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When the Schwarzschild radius is inside the star

  1. Aug 23, 2011 #1

    lavinia

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    What happens to the Schwarzschild metric for an isolated non-rotating body when the horizon radius is inside the body? As I remember from classical physics all of the gravitational pull on an object inside a shell cancels out so it would seem that the horizon radius can not include any mass outside of it.
     
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  3. Aug 23, 2011 #2

    tiny-tim

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    hi lavinia! :smile:
    yes :smile:

    another way of looking at it is that the Schwarzschild metric is only valid in empty space …

    find the Schwarzschild radius from the actual mass, then if that is less than the radius R of the body, the Schwarzschild metric is valid for r > R :wink:

    (eg the metric in the empty space outside the Sun is the Schwarzschild metric for a black hole with the same mass as the Sun)
     
  4. Aug 23, 2011 #3

    K^2

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    Just to make sure it's clear, that is always the case for a star.
     
  5. Aug 25, 2011 #4

    lavinia

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    Thanks Tiny-tim! So the metric breaks down inside the star. That makes sense. What happens to the metric?
     
  6. Aug 25, 2011 #5

    K^2

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    There is a different metric for inside of a spherically symmetric body. It matches Swarzschild metric at the boundary (surface). There's a name for it, but I don't recall what that name is. I'm also not sure if there is an analytic solution for it, or if it's purely numerical.
     
  7. Aug 25, 2011 #6

    WannabeNewton

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    It just isn't valid in the interior of the star. If R > 2M and you come to r = R you won't really see anything happen in the metric but since this metric is a vacuum solution it can only be allowed to describe anything for r > R. If you want to describe the interior you will have to solve the EFEs, possibly with spherical symmetry conditions applied, for a non - vanishing Stress - Energy tensor (such as that of a static fluid to get the solution for the interior of a static, spherically symmetric star).
     
  8. Aug 25, 2011 #7

    WannabeNewton

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    Oh the Tolman - Oppenheimer equation? - [tex]\frac{\mathrm{d} p}{\mathrm{d} r} = \frac{{(\rho + p)(m(r) + 4\pi r^{3}p)}}{r(r - 2m(r))}[/tex] where [itex]m(r) = \frac{1}{2}r(1 - e^{-2\Lambda })[/itex] and you have the equation [tex](\rho + p)\frac{\mathrm{d} \Phi }{\mathrm{d} r} = \frac{\mathrm{d} p}{\mathrm{d} r}[/tex] to get the metric. I also believe there is no analytic solution for it but I could be wrong.
     
  9. Aug 25, 2011 #8

    George Jones

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    An equation of state [itex]p = p \left(\rho \right)[/itex] also is needed. For realistic models (eauations of state), this is done numerically, but Schwarzschild solved Einstein's equation analytically for a constant-density spherical star, with result
    [tex]
    g=\left( \frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2Mr^{2}}{R^{3}}}\right) ^{2}dt^{2}-\left( 1-\frac{2Mr^{2}}{R^{3}}\right) ^{-1}dr^{2}-r^{2}\left( d\theta ^{2}+\sin ^{2}\theta d\phi ^{2}\right).
    [/tex]

    Unfortunately, the term "interior Schwarzschild solution" can refer to either of two very different metrics (solutions to Einstein's equation):

    1) spacetime inside the event horizon of the Schwarzschild black hole solution, where there "is a singularity";

    2) the non-singular solution inside a spherically symmetric, constant density massive object.
     
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