# Geodesics equation in a 2-space with a certain $ds^2$

This is exercise 3.20 of Hobson's general relativity. It's presented as follows:

In the 2-space with line element $$ds^2=\frac{dr^2+r^2d\theta^2}{r^2-a^2}-\frac{r^2dr^2}{(r^2-a^2)^2}$$ Where r>a, show that the differential equation for the geodesics may be written as: $$a^2\left(\frac{dr}{d\theta}\right)^2+a^2r^2=Kr^4$$ Where $$K$$ is a constant such that $$K=1$$ if the geodesic is null.

In my attempt for a solution, I summed the terms with $$dr^2$$ in the expression for the line element in order to get: $$ds^2=-\frac{a^2}{(r^2-a^2)^2}dr^2+\frac{r^2}{r^2-a^2}d\theta^2$$ From this expression I got the components of the metric tensor, being these: $$g_{rr}=-\frac{a^2}{(r^2-a^2)^2}$$ $$g_{\theta\theta}=\frac{r^2}{r^2-a^2}$$ As the metric tensor is diagonal, it's straightforward to get its contravariant components, since they will just be the inverse of their covariant counterparts: $$g^{rr}=g_{rr}^{-1}$$ and $$g^{\theta\theta}=g_{\theta\theta}^{-1}$$. Having calculated the metric tensor and its inverse, now let's compute the connection coefficients via: $$\Gamma^a{}_{bc}=\frac{1}{2}g^{ad}(\partial_bg_{dc}+\partial_cg_{bd}-\partial_dg_{bc})$$ I found out that every connection coefficient vanishes except for the following:

$$\Gamma^{r}{}_{rr}=\frac{-2r}{r^2-a^2}$$ $$\Gamma^{r}{}_{\theta\theta}=-r$$ $$\Gamma^{\theta}{}_{\theta r}=\Gamma^{\theta}{}_{r \theta}=\frac{-a^2}{r(r^2-a^2)}$$

Using the geodesic equations: $$\ddot{x}^a+\Gamma^{a}{}_{bc}\dot{x}^b\dot{x}^c=0$$, I get the two geodesic equations:

$$\ddot{r}+\Gamma^{r}{}_{rr}\dot{r}^2+\Gamma^r{}_{\theta \theta}\dot{\theta}^2=\ddot{r}+r\left( \frac{-2\dot{r}^2}{r^2-a^2}-\dot\theta^2\right)=0$$ $$\ddot\theta + 2\Gamma^\theta{}_{r \theta}\dot{r}\dot{\theta}=\ddot{\theta}-\frac{2a^2}{r(r^2-a^2)}\dot{r}\dot{\theta}=0$$

Where $$\dot x$$ stands for the derivative with respect to the parameter of the geodesic, $$\dot x=\frac{dx}{du}$$.

So far so good, I believe (unless I made a mistake calculating, which is actually possible though I checked my calculations several times before posting), but here I'm stuck. I think working out a bit with both equations and swapping the derivatives with respect to the parameter to derivatives with respect to the coordinates I might be able to get an expression as the one I'm after... But I wasn't able to do it. Any help on how to continue will be much appreciated!

$$\ddot{r}+r\left( \frac{-2\dot{r}^2}{r^2-a^2}-\dot\theta^2\right)=0 \tag{1}$$ $$\ddot{\theta}-\frac{2a^2}{r(r^2-a^2)}\dot{r}\dot{\theta}=0 \tag{2}$$

The Eq.(2) can be solved by separation: $$\ddot{\theta}-\frac{2a^2}{r(r^2-a^2)}\dot{r}\dot{\theta}=0$$ $$\frac{\ddot{\theta}}{\dot{\theta}} = \frac{2a^2}{r(r^2-a^2)}\dot{r} = \frac{a^2}{r^2(r^2-a^2)} 2 r \dot{r} = \left[ \frac{1}{r^2-a^2} - \frac{1}{r^2}\right] \frac{d}{dt}r^2$$

Both sides are integrable:

$$\ln\dot\theta = \ln\frac{r^2 - a^2}{r^2} + constant$$ $$\tag{3} \dot\theta = C \frac{r^2 - a^2}{r^2}$$

Substitue Eq.(3) into Eq.(1)

$$\tag{4} \ddot{r}+r\left\{ \frac{-2\dot{r}^2}{r^2-a^2}- C^2 \left(\frac{r^2 - a^2}{r^2}\right)^2 \right\}=0$$

Scale with $$a$$, $$r \to r / a$$ $$\ddot{r}+r\left\{ \frac{-2\dot{r}^2}{r^2-1}- C^2 \left(\frac{r^2 - 1}{r^2}\right)^2 \right\}=0$$

• I do follow your reasoning, but I can't see how to take that last equation to the one I'm supposed to prove. Could you explain a bit further? Commented Mar 15, 2021 at 14:53
• What is the one that you try to prove?
– ytlu
Commented Mar 15, 2021 at 15:08
• Second one in the question Commented Mar 15, 2021 at 15:08
• "... swapping the derivatives with respect to the parameter to derivatives with respect to the coordinates I might be able to get an expression as the one I'm after"> This words? I cannot follow the text meaning here. Can you elaborate more?
– ytlu
Commented Mar 15, 2021 at 15:12
• Sure, I meant that maybe I could try to use the chain rule to differenciate $r$ with respect to $\theta$ instead of with respect to the parameter (that you assumed to be $t$). Later I tried solving the last equation in your answer, but I didn't get anywhere near the solution... Commented Mar 15, 2021 at 19:35