# Limit of distance between two random points in a unit-radius $n$-sphere

This is a companion contrast to the earlier analogous question for unit $n$-cubes, where the answer (provided by several respondents) is $\infty$ .

What is the limit, as $n \to \infty$, of the expected distance between two points chosen uniformly at random within a unit-radius hypersphere in $\mathbb{R}^n$, i.e., in the unit-radius $n$-ball?

Dividing OEIS A093530 by A093531 I see that it appears to be approaching approximately $1.37$ for odd $n$, but I wonder if the limit is actually known, either exactly or to significant precision? I cannot quite extract an answer from the MathWorld article...

This certainly provides a dramatic contrast between the $n$-cube and the $n$-sphere!

• I'm not sure there's such a contrast between cubes and spheres. In high dimensions a unit cube is vastly larger than a unit sphere, as measured by diameter or volume, so it's not as fair a comparison as the word "unit" suggests. – Henry Cohn Apr 15 '14 at 0:44
• If instead we look at the ratio of expected distance to diameter, then for spheres the limit is asymptotically $\frac{1}{\sqrt{2}}$, while for cubes it is (from Nate Eldredge's answer to the linked question) $\frac{1}{\sqrt{6}}$. Is it known whether Spheres maximize this ratio, whether in finite dimensions or asymptotically? – Kevin P. Costello Apr 15 '14 at 21:15

Both points will be very close to (let's pretend: on) the surface with prob almost 1. Call the first point the north pole. By concentration of measure for the sphere, a randomly chosen second point is almost guaranteed to be almost on the equator, so the limit should be $\sqrt{2}$.

• May I ask: Why can you assume that the first point is the north pole? Concentratoin of measure again? – Joseph O'Rourke Apr 15 '14 at 0:17
• I condition on the first point, and the sphere looks the same from any point. – Christian Remling Apr 15 '14 at 0:18
• Are you reading the question as "two points on the surface of the unit sphere", or ought this be overwhelmingly likely? – usul Apr 15 '14 at 0:57
• Yes, "sphere" (as opposed to "ball"), but even if it's the ball, almost all the volume is near the surface, so it won't matter. – Christian Remling Apr 15 '14 at 1:02
• I should have known, after your first comment (sorry for the fake discussion)... Solution stays the same, though, by also applying the trivial concentration of measure near the surface first. – Christian Remling Apr 15 '14 at 1:24