Re: Distance Between 2 Randomly-chosen Points on a Sphere

From: Ross A. Finlayson (raf_at_tiki-lounge.com)
Date: 10/10/04 

Date: 10 Oct 2004 09:55:23 -0700

Ed Murphy <emurphy42@socal.rr.com> wrote in message news:<pan.2004.10.10.05.51.03.330911@socal.rr.com>...
> On Sat, 09 Oct 2004 21:56:20 +0200, David Kastrup wrote:
>
> > for
> > every point at a distance of x from the pole, there is a point mirrored at
> > the equator at distance pi-x.
>
> I think you're taking distances along the surface of the sphere, rather
> than distances through the interior. The original problem did say
> "straight-line distance". This breaks the mirroring.

I agree.

It seems that of two random points on the unit sphere, one of the
points is deemed fixed at (1, 0, 0). That's because compared to a
random point on the sphere that's random.

One way to consider it is that the randomly selected point is either
the same as the fixed point for a distance of zero, or opposite it
along the diameter for a distance of two.

Then, consider the circle that passes through (0,1,0), (0, -1, 0), (0,
0, 1), and (0, 0, -1). The distance from the fixed point to any point
on that circle is sqrt(2).

Then, I can see why the average might be greater than sqrt(2). For
the circle that is through ( 1/sqrt(2), +- 1/sqrt(2), 0) and (
1/sqrt(2), 0, +- 1/sqrt(2) ), the distance from the fixed point is the
square root of (1/sqrt(2))(1- 1/sqrt(2)) = .4550898.... Then, there
is the corresponding parallel circle of the same radius through (
-1/sqrt(2), +- 1/sqrt(2), 0) and ( -1/sqrt(2), 0, +- 1/sqrt(2) ), and
the distance from the fixed point to that circle is the square root of
(1/sqrt(2))(1+ 1/sqrt(2)) = 1.098684..., and the average of those two
distances is around .776887... , where the square root of two is
1.414, and 1/sqrt(2)=sqrt(2)/2 = .707106781....

I think it's safe to presume that the average is greater than sqrt(2),
and it's interesting to see people have explanations for why it is
4/3, 1.3.... I think it is less than 4/3. With considering, say,
z=0, and x^2 + y^2 = 1 and x=y or 2x^2= 1 and then also that sqrt(
(1/x)(1-1/x)) + sqrt( (1/x)(1+1/x)) / 2 = 4/3, or sqrt(
(1/x)(1-(1/x))) + sqrt( (1/x) (1+(1/x))) = 8/3, solving that for x,
and thus the circle through (x, +-x, 0) and (x, 0, +-x) and the circle
through (-x, +-x, 0) and (-x, 0, +-x) should each be parallel, on the
sphere, and having equal radii, and over those two circles the average
distance is 4/3. I don't think the average distance is 4/3.

Basically I think that two circles of equal radius have the same
number of points, and in the context of two circles on the same sphere
with a question of average distance between the points, that two
circles of differing radius have a different number (amount) of
points, with the larger circle having more points.

Now on the sphere, symmetrical in all axes, divide the sphere into
pairs of circles for each +-x for x from zero to one. Then trivially
the fixed (selected) point on (1,0,0) matches to (-1, 0, 0) are the
poles, and the circle y^2+z^2 = 1 is the equator.

If the average distance between the main pole, (1, 0, 0), and a
"randomly selected point on the sphere" is 4/3, then I expect that the
random selection is not uniform over the sphere.

Where I think that the paired circles have equal numbers of points, it
is probably not uniformly at random to select a random point on the
sphere by selecting a point on [-1,1] at uniformly random and then
selecting a polar coordinate at random. That _would_ give a point at
uniform random on an open cylinder, but not a sphere.

One problem with figuring a random value in three dimensions is that
there is not one coordinate (pair), but rather more than one
independent variable. We can say a polar coordinate from 0 to 360 can
select a point on a circle at random from the center, but because of
the symmetry in all axes of the sphere, up to three dimensions, that
does not work.

It is said there are no knots in four or more dimensions.

Here's my question: is there an analog of polar "co-ordinates" for
three dimensions thus that one scalar value can indicate any
normalized three-vector? The radial coordinate can indicate a
normalized two-vector. Here when I say "normalized" I mean "unit
length" and two-vector is a vector expressing two scalar components
scaling e1e2.

The idea is that if there was that kind of "co-ordinate", particularly
if constant modifications swept constant distances on the sphere, then
applying a uniform random selection to that scalar value would give a
point at uniform random on the sphere.

Where there isn't, invent one, that's specifically what it is.

Please refer me to "Droplet Theory". For the average distance to be
square root of two, it's a different question about the the distance
between the big end of an egg and a random point on the egg that has
max radius one and a given egg parameter. That might look more like a
semisphere or cone than a sphere.

Here's one method: determine the distance between vertices of regular
polyhedra at random, for each finite regular polyhedron "inscribed"
into the unit sphere, extrapolate to regular polyhedra of infinitely
many sides, compare to the result for each unit sphere "inscribed"
into any regular polyhedron.

Warm regards,

Ross F.

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