ED COPELAND: I'm going to introduce pi in a way that I
don't think many people have seen before.
A friend of mine, David Tong, told me about this.
And I think it's really neat.
And it involves rolling balls at each other.
Our thought experiment involves just two balls.
So there's a small ball, and we'll give it a mass little m.
And there's a big ball.
And we'll give this mass a big M. This is actually a 10-pin
bowling ball which I believe you stole.
Now, the idea is that this ball, the big ball, is rolling
with some initial velocity.
And it hits the little ball.
Now the little ball moves off and hits the wall.
And it bounces elastically.
That means that when it bounces it doesn't lose any
kinetic energy or any momentum.
It bounces back.
Meanwhile, the big ball is carrying on.
It hits the big ball.
And then, of course, it bounces back.
The big ball--
it's imparted a little bit of momentum to the big ball,
which is now slowing down.
It goes back, bounces again, comes back, bounces back, goes
back, and just keeps doing this.
Eventually, the big ball slows down enough that it stops and
at the next collision will begin to go back.
OK?
So the system is the ball's coming in, there's a number of
collisions, and then the big ball eventually
bounces and goes back.
What's that got to do with pi?
The question is how many collisions between the big
ball and the little ball before the big ball starts
going the other way.
OK?
It's as simple as that.
Now as long as the mass of this big ball is quite
particular.
In fact, it's given as--
in terms of the little mass, it's given as 16 times 100 to
the power N, where I'm going to let this number N vary.
So if N is 0, the big ball is 16 times the mass of the
little ball.
If n is 1, the big ball is 1,600 times the mass of the
little ball.
If N is 2, the mass is 160,000 times that of the little ball.
So as long as you've got this very particular mass, then it
turns out that the number of collisions between the big
ball and the little ball is the first N
plus 1 digits of pi.
So if N equals 0, then the number of--
so the big ball is 16 times the mass of the little ball,
there'll be three collisions.
If N is 1, so the big ball is 1,600 times that of the little
ball, and so there will be 31 collisions.
If N is 3, there'll be 314 collisions.
If N is 4, there'll be 3,141 collisions.
What's remarkable about this, I think, is
that this is exact.
As long as N is an integer and you let it get bigger and
bigger and bigger, you will get exactly the relevant
number of digits of pi.
So if this was three grams, and this is about 4.8
kilograms, then there would be of order 31 collisions.
So if I scale this up, so this was a boulder of mass 480
kilograms, then there should be 314 collisions before the
boulder begins to go backwards.
BRADY HARAN: I guess these things are round and have
circumferences and pi, so is that where this
is all coming from?
ED COPELAND: It's got nothing to do with
that, which is fantastic.
In fact, if this was a point object, with very massive--
and that was a point object, the same
thing would still apply.
The fact that it's round has nothing to do with it.
What is important is that the collision is elastic, that it
conserves energy and conserves and momentum.
But it could be any shaped object that would do that.
BRADY HARAN: So this has nothing to do with circles.
ED COPELAND: This has nothing at all to do with the circle
of this ball and the circle of this ball.
BRADY HARAN: So if that was a cube or something else--?
ED COPELAND: I think, yes, that would still work.
As long as we're in the collision, you've conserved
energy and momentum.
BRADY HARAN: Ahh, I think you're going to have to
explain that.
ED COPELAND: OK.
Here we go.
Let's try and make sense of two balls hitting each other
and how we get pi out of it.
Let's just draw it again.
And so the idea is I have some surface and then I have some
wall at the end.
And I have a little ball of, mass, little m, and I a big
ball of mass big M.
And initially, the big ball is moving with some velocity.
Let's call it u.
And it hits the little ball.
And then what happens is the little ball begins to go
towards the wall.
Meanwhile, the big ball carries on.
The big ball is still moving.
The little ball hits the wall and comes back.
And then it hits the big ball again.
And so the little ball bounces back, hits this
wall, bounces back.
I mean each time the little ball hits the big ball, it
transfers a bit of its momentum over to the big ball.
