Integrating Geodesic Equations: Kevin Brown

  • Context: Graduate 
  • Thread starter Thread starter exmarine
  • Start date Start date
  • Tags Tags
    Geodesic Integral
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 2K views
exmarine
Messages
241
Reaction score
11
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.
 
Physics news on Phys.org
exmarine said:
[itex]\frac{d^{2}\phi}{ds^{2}}=-\frac{2}{r}\frac{dr}{ds}\frac{d\phi}{ds}[/itex]

[itex]\frac{d\phi}{ds}=\frac{h}{r^{2}}[/itex]
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}\\<br /> \frac{d}{ds}(\ln(\frac{d \phi}{ds})) &=& -2 \frac{d}{ds} \ln(r)\\<br /> \ln(\frac{d \phi}{ds}) &=& -2 \ln(r) + const\\<br /> \frac{d \phi}{ds} &=& \frac{h}{r^2}\end{eqnarray*}[/tex]
 
  • Like
Likes   Reactions: 1 person