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Geodesics in AdS space

  1. Dec 24, 2012 #1
    1. The problem statement, all variables and given/known data

    We consider global AdS given by the coordinates [itex](\rho,\tau, \Omega_i), i=1,\ldots,d[/itex] and the metric
    [itex] ds^2=L^2(-cosh^2\,\rho\,d\tau^2+d\rho^2+sinh^2\,\rho\,d \Omega_i{}^2)[/itex]
    Find the trajectory [itex]\tau(\rho)[/itex], radially-directed geodesics, strating from [itex]\rho=\rho_0[/itex] with proper time [itex]\tau(\rho_0)[/itex].


    2. Relevant equations

    Geodesic equation: [itex]\frac{d^2x^{\mu}}{d \lambda^2}+\Gamma^\mu_{\nu \alpha}\frac{dx^\nu}{d \lambda}\frac{dx^\alpha}{d \lambda}=0[/itex]


    3. The attempt at a solution

    I calculated the Christoffel symbols to obtain the following non-zero components:
    [itex]\Gamma^\rho_{\tau\tau}=sinh\,\rho[/itex]
    [itex]\Gamma^\phi_{ii}=-cosh\,\rho[/itex]
    [itex]\Gamma^\tau_{\rho\tau}=\Gamma^\tau_{\tau\rho}=tanh\,\rho[/itex]
    [itex]\Gamma^i_{\rho i}= \Gamma^i_{i\rho}=coth\,\rho[/itex]

    Using the geodesic equation this yields:
    [itex]\frac{d^2\rho}{d\lambda^2}+sinh\,\rho\,\left(\frac{d\tau}{d\lambda}\right)^2-\sum_i cosh\,\rho\,\left(\frac{d\Omega_i}{d\lambda}\right)^2=0[/itex]
    [itex]\frac{d^2\tau}{d\lambda^2}+2tanh\,\rho\,\frac{d \rho}{d\lambda}\frac{d\tau}{d\lambda}=0[/itex]
    [itex]\frac{d^2\Omega_i}{d\lambda^2}+2coth\,\rho\,\sum_j \frac{d \rho}{d\lambda}\frac{d\Omega_j}{d\lambda}=0[/itex]

    We are looking for radially directed geodesics, i.e. try if there are solutions with [itex]\frac{d \Omega_i}{d\lambda}=0[/itex]. Since this ansatz leads to [itex]\frac{d^2 \Omega_i}{d\lambda^2}=0[/itex] it is consistent with the above equations. Therefore, we can simplify:
    [itex]\frac{d^2 \rho}{d\lambda^2}+sinh\,\rho\,\left(\frac{d\tau}{d\lambda}\right)^2=0[/itex] (*)
    [itex]\frac{d^2\tau}{d\lambda^2}+2tanh\,\rho\,\frac{d \rho}{d\lambda}\frac{d\tau}{d\lambda}=0[/itex]

    Since we look for massless geodesics we require
    [itex]g_{\mu\nu}\frac{dx^\mu}{ \lambda}\frac{dx^\nu}{d \lambda}=0[/itex]
    [itex]\Rightarrow -cosh\,\rho\,\left(\frac{d\tau}{d\lambda}\right)^2+\left(\frac{d\rho}{d\lambda}\right)^2+\sum_i sinh\,\rho\,\left(\frac{d\Omega_i}{d\lambda}\right)^2=0[/itex]
    [itex]\Rightarrow -cosh\,\rho\,\left(\frac{d\tau}{d\lambda}\right)^2+\left(\frac{d\rho}{d\lambda}\right)^2 = 0[/itex]

    We can use this to simplify (*) to get
    [itex] \frac{d^2\rho}{d\lambda^2}+tanh\,\rho\,\left(\frac{d\rho}{d\lambda}\right)^2=0[/itex]

    However, I still don't see a solution. Does anyone have an idea?

    Cheers, physicus
     
  2. jcsd
  3. Dec 25, 2012 #2
    These two Christoffels were wrong. It should be
    [itex]\Gamma^\rho_{\tau\tau}=sinh\,\rho\,cosh\,\rho[/itex]
    [itex]\Gamma^\phi_{ii}=-sinh\,\rho\,cosh\,\rho[/itex]

    However, I end up with the same equation.
     
  4. Dec 25, 2012 #3

    George Jones

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    Try reduction of order. Set

    [tex]u = \frac{d\rho}{d\lambda},[/tex]

    so that

    [tex]\frac{d^2 \rho}{d\lambda^2} = \frac{du}{d\lambda} = \frac{du}{d\rho} \frac{d\rho}{d\lambda} = \frac{du}{d\rho}u.[/tex]
     
  5. Dec 25, 2012 #4
    Why are you looking for massless geodesics? Is this required?
     
  6. Dec 26, 2012 #5
    Sorry, I forgot the "massless" in the problem statement. It is indeed required.

    I try reduction of order [itex]u=\frac{d\rho}{d\lambda}, v=\frac{d\tau}{d\lambda}[/itex].

    This yields
    [itex]\frac{du}{d\lambda}+tanh\,\rho\,u^2=0[/itex]
    Since [itex]\frac{du}{d\lambda}=\frac{du}{d\rho}u[/itex] we get
    [itex]\frac{du}{d\rho}=-tanh\,\rho\,u[/itex]
    [itex]\Rightarrow u=\frac{1}{cosh\,\rho}[/itex]
    [itex]\Rightarrow \frac{d\rho}{d\lambda}=\frac{1}{cosh\,\rho}[/itex]

    Also, from the second equation [itex]\frac{d^2\tau}{d\lambda^2}+2tanh\,\rho\,\frac{d \rho}{d\lambda}\frac{d\tau}{d \lambda}=0[/itex]
    [itex]\Rightarrow \frac{dv}{d\lambda}+2tanh\,\rho\,\frac{d\rho}{d \lambda}v=0[/itex]
    Since [itex]\frac{dv}{d\lambda}=\frac{dv}{d\rho}\frac{d\rho}{d\lambda}[/itex]
    [itex]\Rightarrow \frac{dv}{d\rho}+2tanh\,\rho\,v=0[/itex]
    [itex]\Rightarrow \frac{dv}{v}=-2tanh\,\rho\,d\rho[/itex]
    [itex]\Rightarrow log\,v=-2log\,cosh\,\rho[/itex]
    [itex]\Rightarrow v=\frac{1}{cosh^2\,\rho}[/itex]
    [itex]\Rightarrow \frac{d\tau}{d\lambda}=\frac{1}{cosh^2\,\rho}[/itex]

    This is in agreement with the third equation I found:
    [itex]-cosh^2\,\rho\,\left(\frac{d\tau}{d\lambda}\right)^2+\left(\frac{d\rho}{d\lambda}\right)^2=0[/itex]

    We want to find [itex]\tau(\rho)[/itex]. Can we conclude the following?
    [itex]\frac{d\tau}{d\rho}=\frac{d\tau}{d\lambda}\frac{d \lambda}{d\rho}= \frac{d\tau}{d\lambda}\left(\frac{d\rho}{d\lambda}\right)^{-1}=\frac{1}{cosh\,\rho}[/itex]

    This would yield: [itex]\tau(\rho)=2tan^{-1}(tanh(\rho/2))+\tau(\rho_0)[/itex]
    This seems a bit complicated. Is it right?
     
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