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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

  • wall.

  • 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

  • one?

  • 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 verticaltheta”.

  • 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.

Last video I left you with a puzzle. The setup involves two sliding blocks in a perfectly

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B2 US theta m1 circle coordinate v1 momentum

Why do colliding blocks compute pi?

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    kelly posted on 2020/06/06
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