Solution to Geodesics on a 2-sphere

Question

I have been tasked to finding particle trajectories for a point mass travelling along the surface of the 2-sphere $t=t(\tau)$, $\theta=\theta(\tau)$ and $\phi=\phi(\tau)$. My supervisor gave me the spacetime metric

$ds^2 = -dt^2 +R^2(d\theta^2 +sin^2{\theta}d\phi^2)$

I am finding timelike geodesics, $ 1 = -g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}$.

Here is what I have so far,

$t = E\tau$, $\dot{t} = E$, $sin^2\theta\frac{d\phi}{d\tau} = \frac{k}{R}$ where $\frac{k}{R}$ is some dimensionless constant. I substituted $\dot{\phi}$ and $\dot{t}$ back into the proper time gauge to get.

$\dot{\theta} = \pm\frac{1}{Rsin\theta}\sqrt{E^2sin^2\theta-k^2-sin^2\theta}$

I attempted using the substitution $u=cos\theta$ to eliminate the $sin\theta$ and hopefully get some expression that I could integrate to obtain some inverse trigonometric function, I know that the geodesics should describe great circles along the surface of the sphere. But I cant solve this final equation. Thank you

Answer 1

Your manifold is the product $(\Bbb R, -{\rm d}t^2)\times \Bbb S^2(R)$. By Proposition $38$ in page $208$ (with $f=1$) of O'Neill's Semi-Riemannian Geometry With Applications to Relativity, a curve $\gamma = (\alpha, \beta)$ there is a geodesic if and only if $\alpha$ and $\beta$ are geodesics in $\Bbb R$ and $\Bbb S^2(R)$, respectively.

It is then clear that $\alpha$ must be $\alpha(t) = \pm t+a$ for some $a \in \Bbb R$, and $\beta$ must parametrize a great circle. As for checking that geodesics of the sphere are great circles, there are better ways to do it instead of solving those differential equations. For instance, one can argue that given a tangent vector $v$ at a point $p$, there is a unique maximal geodesic starting at $p$ with velocity $v$, compute directly that great circles are geodesics, and finally that given $v$ and $p$, there is a great circle passing through $p$ with direction $v$.