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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?

• 65 1
kelly posted on 2020/06/06
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