Name: 為什麼碰撞塊會計算pi？ (Why do colliding blocks compute pi?)
Uploaded: 2021-01-14T11:03:56.000Z
Duration: 15 min 16 s

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

在上一支影片當中，我提出了一個問題 有兩個滑動的方塊

idealized world where there's no friction, and all collisions are perfectly elastic,

在沒有摩擦力，且所有的碰撞 皆為完全彈性碰撞的理想情況下

關係子句

meaning no energy is lost. One block is sent towards another smaller one, which starts

也就是沒有任何的能量散失 其中一個大方塊撞向另外一個維持不動的小方塊

off stationary, and there's a wall behind it so that the small one bounces back and

並且左側有面牆 導致小方塊會不斷地反彈回去

forth until it redirects the big block's momentum enough to outpace it away from the

直到小方塊的速度追不到大方塊

如果大方塊的質量比小方塊還要大100的某次方倍時

If that first block has a mass which is some power of 100 times the mass of the second,

例如1百萬倍(100^3) 會有個非常驚人的事實

for example 1,000,000 times as much, an insanely surprising fact popped out: The total number

碰撞的總數，包含了小方塊撞到牆壁的次數

of collisions, including those between the second mass and the wall, has the same starting

會跟圓周率 π 的前幾位數字相同 在這個例子中，發生了3,141次碰撞

digits as pi. In this example, that's 3,141 collisions.

如果大方塊的質量是小方塊的1萬億倍(100^6) 總共會有3,141,592次的碰撞

If it was one trillion times the mass, it would take 3,141,592 collisions before this

幾乎所有的碰撞都在一個爆發中發生

happens, almost all of which happen in one huge burst.

提到了爆發，在上支影片發布後

Speaking of unexpected bursts, in the short time since that video lots of people have

在很短的時間內，有非常多的網友們 分享了他的解法、嘗試和模擬，我覺得很好

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

所以為什麼會這樣？ 為什麼 π 會在這個意想不到的地方出現？

並以這種意想不到的方式出現？

First and foremost this is a lesson about using a phase space, also commonly called

首先，我們要利用相空間(又被稱為位形空間) 來解決這個問題

a configuration space, to solve problems. So rest assured that you're not just learning

所以請放心，你不僅僅是在學 π 的深度算法

about an esoteric algorithm for pi, the tactic here is core to many other fields.

這個手法是其他許多領域的核心 是一個適合牢牢記住的好用工具

To start, when one block hits another, how do you figure out how the velocity of each

一開始，當第一個方塊撞到另外一個方塊時 你要怎麼知道它們碰撞後的速度分別是多少？

one after the collision? The key is to use the conservation of energy, and the conservation

關鍵是要利用「能量守恆」以及「動量守恆」

of momentum. Let's call their masses m1 and m2, and their velocities v1 and v2, which

我們先假設兩個方塊的質量分別是m1和m2 而速度分別是v1及v2

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

無論什麼時候，動能總和都會是 (½)m1(v1)^2 + (½)m2(v2)^2

v1 and v2 will change as the blocks get bumped around, the value of this expression must

所以說，雖然v1和v2會因為方塊的碰撞而改變 但是這個式子的值一定會是一個常數

remain constant. The total momentum of the two blocks is m1*v1 + m2*v2. This also remains

兩個方塊的動量總和是 m1*v1 + m2*v2

constant when the blocks hit each other, but it can change as the second block bounces

當兩個方塊互撞時，這個式子的值也會是一個定值

off the wall. In reality, that second block would transfer its momentum to the wall during

但當小方塊撞到牆壁的時候，這個式子的值就會改變

this collision. Again we're being idealistic, say thinking of the wall as having infinite

在現實中，碰撞到牆壁時 小方塊會將自己的動量傳給牆壁

mass, so such a momentum transfer won't actually move the wall.

