Finding Geodesics/Solving Differential Equations

  • #1
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I want to find geodesics on 2-dimensional Anti de-Sitter space. According to http://www.aei.mpg.de/~gielen/ads.pdf I can write the metric of this as

[tex] ds^2 = - \cosh^2{r} dt^2 + dr^2 [/tex]

Now, we know the geodesic equation is

[tex] \frac{d^2 x^\mu}{d \tau^2} + \Gamma^\mu{}_{\nu \rho} \frac{d x^\nu}{d \tau} \frac{d x^\rho}{d \tau} = 0, \quad \Gamma^\mu{}_{\nu \rho} = \frac{1}{2} g^{\mu \sigma} \left( g_{\nu \sigma, \rho} + g_{\rho\ sigma, \nu} - g_{\nu \rho, \sigma} \right) [/tex]

and where [itex] \tau [/itex] is an affine parameter.

Let's consider the case [itex] \mu = r[/itex], then to work out the Christoffel symbols, I need to know that [itex] g^{rr} = 1[/itex] (using the fact that the metric is diagonal).

Then I compute
[tex] \Gamma^{r}_{tt} = \frac{1}{2} \left( g_{tr,t} + g_{tr,t} - g_{tt, r} \right) = \frac{1}{2} 2 \cosh{r} \sinh{r} = \cosh{r} \sinh{r} [/tex]
[tex] \Gamma^{r}_{tr} = \frac{1}{2} \left( g_{tr,r} + g_{rr,t} -g_{tr,r} \right) = 0 [/tex]
[tex] \Gamma^r{}_{rr} = \frac{1}{2} \left( g_{rr,r} + g_{rr,r} - g_{rr,r} \right) = 0 [/tex]

This means that the Geodesic equation takes the form

[tex] \ddot{r} + \left( \cosh{r} \sinh{r} \right) \dot{t}^2 = 0 [/tex]

Where the dot denotes differentiation with respect to the affine parameter [itex]\tau[/itex].

I don't know how to solve this equation. I can multiply the cosh and sinh to get it to the form [itex] \ddot{r} + \frac{1}{2} \left( \sinh{2r} \right) \dot{t}^2 = 0 [/itex] but that doesn't seem to help. Does anyone have any suggestions? Am I using a poor choice of coordinates - hopefully not as I'd ideally like this in terms of sinh and cosh if possible.

Thanks.
 

Answers and Replies

  • #2
Matterwave
Science Advisor
Gold Member
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You should have 2 geodesic equations though right? Where's your other geodesic equation? You have 2 variables, but only 1 equation.
 
  • #3
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You should have 2 geodesic equations though right? Where's your other geodesic equation? You have 2 variables, but only 1 equation.
So I find [itex] g^{tt} = - \frac{1}{\cosh^2{r}}[/itex] and so

[tex] \Gamma^t_{tt} = \frac{1}{2} g^{tt} g_{tt,t} = - \frac{1}{2 \cosh^2{r}} \left( -2 \cosh{r} \sinh{r} \frac{dr}{dt} \right) = \tanh{r} \frac{\dot{r}}{\dot{t}}[/tex]
[tex] \Gamma^{t}_{tr} = \frac{1}{2} g^{tt} \left( g_{tt,r} + g_{tr,t} -g_{tr,t} \right) = - \frac{1}{2 \cosh^2{r}} \left( -2 \cosh{r} \sinh{r} \right) = \tanh{r} [/tex]
[tex] \Gamma^{t}_{rr} = \frac{1}{2} g^{tt} \left( g_{tr,t} + g_{rt,t} - g_{rr,t} \right) =0 [/tex]

This means the differential equations are
1, [tex] \ddot{t} + \left(2 \tanh{r} \right) \dot{r} \dot{t} =0 [/tex]
2, [tex] \ddot{r} + \cosh{r} \sinh{r} \dot{t}^2 =0 [/tex]

Obviously these are coupled and I cannot see how to solve them - perhaps a suitable change of coordinates will improve things but I really cannot see it. Any ideas?

Thanks.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
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Since the first equation involves only derivatives of t, not t, itself, let [tex]u= \dot{t}[/tex] so that the equation is [tex]\dot{u}+ \left(2\tanh(r)\right)\dot{r}u= 0[/tex] or
[tex]\dot{u}= \left(2\tanh(r)\right)\dot{r}u[/tex]
[tex]\frac{du}{u}= 2\tanh(r) dr[/tex]

That can be integrated for u in terms of r.
 
