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Radial Null Geodesic!

  1. Mar 31, 2014 #1
    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!
     
  2. jcsd
  3. Mar 31, 2014 #2

    Bill_K

    User Avatar
    Science Advisor

    The same integral can have several different forms. You need to think about whether r > rs or r < rs.
     
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