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How Does Gravity Propagate Backwards in Time?

  1. Nov 11, 2009 #1
    If the radius of the event horizon is determined by the mass in the singularity and the singularity is in the future for everything inside the event horizon, how does gravity propagate backwards in time from the singularity to the event horizon?
     
  2. jcsd
  3. Nov 12, 2009 #2
    Because an event horizon cuts off everything inside a black hole from everything outside, thus the gravity we "see" is a frozen version of what the mass evolved into before it became a singularity.

    Sounds good? It's what someone ran past me once when I asked a similar question. A more correct answer is that whole "space becoming time" thing is an over-simplification and not a very good one. Infalling observers, once past the event horizon, inexorably fall towards the singularity in the Schwarzschild metric, which one can liken to space trading places with time... but not really. In other black hole metrics the situation gets more complicated with multiple event horizons, ring-shaped singularities and other sorts of exotica.

    But a point often missed is that we really have no good reason to believe our calculations past the event horizon - it's terra incognita and unobservable.
     
  4. Nov 12, 2009 #3
    Thank you graal for your thoughtful answer even if it is unsatisfying.

    The idea that the gravity we "see" is a frozen version of the mass that crossed the event horizon seems contrived at best. If gravity is to be considered frozen at the event horizon why not consider that it is at the event horizon where the matter resides instead of at the singularity?

    I agree with you about "space becoming time" being an over simplification and have wondered if the only basis for it is the similarity of gamma becoming imaginary inside the event horizon with the orthogonal relationship between time and space.
     
  5. Nov 12, 2009 #4
    The horizon itself is only apparent to one system of observers. A free-falling observer would in-fall all the way to the core. A different co-ordinate system eliminates the apparent singularity at the horizon and let's one compute all the way to the core.
     
  6. Nov 13, 2009 #5
    It's true that other coordinate systems can be devised e.g one based on radially infalling photons such as the Eddington-Finkelstein coordinates but it eliminates the singularity at the event horizon by becoming singular itself.
     
  7. Nov 13, 2009 #6
    Well, I agree with you about "space becoming time" being an over simplification and have wondered if the only basis for it is the similarity of gamma becoming imaginary inside the event horizon with the orthogonal relationship between time and space.
     
    Last edited by a moderator: Nov 13, 2009
  8. Nov 13, 2009 #7

    George Jones

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    What do you mean? Can you be quantitative?
     
  9. Nov 13, 2009 #8

    jambaugh

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    Actually the shape of the local universe becomes (hyper)tubular, [tex] \mathbb{R}\times \mathbb{S}^2[/tex] with circumferential radius shrinking to zero in finite time. And yes you can sort of say all matter inside the event horizon is "simultaneous" but it isn't meaningful.

    Visualize it this way. Imagine a string of observers infalling into a black hole. As they approach the event horizon the surface will seem to stretch flat and as an observer crosses he will see the universe as a hyper-tube, with cross section the circumferential two sphere and extensive forward and backward direction seemingly spatially flat. Observers entering before him will be ahead of him and those who enter later will be behind him. They will accelerate away from each other due to tidal gravitation along the length of the tube. Each observer will also experience a tidal force growing over time.

    If it were not for the shrinking radius he would be able to move laterally "completely around the universe" returning back were he started (equivalent to circumnavigating the black hole) and indeed observers entering at different angles will be displaced laterally w.r.t. each other. I think eventually there would be event horizons fore and aft growing nearer as time passes due to increasing tidal acceleration along this z axis. Eventually the radius will decrease squishing the observer laterally against himself while tidal forces rip him apart in the fore and aft directions. The singularity will be when the "tube" shrinks to a line.

    Schwarzchild metrics are:
    [tex]d\tau^2 = \left(1-\frac{R}{r}\right)dt^2 - \left(1-\frac{R}{r}\right)^{-1}dr^2 - r^2 d\Omega_2^2[/tex]
    which I pluralize because of the coordinate discontinuity at the [tex]r=R[/tex] event horizon so we have two metrics one inside and one outside. (Omega here is the unit sphere coordinates and d Omega^2 the unit sphere metric.)

    Recall that [tex]r[/tex] is defined here as the circumferential radius and for the interior metric r<R this parameter is time-like. The parameter t becomes space-like so it is associated with time only formally. Now given this I would relabel t->z and r->-T and rewrite the metric as:
    [tex]d\tau^2 =\frac{|T|}{R+T}dT^2 -|T|^{-1}\left(R+T\right)dz^2 - T^2 d\Omega^2[/tex]

    (Note: T is negative r so T increases in the forward time direction. I put in |T| so the signs of metric terms would be clear but |T|=-T = r.)

    We can "normalize" the time coordinate with suitable reparameterization t = f(T) so that...
    (pardon me for reusing variable names but bear with me...)

    [tex]d\tau^2 = dt^2 - a(t)^2dz^2 - r^2(t)d\Omega^2[/tex]

    It is not proper to call the inner Schwarzchild solution "stationary" since the metric depends there on a time-like parameter
    [tex] r=T=f^{-1}(t)[/tex].

    It is then clear that you here have a z-axis by 2-sphere spatial component (hyper tube) with time varying but not spatially varying metric (unless you add in the effects of the infalling masses) with the radius going to zero as t approaches some limit value which further analysis will show is finite.

    Now of course other coordinate systems make clearer the continuity across the event horizon but I think this analysis give the best intuitive grasp of what is going on inside. We think of hitting the singularity as an impact. What one would experience is a time at which our local universe shrinks down and stretches out into a straight line. One never runs into the original matter which formed the black hole or any matter which fell in earlier or later if you aren't real close to it in both time and point on the event horizon when you cross. Eventually the tidal forces will pull any earlier in-falling object out of reach (beyond tidal acceleration horizons fore and aft).

    I'm not as clear on whether this line can be considered to exist beyond the crunch event at r=0. Essentially time stops locally in the same sense that time begins at the big-bang event.

    This is a little bit tangential to the OP but I hope instructive.
     
    Last edited: Nov 13, 2009
  10. Nov 13, 2009 #9

    jambaugh

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    Oops! My earlier post seems to have disappeared. I had mentioned in particular that the size of the event horizon isn't precisely "a function of the mass at the singularity" but rather a function of the total mass interior to the event horizon.

    I then mentioned that inside the event horizon it appears rather that one is in a shrinking "universe". I speculated as to the relationship between objects falling into a black hole at different times. The other post was then to clarify further.
     
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