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Can a maximal charged black hole have "flavor" or weak hair?

  1. Jan 24, 2015 #1
    Consider a Maximal (negatively) charged black hole. Can it have "flavor" or weak charge?
    Suppose a muon collided with that hole. Would an electron be emitted?
    What if electron neutrinos collided with that black hole; would electron neutrinos more likely be emitted as Hawking Radiation (assuming that happens at all) than muon neutrinos?
     
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  3. Jan 24, 2015 #2

    PeterDonis

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    Is there any particular reason why you specified a maximal (i.e., charge = mass) hole? The answer to your question would be the same for any charged hole.

    I don't think so. The only "hair" a black hole can have is associated with conserved charges/currents, and AFAIK there are no nonzero conserved charges/currents associated with the weak interaction.
     
  4. Jan 24, 2015 #3
    I specified a maximal charged black hole because of where I am headed with the muon colliding with it. That makes the (EM) charge too much and something gotta give. And if "flavor" or Weak Charge is not conserved at the horizon, then muons will be converted to electrons.
    If it is believed that the Coulomb repulsion will be too great for the muon to be absorbed unless the relativistic energy is enormous, then that seems to be a way to get a (charged) observation apparatus into close orbit with no fear of reaching the horizon.
     
    Last edited: Jan 24, 2015
  5. Jan 24, 2015 #4

    PeterDonis

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    No, it doesn't. As you note later in your post, there will be Coulomb repulsion between the muon and the hole; to overcome that and get the muon to cross the horizon, energy must be added to the system. The result is that the hole's mass must increase by at least enough to keep the hole at or below the maximal threshold.

    The conservation or non-conservation doesn't happen "at the horizon". The black hole's "hair" or lack thereof is not a property of the horizon; it's a property of the spacetime as a whole.

    That happens to muons anyway; there's nothing special about the horizon in this respect.

    The final question in your OP is a better way of approaching what I think you're getting at:

    I think the answer to this is no; but I don't know if anyone has actually done the analysis of Hawking radiation for this case (where multiple quantum fields are assumed to be present).

    It won't be an "orbit" in the usual sense, because the apparatus won't be in free fall; it will require rocket thrust to maintain its trajectory.
     
  6. Jan 25, 2015 #5
    And without the rocket thrust, what will happen to this highly charged apparatus?
     
  7. Jan 25, 2015 #6

    PeterDonis

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    It will fly away from the hole because of electromagnetic repulsion.
     
  8. Jan 25, 2015 #7
    So we would have a bouncy apparatus - we can throw it harder and harder to get as close as we like to the horizon, without fear of losing it (except to tidal forces if the BH is small enough). Or just use less charge on it, to get closer before the bounce. Considering this as a kind of Mirror, it can be seen that charged radiation from such a BH will have extremely high velocity at a distance.

    But surely at some distance it can orbit.
     
    Last edited: Jan 25, 2015
  9. Jan 25, 2015 #8
    Has that been observed already? I mean a neutrinoless conversion?
    http://mu2e.fnal.gov/

    For that matter, a maximal (positively) charged black hole (or a very small black hole) could demonstrate proton conversion to a positron or muon in the same way. THAT has not been observed yet.

    Since we have a problem with the Coulomb repulsion, we could use a maximal rotating black hole and toss in mass off-center (or mass with high angular momentum) to make it briefly super-maximal, in order to get radiation. No bounce, but the protons convert ("decay").
     
  10. Jan 25, 2015 #9

    PeterDonis

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    But there's a limit to how hard you can throw it without it falling in; if you throw it hard enough, it won't bounce before it reaches the horizon.

    What charged radiation are you talking about? Are you talking about Hawking radiation?

    If you mean, at some distance the hole's gravity can counteract the repulsion of the charge, no. Both forces decrease as the inverse square of the distance, so the ratio between them stays the same.

    I didn't say the conversion was neutrinoless. I was talking about the ordinary muon decay reaction. That will happen to muons that fall into a black hole, just as it happens to muons anywhere else.

    I don't understand what you are talking about with this. Particles that fall into a black hole don't get "converted" to anything. If you are talking about Hawking radiation, that is not a "conversion" of particles that fall in; it happens even if the black hole is in vacuum, with nothing falling in at all.

