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D'Inverno derivation of Schwarzschild solution

  1. Dec 29, 2011 #1
    If you happen to have D'Inverno's Introducing Einstein's Relativity, this is on page 187. He has reduced the metric to non-zero components:
    [itex]g_{00}= e^{h(t)}(1-2m/r)[/itex]
    [itex]g_{11}=-(1-2m/r)^{-1}[/itex]
    [itex]g_{22}=-r^2[/itex]
    [itex]g_{33}=-r^2\sin^2\theta[/itex]
    The final step is a time coordinate transformation that reduces [itex]g_{00}[/itex] to [itex]1-2m/r[/itex]. This is achieved by making [itex]e^{h(t')}=1[/itex], so [itex]h(t')=0[/itex]. He does this with the relation
    [itex]t'=\int^t_c e^{\frac{1}{2}h(u)}du[/itex], c is an arbitrary constant
    I suppose that, since c is arbitrary, I can assign whatever value to c to make [itex]h(t')=0[/itex], but why use this particular integral as the relation between t and t'? Is there something special about this integral?
     
  2. jcsd
  3. Dec 30, 2011 #2

    Bill_K

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    This is pretty easy. You want (1 - 2m/r) dt'2 = eh(t)(1 - 2m/r) dt2, so take dt' = eh(t)/2 dt
     
  4. Dec 30, 2011 #3
    Ah, OK...but why would the relation not be [itex]t'=\int e^{h(t)/2}dt[/itex]? Instead, they have it as a definite integral from c to t.
     
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