# Geodesics of the Unit Sphere using Christoffel symbols

I want to derive the geodesics of the unit sphere by using Christoffel symbols.

The metric is the standard round metric:

$ds^2 = d\theta^2 + \sin^2\theta d\phi^2$

ie. $g_{ij} = \begin{pmatrix} 1 & 0 \\ 0 & \sin^2\theta \\\end{pmatrix}$

And the Christoffel symbols:

$\Gamma_{jk}^i := \frac12g^{il}(\partial_j g_{l k} + \partial_k g_{j l} - \partial_\beta g_{j k})$

I worked out to be:

$\Gamma^\theta = \begin{pmatrix} 0 & 0 \\ 0 & -\sin\theta\cos\theta \\\end{pmatrix}$, $\Gamma^\phi = \begin{pmatrix} 0 & \cot\theta \\ \cot\theta & 0 \\\end{pmatrix}$

And we are given that the geodesics parametrised by arclength

$g_{ij}\frac{du^i}{d\tau}\frac{du^j}{d\tau} = \frac{d\theta}{d\tau}^2 + \sin^2\theta \frac{d\phi}{d\tau}^2 = 1$

satisfy the geodesic equation

$\frac{d^2u^i}{d\tau^2} + \Gamma_{jk}^i\frac{du^j}{d\tau}\frac{du^k}{d\tau} = 0 = \frac{d^2\theta}{d\tau^2} - \sin\theta\cos\theta\frac{d\phi}{d\tau}^2 = \frac{d^2\phi}{d\tau^2} + 2\frac{\cos\theta}{\sin\theta}\frac{d\phi}{d\tau}\frac{d\theta}{d\tau}$

Now all of the simple Great Circles satisfy this equation fairly simply; for lines of longitude $(\tau,a)$, and for the equator $(\frac\pi2,\tau)$ all terms vanish.

I also verified that simple lines of latitude fail to satisfy the equations.

So my question is this: do these differential equations characterise great circles? how can I verify them?

$\frac{d^2\theta}{d\tau^2} - \sin\theta\cos\theta\frac{d\phi}{d\tau}^2 = 0$

$\frac{d^2\phi}{d\tau^2} + 2\frac{\cos\theta}{\sin\theta}\frac{d\phi}{d\tau}\frac{d\theta}{d\tau} = 0$

Key idea. You may simplify your target by choosing an appropriate coordinate before calculations.

Preliminaries. You might know that, for a set of ordinary differential equation (ODE) system, it would not be complete unless you specify initial conditions for them (of course, you can still seek for general solutions even without initial conditions, but this does not help especially if you hope to get a proof with mathematical rigor). So for your ODE system, a complete expression would be \begin{align} \ddot{\theta}-\dot{\phi}^2\sin\theta\cos\theta&=0,\\ \ddot{\phi}+2\dot{\theta}\dot{\phi}\cot\theta&=0,\\ \theta|_{\tau=0}&=\theta_0,\\ \dot{\theta}|_{\tau=0}&=u_0,\\ \phi|_{\tau=0}&=\tau_0,\\ \dot{\phi}|_{\tau=0}&=v_0, \end{align} where $\theta_0$ and $\phi_0$ are initial values for $\theta$ and $\phi$, while $u_0$ and $v_0$ are initial values for $\dot{\theta}$ and $\dot{\phi}$.

Geometrical interpretation. The initial value $\left(\theta_0,\phi_0\right)$ indicates the point on the sphere at which the mass point starts moving when $\tau=0$, while $\left(u_0,v_0\right)$, a vector pinned to this point, is the velocity of the mass point at $\tau=0$. So $\left(\theta_0,\phi_0,u_0,v_0\right)$ as a whole gives you a tangential vector somewhere on the sphere.

Now recall the spherical coordinate system, as is shown in the following figure. The tricky part is, we can always reconsider the coordinate after the tuple $\left(\theta_0,\phi_0,u_0,v_0\right)$ is given. That is, as $\left(\theta_0,\phi_0,u_0,v_0\right)$ is specified, we perform a transformation, making this $\left(\theta_0,\phi_0,u_0,v_0\right)$ appear to be a vector at the north pole pointing parallel to the $x$-axis in our new coordinate system. Therefore, in our new system, $\theta_0=0$, $\phi_0=0$, $v_0=0$. Plus, to keep the normality of the velocity, let $u_0=1$.

