# Geodesics on a 2-sphere

I've been doing some work where I need to find the geodesics in a given Riemannian Manifold. Let's take the example of the two sphere, for simplicity, with unitary radius. The distance between two points, say: $p$ and $q$, is given by the length of the geodesic that connect those two points, such that: $$d=\int_0^1 [\dot{\theta}^2+\sin^2(\theta)\dot{\phi}^2]^{1/2}dt,$$ where $c(t)=(\theta (t),\phi(t))$ is the geodesic that connect the two considered points (i.e. $c(0)=p$ and $c(1)=q$).

To find the expression of the geodesic one uses the Variational Principle, which leads to the Euler-Lagrange equations for $\theta$ and $\phi$. To simplify our work one can, instead of searching for the curve that minimizes the distance, search for the curve that minimizes the Energy functional and it can be proven that if the curve is a geodesic then the curve that minimizes the energy is the same as the curve that minimizes the distance (proof in the previous link).

Using what I've stated to search for the geodesics on the 2-sphere we find two differential equations which admit solutions to be the great circles or the meridians.

Now to the actual question(s):

If instead of parametrizing the geodesic with the parameter $t$ one parametrizes it with the coordinate $\theta$ (knowing that one can no longer describe the geodesics that connect two points with the same $\theta$ coordinate), such that $c(\theta)=(\theta,\phi(\theta))$ and $\dot{c}=(1,\dot{\phi})$, is the curve that minimizes the distance the same that minimizes the energy?

If the answer is yes one finds that: $\phi(\theta)=\frac{\phi_2}{\cot(\theta_2)}\cot(\theta)$, assuming the first point's coordinates to be $(\pi/2,0)$ and $(\theta_2,\phi_2)$ for the second point. But that leads to a wierd behaviour for points near the poles and $\phi_2>\pi$... (see picture)

What am I doing wrong?It's the answer to the first question, the coordinate system or something deeper...?

-

I think the mistake you are making is confusing a path and its image.

To make myself clearer, let's have a look at the proof that a minimizer of the energy is a geodesic. Let $\gamma : [a,b] \longrightarrow S$ be a smooth path

$$\int{||\gamma'||} \leq (b-a)^2\int{||\gamma||^2}$$ and equality holds iff $||\gamma'||$ is a constant function.We can suppose $b-a=1$ to have $L(\gamma) \leq E(\gamma)$ without loss of generality. Since making a reparametrisation of $\gamma$ doesn't change the value of the lenght(exercise), take a minimizer of the lenght $\gamma$ and let $\tilde{\gamma}$ its reparametrisation at constant speed $L(\gamma)$

$$L(\gamma) = L(\tilde{\gamma}) \leq E(\gamma)$$

since $\gamma$ minimizes $L(\tilde{\gamma}) = E (\tilde{\gamma}) \geq E(\gamma)$

then $L(\gamma) = E(\tilde{\gamma})$ and this proves that a minimizer of the lenght is a minimizer of the energy iff and only if it is parametrised at constant speed ! Especially energy is not invariant by reparametrisation

But the parametrisation by $\theta$ is not at constant speed, and I think that's were you are wrong

-
Yes you're right. When I first thought about that I thought that the Euler-Lagrange equations were equivalent to the geodesic equations, well they are but for the distance not for the Energy functional (in general). I've worked out explicitly the geodesic equations for this parameterization and the solution I've found verifies one of the equations but fails to verify $\dot{\phi}^2\sin(\theta)\cos(\theta)=0$... Thank you for your answer. – PML Feb 28 '13 at 21:34