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

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To calculate null geodesics for the given Schwarzschild line element, the Euler-Lagrange equations must be solved using the squared Lagrangian derived from the line element. The Lagrangian is expressed in terms of proper time, leading to equations for the derivatives of time and radius. The user derives expressions for the derivatives of time and radius but struggles with the change of variable to express dr/dt. Clarification is sought on the approach taken and any potential errors in the calculations. The discussion emphasizes the importance of correctly applying the Euler-Lagrange equations in this context.
Confused Physicist
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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!
 
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