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.

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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…