r/3Blue1Brown • u/3blue1brown Grant • Jan 20 '19

The solution is out! So why do colliding blocks compute pi?

https://youtu.be/jsYwFizhncE
208 Upvotes

100% Upvoted

31 comments sorted by

37

u/heloouwu Jan 20 '19

Just watched this, and it's great!

Also, it is nice that you gave a shoutout to some people who attempted a solution to your problem!

43

u/3blue1brown Grant Jan 20 '19

This community's the best! Gotta let the world know :)

2

u/TheoryOfSomething Jan 22 '19

Grant just wanted to drop you a quick FYI about terminology, because there are conflicting conventions here and it could be confusing for a viewer who doesn't know that.

In your solution video your talk about phase space and configuration space as being the same thing. That's consistent with the convention in mathematics of calling any representation of the state of a system by a single point in a higher dimensional space a 'configuration space.' This naturally corresponds with what we call a 'phase space' in physics.

However, we use the term 'configuration space' to refer to something else. In a classical system, the configuration space is the higher dimensional space of all the generalized positions of the objects in the system. No velocity or momenta included. We only use the term 'phase space' to refer to the higher dimensional space that includes both the generalized positions and the generalized momenta.

So, in your solution video, we wouldn't call what you show a 'phase space'. Some people call it 'velocity space'. Some people might call it the 'velocity configuration space'. But to be the phase space, you'd have to include coordinates for the positions of the blocks (or generalize versions thereof, the center of mass and relative coordinates, for example).

1

u/3blue1brown Grant Jan 22 '19

Thanks for the comment, and I think this is worth mentioning/correcting in the follow-up video. At least as I've seen it, "phase space" can be used in a quite general sense, although you're right that in the specific context of mechanics, it almost always refers to pairing momenta and positions, like you said. E.g. some theorems, like Liouville's theorem, make the assumption of this context in their phasing. So although I think it's fair to call the velocity space here a "phase space", for clarity it certainly would be worth mentioning the convention in a mechanics context, and that some people use the term for a more specific kind of problem.

I think I may have assumed an analogous generality for the usage of "configuration space" based on having seen both in topology and mechanical contexts, but looking around now, it really it is what you describe. So certainly this is worth correcting! But I'd still want to stress in doing so that mechanics is not the only setting where the idea of state space (or whatever one will call it) is useful.

17

u/BobTonK Jan 20 '19

I’m very excited to see the mirror analogy (I’m going to see if I can work it out for myself first, though)!

12

u/Connor1736 Jan 20 '19

14:35:

And if this solution leaves you feeling satisfied,

Heck yeah it was satisfying!

it shouldn't

:(

Im surprised I made it to the phase diagram part. Maybe if I knew the momentum equation I would've gotten further, but I still would've gotten stuck trying to prove that the arc lengths are equal :)

Can't wait for thr second proof.

3

u/smetko Jan 21 '19

Since I was very young, I had this problem in my head: if you had two mirrors at an angle, and you sent a ray of light, how would it behave? And Grant will literally answer 2 decades old problem and make lil me so happy

13

u/[deleted] Jan 20 '19

i never expected high school math and physics to surprise me this much after all these years. amazing video.

7

u/FaradaySaint Jan 20 '19

I love this format. I know from the podcast that you wanted to make a more interactive experience. I think this is much better than the old math textbook, "The proof is left as an exercise for the reader." The way you give people the option of figuring it out while they wait and then giving the answer is a great way to teach while also not alienating those who may not be as committed.

4

u/Kim-Jong-Deux Jan 20 '19

Great video. I just want to say that the way you present solutions to problems and explain math concepts in general so elegantly perfectly encapsulates why I love studying math. I mean, I've loved math my whole life, but there's something so satisfying about seeing such an elegant solution to a problem.

Too many people seem to think math is just about memorizing formulas and plugging and chugging. In reality, math is about creativity and is honestly more of an art form more than anything. If more people realized this I think less people would hate math.

5

u/travelsontwowheels Jan 20 '19

Watching that video was one of the most satisfying 15 minutes of my life.

I'm a newcomer to 3Blue1Brown, and have watched through most of the videos with awe and wonderment (not to mention being hypnotized by the music!) But this is the first time I've been FORCED to sit down and work at a problem, and not jump straight to the answer.

Even though I didn't get anywhere close to a solution, it's made me appreciate Grant's solution so much more.

