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A Black Holes and Charged Particles

  1. Jun 21, 2016 #1
    What happens when charged particles fall into a black hole?
    Say like N electrons fall in, giving the black hole a net charge of -N.
    Since light cannot escape the event horizon, I imagine electric fields cannot either, since they are mediated by photons.
    So is that charge effectively lost until the black hole decays?

    Or is there some kind of compensation mechanism with the virtual particles near the event horizon such that N positrons are pulled out of the vacuum, into the blackhole to neutralize the net charge, and N electrons are ejected into space? This would conserve charge outside the black hole. Thanks.
     
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  3. Jun 21, 2016 #2

    Nugatory

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    Neither. You end up with a charged black hole, described in the simplest case by the Reissner-Nordstrom solution to the Einstein Field Equations. Google will find many references, although much of it is (unavoidably) fairly math-intensive.
    Although electric fields are mediated by photons, that does not mean that the electric field is carried by photons moving through space or crossing the horizon. The charge is still there and there's still an electric field outside the event horizons (note the plural! R-N black holes have two event horizons).
     
  4. Jun 21, 2016 #3

    BiGyElLoWhAt

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    Well, you can't ever observe something cross the event horizon, you would just see it approach the event horizon and sit there, creating a static field.
     
  5. Jun 21, 2016 #4

    PeterDonis

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    No. One key point to grasp that helps to resolve this apparent difficulty is that "charge" in GR is not a localized property of a particle; it's a property of the spacetime as a whole. A quick and dirty heuristic way to think of this is to imagine a charged particle sending out electric field lines; these lines go all the way out to infinity, and the presence and configuration of the field lines at infinity tells you about the presence of charge somewhere in the spacetime. Those field lines are still there even if there is a black hole present and the source of the charge falls into it.

    Another important point to keep in mind is that, at any event in spacetime, the electromagnetic field at that event is entirely determined by the sources (charges and currents) that are present in the past light cone of the event. (The formal way to demonstrate this is to write everything in terms of retarded potentials, known as the Lienard-Wiechert potentials.) At any event outside the horizon, the past light cone lies entirely outside the horizon; so even if there are "now" no sources outside the hole, the electromagnetic field that you detect at that event is not actually coming from inside the hole. It's coming from the charges and currents that fell into the hole in the past.
     
  6. Jun 21, 2016 #5
    Not true.

    For the sake of argument we could all right now pass some event horizon.
     
  7. Jun 21, 2016 #6

    BiGyElLoWhAt

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    I could see me pass an event horizon, but you couldn't see me pass an event horizon. Once I hit the event horizon, I will stop and sit there, as you see me.
    https://en.wikipedia.org/wiki/Event_horizon
     
  8. Jun 21, 2016 #7

    PeterDonis

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    No, that is not what an outside observer will see. An outside observer will see you get closer and closer to the horizon, moving more and more slowly, but never actually reach it.

    Also, Wikipedia is not a good source to be using for these kinds of discussions, certainly not in an "A" level thread.
     
  9. Jun 21, 2016 #8

    BiGyElLoWhAt

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    Technically, yes, but for all intents and purposes I'll eventually "stop".
     
  10. Jun 21, 2016 #9

    PeterDonis

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    Only if "eventually" means "never as seen by the outside observer". In the idealized model we are talking about here, the outside observer literally never sees you stop, even if he waits for an infinite amount of time by his clock.

    Remember that this is an "A" level thread, and approximate statements that would be OK in a "B" or possibly even an "I" level thread are not appropriate here. We should be giving the most exact, precise, detailed statement of our best current theory, as applied to the situation under discussion, that we can.
     
  11. Jun 21, 2016 #10

    PAllen

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    Also, putting together two of Peter's answers, for any event outside the horizon, the EM field originates from charges in the past positioned just outside the horizon; these are the events in the past light cone of some external event. On the other hand, EM fields exist inside the horizon(s) as well, and these fields may originate from charges both outside and inside the horizon(s) - the past light cone of an interior event includes much of the exterior.
     
  12. Jun 21, 2016 #11
    It all depends on how you define "the outside observer".

    For instance an inertial observer B trailing a person A falling through the event horizon could remain in visual contact provided A's light signal reaches B after B falls through the event horizon.
     
  13. Jun 21, 2016 #12

    PeterDonis

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    An observer that is outside the event horizon.

    And once B falls through the horizon, B is no longer an outside observer.
     
  14. Jun 21, 2016 #13
    True, but B was an outside observer before and when A inside emitted light B cached up to it when he fell through it himself.
     
  15. Jun 21, 2016 #14

    PeterDonis

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    He isn't when he receives the light from A that you are talking about. That is the key point. We have been talking about what the outside observer sees; he only sees things when the light reaches him, not when it is emitted.
     
  16. Jun 21, 2016 #15
    We are actually agreeing.....
     
  17. Jun 21, 2016 #16

    PeterDonis

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    If so, that's good. I couldn't tell.
     
  18. Jun 22, 2016 #17
    Is the diverging coordinate in-fall time for an external observer also true in R-N spacetime, which is different in many respects from the Schwarzschild case ?
    Also, if you let a charged particle fall towards the black hole, then that charge should be affected by its very own electromagnetic field, meaning that in principle it wouldn't trace out a geodesic at all. How does that affect the total coordinate in-fall time as seen by an external observer ?
     
  19. Jun 22, 2016 #18
    With that being said, what would that look like? When I read up on it my understanding was that you'd never see someone cross, as you say. However, they'd still stop having light reach outside observers, due to crossing the event horizon. Rather than seeing a "stop", wouldn't you see a fading approach until you stopped seeing anything (IE it just fades enough that you can't detect it)? This is pretty difficult to visualize for me.
     
  20. Jun 22, 2016 #19

    PeterDonis

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    It is for an uncharged object falling into a R-N hole. The case of a charged object falling into an R-N hole is different, though; see below.

    Yes, that's correct. A particle with the same charge as the hole will, heuristically, feel a force pushing it away from the hole, so it will fall in more slowly than an uncharged object, and will eventually decelerate, stop, and turn around and accelerate away. If this happens before the object reaches the horizon, then of course it will never fall in at all. If it happens after the object crosses the horizon--more precisely the outer horizon (there is also an inner one), things get more complicated, and I won't discuss that case further here; the only thing we would need to know is that the qualitative behavior will be the same as for the Schwarzschild case, but possibly with different details, as for the case of a particle with opposite charge, discussed below.

    A particle with opposite charge from the hole will, heuristically, feel a force pulling it towards the hole, so it will fall in more quickly than an uncharged object. I actually have not seen a computation of how that affects what an observer who remains far away from the hole sees. My quick and dirty intuitive guess is that the faraway observer still sees the object appear to slow down more and more as it gets closer to the horizon, and never quite reach it. That's based on the fact that the horizon (more precisely the outer horizon) is still an outgoing null surface, where radially outgoing light remains at the same radial coordinate forever. The details of exactly how the slowdown is seen, as a function of the faraway observer's clock time, I would expect to be different from the Schwarzschild case; but I would expect the qualitative behavior to be the same.
     
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