How to find the average chord length on a sphere of radius $1$? It is clear that this cannot be done as long as there is no law of distribution of its endpoints. Two probability distributions seem logical:
- Two points uniformly distributed on the sphere are taken and connected by a segment;
- A random direction is taken and parallel to it, uniformly along the projection of the sphere onto a plane perpendicular to this direction, a segment inside the ball is taken.
It turns out that these are completely different models!
In model 1), the average chord length for all $n$ is greater than $\frac{1}{2}$, since the second point with probability $\frac{1}{2}$ is in another hemisphere and the distance in this case is greater than $1$. In model 2), the average chord length tends to zero! Since it is equal to the volume of the $n$-dimensional ball divided by the volume of the $(n-1)$-dimensional ball.
Such nonsense happens in the $n$-dimensional case, I want to figure it out! And how to calculate exactly the average chord length in 1) case?
Best Answer
We can find the average in case 1) by using polar coordinates in $n$ dimensions. As endpoints of the chord we can take $$ \begin{align} &P=(1,0,\dots,0),\\ &Q=(\cos\phi_1, \sin\phi_1\cos\phi_2,\dots, \sin\phi_1\cdots\sin\phi_{n-2}\cos\phi_{n-1}, \sin\phi_1\cdots\sin\phi_{n-2}\sin\phi_{n-1}), \end{align} $$ so that: $$ PQ^2=(1-\cos\phi_1)^2+\sin^2\phi_1=2(1-\cos\phi_1). $$ We can then integrate that over the whole sphere to get the average value of $PQ$: $$ \langle PQ\rangle={ \int_0^\pi\int_0^\pi\cdots\int_0^{2\pi} \sqrt{2(1-\cos\phi_1)} \sin^{n-2}\phi_1\sin^{n-3}\phi_2\cdots \sin\phi_{n-2}\,d\phi_1\,d\phi_2\cdots d\phi_{n-1} \over \int_0^\pi\int_0^\pi\cdots\int_0^{2\pi} \sin^{n-2}\phi_1\sin^{n-3}\phi_2\cdots \sin\phi_{n-2}\,d\phi_1\,d\phi_2\cdots d\phi_{n-1} }. $$ Note that all integrals factor out, with the exception of that in $\phi_1$: $$ \langle PQ\rangle={ \int_0^\pi \sqrt{2(1-\cos\phi_1)} \sin^{n-2}\phi_1\,d\phi_1 \over \int_0^\pi \sin^{n-2}\phi_1\,d\phi_1 } =\frac{2^{n-1} \left(\Gamma \left(\frac{n}{2}\right)\right)^2} {\sqrt{\pi }\,\Gamma\left(n-\frac{1}{2}\right)}. $$