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Null geodesics given line element

  • #1
Hi, I'm the given the following line element:

[tex]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)[/tex]

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:

[tex]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)[/tex]

And then the Lagrangian squared:

[tex]\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)[/tex]

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:

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

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

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

What am I doing wrong?

Thansk!!!
 
Last edited:

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  • #2
stevendaryl
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Post your partial results.
 

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