The metric on $S^2$ is given as $\bar{g} = ds^2 = d\theta^2 + \sin^2\theta d\phi^2$ where $x^1 = \theta$ and $x^2 = \phi$. The only non-zero components of the Christoffel symbols are $\Gamma^{\ 1}_{ \ 2 \ 2}$ and $$\Gamma^{\ 2}_{ \ 2 \ 1} = \Gamma^{\ 2}_{ \ 1 \ 2} = \cot\theta$$

Write down the geodesic equations for the co-ordinates $\theta(t)$ and $\phi(t$)

I know that in local co-ordinates on any smooth manifold $M$ the geodesic equation is given by $$\frac{d^2x^a}{dt^2} + \Gamma^{\ a}_{ \ b \ c}\frac{dx^b}{dt}\frac{dx^c}{dt} = 0$$

on $S^2$ then substituting for $x^1$ and $x^2$ in local co-ordinates we get $$\frac{d^2\theta}{dt^2} -\sin\theta\cos\theta \left(\frac{d\phi}{dt}\right) + \frac{d^2\phi}{dt^2} + 2\cot\theta \frac{d\theta}{dt}\frac{d\phi}{dt} = 0$$

rearranging for $\frac{d^2\theta}{dt}$ we get

$$\frac{d^2\theta}{dt^2} =\sin\theta\cos\theta \left(\frac{d\phi}{dt}\right) - \frac{d^2\phi}{dt^2} - 2\cot\theta \frac{d\theta}{dt}\frac{d\phi}{dt} $$

and similarly we can rearrange and solve for $\frac{d^2\phi}{dt}$. My question is, what exactly do the authors mean by "write down the geodesic equations for the co-ordinates $\theta(t)$ and $\phi(t)$, do I need to solve the above differential equation and then obtain $\theta(t)$? If so how can I go about doing that, what's the best approach?

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    $\begingroup$ the index $a$ is not summed, hence you have the equations for $a=1$ getting $\ddot{\theta}=\ldots$ and for $a=2$ getting $\ddot{\phi}=\ldots$. $\endgroup$ – Baol Oct 6 '18 at 9:49

There are two geodesic equations, one for $x^1 = \theta$ and one for $x^2 = \phi$. Note that the Christoffel symbols should be $\Gamma_{22}^2 = -\sin\theta\cos\theta$ and $\Gamma_{12}^2=\Gamma_{21}^2=2\cot\theta$ (with a factor $2$). Substituting gives $$ \begin{align*} \theta''-\sin\theta\cos\theta \left(\phi'\right)^2&=0 \\ \phi'' + 2\cot\theta \theta'\phi' &=0. \end{align*} $$ where ${}'$ is the derivative w.r.t. to $t$. First integrate the second equation $\frac{ \phi''}{\phi'}=-2\cot \theta \theta'$. This gives $$\phi'=\frac{c}{\sin^2 \theta}.$$ Recall the fact that geodesics have unit speed. In this example this is expressed by $$ (\theta')^2 + \sin^2 \theta (\phi')^2 =1.$$ Substituting $\phi'$ gives $$ \theta' = \sqrt{\frac{\sin^2 \theta -c^2}{\sin^2 \theta}}.$$ Dividing $d\phi$ by $d\theta$ gives us the separable differntial equation $$ \frac{d\phi}{d\theta} = \frac{c}{\sin\theta\sqrt{\sin^2\theta -c^2}}. $$ Integrating this is not a picknick, but essentially, it will give you a not so recognisable parametrisation of the great circles. (An example of a similar calculation can be found here.)

This approach hereabove holds more general. The parametrisation here is a Clairaut parametrisation i.e. the parametrisation is orthogonal $F=0$ and $E$ and $G$ only depend on one parameter (here $E_\phi = G_\phi=0$). If you have a Clairaut parametrisation, you can integrate one of the 2nd order geodesic equations in the fashion as we did hereabove, so you will get a 1st order equation. Or, even better, you just directly use the socalled Clairaut equations (the integrated forms), so that you don't need to redo the integration yourself.


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