Re: [Phys-l] a mathematical curiosity

[John Denker <jsd@av8n.com>]

On 01/09/2012 10:25 AM, Bob Sciamanda wrote:
> Similarly all the surface-area is near the equator. You can see this already
> in D=3, on the earth, where the area of the tropical regions is
> large compared to the area of the arctic regions ... “
>
> I am befuddled.  One could just as arbitrarily choose an equator running 
> through a rotational pole.  Without the earth’s spin and/or its orientation 
> relative to the sun the choice of an equator is completely arbitrary.

Well, we agree that the tilt-angle of the earth's axis is arbitrary
... but once you settle on an angle, the *same* angle determines
the size of the tropic regions and the size of the arctic regions.
  Tropic regions = 2x 23.5 degrees
  Arctic regions = 2x 23.5 degrees

These regions are the same latitude-wise, but very different area-wise.

This is an example of sphere hardening.  Most of the volume is 
near the surface.  Most of the surface-area is near the equator.

> One could just as arbitrarily choose an equator running 
> through a rotational pole.

Hmmmm.  One could choose a _great circle_ running through the pole,
but it wouldn't be called the equator.

The larger point remains:  Suppose you pick two random points on a 
sphere, perhaps a sphere with no rotation and no coordinates (lat/lon 
or otherwise).  Consider the angle those points subtend, as seen 
from the center of the sphere.  That angle is much more likely to
be within a few degrees of 90 than within a few degrees of zero.

The corresponding result is not valid in D=1;  a random point on
the unit interval is just as likely to be near x=0 as x=1.  The
hardening effect is noticeable in D=2 and gets more pronounced as 
D increases.