    This won't work. The hole's mass will always end up increasing by enough to compensate for the increase in angular momentum and keep the hole at or below the maximal threshold.

    I don't understand this either. Do you think that only super-maximal holes emit Hawking radiation? That's not the case.

    Same comment here as above; I don't understand where you're getting this from. You're going to need to give some references, or this thread will be closed.
     
  11. Jan 25, 2015 #10
    There is this from Bekenstein:
    http://arxiv.org/pdf/hep-th/0107045.pdf
    "This 'no hair principle' was overturned in the 1990’s with the appearance of a number of new black hole parameters and properties: skyrmion number, nonabelian magnetic monopole, color, etc."
    Sorry it's not a proper Journal, let me browse some more.
    This thread does seem to be getting off-topic; I'll try to keep it focused and/or start another topic.
     
  12. Jan 25, 2015 #11
    "No bounce, but the protons convert ('decay')."
    I am using the word "convert" in the following sense:
    If you put hydrogen atoms into the black hole, you are more likely to get electrons and positrons out as Hawking radiation, than protons. If you put a LOT of matter into a small black hole, you will get out a lot of positrons but few protons.
    (This also means that ordinary matter, including dark matter, will be "converted" to neutrinos and photons - and to particles and their anti-particles (mostly electrons and positrons) with equal probability. This would seem to be the future makeup of the universe... unless Hawking radiation is largely dark matter.)
     
  13. Jan 25, 2015 #12

    PeterDonis

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    Hm, interesting; it would have been nice if he had given specific references at that point, but he might have considered this statement too commonplace to need references. Which is perfectly understandable for him, as one of the world's leading experts in the field, but doesn't help us very much. :eek:

    Googling did turn up some papers on arxiv about black holes with "hair" of the types he lists--skyrmion hair, monopole hair, and Yang-Mills hair (which is what "color" presumably refers to--but both the weak and strong interactions are mediated by Yang-Mills fields, so this kind of hair could be ). So it does appear that this has been studied. From a quick skim of the papers, though, I can't tell whether, for example, the Yang-Mills hair models would help to answer the questions in your OP.

    In this case that's not an issue; Bekenstein is a reputable source.
     
  14. Jan 25, 2015 #13

    PeterDonis

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    You're more likely to get photons out than anything else, since they're massless. Neutrinos would be next, since, although they are believed to have mass, their masses are extremely small. Electrons and positrons, the next lightest known particles, would be much less likely to be emitted because of their much larger mass. At least, this would be the case for any black hole of stellar mass or larger, since the Hawking temperature is so low.

    (Actually, any such hole in our current universe has a Hawking temperature much lower than the temperature of the CMBR, so it wouldn't actually be losing any mass via Hawking radiation; it would be gaining mass on net from the CMBR even if nothing else ever fell in. So none of the above will happen to any actual black hole in our universe for a very long time.)

    So what you say is correct, but I'm not sure "conversion" is a good word to use, because it implies that the hydrogen atoms falling into the hole somehow turn into photons, etc. by some process involving those specific hydrogen atoms. On our current understanding of Hawking radiation, that's not the case; the energy for the Hawking radiation comes from the spacetime curvature of the hole, and it will happen even if nothing at all is falling in. It is true, though, that the overall process of, say, forming a black hole from a star made of a lot of atoms, and then eventually having that hole evaporate into mostly photons, grossly violates conservation of baryon number, lepton number, etc.--anything that isn't a type of "hair". (Note that even the additional kinds of "hair" that Bekenstein mentions won't mean that baryon number or lepton number are conserved in a black hole.)
     
  15. Jan 25, 2015 #14
    Yes, and even if positrons and electrons are emitted in equal numbers they will eventually annihilate. So rest mass seems to be doomed in the far future.
    But if you put just protons into a black hole, at some point you are gonna get positrons out. As it radiates off its mass, its charge becomes more and more important. If it becomes maximal before it emits positrons, it will stop radiating uncharged particles altogether until a positron is emitted.
     