Consequently, our system becomes \begin{align} \ddot{\theta}-\dot{\phi}^2\sin\theta\cos\theta&=0,\\ \ddot{\phi}+2\dot{\theta}\dot{\phi}\cot\theta&=0,\\ \theta|_{\tau=0}&=0,\\ \dot{\theta}|_{\tau=0}&=1,\\ \phi|_{\tau=0}&=0,\\ \dot{\phi}|_{\tau=0}&=0. \end{align} In addition, to show that the solution corresponds to a great circle, it suffices to show that the above system implies $\phi\equiv 0$ (or more rigorously, $\phi\equiv 0$ when $\theta\in\left[0,\pi\right]$, and $\phi\equiv\pi$ when $\theta\in\left(\pi,2\pi\right)$).

Remark. The symbol "$\equiv$" I adopted here means "always equal to" or "constantly equal to".

Main steps. To begin with, thanks to \begin{align} \ddot{\theta}-\dot{\phi}^2\sin\theta\cos\theta&=0,\\ \ddot{\phi}+2\dot{\theta}\dot{\phi}\cot\theta&=0, \end{align} we have $$\frac{\rm d}{{\rm d}\tau}\left(\dot{\theta}^2+\dot{\phi}^2\sin^2\theta\right)=0,$$ and the initial values yield $$\dot{\theta}^2+\dot{\phi}^2\sin^2\theta\equiv\left(\dot{\theta}^2+\dot{\phi}^2\sin^2\theta\right)\Big|_{\tau=0}=1.$$ (This is exactly your arc-length-parametrization argument. I mention it here just to highlight that it is a natural consequence of the geodesic equations, for which there is no need to impose it as an additional constraint.)

We observe that, due to $\dot{\theta}|_{\tau=0}=1$, $\theta\not\equiv 0$. Hence $\sin\theta\not\equiv 0$. Thanks to this fact, use the last constraint to eliminate $\dot{\phi}^2$ in the first equation, which leads to $$\ddot{\theta}=\left(1-\dot{\theta}^2\right)\cot\theta.$$ This is a decoupled ODE for $\theta$, and it could be dealt with using the following standard trick. We have $$\ddot{\theta}=\frac{\rm d}{{\rm d}\tau}\dot{\theta}=\frac{{\rm d}\theta}{{\rm d}\tau}\frac{{\rm d}\dot{\theta}}{{\rm d}\theta}=\dot{\theta}\frac{{\rm d}\dot{\theta}}{{\rm d}\theta}=\frac{\rm d}{{\rm d}\theta}\left(\frac{1}{2}\dot{\theta}^2\right).$$ As a result, the decoupled ODE becomes $$\frac{\rm d}{{\rm d}\theta}\left(\frac{1}{2}\dot{\theta}^2\right)=\left(1-\dot{\theta}^2\right)\cot\theta\iff-\frac{{\rm d}\left(1-\dot{\theta}^2\right)}{1-\dot{\theta}^2}=2\frac{{\rm d}\sin\theta}{\sin\theta}.$$ This is nothing but $${\rm d}\log\left[\left(1-\dot{\theta}^2\right)\sin^2\theta\right]=0,$$ and we immediately have $$\left(1-\dot{\theta}^2\right)\sin^2\theta\equiv\left[\left(1-\dot{\theta}^2\right)\sin^2\theta\right]\Big|_{\tau=0}=0,$$ meaning that, with $\dot{\theta}|_{\tau=0}=1$, $$\text{either}\quad\dot{\theta}\equiv 1\quad\text{or}\quad\sin\theta\equiv 0$$ must be true. As has been mentioned that $\sin\theta\not\equiv 0$, it is a must that $\dot{\theta}\equiv 1$.

