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Last video I left you with a puzzle. The setup involves two sliding blocks in a perfectly
idealized world where there’s no friction, and all collisions are perfectly elastic,
meaning no energy is lost. One block is sent towards another smaller one, which starts
off stationary, and there’s a wall behind it so that the small one bounces back and
forth until it redirects the big block’s momentum enough to outpace it away from the
If that first block has a mass which is some power of 100 times the mass of the second,
for example 1,000,000 times as much, an insanely surprising fact popped out: The total number
of collisions, including those between the second mass and the wall, has the same starting
digits as pi. In this example, that’s 3,141 collisions.
If it was one trillion times the mass, it would take 3,141,592 collisions before this
happens, almost all of which happen in one huge burst.
Speaking of unexpected bursts, in the short time since that video lots of people have
shared solutions, attempts, and simulations, which is awesome. See the description for
some of my favorites. So why does this happen?! Why should pi show up in such an unexpected
place, and in such an unexpected manner?
First and foremost this is a lesson about using a phase space, also commonly called
a configuration space, to solve problems. So rest assured that you’re not just learning
about an esoteric algorithm for pi, the tactic here is core to many other fields.
To start, when one block hits another, how do you figure out how the velocity of each
one after the collision? The key is to use the conservation of energy, and the conservation
of momentum. Let’s call their masses m1 and m2, and their velocities v1 and v2, which
will be variables changing throughout the process.
At any given moment, the total kinetic energy is (½)m1(v1)^2 + (½)m2(v2)^2. Even though
v1 and v2 will change as the blocks get bumped around, the value of this expression must
remain constant. The total momentum of the two blocks is m1*v1 + m2*v2. This also remains
constant when the blocks hit each other, but it can change as the second block bounces
off the wall. In reality, that second block would transfer its momentum to the wall during
this collision. Again we’re being idealistic, say thinking of the wall as having infinite
mass, so such a momentum transfer won’t actually move the wall.
So we’ve got two equations and two unknowns. To put these to use, try drawing a picture
to represent the equations.
You might start by focusing on this energy equation. Since v1 and v2 are changing, maybe
you think to represent this equation on a coordinate plane where the x-coordinate represents
v1, and the y-coordinate represents v2. So individual points on this plane encode the
pair of velocities of our block. In that case, the energy equation represents an ellipse,
where each point on this ellipse gives you a pair of velocities, and all points of this
ellipse correspond to the same total kinetic energy.
In fact, let’s actually change our coordinates a little to make this a perfect circle, since
we know we’re on a hunt for pi. Instead of having the x-coordinate represent v1, let
it be sqrt(m1)*v1, which for the example shown stretches our figure in the x-direction by
sqrt(10). Likewise, have the y-coordinate represent sqrt(m2)*v2. That way, when you
look at this conservation of energy equation, it’s saying ½(x^2 + y^2) = (some constant),
which is the equation for a circle. Which specific circle depends on the total energy.
At the beginning, when the first block is sliding to the left and the second one is
stationary, we are at this leftmost point on the circle, where the x-coordinate is negative
and the y-coordinate is 0. What about after the collision, how do we know what happens?
Conservation of energy tells us we must jump to some other point on this circle, but which
Well, use the conservation of momentum! This tells us that before and after a collision,
the value m1*v1 + m2*v2 must stay constant. In our rescaled coordinates, that looks like
saying sqrt(m1)*x + sqrt(m2)*y = (some constant), which is the equation for a line with slope
-sqrt(m1/m2). Which specific line depends on what that constant momentum is. But we
know it must pass through our first point, which locks us into place.
Just to be clear what all this is saying: All other pairs of velocities which would
give the same momentum live on this line, just as all other pairs of velocities which
give the same energy live on our circle. So notice, this gives us one and only one other
point that we could jump to. And it should make sense that it’s something where the
x-coordinate gets a little less negative and the y-coordinate is negative, since that corresponds
to our big block slowing down a little while the little block zooms off towards the wall.
When the second block bounces off the wall, it’s speed stays the same, but will go from
negative to positive. In the diagram, this corresponds to reflecting about the x-axis,
since the y-coordinate gets multiplied by -1. Then again, the next collision corresponds
to a jump along a line of slope -sqrt(m1 / m2), since staying on such a line is what conservation
of momentum looks like in this diagram.
This gives us a very satisfying picture of how we hop around on our picture, where you
keep going until the velocity of that smaller block is both positive, and smaller than the
velocity of the big one, meaning they’ll never touch again. That corresponds to this
region of the diagram, so in our process, we keep bouncing until we land in that region.
What we’ve drawn here is called a “phase diagram”, which is a simple but powerful
idea in math where you encode the state of some system, in this case the velocities of
our sliding blocks, as a single point in some abstract space. What’s powerful here is
that it turns questions about dynamics into questions about geometry. In this case, the
dynamical idea of all pairs of velocities that conserve energy corresponds to the geometric
object of a circle, and counting the total number of collisions turns into counting the
number of hops along these lines, alternating between vertical and diagonal.
Specifically, why is it that when the mass ratio is a power of 100, that number of steps
shows the digits of pi?
Well, if you stare at this picture, maybe, just maybe, you might notice that all the
arc-lengths between the points of this circle seem to be about the same. It’s not immediately
obvious that this should be true, but if it is, it means that computing the value of that
one arc length should be enough to figure out how many collisions it takes to get around
the circle to the end zone.
