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*}

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?

  • $\begingroup$ There is a very small typo: $ds^s$ should be $ds^2$. $\endgroup$
    – Ernie060
    Oct 30, 2018 at 18:12
  • $\begingroup$ oh yes a double tap, my bad $\endgroup$
    – MRT
    Oct 30, 2018 at 18:13

1 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.

  • $\begingroup$ @MRT Maybe you might find this question also interesting. $\endgroup$
    – Ernie060
    Oct 30, 2018 at 18:14
  • $\begingroup$ okay so having that equation and getting $\ddot{\theta} = 0$ this answers the question? I'm still a little confused how this is related to $\phi$... If I think about having constant $\phi$ then that makes the first equation $\ddot{\theta} = 0$ as the first $\phi$ derivative would be zero for a constant and the second equation would just be $0=0$ with the first and second derivatives of $\phi$ are zero. So I guess I don't see why $\ddot{\theta} = 0$ shows that constant $\phi$ is a geodesic. $\endgroup$
    – MRT
    Oct 30, 2018 at 18:22
  • $\begingroup$ I think I understand your confusion. I will reformulate my answer. Hopefully this will help a bit. $\endgroup$
    – Ernie060
    Oct 30, 2018 at 20:53
  • $\begingroup$ Thank you for your answer. I was thinking that could you just say$$\large \ddot{\theta}=0\Rightarrow\dot{\theta}=\text{const.}\Rightarrow\theta=a\lambda +\text{const.}$$ where $\theta$ is some linear result in the variable of differentiation $\lambda$. I’m just not sure how you found the reparametrisation $\endgroup$
    – MRT
    Nov 2, 2018 at 8:59
  • $\begingroup$ I used the fact that every regular curve can be reparametrised such that it has constant speed. I first used this reparametrision and then checked that $\ddot \theta =0$ holds. But indeed, so you can first solve $\ddot \theta=0$, and then use this to reparametrise the curve. $\endgroup$
    – Ernie060
    Nov 2, 2018 at 9:19

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