So why do colliding blocks compute pi?

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<div id="description" class="style-scope ytd-video-secondary-info-renderer">Part 1: <a class="yt-simple-endpoint style-scope yt-formatted-string" spellcheck="false" href="https://www.youtube.com/watch?v=HEfHFsfGXjs">https://youtu.be/HEfHFsfGXjs</a> Part 3: <a class="yt-simple-endpoint style-scope yt-formatted-string" spellcheck="false" href="https://www.youtube.com/watch?v=brU5yLm9DZM">https://youtu.be/brU5yLm9DZM</a> Home page: <a class="yt-simple-endpoint style-scope yt-formatted-string" spellcheck="false" href="https://www.youtube.com/redirect?redir_token=3laArFe0s1q4z3Tw8khYB5omUVB8MTU3MDkyMzU5MEAxNTcwODM3MTkw&v=jsYwFizhncE&q=https%3A%2F%2Fwww.3blue1brown.com&event=video_description" target="blank" rel="nofollow">https://www.3blue1brown.com</a> Brought to you by you: <a class="yt-simple-endpoint style-scope yt-formatted-string" spellcheck="false" href="https://www.youtube.com/redirect?redir_token=3laArFe0s1q4z3Tw8khYB5omUVB8MTU3MDkyMzU5MEAxNTcwODM3MTkw&v=jsYwFizhncE&q=http%3A%2F%2F3b1b.co%2Fclacks-thanks&event=video_description" target="blank" rel="nofollow">http://3b1b.co/clacks-thanks</a> Many of you shared solutions, attempts, and simulations with me this last week. I loved it! Y'all are the best. Here are just two of my favorites. By a channel STEM cell: <a class="yt-simple-endpoint style-scope yt-formatted-string" spellcheck="false" href="https://www.youtube.com/watch?v=ils7GZqp_iE">https://youtu.be/ils7GZqp_iE</a> By Doga Kurkcuoglu: <a class="yt-simple-endpoint style-scope yt-formatted-string" spellcheck="false" href="https://www.youtube.com/redirect?redir_token=3laArFe0s1q4z3Tw8khYB5omUVB8MTU3MDkyMzU5MEAxNTcwODM3MTkw&v=jsYwFizhncE&q=http%3A%2F%2Fbilimneguzellan.net%2Fbouncing-cubes-and-%25CF%2580-3blue1brown%2F&event=video_description" target="blank" rel="nofollow">http://bilimneguzellan.net/bouncing-c...</a> And here's a lovely interactive built by GitHub user prajwalsouza after watching this video: <a class="yt-simple-endpoint style-scope yt-formatted-string" spellcheck="false" href="https://www.youtube.com/redirect?redir_token=3laArFe0s1q4z3Tw8khYB5omUVB8MTU3MDkyMzU5MEAxNTcwODM3MTkw&v=jsYwFizhncE&q=https%3A%2F%2Fprajwalsouza.github.io%2FExperiments%2FColliding-Blocks.html&event=video_description" target="blank" rel="nofollow">https://prajwalsouza.github.io/Experi...</a> NY Times blog post about this problem: <a class="yt-simple-endpoint style-scope yt-formatted-string" spellcheck="false" href="https://www.youtube.com/redirect?redir_token=3laArFe0s1q4z3Tw8khYB5omUVB8MTU3MDkyMzU5MEAxNTcwODM3MTkw&v=jsYwFizhncE&q=https%3A%2F%2Fwordplay.blogs.nytimes.com%2F2014%2F03%2F10%2Fpi%2F&event=video_description" target="blank" rel="nofollow">https://wordplay.blogs.nytimes.com/20...</a> The original paper by Gregory Galperin: <a class="yt-simple-endpoint style-scope yt-formatted-string" spellcheck="false" href="https://www.youtube.com/redirect?redir_token=3laArFe0s1q4z3Tw8khYB5omUVB8MTU3MDkyMzU5MEAxNTcwODM3MTkw&v=jsYwFizhncE&q=https%3A%2F%2Fwww.maths.tcd.ie%2F%7Elebed%2FGalperin.%2520Playing%2520pool%2520with%2520pi.pdf&event=video_description" target="blank" rel="nofollow">https://www.maths.tcd.ie/_{lebed/Galpe...</a}> For anyone curious about if the tan(x) ≈ x approximation, being off by only a cubic error term, is actually close enough not to affect the final count, take a look at sections 9 and 10 of Galperin's paper. In short, it could break if there were some point where among the first 2N digits of pi, the last N of them were all 9's. This seems exceedingly unlikely, but it quite hard to disprove. Although I found the approach shown in this video independently, after the fact I found that Gary Antonick, who wrote the Numberplay blog referenced above, was the first to solve it this way. In some ways, I think this