## Galperin’s billiard method of computing pi

For the lack of original pi-related ideas, I describe a billiard experiment invented and popularized by Gregory Galperin (Григорий Гальперин). I read about it in a short note he published in 2001 in the Russian journal “Mathematical Education” (“Математическое Просвещение”). Although the table of contents is in Russian, it’s not hard to find the article with “pi” in the title. It’s clear from the article that the observation precedes its publication by some years; the author gave talks on this subject for a while before committing it to print.

Place two billiard balls (red and blue) on the half-line $[0,\infty)$: the red ball is at rest, the blue ball is to the right of it and is moving the the left. At $x=0$ there is an elastic wall.

Assume perfectly elastic collisions: the kinetic energy is preserved. How many collisions will happen depends on the masses of the ball, or rather their ratio. Suppose they have equal mass (ratio=1). Then we’ll see the following:

1. Blue hits Red and stops; Red begins to move to the left
2. Red hits the wall and bounces back
3. Red hits Blue and stops; Blue begins to move to the right

That’s all, 3 collisions in total. Here is a Scilab-generated illustration: I placed red at $x=1$ and blue at $x=1.5$, with initial velocity $-1$. The time axis is vertical.

Number 3 is also the first digit of $\pi$. Coincidence? No.

Consider the blue/red mass ratio of 100. Now we really need scilab. Count the collisions as you scroll down:

Yes, that’s exactly 31 collisions. The mass ratio of 10000 would yield 314 collisions, and so on.

How does this work? The trick is to consider the two-dimensional configuration space in which the axes are scaled proportional to the square roots of masses. A point on the configuration plane describes the position of two balls at once, but it also behaves just like a billiard ball itself, bouncing off the lines (red=0) and (red=blue). The scaling by square roots of masses is needed to make the (red=blue) bounce work correctly: the angle of incidence is equal to the angle of reflection. (See Update 2 below.)

The opening angle is $\alpha=\tan^{-1}\sqrt{M_r/M_b}$. Now we can straighten the billiard trajectory by repeated reflection: this will help us count the number of collisions.

It remains to count how many angles of size $\alpha=\tan^{-1}\sqrt{M_r/M_b}$ fit inside of angle $\pi$. By taking $\sqrt{M_r/M_b}=10^{-n}$, we should get $10^n \pi$ collisions, rounded to an integer one way or the other. It seems that the rounding is always down, that is, we get exactly the first $n$ digits of $\pi$. I once tried to prove that this always happens, but unsuccessfully. The error of approximation $\tan^{-1} 10^{-n}\approx 10^{-n}$ needs to be compared to the fractional part of $10^n \pi$, which looks like tricky business.

Here’s the scilab source. The plots in this post were obtained with galperin(1) and galperin(100).

function galperin(m)
h=0.001; steps=4000;  x1=zeros(steps); x2=zeros(steps);
x1(1)=1.5;  v1=-1;  x2(1)=1;  v2=0; collision=0; j=1;
while j<steps,
select collision
case 0 then
newx1=x1(j)+h*v1; newx2=x2(j)+h*v2;
if (newx2<0) then collision=1;
elseif (newx1<newx2) then collision=2;
else
x1(j+1)=newx1; x2(j+1)=newx2; j=j+1;
end
case 1 then
v2=-v2; x1(j+1)=x1(j); x2(j+1)=x2(j); collision=0; j=j+1;
case 2 then
newv1=(v1*(m-1)+2*v2)/(m+1);  newv2=(v2*(1-m)+2*m*v1)/(m+1);
v1=newv1; v2=newv2; x1(j+1)=x1(j); x2(j+1)=x2(j);
collision=0; j=j+1;
end
end
t=linspace(0, h*steps, steps); plot(t, x1); plot(t, x2, 'red');
endfunction

Update: this post was referred to from Math StackExchange, with a request for a more intuitive explanation of how the scaling by $\sqrt{M}$ works. Let $V_c$ be the velocity vector of the point in the configuration space. Thanks to the scaling, we have the following:

• The magnitude of $V_c$ is equal to the total kinetic energy of the system, up to a constant factor.
• The projection of $V_c$ onto the collision line $x_r=x_b$ is equal to the total momentum of the system, up to a constant factor.

When the balls collide with each other, both the energy and the momentum are preserved. It follows that the new vector $V_c$ will have the same magnitude and the same projection onto the collision line. By trigonometry, this implies the equality of the incident and reflected angles.