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GR - The horizon a null surface?

  1. Apr 23, 2007 #1

    quasar987

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    1. The problem statement, all variables and given/known data
    In my GR final, there was a question that asked to show that the horizon r=2GM is a null surface, i.e. [itex]d\tau^2=0[/itex]. How does that work, since just plugging r=2GM and dr=0 in the Schwarzschild metric yields

    [tex]d\tau^2=-(2GM)^2d\Omega^2[/tex]

    which is obviously not zero. :confused:
     
  2. jcsd
  3. Apr 23, 2007 #2

    Dick

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    A null surface has the property that its normal vector is null. What's a normal covector to the surface defined by r-2GM=0? Is it null?
     
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