How Do You Calculate Null Geodesics for the Given Schwarzschild Line Element?

[Confused Physicist]

Hi, I'm the given the following line element:

$$ds^2=\Big(1-\frac{2m}{r}\large)d\tau ^2+\Big(1-\frac{2m}{r}\large)^{-1}dr^2+r^2(d\theta ^2+\sin ^2 (\theta)d\phi ^2)$$

And I'm asked to calculate the null geodesics.

I know that in order to do that I have to solve the Euler-Lagrange equations. For this I always do the following. First I calculate the Lagrangian squared in terms of the proper time $\tau$. In this case first I have written the line element as:

$$ds^2=-\Big(1-\frac{2m}{r}\Big)c^2dt ^2+\Big(1-\frac{2m}{r}\Big)^{-1}dr^2+r^2(d\theta ^2+\sin ^2 (\theta)d\phi ^2)$$

And then the Lagrangian squared:

$$\mathcal{L}^2=-\Big(1-\frac{2m}{r}\Big)c^2\dot{t}^2+\Big(1-\frac{2m}{r}\Big)^{-1}\dot{r}^2+r^2(\dot{\theta}^2+\sin^2(\theta)\dot{\phi}^2)$$

Where $\dot{ }$ denotes derivative with respect to proper time: $d/d\tau$.

When I solve the E-L equation for $t$ and $r$ I get:

$$\dot{t}=\frac{k_t}{1-2m/r}$$
$$\dot{r}=k_r\Big(1-2m/r\Big)$$

I'm doing this because I'm looking for a change of variable which I believe has to be:

$$\frac{dr}{dt}=\pm \Big(\frac{1}{1-2m/r}\Big)$$

What am I doing wrong?

Thansk!

 

[stevendaryl]

Post your partial results.