 # Why does a (hyper-)sphere shrink in high dimensions?

I have always liked geometry. I can make sense of a geometrical problem if I can picture it or sketch it on a piece of paper. For example, by drawing a cube box of side $R$, I can immediately convince myself that its surface area is $6 \times R^2$ (as there are six equal sides) and its volume is $R^3$. Things get complicated, though extremely interesting, when the problem cannot be represented so easily, like the question of computing the volume and surface of a $d$-dimensional sphere, also known as hyper-sphere!

(disclaimer: to completely follow the calculations contained in this article you need some knowledge of integral calculus; however, if you don’t have this you can still follow the reasoning, but you will have to blindly trust the formulas!)

## Volume and surface of a sphere in dimensions $d\le 3$

Let’s first start from what we study at school: we can define a sphere as the set of points equidistant a length $R$ from its centre. A two-dimensional sphere of radius $R$ is a sphere that can be represented (embedded) in two dimensions, for example the plane $(x, y)$. Formally, the two-dimensional sphere of radius $R$ would be the set of points $(x_R, y_R)$ whose distance from the centre $(0, 0)$ is R, that is according Pythagorean theorem $\sqrt{x_R^2+y_R^2} = R$. The points inside the two-dimensional sphere are the ones having distance with its centre less or equal than $R$. In what follows I use the notation of surface of a sphere to indicate the points at the boundary, that is having distance $R$, while volume of a sphere to indicate the points also inside it, that is having distance less or equal than $R$. For example, in three dimensions, you can think the surface of the sphere being a football ball (the leather making its exterior, empty inside), while the volume of the sphere being a bowling ball (a solid ball, usually made in plastic or resin).

According to what studied at school, the surface and volume of a two-dimensional are, respectively, $2\pi R$ (the perimeter of the circle) and $\pi R^2$ (the area or the circle). Again, remembering what studied at more advanced geometry at school, the surface and the volume of a three-dimensional sphere are, respectively, $4\pi R^2$ (the external area of the sphere) and $4 \pi R^3 / 3$ (the volume of the sphere). We can summarise all this by writing

\begin{equation} S_1(R) = 2\pi R \, , \quad S_2(R)=4\pi R^2 \, , \quad \textrm{and} \quad V_2(R) = \pi R^2 \, , \quad V_3(R)=\frac{4\pi}{3} R^3 \, , \end{equation}

where $S_d(R)$ and $V_d(R)$ denote the (hyper-)surface and (hyper-)volume and the subscript $d$ denotes the dimensionality of the quantity measured, for example $d=1$ for a length, $d=2$ for an area, and $d=3$ for a volume. Using purely geometrical considerations we can also convince ourselves that a sphere embedded in one dimension has surface $S_0(R) = 2$, as there are two points at the end of its radius $R$, while the volume of sphere is $V_1(R)=2R$. Finally the volume of the sphere embedded in 0 dimensions is simply $V_0(R)=1$, as there is only one point, while its surface is not defined. See the figure below for a summary of all this.

## How to generalise this for any dimension $d$?

We can immediately observe that for both surfaces and volumes, the $R$-dependence is simply given by $R^d$, so that the general values result in

\begin{equation} S_d(R) = S_d R^d \quad \textrm{and} \quad V_d(R) = V_d R^d \, , \end{equation}

where $S_d$ and $V_d$ are merely numbers (without physical units). So far we know that

\begin{equation} \begin{split} & S_0 = 2 \, , \quad S_1 = 2\pi \, , \quad S_2=4\pi \, , \quad \textrm{and} \\ & V_0=1 \, , \quad V_1=2 \, , \quad V_2 = \pi \, , \quad V_3=\frac{4\pi}{3} \, , \end{split} \end{equation}

but what are the values for a generic $d$? Do they keep increasing for $d$ increasing?

Let us try to answer this question using, as much as possible, only geometrical considerations.

## A relation between the volume and the surface

We can intuitively convince ourselves that the volume of a two-dimensional sphere of radius $R$, $V_2(R)$, is given by filling this volume with surfaces of a one-dimensional sphere of radius $r$, taking $0 \le r \le R$, as sketched in the figure below.

The summing up of these (infinitely) many surfaces of one-dimensional spheres is mathematically the integral of the surfaces with $0 \le r \le R$, leading to

\begin{equation} V_2(R) = \int_0^R S_1(r) \, \textrm{d} r  = \int_0^R 2\pi r \, \textrm{d} r = \pi R^2 \quad \Longleftrightarrow \quad  V_2 = \pi \, ,  \end{equation}

a result that we already know. The beauty of this reasoning is that we can extend it to any dimension, leading to the general recursive result between $S_{d-1}$ and $V_d$ given by

\begin{equation} V_d(R) = \int_0^R S_{d-1}(r) \, \textrm{d} r  = \int_0^R S_{d-1} r^{d-1} \, \textrm{d} r = \frac{S_{d-1}}{d} R^{d} \quad \Longleftrightarrow \quad  V_d = \frac{S_{d-1}}{d} \, . \label{eq:surface} \end{equation}

For example, we can check that this gives the correct value for the volume of a three-dimensional sphere $V_3(R) = 4\pi R^3/3$ given the notion of $S_2(R)=4\pi R^2$. However, so far we only know from geometry $S_d$ for $d=0, 1, 2$, so we cannot use this relation for obtaining $V_d$ for $d > 3$.  We therefore need a second (independent) recursive relation to completely solve our problem.

