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

  • $\begingroup$ May I suggest you look at the $k=0$ solutions first? This should result in great circles through the poles... $\endgroup$ – mmeent Feb 19 '18 at 9:04
  • $\begingroup$ Are you sure your differential equations are correct? What happens if you set $E=1$? $\endgroup$ – octonion May 11 '18 at 18:46

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 $v$.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.