is the most natural approach one might take given the problem statement, as corroborated by the fact that many solutions people sent my way in this last week had this flavor. The Galperin solution you will see in the next video, though, involves a wonderfully creative perspective. If you want to contribute translated subtitles or to help review those that have already been made by others and need approval, you can click the gear icon in the video and go to subtitles/cc, then "add subtitles/cc". I really appreciate those who do this, as it helps make the lessons accessible to more people. 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And you know the drill with YouTube, if you want to stay posted on new videos, subscribe: <a class="yt-simple-endpoint style-scope yt-formatted-string" spellcheck="false" href="https://www.youtube.com/redirect?redir_token=3laArFe0s1q4z3Tw8khYB5omUVB8MTU3MDkyMzU5MEAxNTcwODM3MTkw&v=jsYwFizhncE&q=http%3A%2F%2F3b1b.co%2Fsubscribe&event=video_description" target="blank" rel="nofollow">http://3b1b.co/subscribe</a> Various social media stuffs: Website: <a class="yt-simple-endpoint style-scope yt-formatted-string" spellcheck="false" href="https://www.youtube.com/redirect?redir_token=3laArFe0s1q4z3Tw8khYB5omUVB8MTU3MDkyMzU5MEAxNTcwODM3MTkw&v=jsYwFizhncE&q=https%3A%2F%2Fwww.3blue1brown.com&event=video_description" target="blank" rel="nofollow">https://www.3blue1brown.com</a> Twitter: <a class="yt-simple-endpoint style-scope yt-formatted-string" spellcheck="false" href="https://www.youtube.com/redirect?redir_token=3laArFe0s1q4z3Tw8khYB5omUVB8MTU3MDkyMzU5MEAxNTcwODM3MTkw&v=jsYwFizhncE&q=https%3A%2F%2Ftwitter.com%2F3blue1brown&event=video_description" target="blank" rel="nofollow">https://twitter.com/3blue1brown</a> Reddit: <a class="yt-simple-endpoint style-scope yt-formatted-string" spellcheck="false" href="https://www.youtube.com/redirect?redir_token=3laArFe0s1q4z3Tw8khYB5omUVB8MTU3MDkyMzU5MEAxNTcwODM3MTkw&v=jsYwFizhncE&q=https%3A%2F%2Fwww.reddit.com%2Fr%2F3blue1brown&event=video_description" target="blank" rel="nofollow">https://www.reddit.com/r/3blue1brown</a> Instagram: <a class="yt-simple-endpoint style-scope yt-formatted-string" spellcheck="false" href="https://www.youtube.com/redirect?redir_token=3laArFe0s1q4z3Tw8khYB5omUVB8MTU3MDkyMzU5MEAxNTcwODM3MTkw&v=jsYwFizhncE&q=https%3A%2F%2Fwww.instagram.com%2F3blue1brown_animations%2F&event=video_description" target="blank" rel="nofollow">https://www.instagram.com/3blue1brown...</a> Patreon: <a class="yt-simple-endpoint style-scope yt-formatted-string" spellcheck="false" href="https://www.youtube.com/redirect?redir_token=3laArFe0s1q4z3Tw8khYB5omUVB8MTU3MDkyMzU5MEAxNTcwODM3MTkw&v=jsYwFizhncE&q=https%3A%2F%2Fpatreon.com%2F3blue1brown&event=video_description" target="blank" rel="nofollow">https://patreon.com/3blue1brown</a> Facebook: <a class="yt-simple-endpoint style-scope yt-formatted-string" spellcheck="false" href="https://www.youtube.com/redirect?redir_token=3laArFe0s1q4z3Tw8khYB5omUVB8MTU3MDkyMzU5MEAxNTcwODM3MTkw&v=jsYwFizhncE&q=https%3A%2F%2Fwww.facebook.com%2F3blue1brown&event=video_description" target="_blank" rel="nofollow">https://www.facebook.com/3blue1brown</a></div>

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<div class="style-scope ytd-metadata-row-renderer">00:03
Last video I left you with a puzzle. The setup involves two sliding blocks in a perfectly
00:08
idealized world where there’s no friction, and all collisions are perfectly elastic,
00:13
meaning no energy is lost. One block is sent towards another smaller one, which starts
00:17
off stationary, and there’s a wall behind it so that the small one bounces back and
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forth until it redirects the big block’s momentum enough to outpace it away from the
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wall.
