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

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

wall.

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

(點開完整訊息來看我喜歡的)

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？

place, and in such an unexpected manner?

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.

v1和v2會是碰撞過程中的變數

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

to represent the equations.

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.

同理，使y座標為(√m2)*v2

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

此時位於圓的最左側的點

one?

x的座標是負值 y的座標為0

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

此時就要用動量守恆了

-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

of momentum looks like in this diagram.

在座標圖中，就像是對x軸鏡射

This gives us a very satisfying picture of how we hop around on our picture, where you

因為y座標的值乘了(-1)

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

也就是它們不會再發生碰撞

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

shows the digits of pi?

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

這些線在鉛直與斜線兩者交替

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

總碰撞次數會有 π 的數字？

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

the circle to the end zone.

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

這個定理永遠都會是正確的

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

all of these arcs must also be 2*theta.

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

with a previous one.

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

In other words, how many times do you have to add 2*theta to itself before it covers

直到觸碰到停止區域就停止，別忘了停止區域代表 兩個方塊都向右，但小方塊速度較小

more than 2*pi radians? The answer to this is the same as the number of collisions between

不過你也可以把它想像成：再加一個 長 2θ 的圓弧會導致它覆蓋到另外一個圓弧時的停止點

our blocks.

換句話說，在覆蓋超過整個圓 2π 前 你可以加多少個 2θ ？

Or, simplifying things a little, what's the largest integer multiple of theta that

這個問題的答案與方塊的碰撞次數一樣

doesn't surpass pi?

更簡潔的說，在不大於 π 的情況下 θ 的最大整數倍是多少？

For example, if theta was 0.01 radians, then multiplying by 314 would put you a little

舉個例子，如果 θ = 0.01 弧度 那麼將它乘以314會使之小於 π

less than pi, but multiplying by 315 would bring you over that value. So the answer would

但乘以315，則會超過 π

be 314, meaning if our mass ratio were one such that the angle theta in our diagram was

所以答案會是314 也就是如果有某個質量比值使得圖中的 θ = 0.01

0.01, the blocks would collide 314 times.

那麼方塊的碰撞次數將會是314次

In fact, let's go ahead and compute theta, say when the mass ratio is 100 : 1. Remember

現在，你知道該怎麼做了 那麼我們現在來真正的計算 θ 的值

that the rise-over-run slope of this constant momentum line is -sqrt(m1/m2), which in this

假設方塊的質量比為 100：1

example is -10. That would mean the tangent of this angle theta, opposite over adjacent,

別忘了，這條動量守恆線的上升量除以橫移量的 斜率的值為 -√(m1/m2)

is that run over the negative rise, which is 1/10 in this example. So theta = arctan(1/10).

在這個例子中，該值為 -10

In general, it'll be the inverse tangent of the square root of the small mass over

這代表著 tan(θ) 也就是對邊除以鄰邊 將會是上升量除以負的橫移量

the square root of the big mass.

在這例子中，該值為1/10

If you go an plug these into a calculator, you'll notice that the arctan of each such

所以 θ = arctan(1/10)

small value is quite close to the value itself. For example, arctan(1/100), corresponding

更明確的說， θ 會是 小質量除以大質量的平方根的反正切值(arctan)

to a big mass of 10,000 kilograms, is extremely close to 0.01.

如果你用計算機輸入這些數值 你會發現這些反正切值會跟質量比值的平方根很接近

In fact, it's so close that for the sake of our central question, it might as well

舉個例子， arctan(1/100) 對應到了大質量為 10,000 kg

be 0.01. That is, analogous to what we saw a moment ago, adding this to itself 314 times

該值會非常接近 0.01

won't surpass pi, but the 315th time would. Remember, unraveling why we're doing this,

事實上，為了解開我們最主要的問題 可以將它看成 0.01

that's a way of counting how many of our jumps on the phase diagram gets to the end

也就是說，就像剛剛看到的那樣 把它加到314次

zone, which is a way of counting how many times the blocks collide until they're sailing

該值還是不會超過 π 但如果加到315次，則會大於 π

off never to touch again. So that's why a mass ratio of 10,000 gives 314 collisions.

別忘了，我們現在正在做的 是在計算於相圖中，在結束區域前發生了多少次跳躍

Likewise a mass ratio of 1,000,000 to 1 will give an angle of arctan(1/1,000) in our diagram.

也就是在計算方塊發生了多少次的碰撞

This is extremely close to 0.001. And again, if we ask about the largest integer multiple

我的朋友，這就是為什麼當質量比值為10,000時 會發生314次的碰撞

of this theta that doesn't surpass pi, it's the same as it would be for the precise value

同樣方法，當質量比為 1,000,000：1 時 角度 θ 的值會是 arctan(1/1,000)

of 0.001: 3,141. These are the first four digits of pi, because that is by definition

這個值會非常的接近 0.001

what the digits of pi mean. This explains why with a mass ratio of 1,000,000, the number

一樣，我們要找出在不大於 π 的情況下 角度 θ 的最大整數倍是多少

of collisions is 3,141.

