Integral of geodesic equations

1. Apr 9, 2014

exmarine

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

$\frac{d^{2}t}{ds^{2}}=-\frac{2m}{r(r-2m)}\frac{dr}{ds}\frac{dt}{ds}$

$\frac{d^{2}\phi}{ds^{2}}=-\frac{2}{r}\frac{dr}{ds}\frac{d\phi}{ds}$

to get:

$\frac{dt}{ds}=\frac{kr}{(r-2m)}$

$\frac{d\phi}{ds}=\frac{h}{r^{2}}$

Does anyone understand that? I certainly don't.

2. Apr 9, 2014

Bill_K

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

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