## 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

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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
Staff Emeritus
Gold Member
Try reduction of order. Set

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

so that

$$\frac{d^2 \rho}{d\lambda^2} = \frac{du}{d\lambda} = \frac{du}{d\rho} \frac{d\rho}{d\lambda} = \frac{du}{d\rho}u.$$

## 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?

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?