Radial Null Geodesic!

  • Thread starter Ichimaru
  • Start date
  • #1
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Main Question or Discussion Point

Hi,

I've found geodesic equations for the metric:

\begin{equation}

ds^{2} = -c^{2} \alpha dt^{2} + \frac{1}{ \alpha } dr^{2} + d \omega ^{2}

\end{equation}

where

\begin{equation}

\alpha = 1 - \frac{r^{2}}{r_{s}^{2}}

\end{equation}

I have found that for a light ray:

\begin{equation}

\alpha \frac{dt}{d \lambda} = 1 ;
\frac{dr}{d \lambda} = c

\end{equation}

Where lambda is an affine parameter and i have appliedthe conditions:

\begin{equation}


\frac{dt}{d \lambda} = 1 , r = 0

\end{equation}

I am then told to integrate these equations to get:

\begin{equation}

r = r_{S}tanh(\frac{ct}{r_{s}})

\end{equation}

However when I try to integrate I get:

\begin{equation}

\int_{0}^{r} \frac{1}{1-\frac{r^2}{r_{s}^{2}}}dr = ct

\end{equation}

goes to:

\begin{equation}

\frac{r_{s}}{2} ln \frac{r+r_{s}}{r-r_{s}} = ct

\end{equation}

Which rearranges to:

\begin{equation}

r = r_{s}coth( \frac{ct}{r_s})

\end{equation}

Any help would be appreciated, thanks!
 

Answers and Replies

  • #2
Bill_K
Science Advisor
Insights Author
4,155
195
The same integral can have several different forms. You need to think about whether r > rs or r < rs.
 

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