  16. Jan 26, 2015 #15
    I am imagining turning a fire hose onto a small black hole that is vigorously trying to evaporate, putting in just enough mass to keep its mass steady. Eventually you will be putting in a lot more mass than the BH had when you started, and you will have gotten out just as much mass-energy. So it is like a trashmasher chewing up hair. You could in principle run all the mass in the entire current universe through that little BH. That looks a lot like "conversion". This seems to put a new face on the question of where does the information go. The entropy of the black hole does not change, yet a lot of information is apparently being destroyed - since, as you say, the energy for the Hawking radiation comes from the spacetime curvature of the hole, and it will happen even if nothing at all is falling in... thus it has no correlation with what is fueling the BH.
    But "In 2004 Hawking himself conceded a bet he had made, agreeing that black hole evaporation does in fact preserve information." http://en.wikipedia.org/wiki/Black_hole_information_paradox
    Yet how can that tiny BH contain so much information? So little entropy.
     
    Last edited: Jan 26, 2015
  17. Jan 26, 2015 #16
    http://physicsworld.com/cws/article/news/2011/aug/15/information-paradox-simplified
    "Braunstein and Patra say that the event horizon is purely quantum mechanical in nature, with bits of quantum 'Hilbert' space tunnelling through the barrier. The theorists find that even such a heavily simplified tunnelling model can reconstruct the spectrum of radiation that is thought to emanate from black holes. This is unlike Hawking's pair-creation model, which leads to the information loss"
     
  18. Jan 26, 2015 #17
    I only said "radiation". Because Hawking's radiation involves virtual pairs. Tunneling radiation, or some other kind we have not thought of yet, might not be considered Hawking radiation.
     
  19. Jan 26, 2015 #18

    PeterDonis

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    I haven't looked at detailed treatments of Hawking radiation from a charged black hole, so I don't know how, exactly, the charge gets radiated away. What you're saying seems reasonable, but I would want to look at a reference.

    Yes, this would work in principle.

    One key point to keep in mind here is that Hawking radiation in the presence of infalling matter might work differently from Hawking radiation in the absence of infalling matter. Hawking's original derivation assumed that the hole starts in a vacuum state, i.e., no infalling matter. AFAIK there is not a single accepted model of what happens in the presence of infalling matter; that's part of what the current debate about the black hole information paradox is about.

    The information isn't contained in the tiny BH; it's contained in the matter that falls in, and gets transferred to the radiation coming out. So it doesn't have to all be contained in the tiny BH. At least, that's one proposed resolution of the BH information paradox. But not everybody agrees with that picture AFAIK (one obvious question is what mechanism at the horizon transfers the information).

    AFAIK "Hawking radiation" is just a general term for "any kind of radiation coming out of a BH due to quantum effects". All of the mechanisms proposed, including the one in the paper you quoted, fit that definition. And, as I said before, all of this is still a matter of debate; there is no single accepted model of what happens when things fall into a BH, taking quantum effects into account. There isn't even a single accepted model of what happens when nothing falls into a BH, taking quantum effects into account. For that we would need a complete, verified theory of quantum gravity, and we don't have one. Hawking conceded his bet, but that wasn't based on having a complete, verified model; it was just based on his intuition about what such a model will end up looking like.
     
  20. Jan 26, 2015 #19
    You can send in pulses of matter so that, 99.9999% of the time, radiation proceeds without any infalling matter.
    From John Baez's web site:
    http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/info_loss.html
    "Take a quantum system in a pure state and throw it into a black hole. Wait for some amount of time until the hole has evaporated enough to return to its mass previous to throwing anything in. What we start with is a pure state and a black hole of mass M. What we end up with is a thermal state and a black hole of mass M. We have found a process (apparently) that converts a pure state into a thermal state. But, and here's the kicker, a thermal state is a MIXED state (described quantum mechanically by a density matrix rather than a wave function). In transforming between a mixed state and a pure state, one must throw away information. For instance, in our example we took a state described by a set of eigenvalues and coefficients, a large set of numbers, and transformed it into a state described by temperature, one number. All the other structure of the state was lost in the transformation."
     
  21. Jan 26, 2015 #20

    PeterDonis

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    This is a statement of the problem (what I called the "black hole information paradox" earlier), not a statement of the solution. As I said before, there is no generally accepted solution, even for the case where there is no infalling matter, because we don't have a good theory of quantum gravity.
     
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