The key here is to use the ever-helpful inscribed angle theorem, which says that whenever you
form an angle using three points on a circle P1, P2 and P3 like this, it will be exactly
half the angle formed by P1, the circle’s center, and P3. P2 can be anywhere on this
circle, except in that arc between P1 and P3, and this fact will be true.
So now look at our phase space, and focus specifically on three points like these. Remember
this first vertical hop corresponds to the small block bouncing off the wall, and the
second hop along a slope of -sqrt(m1 / m2) corresponds to a momentum-conserving block
collision. Let’s call the angle between this momentum line and the vertical “theta”.
Then using the inscribed angle theorem, the arc length between these bottom two points,
measured in radians, will be 2*theta. Notice, since this momentum line has the same slope
for all of those jumps from the top of the circle to the bottom, the same reasoning means
all of these arcs must also be 2*theta.
So for each hop, if we drop down a new arc, like so, then after each collision we cover
another 2*theta radians of the circle. We stop once we’re in this endzone, corresponding
to both blocks moving to the right, with the smaller one going slower. But you can also
think of this as stopping at the point when adding another arc of 2*theta would overlap
with a previous one.
In other words, how many times do you have to add 2*theta to itself before it covers
more than 2*pi radians? The answer to this is the same as the number of collisions between
our blocks.
Or, simplifying things a little, what’s the largest integer multiple of theta that
doesn’t surpass pi?
For example, if theta was 0.01 radians, then multiplying by 314 would put you a little
less than pi, but multiplying by 315 would bring you over that value. So the answer would
be 314, meaning if our mass ratio were one such that the angle theta in our diagram was
0.01, the blocks would collide 314 times.
In fact, let’s go ahead and compute theta, say when the mass ratio is 100 : 1. Remember
that the rise-over-run slope of this constant momentum line is -sqrt(m1/m2), which in this
example is -10. That would mean the tangent of this angle theta, opposite over adjacent,
is that run over the negative rise, which is 1/10 in this example. So theta = arctan(1/10).
In general, it’ll be the inverse tangent of the square root of the small mass over
the square root of the big mass.
If you go an plug these into a calculator, you’ll notice that the arctan of each such
small value is quite close to the value itself. For example, arctan(1/100), corresponding
to a big mass of 10,000 kilograms, is extremely close to 0.01.
In fact, it’s so close that for the sake of our central question, it might as well
be 0.01. That is, analogous to what we saw a moment ago, adding this to itself 314 times
won’t surpass pi, but the 315th time would. Remember, unraveling why we’re doing this,
that’s a way of counting how many of our jumps on the phase diagram gets to the end
zone, which is a way of counting how many times the blocks collide until they’re sailing
off never to touch again. So that’s why a mass ratio of 10,000 gives 314 collisions.
Likewise a mass ratio of 1,000,000 to 1 will give an angle of arctan(1/1,000) in our diagram.
This is extremely close to 0.001. And again, if we ask about the largest integer multiple
of this theta that doesn’t surpass pi, it’s the same as it would be for the precise value
of 0.001: 3,141. These are the first four digits of pi, because that is by definition
what the digits of pi mean. This explains why with a mass ratio of 1,000,000, the number
of collisions is 3,141.
All this relies on the hope that the arctan of a small value is sufficiently close to
the value itself, which is another way of saying that the tangent of a small value is
approximately that value. Intuitively, there’s a nice reason this is true. Looking at a unit
circle, the tangent of any given angle is the height of this little triangle divided
by its width. When that angle is really small, the width is basically 1, and the height is
basically the same as the arc length along the circle, which by definition is theta.
To be more precise about it, the Taylor series expansion of tan(theta) shows that this approximation
will only have a cubic error term. So for example, tan(1/100) differs from 1/100 by
something on the order of 1/1,000,000. So even if we consider 314 steps with this angle,
the error between the actual value of arctan(1/100) and the approximation of 0.01 won’t have
a chance to accumulate enough to be significant.
So, let’s zoom out and sum up: When blocks collide, you can figure out how their velocities
change by slicing a line through a circle in a velocity phase diagram, each curve representing
a conservation law. Most notably, the conservation of energy plants the circular seed that ultimately
blossoms into the pi we find in the final count.
Specifically, due to some inscribed angle geometry, the points we hit of this circle
are spaced out evenly, separated by the angle we were calling 2*theta. This lets us rephrase
the question of counting collisions as instead asking how many times we must add 2*theta
to itself before it surpasses 2pi.
If theta looks like 0.001, the answer to that question has the same first digits as pi.
And when the mass ratio is some power of 100, because arctan(x) is so well approximated
by x for small values, theta is sufficiently close to this value to give the same final
count. Setup for next video
I’ll emphasizes again what this phase space allowed us to do, because this is a lesson
useful for all sorts of math, like differential equations, chaos theory, and other flavors
of dynamics: By representing the relevant state of your system as a single point in
an abstract space, it lets you translate problems of dynamics into problems of geometry.
I repeat myself because I don’t want you to come away just remembering a neat puzzle
where pi shows up unexpectedly, I want you to think of this surprise appearance as a
distilled remnant of the deeper relationship at play.
And if this solution leaves you feeling satisfied, it shouldn’t. Because there is another perspective,
more clever and pretty than this one, due to Galperin in the original paper on this
phenomenon, which invites us to draw a striking parallel between the dynamics of these blocks,
and that of a beam of light bouncing between two mirrors. Trust me, I’ve saved the best
for last on this topic, so I hope to see you again next video.