1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?

  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted