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Geodesics in AdS space

  • Thread starter physicus
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  • #1
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Homework Statement



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].


Homework 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]


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
 
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Answers and Replies

  • #2
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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]
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.
 
  • #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]
 
  • #4

Homework Statement



Since we look for massless geodesics we require
[itex]g_{\mu\nu}\frac{dx^\mu}{ \lambda}\frac{dx^\nu}{d \lambda}=0[/itex]
Why are you looking for massless geodesics? Is this required?
 
  • #5
55
3
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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