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A 2d Geodesic equation

  1. Jan 10, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider the 2-dim metric [tex]{{\it ds}}^{2}=-{\frac {{a}^{2}{{\it dr}}^{2}}{ \left( {r}^{2}-{a}^{2}\right) ^{2}}}+{\frac {{r}^{2}{d\theta }^{2}}{{r}^{2}-{a}^{2}}}[/tex], where r > a. What is its signature? Show that its geodesics satisfy

    [tex]{\frac {{a}^{2}{{\it dr}}^{2}}{{d\theta }^{2}}}+{a}^{2}{r}^{2}={k}^{2}{r}^{4}[/tex]
    where k is a constant. For which value(s) of k are the geodesics null?

    3. The attempt at a solution

    1- The signature is clearly (-,+).

    2- I can show straightly from the metric itself that if [tex]\tau[/tex] is the proper time, dividing each side of metric by [tex]d\tau^2[/tex] and using [tex](dr/d\tau)/(d\theta /d\tau)=dr/d\theta[/tex] yields the left-hand side of the desired geodesic equation and the other side would be of the form [tex]k^2r^4 [/tex] with [tex]k=\mbox {{\pm}} \left( \sqrt {1-{\frac {{{\it ds}}^{2} \left({r}^{2}-{a}^{2} \right) ^{2}}{{r}^{4}{d\theta }^{2}}}} \right) [/tex]. This can be used to show that if k=+1 or -1 then the geodesics are null. But I don't know anything about how k is supposed to be constant with those irritating r's. What is probably wrong?

    Thanks
    AB
     
  2. jcsd
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