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I'm wondering how to solve the geodesic equation to get a null geodesic. I know the two equations

$$\frac{d^2x^\mu}{ds^2}+{\Gamma^\mu}_{\nu\lambda}\frac{dx^\nu}{ds}\frac{dx^\lambda}{ds}=0$$

and the null one

$$g_{\mu\nu}\frac{dx^\mu}{ds}\frac{dx^\nu}{ds}=0,$$

but how do you do this, given a metric? Consider, for example,

$$ds^2=\cos\sigma\,d\tau^2+d\sigma^2.$$

I don't understand how to plug this in and actually find the null geodesics once I've calculated the Christoffel symbols. thanks!

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  • $\begingroup$ Please use MathJax for all math. $\endgroup$
    – G. Smith
    Mar 4, 2021 at 22:40
  • $\begingroup$ How much experience do you have solving coupled nonlinear differential equations? $\endgroup$
    – G. Smith
    Mar 4, 2021 at 22:43
  • $\begingroup$ i have done some in the past, and i'm sure I can figure it out in general, i'm just generally confused on what this 's' is denoting when taking the derivatives, and overall what the equation should look like plugged in. $\endgroup$ Mar 4, 2021 at 22:44
  • $\begingroup$ Do you understand what a parameterized curve is? $s$ parameterizes the geodesic. When the geodesic is timelike instead of null, it’s the proper time along the geodesic. $\endgroup$
    – G. Smith
    Mar 4, 2021 at 22:47
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    $\begingroup$ your example metric is positive definite and has no null geodesics. $\endgroup$ Mar 5, 2021 at 4:19

1 Answer 1

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Following Ray D'Inverno. Introducing Einstein's Relativity. Page 101.

Assuming u is an affine parameter.

If $2K\equiv g_{ab}\dot{x}^a\dot{x}^b =\dot{x}^a\dot{x}_a $ then you can quickly read off $\Gamma^a _{bc}$ from the geodesic equation in the form

$\frac{\partial K}{\partial x^a } - \frac{d}{du} (\frac{\partial K}{\partial \dot{x}^a})=0$ where $2K=\text{C}=0,+1,-1$.


Example:

$ds^2 = \eta^2 d\tau^2 - d\eta^2 \to K=\frac{1}{2}(\eta^2 \dot\tau^2 - \dot{\eta}^2 ) $

$\frac{\partial K}{\partial x^a } - \frac{d}{ds} (\frac{\partial K}{\partial \dot{x}^a})=0 \to \frac{\partial K}{\partial \tau } - \frac{d}{ds} (\frac{\partial K}{\partial \dot{\tau}})=0\to2\eta \dot{\eta}\dot{\tau}+\eta^2 \ddot{\tau}=0\to$

$\ddot{\tau}+\frac{2}{\eta} \dot{\eta} \dot{\tau}=0\implies\Gamma^\tau _{\eta \tau}=\frac{2}{\eta}=\Gamma^\tau _{\tau\eta}$


$ds^2=\cos σ dτ^2+ dσ^2$

$\to$

$K=\cos σ {\dot{τ}}^2+ {\dot{σ}}^2 =0$

$\to$

$\cos(\sigma){\dot{τ}}^2 = - {\dot{σ}}^2 $

That's one differential equation.

Next calculate,

$\frac{\partial K}{\partial x^a } - \frac{d}{du} (\frac{\partial K}{\partial \dot{x}^a})=0$

$\to$

$ \frac{d}{du} (\frac{\partial K}{\partial \dot{\tau}})=0$

$ \frac{d}{du} (\frac{\partial K}{\partial \dot{\sigma}})=0$

$\to$

$\frac{\partial (\cos(σ) {\dot{τ}})}{\partial u}=0$

$\to$

$ -\sin(σ) \dot{\sigma} \dot{τ} + cos(\sigma) \ddot{\tau}=0$

and

$\ddot{\sigma}=0$.


$\cos(\sigma){\dot{τ}}^2 = - {\dot{σ}}^2 $

$ -\sin(σ) \dot{\sigma} \dot{τ} + cos(\sigma) \ddot{\tau}=0$

$\ddot{\sigma}=0$.

$\to$

$\sigma(u)=\frac{a}{2} u^2 +bu +c$

$\to$

$\dot{\sigma}(u)= au +b $

$\to$

$\dot{\tau}^2 = \frac{-(u+b)^2}{cos(\frac{1}{2} au^2 +ub +c)}$

set $b=c=0$ and $a=2$

$\to$

$\dot{\tau}^2 = \frac{-(u)^2}{cos(u^2)}$

etc. etc... ?

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