# Finding Geodesics/Solving Differential Equations

1. Jun 16, 2014

### latentcorpse

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.

2. Jun 17, 2014

### Matterwave

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

3. Jun 17, 2014

### latentcorpse

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.

4. Jun 17, 2014

### HallsofIvy

Staff Emeritus
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.

5. Jun 17, 2014

### latentcorpse

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!

6. Jun 18, 2014

### latentcorpse

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.

7. Jun 18, 2014

### bloby

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

Last edited: Jun 18, 2014
8. Jun 18, 2014

### latentcorpse

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?

9. Jun 18, 2014

### bloby

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: Jun 18, 2014
10. Jun 18, 2014

### latentcorpse

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?

11. Jun 19, 2014

### bloby

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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