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Does time stop at the singularity?

  1. Feb 11, 2016 #1
    In the schwarzschild metric, inside the event horizon the r coordinate becomes timelike, and when when r=0 the coefficient of dr^2 is also 0. So time stops at the singularity, yet somehow it can move around and emit gravitons, i.e. "evolve" even when it's supposed to be frozen like a video that's paused. How?
     
    Last edited by a moderator: Feb 11, 2016
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  3. Feb 11, 2016 #2

    bcrowell

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    There are two problems with your reasoning. (1) Inside the event horizon there is still time. It's just that the timelike coordinate now happens to be the one that we label "r" in the Schwarzschild coordinates. (2) Your reasoning about what happens inside the event horizon doesn't connect logically to your statement about the singularity. The event horizon and the singularity are two separate things.
     
  4. Feb 11, 2016 #3

    Nugatory

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    The word "singularity" means "a place where the mathematical description that we've been using cannot be used because it doesn't correctly describe the physics that's really happening there". The singularity that appears in the Schwarzchild solution is telling us that the Schwarzschild solution describes spacetime in the region ##r\gt{0}##, not ##r\ge{0}##. This, you can't use the Schwarzschild metric as the starting point for reasoning about what happens at ##r=0##.

    Furthermore, it is unlikely that the Schwarzschild solution is correct in the region where ##r## is non-zero but very small. The problem that is that it is a vacuum solution; it applies to the vacuum outside of a spherically symmetric mass and it is very unlikely that quantum mechanics will allow us to concentrate the mass in an arbitrarily small spherically symmetric volume. Thus, the Schwarzschild solution should only be trusted in regions where ##r## is large enough that we can ignore quantum mechanical effects and model the matter in the black hole as a perfect fluid.

    There's a similar situation in classical physics, one so familiar that you may never have noticed it. You're familiar with Newtonian law that describes the earth's gravitational field acting on a body of mass ##m##: ##F=GmM_E/r^2##. But did you notice that it has a singularity at ##r=0##? The only reason we don't worry about the absurd result of an infinite gravitational force there is that the Newtonian formula is treating the earth as an ideal point mass, and we know that that idealization breaks down for ##r\lt{R}_E## where ##R_E## is the radius of the earth.
     
  5. Feb 11, 2016 #4

    bcrowell

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    This is awfully vague. What is "very small?" Are you referring to the Planck scale?

    In general, you appear to be answering a question that the OP didn't ask.
     
  6. Feb 12, 2016 #5

    PeterDonis

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    No, when r=0 the metric is undefined, because it has 1/r in two coefficients (g_rr and g_tt). However, that in itself doesn't tell you that there is a physical issue, it just tells you that the coordinates you are using break down. (Note also that there are other coordinate charts in which the coefficient of dr^2 in the metric is perfectly well defined at r = 0.) Coordinates are not physics.

    The physical issue is that curvature invariants increase without bound as ##r \rightarrow 0##; that is true independent of your choice of coordinates.
     
  7. Feb 12, 2016 #6

    PeterDonis

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    This is a little bit misleading as you state it. Inside the horizon, the Schwarzschild solution is not static; the spherically symmetric mass, outside of which the vacuum exists that the Schwarzschild metric describes, must be either collapsing or expanding, and only the "collapsing" case is relevant if we are considering the future horizon (the "expanding" case is what we get if we extend through the past horizon). What is (probably) very unlikely is that quantum mechanics will allow the mass to collapse into an arbitrarily small volume without some new physics coming into play. But it is not "very unlikely" that QM will allow a static mass inside a horizon to occupy an arbitrarily small volume; that is already impossible at the classical level, regardless of quantum corrections.
     
  8. Feb 12, 2016 #7

    Nugatory

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    That's true.... I must confess that I was mostly focused on discouraging the idea of taking the Schwarzschild metric coefficients at face value ib that region.
     
  9. Feb 13, 2016 #8
    What do u mean by that? That some region of spacetime has no "time?
     
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