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Flat spacetime at the event horizon of a black hole?

  1. Jun 23, 2015 #1
    I've been working through Leonard Susskind's "The Theoretical Minimum" lecture series (which are a fantastic introduction to the topics covered by the way) and a couple of his comments confused me when he was covering the Kruskal-Szekeres metric/coordinates in General Relativity.

    The end of the relevant section is at 28:48 here:

    He uses the Schwarzchild metric (ignoring the scale factor of Rs)
    [tex]d\tau^2 = (1-1/r) dt^2 - dr^2 / (1-1/r) - r^2 d\Omega^2[/tex]
    and redefines it in terms of the proper distance from the event horizon (ρ) and a new time coordinate (ω = t/2) to arrive at the metric
    [tex]d\tau^2 = F(\rho) \rho^2 d\omega^2 - d\rho^2 - r(\rho)^2 d\Omega^2[/tex]
    and as we approach the event horizon where ρ→0 the metric approaches
    [tex]d\tau^2 = \rho^2 d\omega^2 - d\rho^2 - (1+\rho^2/4)^2 d\Omega^2[/tex]

    It makes sense to me that as ρ→∞ the metric approximates the metric of flat space and I can see that at the event horizon the metric is basically the same as the flat space metric using hyperbolic polar coordinates but I'm not sure how to interpret that information and don't have any intuition. There are both tidal forces and curvature at the event horizon so the spacetime is not flat. Is this just related to the facts that hyperbolic polar coordinates can be used to define hyperbolas of constant r on a spacetime diagram which are equivalent to a constant relativistic acceleration and that an observer in a static position outside a black hole is undergoing a constant acceleration to cancel out the acceleration due to gravity?
     
  2. jcsd
  3. Jun 23, 2015 #2

    bcrowell

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    Spacetime is not flat at the event horizon. Near the event horizon, the metric may look similar to the flat-space metric, but given any point in any spacetime, we can *always* choose coordinates such that the metric has that form near that point. To determine whether spacetime is flat, you need to compute the curvature, which involves taking second derivatives of the metric.
     
  4. Jun 23, 2015 #3
    Oh! That makes perfect sense and actually really helps me tie in the last few lectures with the first half of the series where he covers that topic explicitly. Thank you!
     
  5. Jun 23, 2015 #4

    WannabeNewton

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    The space-time metric can be approximated by the Rindler metric near the horizon.
     
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