但再重申一次，我們是理想主義者 假設牆壁有著無窮大的質量

So we've got two equations and two unknowns. To put these to use, try drawing a picture

因此當動量轉移給牆壁時 並不會讓牆壁有所移動

我們目前有兩個方程式和兩個未知數

You might start by focusing on this energy equation. Since v1 and v2 are changing, maybe

要解聯立的話，可以畫兩個方程式的座標圖

you think to represent this equation on a coordinate plane where the x-coordinate represents

你或許會先從能量守恆的式子開始下手 因為v1和v2是變數

v1, and the y-coordinate represents v2. So individual points on this plane encode the

所以在座標平面上表示該式 其中x座標為v1，y座標代表v2

pair of velocities of our block. In that case, the energy equation represents an ellipse,

因此在這個座標上的各點 代表了兩個方塊的速度

where each point on this ellipse gives you a pair of velocities, and all points of this

在這個情況下，能量守恆的式子會是一個橢圓

ellipse correspond to the same total kinetic energy.

在橢圓上每個點的座標都代表一對速度

In fact, let's actually change our coordinates a little to make this a perfect circle, since

而所有的點都有著相同的總動能

we know we're on a hunt for pi. Instead of having the x-coordinate represent v1, let

事實上，我們把座標稍微改變一點 讓橢圓變成一個圓

it be sqrt(m1)*v1, which for the example shown stretches our figure in the x-direction by

因為我們要知道 π 是從哪裡來的

sqrt(10). Likewise, have the y-coordinate represent sqrt(m2)*v2. That way, when you

因此不要使x座標為v1 而是(√m1)*v1

look at this conservation of energy equation, it's saying ½(x^2 + y^2) = (some constant),

在這個例子中 會在x方向將圖形拉長了√10倍

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

這樣的話，能量守恆的式子就會變成 ½(x^2 + y^2) 等於一個常數

stationary, we are at this leftmost point on the circle, where the x-coordinate is negative

也就是圓的方程式 圓的大小取決於總能量，不過在這個問題中並沒有差別

and the y-coordinate is 0. What about after the collision, how do we know what happens?

一開始，大方塊向左滑動，而小方塊保持靜止

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,

那在碰撞之後呢？ 我們要怎麼知道發生了什麼事？

the value m1*v1 + m2*v2 must stay constant. In our rescaled coordinates, that looks like

能量守恆告訴了我們每個點只能在這個圓上移動 但是是在哪一點？

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

動量守恆告訴我們 m1*v1 + m2*v2 的值一定等於一個常數

know it must pass through our first point, which locks us into place.

在我們重新調整的座標中 這個式子會變成 (√m1)x + (√m2)y 等於一個常數

Just to be clear what all this is saying: All other pairs of velocities which would

也就是直線的方程式 一條斜率為 -√(m1/m2) 的直線

give the same momentum live on this line, just as all other pairs of velocities which

你或許會問那條線的確切位置在哪 這要取決於該方程式的常數決定

give the same energy live on our circle. So notice, this gives us one and only one other

但確定的是，它一定會經過第一個點 所以只剩下一種可能

point that we could jump to. And it should make sense that it's something where the

先將這一切給弄清楚，在這條線上， 每對速度都有著相同的總動量

x-coordinate gets a little less negative and the y-coordinate is negative, since that corresponds

就像是在這個圓上 每對速度都有著相同的總能量

to our big block slowing down a little while the little block zooms off towards the wall.

因此請注意，這裡給了我們一個 也是唯一一個可以跳過去的點

When the second block bounces off the wall, it's speed stays the same, but will go from

應該滿有道理的，因為x座標的值變大了一些 而y座標的值變成了負值

negative to positive. In the diagram, this corresponds to reflecting about the x-axis,

對應了大方塊速度降低了一些 而小方塊往牆壁方向前進

since the y-coordinate gets multiplied by -1. Then again, the next collision corresponds

從這裡開始，看看事情如何發展還滿有趣的 當小方塊撞到牆壁反彈的時候

to a jump along a line of slope -sqrt(m1 / m2), since staying on such a line is what conservation

它的速度量值會維持一樣 但會由負值變成正值

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

接下來，下一次的碰撞對應了 沿著斜率為 -√(m1/m2) 的直線的點跳躍

velocity of the big one, meaning they'll never touch again. That corresponds to this

因為在該線上移動就是動量守恆的式子在圖形中的模樣

region of the diagram, so in our process, we keep bouncing until we land in that region.

