Vanishing Hyperballs

Sphere-in-Cube

Spherical ball contained within a cubic region
[Image from https://grabcad.com ].

We all know that the area of a disk — the interior of a circle — is {\pi r^2} where {r} is the radius. Some of us may also remember that the volume of a ball — the interior of a sphere — is {\frac{4}{3}\pi r^3}.

The unit disk and ball have equations in two and three variables:

Disk:   x^2+y^2<1

Ball:   x^2+y^2+z^2<1.

These are special cases of an {n}-ball, the region {\mathbb{B}_n} of {n}-dimensional space comprising all points whose distance from the origin is less than 1:

n-Ball:   x_1^2+x_2^2+ . . . x_n^2<1.

or {\mathbb{B}_n = \{ x\in \mathbb{R}^n : ||x|| < 1 \}}. For {n>3}, we call {\mathbb{B}_n} a hyperball. A 1-ball is just a line segment. A 2-ball is a disk and a 3-ball is the interior of a sphere. The surface of the n-ball {\mathbb{B}_n} is the (n-1)-sphere {\mathbb{S}_{n-1}}

Volumes and Hyper-volumes

We denote the hyper-volume of the n-ball {\mathbb{B}_n} by {V_n} and its surface area by {S_{n-1}}. These quantities can be defined recursively, starting from {n=0} or {n=1} and iterating to higher dimensions. But there are also closed form solutions. The volume of the {n}-ball is

V_n=\frac{\pi^{n/2} r^n}{\Gamma(\frac{n}{2}+1)}

where {\Gamma} is the gamma function. We recall some properties of {\Gamma(n)}:

\Gamma(\frac{1}{2}) = \sqrt{\pi} ,  \Gamma(1) = 1 \quad and \quad \Gamma(x+1)=x\Gamma(x)

Using these properties, we can express {V_n} without evaluation of integrals using Euler’s definition of the gamma function. We get

Even\ Dimension: \quad V_{2n} = \frac{\pi^n r^{2n}}{n!}

Odd\ Dimension: \quad V_{2n+1} = \frac{2^{2n+1}n!\pi^n r^{2n+1}}{(2n+1)!}

Vanishing Hyper-volumes

Hyperball-Volume

Hyper-volume of B_n for n<= 8 [Table from Wikipedia article “Volume of an n-ball“].}

A table of the values of {V_n} for {1\le 8} is shown here. We note from this table that the volume of {\mathbb{B}_n} is proportional to {r^n}. But the constant of proportionality behaves in a curious way. The coefficient increases with dimension for {n\le 5}, but for higher dimensions it decreases. Indeed, as we shall see, this coefficient tends to zero as {n\rightarrow\infty}.

The expressions above allow us to relate the hyper-volumes for different dimensions:

\frac{V_{2n+2}}{V_{2n}} = \frac{\pi r^2}{n+1}

\frac{V_{2n+1}}{V_{2n-1}} = \frac{\pi r^2}{n+\textstyle{\frac{1}{2}}}

It is a simple matter to confirm that the familiar values for low dimensions satisfy these relationships.

Hyperballs and Hypercubes

We may wonder how we can compare hyper-volumes in different dimensional spaces. But it is straightforward to compare the hyperball in {n} dimensions with a fixed cube in the same space. The hyperball {\mathbb{B}_n} of unit radius, and diameter 2, fits within a cube of edge-length 2, touching it at the centre of each face. The volume of this cube is {2^n}, and we may compare the volume {V_n} of the n-ball to this.

With {r=1}, the above equations become

\frac{V_{2n+2}}{V_{2n}} = \frac{\pi}{n+1}\quad\quad\quad \frac{V_{2n+1}}{V_{2n-1}} = \frac{\pi}{n+\textstyle{\frac{1}{2}}}

Combining these, we see the effect of an increase in dimension by 2 is

V_{n+2} \sim \frac{\pi}{n}\times V_n

It is abundantly clear that, as {n} increases, {V_n} becomes smaller, rapidly approaching zero for large {n}. This is in marked contrast to the surrounding cube, whose volume {2^n} doubles with every additional dimension. For growing {n}, the hyperball occupies a smaller and smaller proportion of the associated cube.


Last 50 Posts

Categories

Archives