- #1

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[tex] ds^2 = - \cosh^2{r} dt^2 + dr^2 [/tex]

Now, we know the geodesic equation is

[tex] \frac{d^2 x^\mu}{d \tau^2} + \Gamma^\mu{}_{\nu \rho} \frac{d x^\nu}{d \tau} \frac{d x^\rho}{d \tau} = 0, \quad \Gamma^\mu{}_{\nu \rho} = \frac{1}{2} g^{\mu \sigma} \left( g_{\nu \sigma, \rho} + g_{\rho\ sigma, \nu} - g_{\nu \rho, \sigma} \right) [/tex]

and where [itex] \tau [/itex] is an affine parameter.

Let's consider the case [itex] \mu = r[/itex], then to work out the Christoffel symbols, I need to know that [itex] g^{rr} = 1[/itex] (using the fact that the metric is diagonal).

Then I compute

[tex] \Gamma^{r}_{tt} = \frac{1}{2} \left( g_{tr,t} + g_{tr,t} - g_{tt, r} \right) = \frac{1}{2} 2 \cosh{r} \sinh{r} = \cosh{r} \sinh{r} [/tex]

[tex] \Gamma^{r}_{tr} = \frac{1}{2} \left( g_{tr,r} + g_{rr,t} -g_{tr,r} \right) = 0 [/tex]

[tex] \Gamma^r{}_{rr} = \frac{1}{2} \left( g_{rr,r} + g_{rr,r} - g_{rr,r} \right) = 0 [/tex]

This means that the Geodesic equation takes the form

[tex] \ddot{r} + \left( \cosh{r} \sinh{r} \right) \dot{t}^2 = 0 [/tex]

Where the dot denotes differentiation with respect to the affine parameter [itex]\tau[/itex].

I don't know how to solve this equation. I can multiply the cosh and sinh to get it to the form [itex] \ddot{r} + \frac{1}{2} \left( \sinh{2r} \right) \dot{t}^2 = 0 [/itex] but that doesn't seem to help. Does anyone have any suggestions? Am I using a poor choice of coordinates - hopefully not as I'd ideally like this in terms of sinh and cosh if possible.

Thanks.