### Even-Dimensional Balls

#### Posted by John Baez

Some of the oddballs on the $n$-Café are interested in odd-dimensional balls, but here’s a nice thing about *even*-dimensional balls: the volume of the $2n$-dimensional ball of radius $r$ is

$\frac{(\pi r^2)^n}{n!}$

Dillon Berger pointed out that summing up over all $n$ we get

$\sum_{n=0}^\infty \frac{(\pi r^2)^n}{n!} = e^{\pi r^2}$

It looks nice. But *what does it mean?*

First, why is it true?

We can show the volume of the $2n$-dimensional ball of radius $r$ is $(\pi r^2)^n/n!$ inductively. How? It’s enough to show that the volume of the unit $2(n+1)$-ball is $\pi /(n+1)$ times the volume of the unit $2n$-ball. Wikipedia gives a proof, and Greg Egan recently summarized the argument in a couple of tweets.

The idea is to map the unit $2(n+1)$-ball to the unit disk in the plane via a projection $\mathbb{R}^{2n} \to \mathbb{R}^2$. Imagine a $2(n+1)$-ball hovering over a disk! I can’t draw it, because the first interesting example is 4-dimensional. But imagine it.

The points over any point in the disk at distance $r$ from the origin form a $2n$-ball of radius $\sqrt{1 - r^2}$. We can get the volume of the $2(n+1)$-ball by integrating the volumes of these $2n$-balls over the disk.

That’s the idea. The rest is just calculus. Let the volume of the unit $2n$-ball be $V_{2n}$. We do the integral using polar coordinates:

$V_{2(n+1)} = \int_0^1 V_{2n} (1 - r^2)^n \; 2 \pi r d r$

Make the substitution $u = 1 - r^2$, so $d u = -2 r d r$. Some minus signs and 2’s cancel and we get

$V_{2(n+1)} = \pi V_{2n} \int_0^1 u^n \, d u = \frac{\pi}{n+1} V_{2n}$

as desired!

So that’s simple enough: since $V_0 = 1$ we get

$V_{2n} = \frac{\pi^n}{n!}$

And maybe that’s all that needs to be said. But suppose we follow Dillon Berger and sum the volumes of all even-dimensional balls of radius $r$. We get

$\sum_{n=0}^\infty V_{2n} r^{2n} = \sum_{n=0}^\infty \frac{(\pi r^2)^n}{n!} = e^{\pi r^2}$

This looks cute, but what does it mean?

It seems odd to sum the volumes of shapes that have different dimensions. But we do it in classical statistical mechanics when considering a system of particles with a *variable* numbers of particles: a so-called ‘grand canonical ensemble’, like an open container of gas, where molecules can flow in and out.

If we have a single classical harmonic oscillator, with momentum $p$ and position $q$, its energy is

$\frac{1}{2}(p^2 + q^2)$

where I’m choosing units that get rid of distracting constants.

If we impose the constraint that the energy is $\le E$ for some number $E$, the state of the oscillator is described by a point $(p,q)$ with

$p^2 + q^2 \le 2 E$

This is a point in the disk of radius $r = \sqrt{2E}$.

Now suppose we have an arbitrary collection of such oscillators—and we don’t know how many there are! Then the state of the system is described first by a natural number $n$, the number of oscillators, and then momenta $p_1, \dots, p_n$ and positions $q_1, \dots, q_n$ obeying

$p_1^2 + q_1^2 + \cdots + p_n^2 + q_n^2 \le 2 E$

In other words, it’s a point in the *union* of all even-dimensional balls of radius $r = \sqrt{2E}$.

Physicists would call this union the ‘phase space’ of our collection of oscillators. Its volume is the sum of the volumes of the balls:

$\sum_{n=0}^\infty V_{2n} r^{2n} = e^{\pi r^2} = e^{2 \pi E}$

Actually physicists would include the constants that I’m hiding, and this is actually somewhat interesting. For example, to make the area 2-form $d p \wedge d q$ dimensionless, they divide it by Planck’s ‘quantum of action’. This is what allows them to add up volumes of balls (or other symplectic manifolds) of different dimensions and get something that doesn’t break the rules of dimensional analysis. But this also brings Planck’s constant into a question of *classical* statistical mechanics—a surprise, but a surprise that’s well-known among experts: it shows up in computations such as the entropy of an ideal gas.

And indeed, entropy is exactly what I’m interested in now! Why would a physicist care about the volume of phase space for a collection of arbitrarily many oscillators with total energy $\le E$? Here’s why: if *all we know* about this collection of oscillators is that it’s energy is $\le E$, its entropy is just the logarithm of this phase space volume! So, it’s

$\ln (e^{2 \pi E}) = 2 \pi E$

Nice!

Physicists will not like how I tucked all the constants under the rug, getting a formula for entropy that’s missing things like Boltzmann’s constant and Planck’s constant, as well as the mass of the particles in our harmonic oscillators, and their spring constant. I leave it as a challenge to restore all these constants correctly. I suspect that the $2\pi$ will get eaten by the $2 \pi$ hiding in Planck’s constant $\hbar = h / 2\pi$. This I know: we will get something proportional to $E$, but with units of entropy.

This is what I’ve got so far. It’s not yet a new *proof* that the volumes of all even-dimensional balls of radius $r$ sum to

$e^{\pi r^2}$

but rather a physical interpretation of this fact. Maybe someone can keep the ball rolling….

## Re: Even-Dimensional Balls

This is a bit of a diversion that’s orthogonal to the physical interpretation, but the basic formula generalises in a nice way to the $p$-norm.

The volume of a ball in $\mathbb{R}^n$ under the $p$-norm is:

$V_{p,n}(r) = 2^n\frac{\Gamma(1+1/p)^n}{\Gamma(1+n/p)} r^n$

This formula comes from “Volumes of Generalized Unit Balls” by Xianfu Wang, Mathematics Magazine Vol. 78, No. 5 (Dec., 2005), pp. 390-395, via Anand Sarwate and John D Cook.

We then have:

$V_{p, k p}(r) = \frac{V_{p,p}(r)^k}{k!}$

and hence:

$\sum_{k=0}^{\infty} V_{p,k p}(r) = e^{V_{p,p}(r)}$

If you put $p=1$ and take out the factor of $2^n$, this includes the fact that the sum of the volumes of the $n$-dimensional simplices where $n$ orthogonal edges of length $L$ meet at one vertex is just $e^L$.