Intuition why the volume and surface area of the unit sphere eventually decrease

Solution 1:

One should consider that the formula for a sphere can be presented as a product of fractions of the form for even and odd dimensions

  1     3     5     7     9                0     2    4    6    8
 ---------------------------              -------------------------
  2    2pi   2pi   2pi   2pi      and      1    2pi  2pi  2pi  2pi
       --    --    ---   ---                    --  ---  ---   ---
        3     5     7     9                      2    4    6    8

One sees that the numerator is stuck on $2\pi$, but the denominator accelerates away. The intuitive answer is not so much that the sphere becomes smaller, but that the cube goes very large. The intuition follows exactly, a square, cube, etc that has a specific margin (ie 'dihedral') angle in hyperbolic space. The body stays around, but the vertices go to infinity etc. This tells us that it's the cube's fault, not the sphere's, that the ratio runs away.

Volume is measured as the moment of surface. What this means is that we construct a vector pointing out from the solid, and then integrate the dot product of a ray from a point or plane, over the element of the surface, that is this integrated over the whole surface. thus:

$d \text{ volume} = \text{coordinate}\cdot d\text{ surface}$

When the coordinate is radial, we need to divide the value by $n$. When it is done from a plane (eg relative to the x-axis), there is no need to divide by $n$. The sum by coordinate by radial, is exactly the sum of doing it from the $X$ axis, and the $Y$ axis etc. That's where the division by $n$ comes in.

When one uses coordinates in radial terms, and uses the above as a definition of volume, one gets measures in tegmic units as in my Polygloss. The ordinary measure-polytope units (square, cube, ...), derive from considering just the x-coordinate. The radial scheme is equivilant to considering all of the perpendiculars, so the tegmic volume is n times the prismic volume for the same surface. And since surface is a volume of N-1 dimensions, one can prove by induction, that the prismic unit is $n!$ of the tegmic one.

The tegmic radian, which is $1/n!$ of the spheric one, is slightly les than the solid angle of the simplex of the same dimension.

So after roughly 19 dimensions, one can no longer crowd $n!$ simplexes at a point.

Another approach

One should remember that a sphere will sit in any prism of lesser spheres. So, for example, a sphere sits in a circle-prism (or cylinder). This means that the maximum is by steps of 2d, $\pi^{n/2}$. But this is a maximum. The sphere in four dimensions, occupies just half of this space (ie $\pi^2/2$). If we suppose not $\pi$ in the numerator, but $2\pi$, then the divisor corresponds to $n$: the sphere of $2\pi$ is $1/n$ of the cylinder formed by a circle, and an N-2 sphere, which leads to the resluts above. In general, for the even dimensions, it might be writen as $\pi^{n/2}/(n/2)!$, and with odd dimensions, this relation is used to set eg $(1/2)!$ so that $2 = \pi^{1/2}/(1/2)!$, gives $(1/2)! = \pi^{1/2}/2$.