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…

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Davide is an associate professor in applied maths and physics at UEA. His research focusses mainly on vortex dynamics and turbulence in quantum fluids, but he also works on out-of-equilibrium regimes of other physical systems like ocean waves, light propagating in nonlinear media, and nonlinear chains in solid state physics. He likes cooking, growing his own vegetables and fruits, travelling, and playing board games.