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Formation of Naked Singularities

  1. Jul 25, 2009 #1
    Hi!
    I've been trying to find explanations about the theoretical formation of naked singularities,
    and all I could come up with was Wikipedia, and frankly, the explanation didn't "set" in my mind, I couldn't really visualize it. It reads:

    "From concepts drawn of rotating black holes, it is shown that a singularity, spinning rapidly, can become a ring-shaped object. This results in two event horizons, as well as an ergosphere, which draw closer together as the spin of the singularity increases. When the outer and inner event horizons merge, they shrink toward the rotating singularity and eventually expose it to the rest of the universe."

    When they say the two event horizons merge, that's the part that I can't get right, I mean, why do they shrink towards the center? why do they draw closer? Is it because of the geometric transformations to the singularity? Why two, actually? wouldn't one conceal it from the entire universe? I seriously need clarification on this...

    -Itai
     
  2. jcsd
  3. Jul 26, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Itai! Welcome to PF! :smile:

    They don't shrink towards the centre, they shrink towards each other, at r = M.
    They draw closer because that's what the maths says.

    The signature of ordinary space is +--- but the signature changes when you go through an event horizon.

    The Kerr solution (rotating black hole) has one signature change (angular momentum = 0), two signature changes (0 < angular momentum < M), or none (angular momentum ≥ M).

    To be more specific: for a rotating black hole with mass M and angular momentum aM, the coefficient of dr2 in the metric has a factor 1/(r2 - 2Mr + a2), and therefore changes sign at r = M ± √(M2 - a2), and these sign changes are at r = 2M (and 0) for a = 0, they converge as a increases, eventually disappearing when they merge at r = M for a = 1. :wink:
    I don't know what you mean by "geometric transformations to the singularity" :redface:

    The singularity is always at r = 0, nowhere near where the event horizons merge at r = M.

    It is geometrically a point when a = 0, and a ring when a > 0, but it's always at r = 0. This apparent contradiction (a ring of zero radius) is because r isn't actually a very sensible coordinate. :rolleyes:
     
    Last edited: Jul 26, 2009
  4. Jul 26, 2009 #3
    Ok, thanks.
    But still, why do two event horizons form in the first place? I mean, how does the fact that the black hole is shaped as a ring make that happen?
    It says there "...they shrink toward the rotating singularity, eventually exposing it to the rest of the universe"
    Is that just incorrect?
     
    Last edited: Jul 26, 2009
  5. Jul 26, 2009 #4

    tiny-tim

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    It doesn't.

    The event horizons and the ring singularity don't cause each other, they are both caused by the metric.

    (and one event horizon, with a non-ring singularity, virtually never happens … it would require the black hole to have exactly zero angular momentum, which is virtually impossible)
     
  6. Jul 26, 2009 #5
    Oh, ok, thanks, clearly I have to learn a bit more about this whole thing.
    I guess the explanation in Wikipedia wasn't quite accurate, because it differed from what you've told me.

    Thanks anyway :)
     
  7. Jul 26, 2009 #6
    I just thought of another issue.. If there is no event horizon, It means an object can escape he singularity, right? So if the escape velocity is now less than the speed of light, by what definition is the singularity a singularity?
     
  8. Jul 26, 2009 #7
    I just thought of another issue.. If there is no event horizon, It means an object, or say, a photon, can escape the singularity, right ? So if the escape velocity is now less than the speed of light, by what definition is the singularity a singularity?

    I thought it may mean that the event horizons are still there, only below the surface of the formerly-black hole... But that would make the singularity not really naked, would it? Although, the effects in the immediate area of the singularity could be observed...
    Am I making ANY sense at all?
     
  9. Jul 26, 2009 #8
    Its a singularity because the cuvature diverges to infinity.

    Naked singularities are non-physical in general. In terms of the solution for a rotating black hole i think the naked siguarity wound repel matter that fell towards it because it has such a high angular momentum. But by the same logic such a singularity couldn't form from a gravitional collapse because the angular momentum would prevent it.
     
  10. Jul 26, 2009 #9

    tiny-tim

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    There's no such thing as escape velocity (for a black hole) … a black hole isn't like a waterfall (and light isn't like fish!).

    The definition of a black hole singularity is that any line going to the singularity ends there.

    Lines not going to the singularity (however close they may get) don't have to end.
    Nope. :redface: Sorry! :smile:
     
  11. Jul 26, 2009 #10
    If it's not too much trouble, could you try to explain the whole process of the creation of a naked singularity? Maybe a complete explanation will do better than asking about specific parts..
     
  12. Jul 27, 2009 #11

    tiny-tim

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    (in post #2 above, I left out a "squared", and it should read "two signature changes (0 < angular momentum < M2), or none (angular momentum ≥ M2)" :redface:)
    It's very simple …

    a naked singularity is created the same way as an ordinary one …

    the only difference is that if the angular momentum J is greater than the mass squared (M2), then the event horizons (which would be at r = M(1 ± √(1 - (J/M2)2)) cannot exist. :smile:
     
  13. Jul 27, 2009 #12
    Ok thanks.. that just about does it for now...
    I'll understand it better in a few years after receiving
    formal university education :D...

    Just one last thing though...
    The curvature of spacetime, is, essentially what we call gravity.
    So if the curvature goes to infinity at the singularity, It's still not clear to me
    how light can be emitted from it...
    Does the fast rate of rotation change the curvature around it?

    If the answer is no, please do try to give the actual reason, and please bare with me,
    It's not like I've studied this for years, I've only recently become aware of this theoretical phenomenon.
     
    Last edited: Jul 27, 2009
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