00:28
If that first block has a mass which is some power of 100 times the mass of the second,
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for example 1,000,000 times as much, an insanely surprising fact popped out: The total number
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of collisions, including those between the second mass and the wall, has the same starting
00:42
digits as pi. In this example, that’s 3,141 collisions.
00:48
If it was one trillion times the mass, it would take 3,141,592 collisions before this
00:56
happens, almost all of which happen in one huge burst.
01:00
Speaking of unexpected bursts, in the short time since that video lots of people have
01:05
shared solutions, attempts, and simulations, which is awesome. See the description for
01:08
some of my favorites. So why does this happen?! Why should pi show up in such an unexpected
01:13
place, and in such an unexpected manner?
01:15
First and foremost this is a lesson about using a phase space, also commonly called
01:20
a configuration space, to solve problems. So rest assured that you’re not just learning
01:24
about an esoteric algorithm for pi, the tactic here is core to many other fields.
01:32
To start, when one block hits another, how do you figure out how the velocity of each
01:35
one after the collision? The key is to use the conservation of energy, and the conservation
01:41
of momentum. Let’s call their masses m1 and m2, and their velocities v1 and v2, which
01:48
will be variables changing throughout the process.
01:52
At any given moment, the total kinetic energy is (½)m1(v1)² + (½)m2(v2)². Even though
02:01
v1 and v2 will change as the blocks get bumped around, the value of this expression must
02:06
remain constant. The total momentum of the two blocks is m1v1 + m2v2. This also remains
02:16
constant when the blocks hit each other, but it can change as the second block bounces
02:20
off the wall. In reality, that second block would transfer its momentum to the wall during
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this collision. Again we’re being idealistic, say thinking of the wall as having infinite
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mass, so such a momentum transfer won’t actually move the wall.
02:35
So we’ve got two equations and two unknowns. To put these to use, try drawing a picture
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to represent the equations.
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You might start by focusing on this energy equation. Since v1 and v2 are changing, maybe
02:47
you think to represent this equation on a coordinate plane where the x-coordinate represents
02:51
v1, and the y-coordinate represents v2. So individual points on this plane encode the
02:57
pair of velocities of our block. In that case, the energy equation represents an ellipse,
03:02
where each point on this ellipse gives you a pair of velocities, and all points of this
03:06
ellipse correspond to the same total kinetic energy.
03:09
In fact, let’s actually change our coordinates a little to make this a perfect circle, since
03:13
we know we’re on a hunt for pi. Instead of having the x-coordinate represent v1, let
03:19
it be sqrt(m1)v1, which for the example shown stretches our figure in the x-direction by
03:25
sqrt(10). Likewise, have the y-coordinate represent sqrt(m2)v2. That way, when you
03:32
look at this conservation of energy equation, it’s saying ½(x² + y²) = (some constant),
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which is the equation for a circle. Which specific circle depends on the total energy.
03:45
At the beginning, when the first block is sliding to the left and the second one is
03:50
stationary, we are at this leftmost point on the circle, where the x-coordinate is negative
03:56
and the y-coordinate is 0. What about after the collision, how do we know what happens?
04:01
Conservation of energy tells us we must jump to some other point on this circle, but which
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one?
04:06
Well, use the conservation of momentum! This tells us that before and after a collision,
04:11
the value m1v1 + m2v2 must stay constant. In our rescaled coordinates, that looks like
04:19
saying sqrt(m1)x + sqrt(m2)y = (some constant), which is the equation for a line with slope
04:29
-sqrt(m1/m2). Which specific line depends on what that constant momentum is. But we
04:37
know it must pass through our first point, which locks us into place.
