1
$\begingroup$

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!

$\endgroup$

1 Answer 1

0
$\begingroup$

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

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

$\endgroup$
6
  • $\begingroup$ 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? $\endgroup$ Commented Mar 15, 2021 at 14:53
  • 1
    $\begingroup$ What is the one that you try to prove? $\endgroup$
    – ytlu
    Commented Mar 15, 2021 at 15:08
  • $\begingroup$ Second one in the question $\endgroup$ Commented Mar 15, 2021 at 15:08
  • 1
    $\begingroup$ "... 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? $\endgroup$
    – ytlu
    Commented Mar 15, 2021 at 15:12
  • $\begingroup$ 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... $\endgroup$ Commented Mar 15, 2021 at 19:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.