I'm trying to find the average arc length between two random points on a unit sphere. The solution I've come up with is rather ugly. Consider a parametric surface:
$$X(u,v)=\sin u\cos v\\Y(u,v)=\cos u\cos v\\Z(u,v)=\sin v$$
(For a sphere, $u\in[0,2\pi]$, $v\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$). Then, integrate the distance formula for two points $(x_1,y_1,z_1),(x_2,y_2,z_2)$along all three axes. This turns into a nasty integral, though.
$$F(\ldots)=\int_0^{2\pi}
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
D(X_1,Y_1,Z_1,X_2,Y_2,Z_2)dv\mbox{ }du$$
$$\int_0^{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}F(\ldots)\mbox{ }dv\mbox{ }du$$
(Where $D(\ldots)$ is the arc length formula.) I could plug this into mathematica or something, but I have a feeling this would
- take too long to compute
- be too complex
- probably also be wrong
What is a more efficient way to do this? How should I go about solving this problem?
Best Answer
You can rotate the coordinates so the first point is at the pole and measure $\theta$ from $0$ at that pole to $\pi$ at the other one. The distance to the other point is then $\theta$. The probability distribution of $\theta$ is proportional to $\sin \theta$ as that is the radius of the small circle the second point is on. Your average distance is then $$\frac {\int_0^\pi \theta \sin(\theta) d\theta}{\int_0^\pi \sin(\theta) d\theta}=\frac {\pi}{2}$$ We can see this is true by matching up points at $\theta$ with points at $\pi - \theta$ The average of each pair is $\frac \pi 2$