## A relation between the volumes of spheres of different dimensions

Let us focus on the three-dimensional sphere and slice it in two by considering the two-dimensional sphere passing by the $(x, y)$ plane at $z=0$, as shown in the figure below.

Given a generic point $(x_r, y_r)$ contained inside the two-dimensional sphere having distance $r=\sqrt{x_r^2+y_r^2}$ with $0 \le r \le R$, all the points contained in the three-dimensional sphere of radius $R$ have $z$-coordinate $z_{r’}$ contained in the one-dimensional sphere of radius $r’=\sqrt{R^2-r^2}$. By summing up of these one-dimensional spheres by considering all the (infinitely many) points $(x_r, y_r)$ we can therefore express the volume of the three-dimensional sphere $V_2(R)$ versus the volume of the one dimensional sphere $V_0(r’)$. This, mathematically, is expressed by the two-dimensional integral

\begin{equation} \begin{split} V_3(R) & = \int_{x^2+y^2 \le R} V_1(\sqrt{R^2-r^2}) \, \textrm{d} x \textrm{d} y \,,  \quad \textrm{then moving to polar coordinates} \\ & = \int_0^R \int_0^{2\pi} 2 \times \left(R^2-r^2\right)^{1/2} \, r \textrm{d} \theta \textrm{d} r  \\ & = 4\pi \frac{R^3}{3} \quad \Longleftrightarrow \quad  V_3 =  \frac{4\pi}{3} \, , \end{split} \end{equation}

confirming the result of $V_3$ we already know from geometry. We can however generalise this reasoning to express the volume of a $d$-dimensional sphere versus the volume of a $(d-2)$-dimensional sphere, leading to the integral

\begin{equation} \begin{split} V_d(R) & = \int_{x^2+y^2 \le R} V_{d-2}(\sqrt{R^2-r^2}) \, \textrm{d} x \textrm{d} y \,,  \quad \textrm{then moving to polar coordinates} \\ & = \int_0^R \int_0^{2\pi} V_{d-2} \times \left(R^2-r^2\right)^{d/2} \, r \textrm{d} \theta \textrm{d} r  \\ & = 2\pi V_{d-2} \frac{R^d}{d} \quad \Longleftrightarrow \quad  V_d =  \frac{2\pi}{d} V_{d-2} \, . \end{split} \end{equation}

This last result is great, as it let us completely solve our problem! Indeed, knowing from elementary geometry that $V_0 = 1$ and $V_1 = 2$, we can recursively obtain $V_d$ for any $d \ge 2$.

## Obtaining the surface and volume for a generic dimension $d$, an unexpected result!

After a bit of tedious algebra (not shown here) that simplifies the recursion relations, we can express $V_d$ versus $d$, resulting in

\begin{equation} V_0=1 \, , \quad V_1=2 \, , \quad V_d = \left\{ \begin{split} & \frac{\pi^{d/2}}{(d/2)!} \,, \quad \text{for $d \ge 2$ even} \\ & \frac{2 \times [(d-1)/2]! \times (4\pi)^{(d-1)/2}}{d!}  \,,  \quad \text{for $d \ge 3$ odd} \end{split} \right. \end{equation}

Then, combining this last result with the equation relating the surface and the volume, we get also the hyper-sphere surface constants $S_d$ versus $d$, being

\begin{equation} S_0=2 \, , \quad S_d = \left\{ \begin{split} & \frac{(d+1) \times \pi^{(d+1)/2}}{[(d+1)/2]!} \,, \quad \text{for $d \ge 1$ odd} \\ & \frac{(d+1) \times 2 \times (d/2)! \times (4\pi)^{d/2}}{(d+1)!}  \,,  \quad \text{for $d \ge 2$ even} \end{split} \right. \end{equation}

Let us plot the values obtained from the two formulas above versus $d$ in the figure below.

We immediately notice that both $S_d$ and $V_d$ have a maximum, and then decay to zero for large dimension $d$ (one can formally prove that these limits are indeed $0$). This means effectively that both the surface and the volume of an hyper-sphere will shrink to zero for $d$ tending to infinity — very very weird

## How to make sense of all this

Does this make sense? Well it does if, for example, we compare the surface and volume of an hyper-sphere with the surface and volume of other hyper-objects, dimension by dimension. Take for instance the volume of a hyper-cube of side $R$ embedded in $d$ dimensions: the volume of a one-dimensional cube is simply the length of its side $V_1^{\textrm{box}}(R)=R$, the volume of a two-dimensional cube is the area of a square $V_2^{\textrm{box}}(R)=R^2$, and so on, being the volume of a $d$-dimensional cube simply $V_d^{\textrm{box}}=R^d$.

If we compare the volumes of the hyper-sphere and the hyper-cube for a given $d$ by taking the ratio of them, we have this ratio increasing for lower values of $d$, reaching a maximum and then decreasing. This means that for low dimensions, the hyper-cube is always contained inside the hyper-sphere, see the figure below to convince yourself for $d=1, 2, 3$.

However for large dimensions, is the other way around: the (infinite) number of points contained in the hyper-sphere will always be lower than the number of points contained in the hyper-cube, and the hyper-sphere will eventually shrink and become infinitesimal compared to the hyper-cube for $d$ tending to infinity!

A similar reasoning can be done by comparing the surfaces of the hyper-sphere and the hyper-cube for a given $d$, but to do so one needs first to understand how many hyper-sides a hyper-cube has: I leave you with this problem to solve…