06 '12 at 16:03
  • @JerrySchirmer: If you read the linked post, the CTC's are in an unphysical skin which is excluded in quantum gravity. This is why you are allowed to leave the same black hole later--- the cut-off to how much later you must leave to avoid real CTC's is provided by quantum gravity. This I knew for a long time, that it is possible, but only now can I calculate how long it takes to leave in exterior time in a reasonable picture. The CTC thing is why nobody noticed before. – Ron Maimon Sep 06 '12 at 16:07