# Finding Geodesics/Solving Differential Equations

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

$$ds^2 = - \cosh^2{r} dt^2 + dr^2$$

Now, we know the geodesic equation is

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

and where $\tau$ is an affine parameter.

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

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

This means that the Geodesic equation takes the form

$$\ddot{r} + \left( \cosh{r} \sinh{r} \right) \dot{t}^2 = 0$$

Where the dot denotes differentiation with respect to the affine parameter $\tau$.

I don't know how to solve this equation. I can multiply the cosh and sinh to get it to the form $\ddot{r} + \frac{1}{2} \left( \sinh{2r} \right) \dot{t}^2 = 0$ 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.

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Matterwave
Gold Member
You should have 2 geodesic equations though right? Where's your other geodesic equation? You have 2 variables, but only 1 equation.

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 $g^{tt} = - \frac{1}{\cosh^2{r}}$ and so

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

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

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.

HallsofIvy
Homework Helper
Since the first equation involves only derivatives of t, not t, itself, let $$u= \dot{t}$$ so that the equation is $$\dot{u}+ \left(2\tanh(r)\right)\dot{r}u= 0$$ or
$$\dot{u}= \left(2\tanh(r)\right)\dot{r}u$$
$$\frac{du}{u}= 2\tanh(r) dr$$

That can be integrated for u in terms of r.

Since the first equation involves only derivatives of t, not t, itself, let $$u= \dot{t}$$ so that the equation is $$\dot{u}+ \left(2\tanh(r)\right)\dot{r}u= 0$$ or
$$\dot{u}= \left(2\tanh(r)\right)\dot{r}u$$
$$\frac{du}{u}= 2\tanh(r) dr$$

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

Exponentiating we get $$u = -A \cosh{r}$$

This becomes $$dt = -A \cosh{(r(\tau))} d \tau$$

So we get $$t(\tau) = -A \sinh{(r(\tau))} + C$$

Is that correct? I seem unable to use this to satisfy the other geodesic equation!

Since the first equation involves only derivatives of t, not t, itself, let $$u= \dot{t}$$ so that the equation is $$\dot{u}+ \left(2\tanh(r)\right)\dot{r}u= 0$$ or
$$\dot{u}= \left(2\tanh(r)\right)\dot{r}u$$
$$\frac{du}{u}= 2\tanh(r) dr$$

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

Substituting into the other geodesic equation gives

$$\ddot{r} + A^2 \frac{\sinh{r}}{\cosh^3{r}} =0$$

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.

##g_{tt,t}## =0? Then no factor 2.
##\int cosh(r(\tau))\,d\tau \neq sinh(r(\tau))## since it's a composite function.

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##g_{tt,t}## =0? Then no factor 2.
I disagree. I find

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

Does that make sense?

It's confusing but "##,_t##" means partial derivative along the coordinate t and Christoffel's symbols are defined independently of geodesics.
I get ##u = \frac{A}{cosh(r)}## and ##\ddot{r}+B\, tanh(r)=0##

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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.
Your eqns are lacking any first order derivatives. How did you get them? Even accepting your dispute with my $\ddot{t}$ equation, I don't see how you get a $\tanh{r}$ in the $\ddot{r}$ equation?