# Finding the geodesic equations on $S^2$

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?

• the index $a$ is not summed, hence you have the equations for $a=1$ getting $\ddot{\theta}=\ldots$ and for $a=2$ getting $\ddot{\phi}=\ldots$. – 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.