而該角度 θ 可以視為 0.001

All this relies on the hope that the arctan of a small value is sufficiently close to

最大整數倍會是 3,141 也就是定義上 π 的前四個數字

the value itself, which is another way of saying that the tangent of a small value is

這解釋了為什麼當質量比值是1,000,000時 碰撞次數是 3,141 次

approximately that value. Intuitively, there's a nice reason this is true. Looking at a unit

你或許會發現，這一切都是因為小數值的反正切值 會跟它本身非常接近

circle, the tangent of any given angle is the height of this little triangle divided

也可以講成，小數值的正切值(tan) 大約等於它本身

by its width. When that angle is really small, the width is basically 1, and the height is

有個非常直觀的原因來解釋為何會這樣

basically the same as the arc length along the circle, which by definition is theta.

如果有個單位圓 某個角度的正切值會等於這個小三角形的高除以底

To be more precise about it, the Taylor series expansion of tan(theta) shows that this approximation

當角度非常小的時候 底的長度會是 1 ，也就是圓的半徑

will only have a cubic error term. So for example, tan(1/100) differs from 1/100 by

而高的長度基本上會是圓弧的弧長 根據定義該弧長等於 θ

something on the order of 1/1,000,000. So even if we consider 314 steps with this angle,

更精確點的話 tan(θ) 的泰勒級數展開式當中

the error between the actual value of arctan(1/100) and the approximation of 0.01 won't have

只會有立方項的誤差項 舉個例子來說

a chance to accumulate enough to be significant.

tan(1/100) 與 1/100 只差了約 1/1,000,000 所以當我們在算該角度的跳躍數時

So, let's zoom out and sum up: When blocks collide, you can figure out how their velocities

實際值 arctan(1/100) 與大約值 1/100 之間的誤差 不會大到影響該最大整數倍

change by slicing a line through a circle in a velocity phase diagram, each curve representing

好的，先總結一下 當方塊碰撞的時候

a conservation law. Most notably, the conservation of energy plants the circular seed that ultimately

你可以在速度的相圖中的圓上畫一條直線 來了解碰撞後的速度情形

blossoms into the pi we find in the final count.

而這個圓和直線都代表著不同的守恆定律

Specifically, due to some inscribed angle geometry, the points we hit of this circle

特別要注意的是，能量守恆定律的圓

are spaced out evenly, separated by the angle we were calling 2*theta. This lets us rephrase

是為什麼最後會有 π 的主要原因

the question of counting collisions as instead asking how many times we must add 2*theta

明確來說，由於圓周角定理 圓上的每個點都將圓平均的分割

to itself before it surpasses 2pi.

都以 2θ 的角度分隔開

If theta looks like 0.001, the answer to that question has the same first digits as pi.

因為這點，我們的問題從計算碰撞次數

And when the mass ratio is some power of 100, because arctan(x) is so well approximated

變成了我們可以加幾次 2θ 而都不會超過 2π

by x for small values, theta is sufficiently close to this value to give the same final

如果 θ 是 0.001 則該問題的解答會與 π 的前4位數字相同

count. Setup for next video

如果當質量的比值是100的某次方時 由於 x 極小的情況下 arctan(x) 會約等於 x

I'll emphasizes again what this phase space allowed us to do, because this is a lesson

因此 θ 會大約等於 arctan(θ) 所以答案不會影響

useful for all sorts of math, like differential equations, chaos theory, and other flavors

我要再次強調相空間可以讓我們做到哪些事情

of dynamics: By representing the relevant state of your system as a single point in

因為像我剛說的，相空間在各種的數學領域中都用的到 像是微分方程、混沌理論，以及其他種類的動力學

an abstract space, it lets you translate problems of dynamics into problems of geometry.

它將一個系統中的相關狀態轉化在抽象空間成單一的點 讓問題從動力學問題變成了幾何學問題

I repeat myself because I don't want you to come away just remembering a neat puzzle

我再重複一次是因為我不希望你離開時，只記得 「為什麼 π 會出現在令人意想不到的地方」的美麗題目

where pi shows up unexpectedly, I want you to think of this surprise appearance as a

我希望你也不要忘記 這道題目背後更深層的關係

distilled remnant of the deeper relationship at play.

如果這個解答讓你很滿意的話 那就不應該了，因為還有另外一種解法

And if this solution leaves you feeling satisfied, it shouldn't. Because there is another perspective,

比這個解法還要更聰明、更漂亮 Galperin 在他的原始論文中

more clever and pretty than this one, due to Galperin in the original paper on this

提到了要我們在「方塊的運動情形」 與「光束在兩個鏡子間的反射情況」兩者做比較

phenomenon, which invites us to draw a striking parallel between the dynamics of these blocks,

相信我，我已經省略很多了

and that of a beam of light bouncing between two mirrors. Trust me, I've saved the best

所以希望能在下一個影片再次看到你

for last on this topic, so I hope to see you again next video.