Brilliant billiard balls ! Pi from the pool.


We all know that mathematics helps us to understand physics or many natural phenomena. But what about using physics to understanding mathematics, specifically a mathematical constant, say the most famous one: \pi?

Well, before I start I must acknowledge my friend Devdatta and give him the full credit for sharing the youtube video that I am posting here right now.

Now you probably have understood what the video tells about. It’s just about exploiting the momentum conservation principle of our everyday physics and the energy conservation property of an elastic collision. What will be the easy pieces to realize the property of an elastic collision? The billiard balls, I guess, come in your mind first. If you hit a billiard motionless ball with another one, you often see that after hitting the ball stops, whereas the motionless ball starts moving. In an ideal situation (perfectly elastic collision) the moving ball transfers its all kinetic energy to the standing one and hence the latter starts moving with the same speed that the first ball initially had.

Now how do we get all the digits \pi has (Note that it was a long chased effort by many mathematicians including the Indian genius Srinivas Ramanujan)? The idea was published in a paper by G. Galperin of the Dept. of Mathematics, Easter Illinois University, USA, with an interesting title : PLAYING POOL WITH \pi (THE NUMBER \pi FROM A BILLIARD POINT OF VIEW).

The paper briefly states the experimental setup, that you might have already seen in the video above, i.e. we hit a billiard ball of mass m with another ball of larger mass M=100^N\, m with a speed fast enough that the heavier ball hits the lighter ball and sends it to the wall. Then the lighter ball gets reflected by the wall and collides the heavier ball again and gets reflected back towards the wall. Under the assumption that all the collisions are elastic (between the balls and between the ball and the wall), there arises a situation when the heavier ball starts moving in the opposite direction (i.e. the last ball-to-ball collision). The paper proves that the number of hits will be always finite and it will be first N+1 digits of \pi ignoring the decimal point.

The way the paper derives the proof is using a configuration space representation for the positions of the two balls: say, if the lighter and heavier balls are at positions x(t) and y(t) respectively at time t, then we can denote the configuration by a point P(t)=(x(t),y(t)) or a position vector \vec P(t)=x(t)\hat x+y(t) \hat y. We also note that collisions happen only when x=y, and y>x\ge 0 at all time before the final hit, if we consider the wall position to be at the origin.

Thus it turns out that we can plot the trajectory of P(t) as multiple reflections inside an angle \alpha=AOB, and we can count the number of hits by counting the number of hits on the line OA and OB.


Galperin showed the number is the nearest integer equal or above \pi/\alpha-1, denoted as the ceiling function: \lceil{ \pi/\alpha}-1\rceil. Galperin showed that \alpha=\arctan \sqrt{m/M}=\arctan\sqrt{10^{-N}}. So the number of hits becomes

\lceil\pi/\text{arctan}\sqrt{10^{-N}}\rceil-1(\simeq 10^N \pi when \gg 1).

One initial check of the formula could be considering the N=0 case, i.e. when
the two balls have the same mass. So we can recall the example of elastic collision that I mentioned in the beginning, an extra piece comes due to hitting the wall. So (i) ball 1 hits the standing ball 2 and stops , (ii) ball 2 hits the wall and reflects back, (iii) ball 2 hits the ball 1 and stops where ball 1 starts moving in the opposite direction: number of hits is 3 and we can verify that indeed \lceil \pi/\text{arctan}-1\rceil=4-1=3.

However, in the video, Prof. Ed Copeland reaches to a similar conclusion after deriving an expression
for the speed of the heavier ball, starting with an initial speed U_0, after n collisions:
U_n =(1+m/M)\sqrt{m/M}U_0 \cos(n\theta), where \cos\theta=(M-m)/(M+m). Now after the last collision (say, when n=p), the heavier ball (see Copeland’s notes)
moves in the opposite direction, U_n changes its sign. But before it changes sign it has to pass through zero, which happens when n\theta=\pi/2 so that \cos(n\theta)=0, thus \pi/2 > p\theta and (p+1)\theta > \pi/2.

However, I couldn’t figure out any published reference of Copeland’s derivation. But do you see any mismatch or difference between Galperin and Copeland? There’s an extra factor 16 in M in Copeland’s work and he also needed the M \gg m assumption too. (I’ll probably try to attach a note in future if I can make a connection between the two approaches successfully).

So if this sounds cool, then you may be ready to say, physics is not all about physics only!

Useful links:

Galperin’s paper.

Galperin’s homepage.

Numberphile’s youtube channel

Devdatta’s blog.

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One thought on “Brilliant billiard balls ! Pi from the pool.

  1. Pingback: Happy Pi Day 2016! (Galperin’s method) | Dave's programming blog

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