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Vacuum Einstein Equation

  1. Oct 31, 2012 #1
    Sorry if I am asking a too trivial question! I am having a confusion regarding the following-The solution to Einstein equation in vacuum is given by the Schwarzschild metric. However, what does the mass represent in the metric in Schwarzschild coordinate? Whose mass is it and how does it enter although we are looking for vacuum solutions, meaning T_{ab}=0
     
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  3. Oct 31, 2012 #2

    Matterwave

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    The Schwarzschild metric is one particular solution to the vacuum Einstein equations for the particular case of one spherically symmetric, stationary (non-rotating), uncharged mass. There are other solutions to the vacuum field equations (e.g. Kerr metric). The mass term is a boundary value representing the total mass-energy of the central object.
     
  4. Oct 31, 2012 #3

    haushofer

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    To be more precise, the Schwarzschild solution is a solution of empty space "up to one point". You could see it as the solution of which the energy momentum tensor is given by

    [tex]
    T_{\mu\nu} = \rho \delta_{\mu}^0 \delta_{\nu}^0 \delta^4 (r)
    [/tex]

    where rho is some mass density. With other words: if [itex] r \neq 0 [/itex] then [itex]T_{\mu\nu} = 0 [/itex], whereas for r=0 one has that [itex]T_{\mu\nu} \rightarrow \infty[/itex] indicating a singularity.
     
  5. Oct 31, 2012 #4
    Or rather, the Schwarzchild solution is valid outside any massive, spherically symmetric object. A good analogy would be to the electric field outside a ball of charge. Outside the ball, there is no charge density, but there is an electric field from the ball. Similarly, outside a massive, spherically symmetric object, there are still effects from gravity, but the stress-energy tensor is zero.
     
  6. Oct 31, 2012 #5

    tom.stoer

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    No.

    As Muphrid says
    the Schwarzschild solution is applicable for all for spherically symmetric massive bodies; the the singularity is only a special case
     
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