Today I did an exercise where I found the angle between two curves $\mathbf{x}(\lambda_1)$ and $\mathbf{x}(\lambda_2)$ on the surface of a unit sphere with line element $ds^2 = d \theta^2 + sin^2 \theta d \phi^2$. $\mathbf{x}(\lambda_1)$ and $\mathbf{x}(\lambda_2)$ are different functions. In my exercise they were $\theta = \theta_0 + \lambda_1$, $\phi = \phi_0 + \lambda_1$ and $\theta = \theta_0 - 2 \lambda_2$, $\phi = \phi_0 + \lambda_2. $ The method was something along these lines:
- Parametrize a 'circle' $\mathbf{x}(\lambda_c)$ of very small radius $\epsilon$ centered about the point of intersection of the two curves $(\theta_0 , \phi_0)$. Since $\epsilon$ is small, assume $\sin \theta(\lambda_c) = \sin \theta_0$. This can be done using the line element. A 'circle' centered about $(\theta_0 , \phi_0)$ with radius $\epsilon$ should be given by the equation $$\epsilon^2 = (\theta - \theta_0)^2 + sin^2 \theta_0 (\phi - \phi_0)^2$$ Then I parametrize this equation with parameter $\lambda_c$.
- Find the values of $\lambda_c$ for which the two curves intersect the circle.
- Find the arc length by integrating ds with limits given by the two values of $\lambda_c$ in the previous step
- Divide the arc length by the radius $\epsilon$ to get the angle between the two curves
But I am not sure $s = r \theta$ holds in curved space. I'm assuming this works since the space is locally Euclidean. But then can't I just find the tangent vectors $d \mathbf{x}/d \lambda_1$ and $d \mathbf{x} / d \lambda_2$ and find the angle between these vectors like I would in a 2d Euclidean space to find the angle between the curves?
Short answer: For this particular surface you are considering, the relation is true.
From page-19 of Visual Differential Geometry by Tristan Needham
You can pretty easily see that the geodesic circle of radius $r$ centered at the point $c$, has the same linear relation of $s = r \theta$ going on i.e: sectors of equal angle are have same arclength.