Showing a Geodesic Result

Question

Given the surface of the unit sphere with usual metric $$\large ds^2 = d\theta^2 + \sin^2(\theta)\,d\phi^2$$ I have calculated the Euler-Lagrange equations \begin{align*}\large \ddot{\theta} - \sin\theta\cos\theta\dot{\phi}^2 &= 0\\ \large \ddot{\phi} + 2\cot\theta\dot{\phi}\dot{\theta} &= 0 \end{align*} Giving the Christoffel symbol components as \begin{align*}\large \Gamma^\theta_{\phi\phi} = -\sin\theta\cos\theta,\quad\Gamma^\theta_{\theta\theta} = \Gamma^\theta_{\theta\phi} &= 0\\ \large \Gamma^\phi_{\theta\phi} = \Gamma^\phi_{\phi\theta} = \cot\theta,\quad\Gamma^\phi_{\theta\theta} = \Gamma^\phi_{\phi\phi} &= 0 \end{align*}

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Now there is the question to show that a line of constant $\phi$ is a geodesic.

How would one go about this, given what I have done so far?

Answer 1

You have to show that curves with constant $\phi$ satisfy the geodesic equations. Geometrically you are showing that the great circles are geodesics.

We substitute $\dot \phi=0$ in the two geodesic equations above. The second geodesic equation holds trivially; the first equation becomes $\ddot \theta =0$. Now, note that we can always reparametrise a (regular) curve such that it has constant speed $c$. For a curve on the sphere, this means that $$ (\dot\theta)^2+\sin^2\theta (\dot \phi)^2= c^2.$$ If we now substitute $\dot \phi=0$, we see that $\dot \theta$ is constant. It follows that $\ddot \theta=0$. Thus the two geodesic equations are satisfied, so the curves with constant $\phi$ (with constant speed parametrisations) are geodesics.