Why does $\frac{V_n}{A_n}$ represent the concentration of volume around the boundary?

Question

I read this answer about the volume of a ball in $\mathbb{R}^n$ being concentrated around the boundary.

I don't understand why:

1. $$\frac{V_n}{A_n}$$ is a measure of concentration around the boundary?
2. $$\frac{V_n}{V_{n-1}}$$ is a measure of concentration around the equator?

$V_n$ and $A_n$ denotes the volume and surface area of a (unit) ball in $\mathbb{R}^n$. We know that $A_n = nV_n$.

Also what exactly is the difference between the equator and the boundary in this context? Is there a diagram that can help me?

Thanks a lot!

Answer 1

I'll try to answer the question in your last paragraph without drawing pictures.

The surface of a disk is a circle; the surface of a ball is a sphere. An $n$-dimensional ball of radius $R$ (centered at the origin) is the set $\{(x_1,x_2,\ldots,x_n)\mid x_1^2+x_2^2+\ldots+x_n^2\le R^2\}$; its surface is the set $\{(x_1,x_2,\ldots,x_n)\mid x_1^2+x_2^2+\ldots+x_n^2=R^2\}$.

In the context of the answer you link to, an equator of a disk is a line segment that cuts it in half (a diameter). An equator of a ball is the cross section you get when you slice the ball in half with a plane. This cross section is a disk—a disk of largest diameter that fits inside the ball. For an $n$-dimensional ball, an equator is the cross section (of dimension $n-1$) you get when you slice the ball into two equal halves. This cross section is a ball of dimension $n-1$ and radius equal to that of the original ball.

Equivalently, an equator is what you get when you do an orthogonal projection of the $n$-dimensional ball onto an $(n-1)$-dimensional hyperplane through its center. An easy way to produce an example is to set one coordinate equal to zero. An equator of the disk $x^2+y^2\le R^2$ is the line segment $x^2\le R^2$ (by setting $y=0$); an equator of the ball $x^2+y^2+z^2\le R^2$ is the disk $x^2+y^2\le R^2$ (by setting $z=0$).

Notice that the equator of a ball is also a ball (of dimension one less). The ratio $V_n/V_{n-1}$ is a length. It tells you how a tall a "cylinder" with an $(n-1)$-dimensional ball as its base would have to be to have the same volume as an $n$-dimensional ball of the same radius. So $V_2/V_1$ is the height a rectangle of base $2R$ would have to have to equal the area of a disk of radius $R$. It is $\pi R^2/(2R)=\frac{\pi}{2}R$. And $V_3/V_2$ is the height a cylinder with base $\pi R^2$ would have to have to equal a ball of radius $R$. It is $\frac{4}{3}\pi R^3/(\pi R^2)=\frac{4}{3}R$.

If $V_n/V_{n-1}$ is a small fraction of $R$, it means that the cylinder doesn't have to be very tall to equal the volume $V_n$; if it is large, then the cylinder does have to be tall. Now if instead of slicing the $n$-dimensional ball in half, we were to slice slightly above and slightly below the midpoint, that is, instead of slicing at, say, $x_n=0$, we sliced at $x_n=\epsilon$ and $x_n=-\epsilon$, where $\epsilon$ is a small positive number, the volume between the slices would be approximately a cylinder with the equator as base. So if $V_n/V_{n-1}$ were small, we could set $\epsilon$ equal to half this ratio, and almost all of the volume $V_n$ would lie between the slices.

When $n=2$ or $3$, the ratio is not a small fraction of $R$. (Both $\pi/2$ and $4/3$ are greater than $1$.) But as $n$ grows, it can be shown that $\sqrt{n}V_n/V_{n-1}$ approaches $\sqrt{2\pi}R$; the coefficient of $R$ in $V_n/V_{n-1}$ approaches $\sqrt{2\pi/n}$, which is indeed small when $n$ is large. So most of the volume $V_n$ lies in a narrow slice containing the equator.

The surface calculation is similar. The ratio $V_n/A_n$ is again a length. The difference $V_n(R)-V_n((1-\epsilon)R)$ (where the argument of $V_n$ is the radius of the ball) is the volume of the "spherical shell" of thickness $\epsilon R$. Roughly speaking, it's the part of the volume of the onion that is contained in the onion peel. Since $V_n(R)=C_nR^n$, where $C_n$ does not depend on $R$, we have, using that $A_n(R)=\frac{d}{dR}V_n(R)$, $$ V_n(R)-V_n((1-\epsilon)R)=C_nR^n(1-(1-\epsilon)^n)\approx C_nR^n\cdot n\epsilon=nC_nR^{n-1}\cdot\epsilon R=A_n(R)\cdot\epsilon R. $$ So if $V_n(R)/A_n(R)$ is a small fraction times $R$, we can set $\epsilon$ equal to this small fraction, and conclude that the volume of the onion peel is approximately that of the whole onion.

For $n=2$ and $n=3$ we get $\pi R^2/(2\pi R)=\frac{1}{2}R$ and $\frac{4}{3}\pi R^3/(4\pi R^2)=\frac{1}{3}R$. In fact, from the previous paragraph you can see that $V_n(R)/A_n(R)=\frac{1}{n}R$ in general. When $n$ is big, this is indeed a small fraction of $R$.

Answer 2

Intuitively, A$_n$ * r should be equal to volume of the shell V$_n$(1) - V$_n$(1-r). Here V$_n$(x) stands for volume of n-dimensional ball with radius x; also, V$_n$ is just V$_n$(1). In words: area of base * height = volume. The strange thing here is that one can take small r (= 1/n) to get almost whole unit ball volume, hence the claim that volume is concentrated around surface. Similarly, one can cut V$_n$ around equator (to get "base" with area V$_{n-1}$) and ask about height of a cylinder with base V$_{n-1}$ so that its volume is equal to volume of V$_n$. The answer is 1/sqrt(n), i.e. quantity much smaller than radius of the ball (=1). These things entered modern (graduate level) analysis and probability under the name "concentration of measure" in the last few decades.