Playing pool with pi

This is a short note based on the paper

Playing pool with π (the number π from a billiard point of view) by G. Galperin in 2003.

Let’s start with two hard balls,  denoted by $B_1$ and $B_2$, of masses $0 on the positive real axis with position $0, and a rigid wall at the origin. Without loss of generality we assume $m=1$. Then push the ball $B_2$ towards $B_1$, and count the total number $N(M)$ of collisions (ball-ball and ball-wall) till the $B_2$ escapes to $\infty$ faster than $B_1$.

Case. $M=1$: first collision at $y(t)=x$, then $B_2$ rests, and $B_1$ move towards the wall; second collision at $x(t)=0$, then $B_1$ gains the opposite velocity and moves back to $B_2$; third collision at $x(t)=x$, then $B_1$ rests, and $B_2$ move towards $\infty$.

Total counts $N(1)=3$, which happens to be first integral part of $\pi$. Well, this must be coincidence, one might wonder.

However, Galperin proved that, if we set $M=10^{2k}$, then $N(M)$ gives the integral part of $10^k\pi$. For example, $N(10^2)=31$; and  $N(10^4)=314$.

Now let’s consider the coordinates of the two balls $X=\{(x,y):0\le x\le y\}$. Between collision, $(x_t,y_t)$ are moving along straight lines in the configuration space $X$,

– until the two balls collide, which induces a folding of the straight trajectories along the diagonal $\{x=y\}$;

– or until $B_1$ hit the wall, which induces a reflection at the $y$-axis $\{x=0\}$.

To count the total number of foldings, we need to specify how the trajectory folds at the diagonal. Let’s illustrate the special case when $M=m=1$:

The dash lines indicates the trajectory moves back in the same way it comes in. After three collisions,  the trajectory escapes to $\infty$. In the general cases, the trajectory escapes to $\infty$ when the direction lies in  $[\pi/4, \pi/2]$.

Galperin also list an interesting footnote in that paper:  consider two series of identical balls, $i$ of them are on the negative axis with velocity $1$,  and $j$ of them are on the positive axis with velocity $-1$. Then after a few collisions, things are settled that $i$ of them are on the positive axis with velocity $1$,  and $j$ of them are on the negative axis with velocity $-1$. The total number of collisions in this process is $i\cdot j$. The trick is to treat teach collision as a jump. This applies to some more general cases.

Consider two moving balls of masses $m_i$ with velocities $v_i$. At some moment they have a elastic collision and end up with velocities $V_i$. Then the law of conservation of momentum and the law of conservation of energy tell us that

$m_1V_1+m_2V_2=m_1v_1+m_2v_2$,
$m_1V_1^2+m_2V_2^2=m_1v_1^2+m_2v_2^2$

Let $M=m_1 + m_2$ be the total mass, and $P=m_1v_1+m_1v_2$ be the total momentum.
Firstly note that by the translation invariance of the masses, the law of conservation of momentum and the law of conservation of energy are preserved under the transformation $u_i=v_i+a$:
(a) $m_1U_1+m_2U_2=m_1V_1+m_2V_2+Ma$
$=m_1v_1+m_2v_2+Ma=m_1u_1+m_2u_2$;

(b) $m_1U_1^2+m_2U_2^2=m_1V_1^2+m_2V_2^2+2Pa+Ma^2$
$=m_1v_1^2+m_2v_2^2+2Pa+Ma^2=m_1u_1^2+m_2u_2^2$.

A special case when $m_1=m_2$: $V_1=v_2$ and $V_2=v_1$.

To get the general solution, we first change to the coordinate $u_i=v_i-v_1$, and rewrite the following

(1) $m_1U_1+m_2U_2=m_2u_2$,
(2) $m_1U_1^2+m_2U_2^2=m_2u_2^2$.

Then one can easily gets $U_2=u_2-\frac{m_1}{m_2}U_1$ from (1); pluggin to (2), one gets $U_1=\frac{2m_1}{m_1+m_2}u_2$, and hence $V_1=v_1+U_1=v_1+\frac{2m_1}{m_1+m_2}(v_2-v_1)$. Similarly solve $V_2$. Note $1=\frac{m_1 - m_2}{m_1 + m_2}+ \frac{2 m_2}{m_1 + m_2}$. So we can rewrite the solutions as:

$V_{1} = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) v_{1} + \left( \frac{2 m_2}{m_1 + m_2} \right) v_{2}$;
$V_{2} = \left( \frac{m_2 - m_1}{m_1 + m_2} \right) v_{2} + \left( \frac{2 m_1}{m_1 + m_2} \right) v_{1}$.

Let $k=\frac{v_2}{v_1}$ (and $K=\frac{V_2}{V_1}$) be the direction that the trajectory is moving to right before (and after) the ball-ball colision. Then
$K=\frac{2m_1+(m_2-m_1)k}{2m_2 k+(m_1-m_2)}$. Let $\theta$ and $\Theta$ be the angles with the normal direction (the diagonal). So (with some orientation)

$\tan \theta=\frac{k+1}{1-k}$, or $k=\frac{\tan \theta-1}{\tan \theta+1}$ and

$\tan \Theta=\frac{-1-K}{1-K}=1+\frac{2}{K-1}$
$=1+\frac{2}{m_1+m_2}\cdot\frac{2m_2 k+(m_1-m_2)}{1-k}$
$=1+\frac{4m_2}{m_1+m_2}+ \frac{(m_1-3m_2)}{m_1+m_2}\cdot(\tan \theta+1)$.
Tise direct approach sees to be difficult to carry on from here.

In his paper, Galperin used a linear transform that makes the reflection law at the diagonal an elastic one. More precisely, consider the new coordinates $X=\sqrt{m}x, Y=\sqrt{M}y$. That is, we use a more accurate ruler to measure the slow movement of the heavier ball.
Then he checked that the reflections at the two boundaries are elastic:

(1). Ball-Wall: $\sqrt{M}\,V$ stays the same, just $\sqrt{m}\,v\to -\sqrt{m}\,v$. Clearly elastic;

(2). Ball-Ball: note that $\vec{m}=(\sqrt{m},\sqrt{M})$ represents the direction of new boundary, $\vec{v}=(\sqrt{m}\,u,\sqrt{M}\,v)$ represents the direction of the new trajectory; the conservation laws can be rewritten as $P=\vec{m}\cdot\vec{v}$, $\sqrt{2E}=|\vec{v}|$ (before and after collision). Note that $P=\vec{m}^{\pm}\cdot\vec{v}^{\pm}=|\vec{m}^{\pm}|\times|\vec{v}^{\pm}|\cos\phi^{\pm}$: therefore $\phi^+=\phi^-$ does not change after the collision.

The final step is to notice that, for elastic collisions, the foldings of trajectory are the same as the mirror extensions along the boundaries. So we just count the number of rollovers of new space to pass the negative $y$-axis. The new cusp angle is given by $\tan \alpha=\frac{\sqrt{m}}{\sqrt{M}}$. In the case $M=10^{2k}m$ we have $\alpha=\arctan \frac{1}{10^k}\simeq 10^{-k}$ and $N(M)=\left[\frac{\pi}{\arctan 10^{-k}}\right]\simeq [10^k \pi]$.

As remarked by Galperin, the space with coordinates $X=\sqrt{m}x, Y=\sqrt{M}y$ is the corrected, elastic billiard configration space. This supports Arnold’s claim that The notion of the configuration space alone let us solve a very difficult mathematical problem.