Then the big ball will slow down.
And in fact, eventually, the big ball will stop and will
begin to go the other way.
And what we want to know is if that happens after, say, p
steps, then how does p--
the number of collisions--
depend upon the mass of the big ball?
Now what's important here-- the only bit that I've had to
throw in that I must say Brady wasn't that impressed with
when I told him, but still I think it's neat--
is that there's a very particular value for the big
ball that we're interested in.
The mass of the big ball is 16 times 100 to the power N times
the mass of the little ball.
So if N equals 0, then the big ball is just 16 times the mass
of the little ball.
If N equals 1, then the big ball is 1,600 times the mass
of the little ball.
Then you get the idea.
If N equals 2, then the big ball is 160,000 times the mass
of the little ball.
So rapidly--
BRADY HARAN: It jumps up in hundreds.
ED COPELAND: It jumps up in hundreds.
That's right.
BRADY HARAN: Not tens.
ED COPELAND: Not tens.
Now, in each case, you can ask how many steps, how many
collisions of the big ball and the little ball are there
before the big ball starts going in the opposite
direction, in other words, before the big ball starts
going backwards.
And the result is remarkable.
So the number of steps when N equals 0 is equal to 3.
So the number of collisions is 3.
The number of collisions when N equals 1 is 31.
And the number of collisions when N equals 2 is 314.
And it goes on.
It goes on so that for the case of a general N, the
number of collisions is the first N plus 1 digits of pi,
pi being 3.141 dot dot dot dot.
And it's true for any N. So as N goes towards infinity, you
just keep recovering the first N plus 1 digits of pi.
It's not an expansion or anything like that.
It's just the expression that you get.
BRADY HARAN: And nothing to do with circles.
ED COPELAND: Nothing to do with circles.
You can work out this, and the key ingredients that you need,
which we do at school, is you need to remember that we have
conservation of energy in the collision and
conservation of momentum.
That's all we need to make use of.
And when you've got that, you can work out the velocity of
the part of the big ball after N collisions.
It's possible to do that.
And then the num-- which I'll call Un.
OK?
So after n collisions, the big ball velocity is Un.
And it turns out I can work out what Un is.
Un, after a bit of algebra, Un, in terms of this initial
velocity, U0 is given by the following.
It's given by this fairly simple expression--
I'll tell you what x is in a minute, so don't panic, Brady.
it's given by--
this bit's kind of irrelevant, but the bit that's important
is the following.
There's a cosine of this angle theta.
So x is just the ratio of the little mass to the big mass.
So it's generally going to be very small.
You can see that the largest it's going to be is 1 over 16.
And then it just gets really, really small very quickly.
So x is generally much less than unity.
So cos theta is equal to 1 minus x over 1 plus x.
So everything is in terms of x, this ratio.
So how can I see when this ball is going to start going
in the opposite direction?
And all I need to do that is to remember what--
sorry, that's calls cos n theta--
I missed that one important n there--
that's cos of n theta, not cos theta.
All I need to do is look at this cosine function.
We know what the cosine looks like from our school days.
Cosine looks like the following.
It's got, as a function of some angle, alpha.
The cosine of alpha looks like this.
It goes between 1 and minus 1.
And it changes when alpha is pi by 2.
It goes from positive values to negative values when alpha
is pi by 2.
BRADY HARAN: That's our smoking gun!
ED COPELAND: There it is, Brady.
Pis are coming in.
So when n theta or alpha is less than pi by
2, cosine is positive.
So Un is positive.
That means it's going in that direction.
But when n theta gets bigger than pi by 2, cos
n theta goes negative.
So Un goes negative.
And it's beginning to go in that direction.
So when N is 0, it's the integer part of 10 to the 0
times pi, which is just pi.
And so the integer part of that is 3.
When N is 1, the number of collisions is the integer part
of 10 times pi, which is 31.
And when N is 2, it's the integer part of 100 times pi,
which is 314.
And this is true for any value of N.
BRADY HARAN: And the light went off.
Perfect.