Computing $\pi$ using atoms

Everyone knows it's possible to compute $\pi$ by throwing darts at random but it's not so well known that you can calculate it by simulating an atom in a periodic box.

Remember Pac-Man? You could move the character all the way to the right of the maze and pop up at the left or down to the bottom and reappear at the top. The maze is akin to the flat torus $\T = \R^2 /\!\sim$, where $x \sim y$ if and only if $x_i = y_i \! \mod 2 \pi$ for $i = 1, 2$. Layout of Pac-Man

If we fix a velocity $v \in \R^2$ and set in motion a particle originally at $x_0 \in \T$, then our particle follows the trajectory $x(t) = x_0 + t v$ forever.

A point (shown as a black circle) and a velocity vector (represented as an arrow with its tail at the point) on the flat torus (represented as a square).

Something interesting happens depending on how we choose $v = (v_1, v_2)$. If $\frac{v_1}{v_2}$ is a rational number, then the trajectory $x(t)$ is periodic.

Example of a periodic trajectory (represented as black tube) on the torus.

However, if $\frac{v_1}{v_2}$ is irrational, then the set of visited points $\{ x(t) \, | \, t \ge 0 \}$ is dense in $\T$. In other words, the particle eventually gets arbitrarily close to any point of the torus without ever visiting twice the same spot.

Left slider: Length of the curve.
Right slider: Value of $\frac{v_1}{v_2}$ (default is $\sqrt{2}$).

We can now use this to measure $\pi$! Indeed, if $S \subset \T$ is a circle of radius $r = 1$, then \begin{equation*} \bar{A} = \lim_{T \to \infty} \frac{1}{T} \int_0^T \mathbb{1}_{S}(x(t)) \, \d t \end{equation*} is the fraction of the time that the particle spends inside $S$. It turns out that \begin{equation*} \pi = (4 \bar{A})^{-1} \end{equation*} as long as our initial velocity satisfies $\boldsymbol{\frac{v_1}{v_2} \not\in \mathbb{Q}}$.

Indeed, our approximations of $(4 \bar{A})^{-1}$ converge to $\pi$ as $T \to \infty$.

Graph showing convergence of the estimate of π as T increases

This works thanks to the ergodic theorem which, in the above setting, tells us that the time average $\bar{A}$ is equal to the expected value \begin{equation*} \mathbb{E}[\mathbb{1}_S] = \mathbb{P}(S), \end{equation*} with respect to the uniform probability distribution $\mathbb{P}$ on $\T$. The latter is simply the ratio of area of the circle $S$ to the area of the flat torus $\T$.

This can be seen as an instance of (Hamiltonian) Monte Carlo, but without randomness. Alternatively, we can regard it as an application of molecular dynamics to a gas particle in a 2D periodic box.

Published: November 15, 2022.