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Maximum angular momentum for charged black hole?

  1. Oct 30, 2014 #1
    Hi all

    I gather a normal black hole has maximum angular velocity at the point that the event horizon is moving at The speed of light.

    However what would be the maximum rotational velocity for a maximally charged black hole- for example one made purely of electrons?

    Thanks
     
  2. jcsd
  3. Oct 30, 2014 #2

    PeterDonis

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    No, this is not correct.

    First, the event horizon is a null surface, so it is always "moving at the speed of light"--but you have to be careful to define what it is "moving" relative to (the answer is, relative to observers free-falling into the hole as they pass the horizon).

    Second, the maximum angular velocity is not really a "maximum" as in "impossible to exceed"; we don't know that. But there is a particular angular velocity for which the hole's angular momentum per unit mass, ##a = J / M##, is equal to its mass ##M## (all this is in geometric units, so ##M## and ##a## both have units of length). If the angular velocity is smaller than this value, call it ##\Omega_{crit}##, then there is no way to make the angular velocity equal to ##\Omega_{crit}## by adding angular momentum to the hole; the amount of mass that has to be added is always sufficient to keep ##\Omega## smaller than ##\Omega_{crit}##. That is the sense in which ##\Omega_{crit}## is a "maximum".

    The angular velocity ##\Omega_{crit}## also represents a "break point" in the type of spacetime geometry present. For ##\Omega < \Omega_{crit}##, the geometry is the one usually described as a Kerr black hole, with two horizons, inner and outer, and a ring singularity (with other features that I won't go into). For ##\Omega = \Omega_{crit}##, there is only a single horizon. For ##\Omega > \Omega_{crit}##, there is no horizon at all, and the singularity is a naked singularity, visible from infinity.

    Adding charge to the rotating black hole doesn't change any of the above qualitatively; it just changes the specific value of ##\Omega_{crit}## (because it now involves the charge ##Q## as well as ##M## and ##a##).
     
  4. Oct 31, 2014 #3
    I've been meaning to ask something related to this point. It is commonly explained that an observer free-falling into a black hole sees no horizon, therefore they must see this naked singularity? OK they will soon be dead and can't tell anyone else about it, but they should still see it, shouldn't they?

    This is related to some thoughts I have had on the Rindler horizon, which only exists whilst an observer is accelerating, and the equivalence principle. Does this mean that a black hole horizon only exists to an observer who is holding station at a fixed point outside it (but possibly at infinity)?
     
  5. Oct 31, 2014 #4

    PeterDonis

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    Where is this "commonly explained"? Do you have a reference?

    No. The horizon is a global feature of the spacetime; it's not observer-dependent. (In this respect it is different from the Rindler horizon, which is observer-dependent.)

    However, it is true that the horizon (more precisely, the absolute horizon--see below) is not detectable locally--regardless of whether an observer is free-falling into the hole or hovering outside it. (Strictly speaking, an observer hovering outside the horizon can't detect it because no light signals can escape outside the horizon. But we'll overlook that technical point.) That's because the horizon is the boundary of the region (the black hole) that can't send light signals to infinity. That's an inherently global definition: in fact, to know where exactly the horizon is, you would have to know the entire future of the spacetime.

    There is a related kind of horizon, called the apparent horizon, which is detectable locally; roughly speaking, an apparent horizon is present if outgoing light no longer moves outward. For a stationary, eternal black hole (i.e., one that has always existed and will always exist, and never gains or loses any mass--i.e., an idealization), the apparent horizon coincides with the absolute horizon, so if you detect the apparent horizon, you know where the absolute horizon is. For a real black hole, however, which can gain mass by things falling in (and lose mass by Hawking radiation), the two do not coincide (although they will be very close for a hole that is gaining or losing mass very slowly).
     
  6. Nov 1, 2014 #5
    By "seeing no horizon" I was merely referring to the finite intrinsic curvature at the horizon (and by implication the complete lack of any indication of a horizon to the observer), which is mentioned pretty much universally eg. in MTW (which goes into painful detail on the subject!), that's all.

    As to seeing the singularity, that is my question. I am assuming that I can see what is in front of me (both outside and inside the horizon coordinates), as I am hurtling in that direction.

    Hope this clarifies what I was asking (most of which I think you have answered, still digesting).
     
  7. Nov 1, 2014 #6

    PeterDonis

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    Ok, that makes it clearer what you meant. I was confused because this has nothing to do with either of the things you were asking about--it has nothing to do with "seeing a naked singularity" (the definition of "naked singularity" refers to the presence or absence of a global horizon, not a locally observed one), and it has nothing to do with the horizon "existing" only for a hovering observer (intrinsic curvature at the horizon is an invariant, the same for all observers).

    The singularity is not "in front of you". It is in your future. You can't see light coming from your future.

    That is, the singularity is not a place in space; it is a moment of time. You can't "see" a moment of time that you haven't reached yet.
     
  8. Nov 1, 2014 #7
    Ah, I think I see your objection now; I believe you are talking in terms of global t and r coordinates. To explain my POV a bit I have recently become used to doing (geodesic) simulations from the perspective of an orbiting/infalling observer, so I am a bit stuck in that mindset ATM. As a bit of fun, recently I plotted t and r as I fell through the horizon and saw t "spike" to infinity then come back down again (I'm sure we've all seen those diagrams in discussions of the Schwarzschild metric). But I did not observe the "switching" of t and r coordinates that I think you are referring to. That simply does not seem to happen in the equations of motion I am using (MTW eqs.33.32).

    I've also recently become a little bit obsessive about what an observer really sees in relativity, so the coordinate perspective is less relevant to me than to some others I guess.

    Getting to the point, can you convince me that the coordinate view trumps the "proper" view in this case? I won't be offended if you disagree, after all we are pretty much discussing a science-free zone when we talk about what happens within the horizon ;)
     
  9. Nov 1, 2014 #8

    PeterDonis

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    No, I'm not. What I've said about the singularity has nothing to do with coordinates; it's a statement about the invariant geometry of Schwarzschild spacetime.

    I don't have to; what I've said is true in the "proper" view.

    No, we aren't; we are using a perfectly consistent physical model. We are extrapolating it beyond the domain in which it's currently been validated experimentally, but we do that in science all the time; if we didn't, science would never be able to make predictions about experiments that haven't yet been done.

    If we were really in a "science-free zone", we could make any claim we wanted and there would be no way to refute it, which would make it pointless to make a claim in the first place. You don't seem to be taking that position.
     
  10. Nov 1, 2014 #9
    I meant science-free only as in the results cannot be observed by anyone outside the horizon. I am still trying to understand what you said above about horizons though, let's see if I can get this straight in my head before I pursue it further, and thanks for replying!
     
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