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Does a Schwartzchild blackhole have a singularity?

  1. Sep 21, 2008 #1
    Does a Schwartzchild blackhole have a singularity at its center?
     
  2. jcsd
  3. Sep 21, 2008 #2

    mathman

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    That is one of the big problems of current physics. General relativity predicts there is one, Quantum theory won't allow it. The reconciliation of these two main theories of modern physics is the impetus (not necessarily the only one) for theories such as string theory and loop quantum gravity.
     
  4. Sep 21, 2008 #3
    Thanks mathman, but I was actually thinking along the lines of general relativity, alone.

    The usual answer is 'yes', but I don't see how a singularity ever forms from a extended mass such as a star.

    I suppose I should post the metric.

    [tex] ds^2 = - \left( 1 - \frac{2Gm}{r} \right) dt^2 + \left( 1 - \frac{2Gm}{r} \right)^{-1} dr^2 + \ r^2 \left( d \theta^2 + sin^2 \theta d \phi^2 \right) [/tex]

    where [tex]\ 2Gm=r_0 [/tex] is the radius of the event horizon.

    This is the metric for the vacuum region surrounding a spherically symmetric mass, valid up to the surface of the mass centered at the origin.

    With an extended mass, where's the central singularity?
     
    Last edited: Sep 21, 2008
  5. Sep 22, 2008 #4

    Jonathan Scott

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    Once an event horizon has formed (which admittedly takes an infinite time as seen and calculated from outside), anything inside it cannot help collapsing to a mathematical point singularity within a strictly limited time as observed locally. There can be no extended mass inside the event horizon.

    Note that General Relativity and Einstein's field equations do NOT actually directly predict the existence or otherwise of black holes, contrary to popular opinion. In the solutions of those equations described by the Schwarzschild solution and later extensions such as the Kerr solution for rotating central bodies, the radial coordinate involves a constant of integration which is determined by assumed boundary conditions. Karl Schwarzschild himself derived his solution using a Euclidean radial coordinate, making the assumption that a point mass was located at the origin, but then pointed out that the mathematics for describing the solution could be simplified by using a different radial coordinate which had a value of 2GM at the physical origin. (See arXiv:physics/9905030 for a translation of Schwarzschild's original paper). When Hilbert later described Schwarzschild's solution, he did it directly in terms of the simplified coordinate system, but assumed that the physical origin was where Schwarzschild's modified radial coordinate was zero rather than 2GM.

    It is this assumption which gives rise to the possibility of black holes, and for a long time it seems that no-one noticed that any assumption had been made, until Leonard S Abrams dared to point this out in 1989 (arXiv:gr-qc/0102055). Unfortunately, as there doesn't seem to be any obvious way to prove which boundary assumption is better, this has given rise to various surprisingly fierce battles based on somewhat unsatisfactory arguments which appear to boil down to "my assumptions are better than your assumptions". My personal opinion at present, on the grounds of simplicity, is that Schwarzschild's assumption makes a lot more sense than Hilbert's, in which case black holes don't happen, although it's difficult to distinguish an exceedingly dense object which hasn't quite collapsed into a black hole from one which has. Experimental evidence suggests that at least some extremely dense objects involved in quasars have strong intrinsic magnetic fields, which means they cannot be black holes because of what is known as the "no hair" theorem, and this supports Schwarzschild's original assumption.
     
  6. Sep 22, 2008 #5

    George Jones

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    These ideas have been thoroughly discredited; see

    https://www.physicsforums.com/showthread.php?t=141985&highlight=Abrams

    http://arxiv.org/abs/gr-qc/0608033.
     
  7. Sep 22, 2008 #6

    Jonathan Scott

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  8. Sep 25, 2008 #7
    George, it seems to me to me, as plotted in assymtotically flat coordinates exterior to the blackhole, if the world line of an infalling particle never stalls on at the coordinate singularity of the blackhole horizon as it recedes with Hawking radiation, then in no coordinate system does it stall on the event horizon, or cross. In other words, there is a sort of race going on with the Schwarzchild radius decreasing in the proper time of a distant observer. If the distance of a test particle from the horizon is always finite in the coordinate system of a distant observer at rest with the black hole, then it is finite in any coordinate system.

    In a vernacular way, it would be rather scitzoid if the particle crossed the event horizon in one coordinate system, and not another.
     
    Last edited: Sep 25, 2008
  9. Sep 26, 2008 #8

    atyy

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    One of the things I found helpful for thinking about the Schwarzschild horizon is the horizon of the relativistic rocket, which only requires special relativity, but is just as counter-intuitive: http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html
     
  10. Sep 26, 2008 #9

    George Jones

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    The particle crosses the event horizon in all coordinate systems that cover the event horizon; Schwarzschild coordinates do not cover the even horizon.
     
  11. Sep 26, 2008 #10
    "Cover" is the keywords I needed. thanks!

    Somehow I confused the two threads I started. The other was on crossing the event horizon of a blackhole shrinking under Hawking radiation--but, no matter.

    If, in this senario--big If-- the trajectory of a particle does not include the coordinate singularity on coordinate map, then it shouldn't include the event horizon in any other.

    I think the approach to solve the problem is to set the radius to a function of time per Hawking, and integrate the proper time over the geodesic of a freely falling particle that follows the radial coordinate. The outstanding issue being: what temperal coordinates did Hawking use? Does this sound correct?
     
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