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The no hair theorem says that a black hole can be characterized by a small number of parameters that are visible from distance - mass, angular momentum and electric charge.

For me it is puzzling why local quantities are not included, i.e. quantum numbers different from electrical charge. Lack of such parameters means breaking of the conservations laws (for a black hole made of baryons, Hawking radiation then is 50% baryonic and 50% anti-baryonic).

The question is:

  • If lack of baryonic number as a black hole parameter is a well established relation?

OR

  • It is (or may be) only an artifact of lack of unification between QFT and GR?
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You might want to see physics.stackexchange.com/questions/4908/… –  Matt Reece Mar 20 '11 at 15:54

3 Answers 3

up vote 14 down vote accepted

The no hair theorem is proven in classical gravity, in asymptotically flat 4 dimensional spacetimes, and with particular matter content. When looking at more general circumstances, we are starting to see that variations of the original assumptions give the black hole more hair. For example, for asymptotically AdS one can have scalar hair (a fact which is used to build holographic superconductors). For five dimensional spaces black holes (and black rings) can have dipole moments of gauge charges. Maybe there are more surprises.

But, the basic intuition behind the no hair theorem is still valid. The basic fact used in all these constructions is that when the object falls into a black hole, it can imprint its existence on the black hole exterior only if it associated with a long ranged field. So for example an electron will change the charge of the black hole which means that black hole will have a Coulomb field. You'd be able to measure the total charge by an appropriate Gauss surface. Note that gravity has no conserved local currents (see this discussion), the only thing you'd be able to measure is the total charge.

As for baryon number, it is not associated with long ranged force, when it falls into black hole there is nothing to remember that fact, and the baryon number is not conserved. This is just one of the reasons there is a general belief that global charges (those quantities which are not accompanied by long ranged forces) are not really conserved. For the Baryon number we know that for a fact: our world has more baryons than anti baryons, so the observed baryon number symmetry must only be approximate. It must have been violated in the early universe when all baryons were generated (look for a related discussion here), a process which is referred to as baryogenesis.

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Great if one feels that his answers are not needed at all, Moshe. ;-) +1 –  Luboš Motl Mar 20 '11 at 18:10
    
I know the feeling. I routinely let things go when I don't have the time since I know you'd be giving a good answer. –  user566 Mar 20 '11 at 18:13

One way out of this--if you believe that Hawking radiation is real, and that black holes completely evaporate, then:

1) the Hawking radiation violates the positive energy conditions. These are assumptions in the no-hair theorem. So, we don't expect the no-hair theorem to be valid for black holes if Hawking radiation is important.

2) If the evaporation profile evolves in a certain way, you can prevent the formation of a true event horizon, only having an apparent horizon appear for a finite amount of time. This will enable the information about the input baryon number to escape the black hole.

But classically, you are right. It is well-established that you lose the information about what you put into the black hole--classically, a black hole formed from neutrino collapse is identical to a black hole formed from the collapse of a neutron star or photons or gravitational waves or whatever.

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4  
Just because the information might be preserved, doesn't mean that the conservation laws apply. I believe we know enough about Hawking radiation to know that, even if the information escapes, the actual baryon number is still not preserved, which I believe was the OP's real question. –  Peter Shor Mar 20 '11 at 16:50
    
@Peter Shor--but without the mechanism of a particle colliding with the singularity, merging with it, and then being radiated as another particle, I'm not 100% clear on what the mechanism of a baryon number-destroying interaction would be. –  Jerry Schirmer Mar 20 '11 at 17:17
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Baryons being accelerated to high energy, in which case they are involved in non-baryon-number-conserving interactions, which we already know exist. Geometry is nice and all, but it is not the only explanation for everything. –  user566 Mar 20 '11 at 17:25
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@Jerry: We don't know what the baryon-number destroying interaction is, but we know it must exist. Create a huge black hole using only baryons. The first radiation from it has to be very long wavelength radiation (unless you believe that this is wrong), which means it won't radiate any baryons at all until it's mass is fairly small. At that point, if the baryon/mass ratio is larger than the proton, it can't possibly radiate enough mass to conserve baryon number. So we can conclude that Planck-scale interactions must not conserve baryon number from this thought experiment. –  Peter Shor Mar 20 '11 at 17:29
    
@Peter I have never heard it argued that way, very cool. –  Columbia Mar 21 '11 at 3:17

A standard answer is that the nuclear force obeys a Yukawa potential $V(r)~=~-e^{-ar}/r$ which drops off with a very short range. A charge dropped on a black hole will have a long range electric field which connects the horizon with conformal infinity. The electric charge becomes distributed on the horizon and remains apparent. The Yukawa type of force that drops off rapidly does not appear in this manner.

There is the holographic content of quantum fields or strings which fall onto a black hole. The baryon number, and in fact all quantum numbers, are preserved on the stretched horizon. They are annihilated in the far future as they are replaced by Hawking radiation modes. However, this Hawking radiation is just some “recoding” of the quantum information. Without the master decyption system what emerges appears to be noise. Yet if quantum information is preserved then baryon number, or more fundamentally quark family types and color charges, is simply transformed into some other form of quantum information.

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Interesting. I wonder what would happen if you dropped something like one quark into a black hole, so you weren't color neutral. –  Carl Brannen Mar 21 '11 at 4:19
    
If you watch a baryon approach a black hole, the system becomes delayed or frozen above the event horizon. There are then the X fields in GUTs which rotate a quark into a lepton. By watching a baryon approach a black hole, and if one can witness the highly redshifted (IR) frequencies, this amounts to witnessing the demolition of a baryon. The correspondence between AdS and CFT, and AdS has a correspondence with fields on its boundary and the black hole horizon is holds there then an equivalency between a GUT proton decay and baryon violation in black holes. –  Lawrence B. Crowell Mar 21 '11 at 13:03

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