# proving a given curve is a geodesic

I am trying to solve the following problem from Lee's Riemannian Manifolds:

where the curve $\gamma : I \to \mathbb{R}^2$ is given by $\gamma(t) = (a(t),b(t))$ so that $M$ is parametrized as $\varphi(\theta,t) = (a(t) \cos \theta, a(t) \sin \theta, b(t))$.

I completed part (a), and got the following: $$\begin{array}{cccc} \Gamma_{\theta \theta}^{\theta} = 0, & \Gamma_{\theta \theta}^t = - a(t) \dot{a}(t), & \Gamma_{\theta t}^{\theta} = \dfrac{\dot{a}(t)}{a(t)}, & \Gamma_{\theta t}^t = 0, \\ \Gamma_{t \theta}^{\theta} = \dfrac{\dot{a}(t)}{a(t)}, & \Gamma_{t \theta}^t = 0, & \Gamma_{tt}^{\theta} = 0, & \text{and } \Gamma_{tt}^t = 0. \end{array}$$

For part (b), I think I should be showing that the curve $\alpha : I \to M$ defined by $\alpha(t) = \varphi(\theta_0,t) = (a(t) \cos \theta_0, a(t) \sin \theta_0,b(t))$ satisfies the geodesic equation $$\ddot{\alpha}^k(t) + \dot{\alpha}^i(t) \dot{\alpha}^j(t) \Gamma_{ij}^k (\alpha(t)) = 0.$$ However, I'm confused as to how to perform the calculation when my curve is in standard coordinates and my Christoffel symbols are in $(\theta,t)$ coordinates. Can anyone tell me where I'm going wrong in trying to do the calculation and what I should be doing instead?

Also, I'm assuming that for part (c), I should again be using the geodesic equation, but that I should be able to derive a condition on the "latitude circle" after performing the calculation.

## 1 Answer

The whole point in calculating the Christoffel symbols in $(\theta,t)$ coordinates, is that having done that, you no longer need to use the standard Cartesian coordinates. Hence, just think of the curve $\{\theta=\theta_0\}$ as $\alpha(t)=(\theta_0,t)$. The vector speed is then just $\dot{\alpha}=\partial/\partial t$, and part (b) is immediate from your calculations. Part (c) shouldn't be too hard either.