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Just to be clear what all this is saying: All other pairs of velocities which would
04:47
give the same momentum live on this line, just as all other pairs of velocities which
04:52
give the same energy live on our circle. So notice, this gives us one and only one other
04:57
point that we could jump to. And it should make sense that it’s something where the
05:01
x-coordinate gets a little less negative and the y-coordinate is negative, since that corresponds
05:06
to our big block slowing down a little while the little block zooms off towards the wall.
05:14
When the second block bounces off the wall, it’s speed stays the same, but will go from
05:17
negative to positive. In the diagram, this corresponds to reflecting about the x-axis,
05:26
since the y-coordinate gets multiplied by -1. Then again, the next collision corresponds
05:32
to a jump along a line of slope -sqrt(m1 / m2), since staying on such a line is what conservation
05:39
of momentum looks like in this diagram.
05:41
This gives us a very satisfying picture of how we hop around on our picture, where you
05:48
keep going until the velocity of that smaller block is both positive, and smaller than the
05:54
velocity of the big one, meaning they’ll never touch again. That corresponds to this
05:59
region of the diagram, so in our process, we keep bouncing until we land in that region.
06:07
What we’ve drawn here is called a “phase diagram”, which is a simple but powerful
06:11
idea in math where you encode the state of some system, in this case the velocities of
06:17
our sliding blocks, as a single point in some abstract space. What’s powerful here is
06:22
that it turns questions about dynamics into questions about geometry. In this case, the
06:27
dynamical idea of all pairs of velocities that conserve energy corresponds to the geometric
06:32
object of a circle, and counting the total number of collisions turns into counting the
06:37
number of hops along these lines, alternating between vertical and diagonal.
06:44
Specifically, why is it that when the mass ratio is a power of 100, that number of steps
06:52
shows the digits of pi?
06:54
Well, if you stare at this picture, maybe, just maybe, you might notice that all the
07:00
arc-lengths between the points of this circle seem to be about the same. It’s not immediately
07:05
obvious that this should be true, but if it is, it means that computing the value of that
07:09
one arc length should be enough to figure out how many collisions it takes to get around
07:14
the circle to the end zone.
07:16
The key here is to use the ever-helpful inscribed angle theorem, which says that whenever you
07:21
form an angle using three points on a circle P1, P2 and P3 like this, it will be exactly
07:27
half the angle formed by P1, the circle’s center, and P3. P2 can be anywhere on this
07:34
circle, except in that arc between P1 and P3, and this fact will be true.
07:40
So now look at our phase space, and focus specifically on three points like these. Remember
07:45
this first vertical hop corresponds to the small block bouncing off the wall, and the
07:49
second hop along a slope of -sqrt(m1 / m2) corresponds to a momentum-conserving block
07:56
collision. Let’s call the angle between this momentum line and the vertical “theta”.
08:01
Then using the inscribed angle theorem, the arc length between these bottom two points,
08:07
measured in radians, will be 2theta. Notice, since this momentum line has the same slope
08:13
for all of those jumps from the top of the circle to the bottom, the same reasoning means
08:18
all of these arcs must also be 2theta.
08:24
So for each hop, if we drop down a new arc, like so, then after each collision we cover
08:29
another 2theta radians of the circle. We stop once we’re in this endzone, corresponding
08:35
to both blocks moving to the right, with the smaller one going slower. But you can also
08:39
think of this as stopping at the point when adding another arc of 2theta would overlap
08:45
with a previous one.
08:46
In other words, how many times do you have to add 2theta to itself before it covers
08:52
more than 2pi radians? The answer to this is the same as the number of collisions between
08:58
our blocks.
08:59
Or, simplifying things a little, what’s the largest integer multiple of theta that
09:05
doesn’t surpass pi?
09:07
For example, if theta was 0.01 radians, then multiplying by 314 would put you a little
09:15
less than pi, but multiplying by 315 would bring you over that value. So the answer would
09:21
be 314, meaning if our mass ratio were one such that the angle theta in our diagram was
09:28
0.01, the blocks would collide 314 times.
09:33
In fact, let’s go ahead and compute theta, say when the mass ratio is 100 : 1. Remember
09:43
that the rise-over-run slope of this constant momentum line is -sqrt(m1/m2), which in this
09:51
example is -10. That would mean the tangent of this angle theta, opposite over adjacent,
09:59
is that run over the negative rise, which is 1/10 in this example. So theta = arctan(1/10).
