# I Geodesics on surfaces

Tags:
1. Nov 1, 2016

### Frank Castle

Consider the metric of $S^{2}$: $$ds^{2}=d\theta^{2}+\sin^{2}(\theta)d\phi^{2}$$ Then in order to determine the geodesics on this surface one can minimise the integral $$s=\int_{l_{1}}^{l_{2}}\sqrt{\left(\frac{d\theta}{dl}\right)^{2}+\sin^{2}(\theta)\left(\frac{d\phi}{dl}\right)^{2}}dl$$ where $l$ parametrises a path connecting two points on the surface. We can identify the Lagrangian as $L=\sqrt{\left(\frac{d\theta}{dl}\right)^{2}+\sin^{2}(\theta)\left(\frac{d\phi}{dl}\right)^{2}}$.

If one parametrises the path by its arc-length $s$ then $L=1$. Now, in the case of space-time the arc-length of a time-like path is equal to the proper time between the two endpoints of the path, and hence an arc-length parametrisation corresponds in this case to choosing proper time to parametrise the path.

My question is, is the analogue of this for a curved surface (such as $S^{2}$) the proper distance between two points on the surface, defined as $dl=\sqrt{ds^{2}}$, such that, in the case of $S^{2}$, the equations of motion are: $$\frac{d^{2}\theta}{dl^{2}}-\sin(\theta)\cos(\theta)\left(\frac{d\phi}{dl}\right)^{2}=0\\ \frac{d^{2}\phi}{dl^{2}}+2\cot(\theta)\frac{d\theta}{dl}\frac{d\phi}{dl}=0$$

2. Nov 6, 2016

### Greg Bernhardt

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Nov 7, 2016

### vanhees71

We had a discussion about this in the context of GR. Despite the fact that in GR you have a pseudo-Riemannian manifold, everything said there is valid here. Instead of using the "square-root form" of the action, use
$$L_2=\frac{1}{2} g_{ab} \dot{x}^a \dot{x}^b.$$
Then automatically the parameter of the geodesic becomes an affine parameter. See the mentioned thread:

$$\mathrm{D}_{l}^2 x^{a}=\ddot{x}^a + {\Gamma^{a}}_{cd} \dot{x}^c \dot{x}^d=0.$$
You can of course scale your parameter $l$ such that $\mathrm{d} s^2=g_{ab} \dot{x}^{a} \dot{x}^B \mathrm{d} l^2=\mathrm{d} l^2$, i.e., you can always impose the constraint
$$g_{ab} \dot{x}^a \dot{x}^b=1.$$