Understanding Geodesic Parametrization on a Sphere

[Jufa]

Let us consider a sphere of a unit radius . Therefore, by choosing the canonical spherical coordinates $\theta$ and $\phi$ we have, for the differential length element:

$$dl = \sqrt{\dot{\theta}^2+sin^2(\theta)\dot{\phi}^2} $$

In order to find the geodesic we need to extremize the following:

$$ \int_{\lambda_0}^{\lambda_f} {\sqrt{\dot{\theta}^2+sin^2(\theta)\dot{\phi}^2}} d\lambda $$

We can do it so by imposing the Euler-Lagrange equations for the lagrangian $L = \sqrt{\dot{\theta}^2+sin^2(\theta)\dot{\phi}^2}$ or equivalently, for the lagrangian $L' = L^2$. These equations for L' look like:

$$ \ddot{\theta} +sin(\theta)cos(\theta)\dot{\phi}^2 = 0 $$

$$ \ddot{\phi}+ cot(\theta) \dot{\phi}\dot{\theta}=0 $$

I am pretty sure that I am right until here.
What does not make sense to me is the following:
Suppose you choose a curve in which $\theta = \pi/2$ i.e. both its first and second derivative vanish.
Then we get the following condition:

$$ \ddot{\phi} = 0 $$

But why does that happen? Once we have fixed the angle $\theta$ we have also fixed the curve (geodesic) and the only condition on $\phi(\lambda)$ should be that it is continuous and injective (i.e., it does not make $\phi$ go back and forth).
For example, the parametrization $\phi(\lambda) = \frac{\lambda^2}{2\pi}$ (which does not have a null second derivative) from $\lambda_0 = 0$ to $\lambda_f =2 \pi$ should be as valid as the parametrization $\phi(\lambda) = \lambda$ from $\lambda_0 = 0$ to $\lambda_f = 2\pi$.Thanks in advance.

 

[Orodruin]

To extremise with Lagrangian $L$ is not equivalent to extremising with $L^2$. If you use $L^2$ it is the same as using $L$ with the additional requirement that $L$ is constant and doing so will therefore give you an affinely parametrised geodesic - ie, a geodesic with a constant length tangent. This coincides with the geodesic concept introduced by a connection.

 

[Jufa]

To extremise with Lagrangian $L$ is not equivalent to extremising with $L^2$. If you use $L^2$ it is the same as using $L$ with the additional requirement that $L$ is constant and doing so will therefore give you an affinely parametrised geodesic - ie, a geodesic with a constant length tangent. This coincides with the geodesic concept introduced by a connection.

Great answer. Many thanks.

 

[stevendaryl]

It's worth going through the exercise. In general, if $L$ is a "lagrangian" of the form:

$L = \sqrt{g_{ij} \dfrac{dx^i}{d\lambda} \dfrac{dx^j}{d\lambda}}$

then extremizing the action gives:

$\dfrac{d}{d\lambda} (g_{i j} \dfrac{dx^j}{d\lambda}) - \dfrac{1}{2} (\dfrac{\partial}{\partial x^i} g_{kj}) \dfrac{dx^k}{d\lambda} \dfrac{dx^j}{d\lambda} = g_{ij} \dfrac{dx^j}{d\lambda} \dfrac{\frac{dL}{d\lambda}}{L}$

The additional requirement that $L$ is unchanging as a function of $\lambda$ gives the usual form of the geodesic equation.