10:07
In general, it’ll be the inverse tangent of the square root of the small mass over
10:16
the square root of the big mass.
10:19
If you go an plug these into a calculator, you’ll notice that the arctan of each such
10:23
small value is quite close to the value itself. For example, arctan(1/100), corresponding
10:30
to a big mass of 10,000 kilograms, is extremely close to 0.01.
10:36
In fact, it’s so close that for the sake of our central question, it might as well
10:40
be 0.01. That is, analogous to what we saw a moment ago, adding this to itself 314 times
10:49
won’t surpass pi, but the 315th time would. Remember, unraveling why we’re doing this,
10:56
that’s a way of counting how many of our jumps on the phase diagram gets to the end
11:01
zone, which is a way of counting how many times the blocks collide until they’re sailing
11:06
off never to touch again. So that’s why a mass ratio of 10,000 gives 314 collisions.
11:14
Likewise a mass ratio of 1,000,000 to 1 will give an angle of arctan(1/1,000) in our diagram.
11:21
This is extremely close to 0.001. And again, if we ask about the largest integer multiple
11:28
of this theta that doesn’t surpass pi, it’s the same as it would be for the precise value
11:34
of 0.001: 3,141. These are the first four digits of pi, because that is by definition
11:43
what the digits of pi mean. This explains why with a mass ratio of 1,000,000, the number
11:48
of collisions is 3,141.
11:57
All this relies on the hope that the arctan of a small value is sufficiently close to
12:02
the value itself, which is another way of saying that the tangent of a small value is
12:06
approximately that value. Intuitively, there’s a nice reason this is true. Looking at a unit
12:12
circle, the tangent of any given angle is the height of this little triangle divided
12:18
by its width. When that angle is really small, the width is basically 1, and the height is
12:25
basically the same as the arc length along the circle, which by definition is theta.
12:32
To be more precise about it, the Taylor series expansion of tan(theta) shows that this approximation
12:38
will only have a cubic error term. So for example, tan(1/100) differs from 1/100 by
12:45
something on the order of 1/1,000,000. So even if we consider 314 steps with this angle,
12:52
the error between the actual value of arctan(1/100) and the approximation of 0.01 won’t have
13:00
a chance to accumulate enough to be significant.
13:03
So, let’s zoom out and sum up: When blocks collide, you can figure out how their velocities
13:09
change by slicing a line through a circle in a velocity phase diagram, each curve representing
13:16
a conservation law. Most notably, the conservation of energy plants the circular seed that ultimately
13:22
blossoms into the pi we find in the final count.
13:25
Specifically, due to some inscribed angle geometry, the points we hit of this circle
13:30
are spaced out evenly, separated by the angle we were calling 2theta. This lets us rephrase
13:36
the question of counting collisions as instead asking how many times we must add 2theta
13:41
to itself before it surpasses 2pi.
13:45
If theta looks like 0.001, the answer to that question has the same first digits as pi.
13:51
And when the mass ratio is some power of 100, because arctan(x) is so well approximated
13:56
by x for small values, theta is sufficiently close to this value to give the same final
14:02
count. Setup for next video
14:03
I’ll emphasizes again what this phase space allowed us to do, because this is a lesson
14:07
useful for all sorts of math, like differential equations, chaos theory, and other flavors
14:12
of dynamics: By representing the relevant state of your system as a single point in
14:17
an abstract space, it lets you translate problems of dynamics into problems of geometry.
14:23
I repeat myself because I don’t want you to come away just remembering a neat puzzle
14:27
where pi shows up unexpectedly, I want you to think of this surprise appearance as a
14:32
distilled remnant of the deeper relationship at play.
14:35
And if this solution leaves you feeling satisfied, it shouldn’t. Because there is another perspective,
14:41
more clever and pretty than this one, due to Galperin in the original paper on this
14:45
phenomenon, which invites us to draw a striking parallel between the dynamics of these blocks,
14:50
and that of a beam of light bouncing between two mirrors. Trust me, I’ve saved the best
14:54
for last on this topic, so I hope to see you again next video.</div>

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So why do colliding blocks compute pi?

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