28

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?
Qmechanic
  • 201,751
Piotr Migdal
  • 6,430

3 Answers3

30

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.

Community
  • 1
  • 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. –  Mar 20 '11 at 18:13
5

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.

Zo the Relativist
  • 41,373
  • 2
  • 74
  • 143
  • 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. – Zo the Relativist Mar 20 '11 at 17:17
  • 3
    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. –  Mar 20 '11 at 17:25
  • 8
    @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 03:17
  • @Peter: that process won't violate baryon number, though--the long wavelength particles will contain equal numbers of, say, anti-neutrions and neutrinos (being a thermal bath), and will not be able to produce a net number of leptons/antileptons or baryons/antibaryons. Then, allow sufficient time for the fields containing the initial degrees of freedom for baryon number or whatever to propogate back out, and you're out to where you started, plus some thermal radiation containing equal numbers of particles and antiparticles. – Zo the Relativist Mar 21 '11 at 05:50
  • @Moshe: last I checked, proton decay hadn't been observed. – Zo the Relativist Mar 21 '11 at 05:51
  • @Jerry, last I checked we see pretty much only baryons around us, so baryon number could not be exactly conserved at all energy scales in our universe. But, I like @Peter’s answer better because it tells you why it had to be the case. I think your reply to Peter with the “black holes don’t really form” scenario has its own problems, but this is not the place to get into this discussion. –  Mar 21 '11 at 06:01
  • @Moshe:If you believe that the black hole completely evaporates, there is no event horizon-that is simply a fact arising from the lack of a singularity at late times.Personally,I'm fine with saying that the black hole might not evaporate completely,since we know so little about Hawking radiation back-reaction anyway, but I seem in the minority on that one.In any regard, you need a process to destroy baryons inside the black hole in order to cause them to not be conserved-if this is a part of singularity resolution,fine-that's not my field.And who said that all quantum numbers were zero at t=0? – Zo the Relativist Mar 21 '11 at 06:24
  • @Jerry The issue is that of time scales, it is not likely that macroscopic number of baryons will be emitted in the last quanta evaporates, just as it is unlikely the universe just started with tiny imbalance of baryons over anti baryons. The relevant physics is that of the horizon, singularity resolution is not the issue in black hole physics - that would be an easy way out. One has to understand what happens during the long lifetime of the BH for which the horizon is real. –  Mar 21 '11 at 06:38
  • @Moshe: then, what on earth is your issue with my answer--that's settled in the third paragraph of my OP. Hawking radiation is a distractor from this, then, because it's just a thermal bath of particles and antiparticles in equal proportion. The black holes are identical no matter what made them. The only thing that happens is that negative energy Hawking radiation allows the classical GR black hole theorems (most importantly here, the area increase and no-hair theorems)to be violated, and thus allows the interior constituents of the BH to escape to infinity. – Zo the Relativist Mar 21 '11 at 06:58
  • I don't really understand your scenario, when the black hole is completely gone, everything in it has turned into Hawking radiation. If the BH was made from baryons initially they all ended up as radiation of mostly neutrinos or whatever. If you are not saying that the baryons are radiating out in the last Planck time of evaporation, when are they getting out? I really think that Peter's argument settles the issue. –  Mar 21 '11 at 07:10
  • @Moshe: I'm saying that all of the null and timelike generators that enter the horizon eventually leave the horizon, aside from those that would actually intersect the singularity itself (and in the case where $a$ or $q \neq 0$, it is not necessary for inbound geodesics to intersect the singularity). Thus, unless an interaction with the singularity doesn't conserve baryon number, the baryons that enter the hole eventually leave the hole. You can see this yourself with a charged Vaiyda model with mass profile looking something like $M(v)=v^{4}(v-v_{0})^{4}$. and zero for $0\leq v \leq v_{0}$ – Zo the Relativist Mar 21 '11 at 07:21
  • It's a theorem of GR that shrinking black hole horizons are transversable, is basically what my argument is. If the black hole completely evaporates, every interior point is in the past of future null infinity or future timelike infinity. Thus, something in the interior has to destroy baryons, or the black hole collapse cannot remove net baryon number, as hawking radiation only will create particle-antiparticle pairs. – Zo the Relativist Mar 21 '11 at 07:26
  • @Jerry: Do you have a citation somewhere to the fact that shrinking black hole horizons are transversable? I haven't seen this theorem anywhere. – Peter Shor Mar 21 '11 at 12:40
  • @Peter: I'll look around. I know that it's in Hayward's trapping horizon stuff, but that is a lot of papers for someone to wade through. You can picture it if you look at a Penrose diagram of a black hole that's shrinking, though. – Zo the Relativist Mar 21 '11 at 14:07
  • Just to sharpen the issue - keep the black hole in a steady state by continuously feeding it baryons. You will never see the baryons coming back. The causal structure of an evaporating black hole is interesting distraction, but I don't think it can explain away the issue. –  Mar 21 '11 at 15:20
  • @Jerry: If the proton and neutron are the elementary particles with the highest baryon number/mass ratio, than a black hole with baryon number $2N$ and the mass of $N$ protons cannot decay into elementary particles and preserve baryon number. And if there were a particle with higher baryon number/mass ratio, it would be stable and we presumably would have seen it. – Peter Shor Mar 21 '11 at 21:16
  • @Moshe @Peter: working out conservation of mass/energy for a system like this is non-trivial, as Moshe's suggestion will destroy asymptotic flatness, and Hawking radiation has negative energy relative to asymptotic infinity. I'm not saying that you guys are wrong, I'm more saying that I'm unconvinced. Which is why my initial answer was "Here's a way you might be able to get out of this" in tone, more than "this is right". – Zo the Relativist Mar 21 '11 at 22:54
  • @Jerry: Incoming flux of baryons with finite energy will certainly not destroy asymptotic flatness, it just corresponds to having a different state at the horizon, whose influence falls off fast enough as to preserve asymptotic flatness. As for the evaporating black hole case, if I understand your scenario correctly then when you discuss quantum fields instead of classical particles (as you should) it comes down to the suggestion that the baryon number is restored “at the last minute”, to which Peter provided a nice counter-argument. –  Mar 21 '11 at 23:06
  • @Moshe: a steady, neverending flux of particles certainly will destroy asymptotic flatness. And if we're talking about Quantum Fields, we have to measure the field at late times anyway, where asymptotic flatness at future null infinity allows us to restore the Poincaré group and get a notion of what 'particle' means, anyway. Any of the derivations of this stuff that I've looked at have required a definition of 'in' and 'out' states at past and future null infinity, anyway. – Zo the Relativist Mar 23 '11 at 14:35
2

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.

Lawrence B. Crowell
  • 9,228
  • 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 04:19
  • 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