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How can EM interaction escape a black hole?

  1. Feb 1, 2016 #1
    Hi.

    Charge is one of the few properties a black hole can have. EM interaction is mediated by photons. How can they escape the BH?
     
  2. jcsd
  3. Feb 1, 2016 #2

    Dale

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    Static fields are not mediated by real photons. They are mediated by virtual photons, which can go faster than c.
     
  4. Feb 1, 2016 #3

    fresh_42

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    Having the long and heated debate about virtual particles on PF in mind, could you explain this process? I assume we are talking about Hawking radiation here. What can be found on, e.g. Wikipedia, left me thinking that it is something like "vacuum energy → pair production → one particle to the BH and one to the jet". I'm completely insecure how far away this naive picture is from the truth. Are there short answers at all?
     
  5. Feb 1, 2016 #4

    PeterDonis

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    This Usenet Physics FAQ article is relevant:

    http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html

    It basically gives the same viewpoint Dale gave, but it also answers the analogous question about gravity, not just EM fields.

    There is also another way to look at it: the observed EM field at any event ultimately depends on the charge-current distribution that is present in the past light cone of that event. Any real charged black hole will have been formed by the collapse of some charged object in the past. At least a portion of that collapse will be in the past light cone of any event outside the hole's horizon at which we are trying to predict what the EM field will be. We can make a correct prediction by starting with the charge-current distribution during the collapse as initial conditions, and then integrating Maxwell's Equations forward in time to the event we are interested in. (Actually we have to integrate the coupled Einstein-Maxwell equations, since we also need to solve for the spacetime geometry in order to integrate correctly. But that's no problem, since those equations have a well-posed initial value formulation when coupled just as they do separately.)

    So on this view, the EM field you feel outside a charged black hole is not coming from inside the horizon; it's coming from the charged matter that originally collapsed to form the hole. It just happens that that collapse process was such that it can be thought of as leaving behind a static field that maintains itself unchanged once formed. Similar remarks apply to the hole's gravity.
     
  6. Feb 1, 2016 #5

    Dale

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    Hawking radiation is real photons.

    I don't think that I can give a good explanation about virtual photons, let alone a short one. My preference is always to avoid quantum complexities when discussing classical concepts. I think it is a little masochistic to do otherwise.
     
  7. Feb 1, 2016 #6

    PeterDonis

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    No. Even if we view the EM field of a black hole (or its gravity) as being mediated by virtual particles, they are not the same as the particles that are postulated in the common heuristic explanation of Hawking radiation that you mention. (Btw, that explanation itself is, as I just said, heuristic; it bears little if any resemblance to the actual underlying math.) But, as my previous post makes clear, you don't actually need to invoke virtual particles to explain the EM field of a black hole (or its gravity); you can explain it purely in terms of classical concepts. I agree with Dale that the latter is always advisable when it can be done.
     
  8. Feb 2, 2016 #7

    fresh_42

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    Thank you both. The only point that left the links open to me is the EM-field transfers momentum but does not transfer information. I am not very certain how to resolve this apparent contradiction.
     
  9. Feb 2, 2016 #8

    PeterDonis

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    I'm not sure what you mean by that. The momentum transfer can be explained classically by looking at the past light cone, as I described, so there's no inconsistency between that and information transfer.
     
  10. Feb 2, 2016 #9

    fresh_42

    Staff: Mentor

    I've meant the link about virtual particles on your linked page.
    Does it mean all we observe about the EM-field are basically the "leftovers" due to relativity. If so, then I think I understood it.
     
  11. Feb 2, 2016 #10

    PeterDonis

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    All that is really saying is that, if you insist on viewing a static EM field as being mediated by virtual particles, then you have to accept that virtual particles can travel outside the light cones, so they can escape from the event horizon. But this view of an EM field does not explain how it can transfer information. To explain that, you need to adopt the classical view I described.

    No. A static EM field is not a "relativistic correction" to anything. (Magnetic fields are sometimes viewed this way, but even that view has limitations.)
     
  12. Feb 3, 2016 #11

    stevendaryl

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    I'm not sure if this helps, but rather than thinking of a field (such as the electromagnetic field) as being created by the exchange of the corresponding particles (photons), it's better to think of it as the photons as reflecting perturbations, or changes, in the field. The field exists in the absence of any photons, and a static field (such as the field of a charged point mass at rest) doesn't involve photons at all. (Virtual photons might be used in calculations, but they shouldn't be taken seriously as a physically meaningful description of what's going on.)

    The following is completely non-rigorous, but I hope it's not blatantly wrong. (Somebody will correct it if it is.)

    If you have a large object such as a star that has a nonzero charge, the electric field can be described in terms of "moments". Far away from the star, it will look approximately like a point-charge. It will have a field that is approximately of magnitude [itex]\frac{Q}{r^2}[/itex], where [itex]Q[/itex] is the total charge, and [itex]r[/itex] is the distance to the center of the star. That's called the "monopole moment". But since a star is not completely spherically symmetric, there will be "corrections" that will have magnitude proportional to [itex]1/r^4[/itex], [itex]1/r^6[/itex] etc. Those corrections are due to the dipole moment, the quadrupole moment, etc. Stuff going on inside the star (matter moving around) will have no effect on the total charge, so that part of the star's electric field is static (at least in a frame in which the star is at rest). But the "corrections" will be time-dependent. Those time-dependent corrections are propagated by electromagnetic waves.

    If the star collapses into a black hole, there can be no propagation of information from inside the star to distant locations. So the "corrections" disappear. The star will look like a perfect sphere, with an electric field exactly [itex]\frac{Q}{r^2}[/itex] in magnitude.
     
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