  • #5
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Since the first equation involves only derivatives of t, not t, itself, let [tex]u= \dot{t}[/tex] so that the equation is [tex]\dot{u}+ \left(2\tanh(r)\right)\dot{r}u= 0[/tex] or
[tex]\dot{u}= \left(2\tanh(r)\right)\dot{r}u[/tex]
[tex]\frac{du}{u}= 2\tanh(r) dr[/tex]

That can be integrated for u in terms of r.
So [tex] \ln{u} = -\ln{\cosh{r}} + c[/tex]

Exponentiating we get [tex] u = -A \cosh{r} [/tex]

This becomes [tex] dt = -A \cosh{(r(\tau))} d \tau[/tex]

So we get [tex] t(\tau) = -A \sinh{(r(\tau))} + C[/tex]

Is that correct? I seem unable to use this to satisfy the other geodesic equation!
 
  • #6
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Since the first equation involves only derivatives of t, not t, itself, let [tex]u= \dot{t}[/tex] so that the equation is [tex]\dot{u}+ \left(2\tanh(r)\right)\dot{r}u= 0[/tex] or
[tex]\dot{u}= \left(2\tanh(r)\right)\dot{r}u[/tex]
[tex]\frac{du}{u}= 2\tanh(r) dr[/tex]

That can be integrated for u in terms of r.
Actually I realise I made a mistake in my above answer. I now find [itex]\ln{u} = -2 \ln{\cosh{r}} + C \Rightarrow u = A ( \cosh{r})^{-2}[/itex]

Substituting into the other geodesic equation gives

[tex] \ddot{r} + A^2 \frac{\sinh{r}}{\cosh^3{r}} =0[/tex]

I don't know how to solve this and am now thinking that maybe I made a mistake earlier on - perhaps with the factor of 2 in the t geodesic equation as if this were a 1 then I would get a tanh in the above equation which would be easier to deal with. However, since we sum over the tr and rt contributions in the geodesic equation, I think it's correct that it should double up.

What do you reckon?

Thanks.
 
  • #7
112
8
##g_{tt,t}## =0? Then no factor 2.
##\int cosh(r(\tau))\,d\tau \neq sinh(r(\tau))## since it's a composite function.
 
Last edited:
  • #8
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##g_{tt,t}## =0? Then no factor 2.
I disagree. I find

[tex] \Gamma^t_{tt} = \frac{1}{2} g^{tt} g_{tt,t} = - \frac{1}{2 \cosh^2{(r(\tau))}} \left( -2 \cosh{r} \sinh{r} \frac{dr(\tau)}{dt} \right) = \tanh{r} \frac{dr}{d \tau} \frac{d \tau}{dt} = \tanh{r} \frac{\dot{r}}{\dot{t}}[/tex]

Does that make sense?
 
  • #9
112
8
It's confusing but "##,_t##" means partial derivative along the coordinate t and Christoffel's symbols are defined independently of geodesics.
Maybe not very helpful..
I get ##u = \frac{A}{cosh(r)}## and ##\ddot{r}+B\, tanh(r)=0##
 
Last edited:
  • #10
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It's confusing but "##,_t##" means partial derivative along the coordinate t and Christoffel's symbols are defined independently of geodesics.
Maybe not very helpful..
I get ##u = \frac{A}{cosh(r)}## and ##\ddot{r}+B\, tanh(r)=0##
Your eqns are lacking any first order derivatives. How did you get them? Even accepting your dispute with my [itex]\ddot{t}[/itex] equation, I don't see how you get a [itex]\tanh{r}[/itex] in the [itex]\ddot{r}[/itex] equation?
 
  • #11
112
8
Trying to help without self confidence: not a good idea...
I find ##\Gamma^t_{tt} = 0 = \Gamma^t_{rr}## and##\Gamma^t_{rt} = tanh(r) = \Gamma^t_{tr}##.
Then ##\ddot{t} + tanh(r) \dot{t}\dot{r}+tanh(r)\dot{r}\dot{t} =0## where I missed the sum over the two gammas. Then as Hallsoflvy ##u=\dot{t}## that gives ##\frac{du}{u}=-2\,tanh(r)\,dr## and like you did ##ln(u)=-2\,ln(cosh(r))+A## then ##u=\frac{A}{cosh^2 (r)}## wich I put in the second equation(##u^2=\dot{t}^2##): ##\ddot{r}+tanh(r)\frac{B}{cosh^2(r)}=0##. That is ##\ddot{r}=C\, tanh(r)\,tanh'(r)## and multiplying both sides with ##\dot{r}## I get ##\ddot{r}\dot{r}=D\frac{d\,tanh(r)}{d\tau}\,tanh(r)##. This is ok with ##\dot{r}=thanh(r)##. Then ##\frac{dr}{tanh(r)}=cst\,d\tau## and ##r=arcsinh(exp(cst\,\tau))## and ##\dot{t}=\frac{A}{1+sinh^2(arcsinh(exp(\alpha\tau)))}## and so on
 

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