從這裡，你可以藉由方塊的碰撞情形 將圓裡剩餘的部分給填滿

What we've drawn here is called a “phase diagram”, which is a simple but powerful

持續這個步驟，直到小方塊的速度是正值 並且小於大方塊的速度

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

這對應到了座標圖中，右上角這個三角形的區域 因此在過程中，點會不斷地跳躍，直到落到該區域內

that it turns questions about dynamics into questions about geometry. In this case, the

我們畫的這張圖稱做「相圖」 它是一個簡單但強大的數學概念

dynamical idea of all pairs of velocities that conserve energy corresponds to the geometric

你可以將某個系統的狀態給繪製出來 在這個例子中，也就是滑動方塊的速度

object of a circle, and counting the total number of collisions turns into counting the

並在一個抽象的空間中以點的形式出現

number of hops along these lines, alternating between vertical and diagonal.

強大的地方在於，它將動力學的問題 變成了幾何的問題

Specifically, why is it that when the mass ratio is a power of 100, that number of steps

在這個情況下，在動力學中所有可能的每對速度 都必須維持能量守恆，對應到了在幾何中的圓

計算方塊碰撞的次數變成了計算 點沿著這些線的跳躍次數

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

不過問題還是沒有解答 為什麼當質量比值為100時

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

如果你看著這張圖，或許，只是或許

你會發現在圓上每個點跟點之間的圓弧的長度 都會是一樣的

The key here is to use the ever-helpful inscribed angle theorem, which says that whenever you

雖然這件事情並非顯而易見的 但這件事情代表了

form an angle using three points on a circle P1, P2 and P3 like this, it will be exactly

只要計算了一個圓弧的長度 就可以知道需要幾次的碰撞才能夠到達結束區域

half the angle formed by P1, the circle's center, and P3. P2 can be anywhere on this

這裡的關鍵就是利用永遠有用的「圓周角定理」

circle, except in that arc between P1 and P3, and this fact will be true.

圓周角定理表示，當你用三個在圓上的點 P1、P2、P3形成一個角度時

So now look at our phase space, and focus specifically on three points like these. Remember

該角度剛好會是由 P1、圓心、P3 所形成的角度

this first vertical hop corresponds to the small block bouncing off the wall, and the

P2 可以在圓上任何一點 除了 P1 與 P3 之間

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”.

現在回到我們的相空間 並專注於這三個點

Then using the inscribed angle theorem, the arc length between these bottom two points,

第一個鉛直跳躍代表了小方塊撞到牆壁反彈

measured in radians, will be 2*theta. Notice, since this momentum line has the same slope

第二個沿著斜率為 -√(m1/m2) 的斜線的跳躍 代表了方塊碰撞時的動量守恆

for all of those jumps from the top of the circle to the bottom, the same reasoning means

我們把動量守恆線和鉛直線之間所形成的角度稱做 θ

現在你大概發現了 利用圓周角定理

So for each hop, if we drop down a new arc, like so, then after each collision we cover

下方兩點所形成的圓弧的長度(單位為弧度) 將會是 2θ

another 2*theta radians of the circle. We stop once we're in this endzone, corresponding

很重要的是，所有從圓上方跳到圓底部的跳躍 都沿著相同斜率的動量守恆線

to both blocks moving to the right, with the smaller one going slower. But you can also

因此，這些圓弧的長度都會是 2θ

think of this as stopping at the point when adding another arc of 2*theta would overlap

所以，對於每個跳躍 如果我們放了一個新的弧，相這樣

那麼在每次的碰撞後，我們在圓上 再放另外一個長 2θ 的圓弧