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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:

  1. Two points uniformly distributed on the sphere are taken and connected by a segment;
  2. 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?

QLimbo
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1 Answers1

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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 $\phi_1$: $$ \langle PQ\rangle={ \int_0^\pi \sqrt{2(1-\cos\phi_1)} \sin^{n-2}\phi_1\,d\phi_1 \over \int_0^\pi \sin^{n-2}\phi_1\,d\phi_1 } =\frac{2^{n-1} \left(\Gamma \left(\frac{n}{2}\right)\right)^2} {\sqrt{\pi }\,\Gamma\left(n-\frac{1}{2}\right)}. $$

Intelligenti pauca
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  • Point $Q$, in my opinion, is not uniformly distributed on the sphere, if the $\phi$ are evenly distributed over the parallelepiped. Consider the case of an ordinary sphere in three-dimensional space. The answer is $\frac{4}{3}$. – QLimbo Nov 08 '21 at 19:08
  • @Cornifer And in fact, plugging $n=3$ in my formula you get $4/3$. – Intelligenti pauca Nov 08 '21 at 19:11
  • Yes, you are right, I jumped to conclusions! :) – QLimbo Nov 08 '21 at 19:11
  • If $Q$ weren't uniformly distributed on the sphere you couldn't use the integral at denominator to compute the area of the spherical surface. – Intelligenti pauca Nov 08 '21 at 19:18
  • Yup, everything is fine. I even calculated the limit. As $n$ tends to infinity $\sqrt{2}$. But in your integrals, the last integrand factor seems to be $\sin \varphi_{n-2}$. I also have a question about $n=1$. What's the point of chord length in this case? – QLimbo Nov 08 '21 at 19:35
  • You are right about $\sin\phi_{n-2}$, I corrected the typo. For the limit $\sqrt2$ see the comment by David K above. For $n=1$ the integrals are meaningless, I think. – Intelligenti pauca Nov 08 '21 at 21:17
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    In case 2) the chord length is obtained by dividing the volume of an $n$-dimensional ball by the volume of an $(n-1)$-dimensional ball, i.e. $\frac{\Gamma(n / 2+1 / 2)}{\Gamma(n / 2+1)} \sqrt{\pi}$ and this tends to zero. – QLimbo Nov 09 '21 at 17:13