Solving Radially-Directed Geodesics in AdS Space

[physicus]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

Using the geodesic equation this yields:
\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
\frac{d^2\tau}{d\lambda^2}+2tanh\,\rho\,\frac{d \rho}{d\lambda}\frac{d\tau}{d\lambda}=0
\frac{d^2\Omega_i}{d\lambda^2}+2coth\,\rho\,\sum_j \frac{d \rho}{d\lambda}\frac{d\Omega_j}{d\lambda}=0

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

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

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

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

Cheers, physicus

 

[physicus]

I calculated the Christoffel symbols to obtain the following non-zero components:
\Gamma^\rho_{\tau\tau}=sinh\,\rho
\Gamma^\phi_{ii}=-cosh\,\rho

These two Christoffels were wrong. It should be
\Gamma^\rho_{\tau\tau}=sinh\,\rho\,cosh\,\rho
\Gamma^\phi_{ii}=-sinh\,\rho\,cosh\,\rho

However, I end up with the same equation.

 

[George Jones]

Try reduction of order. Set

u = \frac{d\rho}{d\lambda},

so that

\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.

 

[kevinferreira]

Homework Statement

Since we look for massless geodesics we require
g_{\mu\nu}\frac{dx^\mu}{ \lambda}\frac{dx^\nu}{d \lambda}=0

Why are you looking for massless geodesics? Is this required?

 

[physicus]

Sorry, I forgot the "massless" in the problem statement. It is indeed required.

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

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

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

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

We want to find \tau(\rho). Can we conclude the following?
\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}

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