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Geodesic in 2D space

  1. May 13, 2013 #1
    1. The problem statement, all variables and given/known data
    I am having trouble understanding how the following statement (taken from some old notes) is true:

    >For a 2 dimensional space such that [tex]ds^2=\frac{1}{u^2}(-du^2+dv^2)[/tex]
    the timelike geodesics are given by [tex]u^2=v^2+av+b[/tex] where [itex]a,b[/itex] are constants.




    2. Relevant equations
    Euler-Lagrange, Normalisation condition


    3. The attempt at a solution

    When I see "geodesics" I jump to the Euler-Lagrange equations. They give me
    [tex]\frac{d}{d\lambda}(-2\frac{\dot u}{u^2})=(-\dot u^2+\dot v^2)(-\frac{2}{u^3})\\
    \implies \frac{\ddot u}{u^2}-2\frac{\dot u^2}{u^3}=\frac{1}{u^3}(-\dot u^2+\dot v^2)\\
    \implies u\ddot u-\dot u^2-\dot v^2=0[/tex]
    and
    [tex]\frac{d}{d\lambda}(2\frac{\dot v}{u^2})=0\\
    \implies \dot v=cu^2[/tex]
    where [itex]c[/itex] is some constant.

    Timelike implies [tex]\dot x^a\dot x_a=-1[/tex] where I have adopted the (-+++) signature.

    I can't for the life of me see how the statement results from these. Would someone mind explaining? Thanks.
     
  2. jcsd
  3. May 15, 2013 #2
    Do not expand the Euler-Lagrange equations. But do a trick like this:

    The Lagrangian is:

    [tex] L = \frac{1}{u}\sqrt{\left(\frac{dv}{du}\right)^2-1}= \frac{1}{u}\sqrt{v'^2-1} [/tex]

    Now you see this doesnt depend on $v$. The Euler Lagrange equations then give:

    [tex]0=\frac{\partial L}{\partial v}=\frac{d}{du}\frac{\partial L}{\partial v'} \Longrightarrow \frac{\partial L}{\partial v'}=C [/tex]

    Now calculate [tex]\frac{\partial L}{\partial v'}[/tex] from the Lagrangian and put it equal to the constant C. This will lead to a differential equation which gives the solution you were looking for.
     
    Last edited: May 15, 2013
  4. May 15, 2013 #3
    Thanks, Thaakisfox, I've got it now!
     
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