Finally, substitute $\theta=\tau$ (obtained from $\dot{\theta}=1$ as well as $\theta|_{\tau=0}=0$) into the second equation, and we have another decoupled system \begin{align} \ddot{\phi}+2\dot{\phi}\cot\tau&=0,\\ \phi|_{\tau=0}&=0,\\ \dot{\phi}|_{\tau=0}&=0. \end{align} Obviously, $\dot{\phi}\equiv 0$ solves this system. We need to show that this is the only solution this system admits. Suppose for contradiction that there exists some $\dot{\phi}\not\equiv 0$. As such, the equation could be re-expressed as $$\frac{{\rm d}\dot{\phi}}{\dot{\phi}}=-2\frac{{\rm d}\sin\tau}{\sin{\tau}}\iff{\rm d}\log\left(\dot{\phi}\sin^2\tau\right)=0,$$ i.e., $$\dot{\phi}\sin^2\tau=\left(\dot{\phi}\sin^2\tau\right)\Big|_{\tau=0}=0\Longrightarrow\dot{\phi}\equiv 0,$$ which contradicts our assumption that $\dot{\phi}\not\equiv 0$. As a consequence, $\dot{\phi}\equiv 0$, and thus $\phi\equiv 0$ as per its initial condition, is the only possible solution.

To sum up, we obtain that \begin{align} \theta&=\tau,\\ \phi&=0 \end{align} is the only solution that satisfies the given geodesic equations as well as its specified initial conditions. Obviously, this solution corresponds to a great circle on the sphere.

Therefore, we may conclude that any curve on the unit sphere that satisfies the geodesic equations must be (part of) a great circle.

• Thanks for the comprehensive answer. The bit I don't understand still is the transformation of coordinates to the pole. It can't be true that for any differential equation in $\phi$ and $\theta$ there is a transformation $\phi \to \phi '$,$\theta \to \theta '$ such that the same differential equation is true for $\phi '$ and $\theta '$, for example, take $\frac{d}{d\tau} \phi = 1$. In this case it matters where the initial condition is; you cannot move it to the pole without changing the differential equation. Could you elaborate on what transformation we make and why it preserves the DE? – Christian Fieldhouse Apr 9 '18 at 10:20
• @ChristianFieldhouse: Hi Christian: I got your point. You wish to fix a coordinate frame in the first place. For different initial conditions, you hope to get different solutions and check that they are all great circles. But these solutions may well be expression-unfriendly, and this is why I suggested to choose the coordinate after a set of initial conditions is given. For a given velocity vector pinned at a point on the sphere, you choose your coordinate such that the point is exactly the north pole of your frame and the vector is parallel to your $x$-axis. This simplifies calculations. – hypernova Apr 9 '18 at 17:07
• @ChristianFieldhouse: You are right that the expression of equations does not preserve under general coordinate transformations. The point is: are we able to find some specific transform under which the expression preserves? We can but it is involved. If it is a must, we would rather deal with the arbitrarily given initial conditions directly than try to figure out such a transform. You may start from $\left(1-\dot{\theta}^2\right)\sin^2\theta\equiv \left(1-u_0^2\right)\sin^2\theta_0$, working out the expression for $\theta$ and then for $\phi$, following the same technique. – hypernova Apr 9 '18 at 17:18
• @ChristianFieldhouse: In general, $\left(1-\dot{\theta}^2\right)\sin^2\theta=K\in\left[0,1\right)$ with $\theta\in\left(0,\pi\right)$ implies $\dot{\theta}=\pm\sqrt{\sin^2\theta-K}/\sin\theta$, or $-\dot{\theta}\sin\theta=\pm\sqrt{1-K-\cos^2\theta}$, or $-{\rm d}\cos\theta=\pm\sqrt{1-K-\cos^2\theta}{\rm d}\tau$. This can be integrated directly, and gives you ${\rm d}\left(\arctan\frac{\cos\theta}{1-K-\cos^2\theta}\pm\tau\right)=0$, which reduces to $\cos^2\theta=\left(1-K\right)\sin^2\left(C\pm\tau\right)$ with some constant $C$. See, though general, it does not help that much with your goal. – hypernova Apr 9 '18 at 17:34
• Correct me if I'm wrong: For a given initial condition, you choose the most convenient coordinates and derive (As I did above) the geodesic equations, then you check (As you've done) that the only curves satisfying all conditions are indeed great circles. It seems though that even if I had got my geodesic equations wrong and arrived at $\frac{d\phi}{d\tau} = 1, \frac{d\theta}{d\tau} = 0$, If I can just choose the coordinates after I get initial conditions so that the IC's are on the equator, then I still get only great circles. Thus there must be something more to check. – Christian Fieldhouse Apr 9 '18 at 18:20