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Integral of geodesic equations

  1. Apr 9, 2014 #1
    Kevin Brown, in his excellent book "Reflections on Relativity" p. 409, "immediately" integrates 2 geodesic equations:

    [itex]\frac{d^{2}t}{ds^{2}}=-\frac{2m}{r(r-2m)}\frac{dr}{ds}\frac{dt}{ds}[/itex]

    [itex]\frac{d^{2}\phi}{ds^{2}}=-\frac{2}{r}\frac{dr}{ds}\frac{d\phi}{ds}[/itex]

    to get:

    [itex]\frac{dt}{ds}=\frac{kr}{(r-2m)}[/itex]

    [itex]\frac{d\phi}{ds}=\frac{h}{r^{2}}[/itex]

    Does anyone understand that? I certainly don't.
     
  2. jcsd
  3. Apr 9, 2014 #2

    Bill_K

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    Science Advisor

    They both go pretty much the same way. For the second one,

    [tex]\begin{eqnarray*}\frac{\frac{d^2 \phi}{ds^2}}{\frac{d \phi}{ds}} &=& - \frac{2}{r}\frac{dr}{ds}\\
    \frac{d}{ds}(\ln(\frac{d \phi}{ds})) &=& -2 \frac{d}{ds} \ln(r)\\
    \ln(\frac{d \phi}{ds}) &=& -2 \ln(r) + const\\
    \frac{d \phi}{ds} &=& \frac{h}{r^2}\end{eqnarray*}[/tex]
     
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