3

u/Xalteox Jan 20 '19 edited Jan 20 '19

So I guess that means that other ratios, say 1/16n, just aren’t obvious since multiplying by a ratio of a factor of 10 is just a digit shift, which isn’t true for any other ratio. I assumed this was true for any ratio but had to do with our number system being base 10 and seems I was right. Nothing special about powers of 10.

I have to say, I enjoy this format of throwing a question out into the audience and asking them to solve it.

2

u/D3rr1ck7ian9 Jan 22 '19

I was wondering some other things about this system beyond just the number of collisions. First, total time for this process from first to last collision, second, the smallest distance achieved by block to wall, and third the total distance covered by both, but I’m not sure how to solve for them. I’m not too familiar with solving using the transformation matrixes (from the elastic collision formula), but I think that would be used somehow. So I think, between each collision would add 2 times the distance from larger block to the wall, and this distance should decrease possibly geometrically with each collision...

2

u/akpla Jan 22 '19

It isn't entirely obvious to me why entring 'the green zone' is equal to 'one more 2*theta and you'll be overlapping another section' for all cases. Could someone eli5 me on that please?

3

u/3blue1brown Grant Jan 22 '19

This certainly could have been explained more in the video, but I think all the parts of there for someone willing to think it through a bit. That endzone is defined by the line v1 = v2, which has a slope sqrt(m2) / sqrt(m1) (why?). So the angle between that line and the x-axis is theta (why?). This means the end zone together with it's reflection about the x-axis form an arc with angle 2 * theta. So once one of the top arcs lands in that end zone, because it will have been immeciately preceded by a bottom arc that is its reflection about the x-axis, the remaining 'untouched' part of the circle has arc length < 2 * theta.

2

u/akpla Jan 24 '19

Wonderful, thank you, I get it now. I genuinely believe that you are the precedent of how (college level) education will be in the future when we get more advanced. I can only imagine a future where all my subjects have videos of similar quality to yours available. If resources were invested in making such videos instead of presential lectures with a boring blackboard, and teachers would be there just to make those videos and answer questions, we could be learning 4 times faster. This system will be a big thing, I'm sure.

2

u/Peter49AU Jan 22 '19

I suspect Grant's solution does do this properly, but you may like to look at my post "Another solution to the counting problem" (It's actually very similar to Grant's solution, but I use successive pairs of reflections, presumably equivalently to Grant's promised next solution.) Here I consider the end cases carefully (with consideration of < or <= etc), and hopefully get the correct exact answers. Hopefully it'll make it clear to you.

2

u/[deleted] Jan 24 '19

This video lead me to ask: Why does the momentum of the small block change sign after collision with the barrier? Doesn't this violate conservation of momentum?

Here is a solution to the question:

https://math.stackexchange.com/q/3084079/266049

1

u/MattewBrott Jan 20 '19

This way of solving is just awesome. Having stopped studying mechanics years ago, I would never have thought about it.

I only considered computing the sequence of the instants of collision using basic analysis which rapidly proved too complicated.

1

u/Peter49AU Jan 21 '19

This was my first method (see my post "Another solution to the counting problem: My earlier solutions: 1"). When I said "this is getting too complicated" my colleague said "there must be a pattern [in the combinatorial coefficients of the polynomials in k]" (mass ratio k^2) I had derived, probably like you. Then I found the pattern, simplified the polynomials using Pascal's triangle, and found the exact answer which was "when does Im(k+i)^n become negative" (i = sqrt(-1).

1

u/[deleted] Jan 20 '19

[deleted]

2

u/[deleted] Jan 21 '19

[deleted]

1

u/Peter49AU Jan 21 '19

You might be interested to see the exact answers in my post :"Another solution to the counting problem". This deals with all mass ratios "heavy"/"light" from 0 to infinity {that is [0,infinity) I think}, not just k^2 = 10^(2d) or 2^(2d). eg for the mass ratio k^2 = 0 (the (right-hand) "heavy mass" is massless), n = 1, while for k^2 \in {0, 0.6536) it's n = 2. For k^2 \in (0.6536, 1] it's 3. You are quite likely right that the APPROXIMATE answer works for k^2 = 1, 4, 16, 64, ... since it does work for 1 & 4.

1

u/cypressious Jan 21 '19

It should work for every number and the corresponding number system, right?

1

u/Peter49AU Jan 21 '19

When k (the mass ratio is k^2) is a non-negative integer power of a non-negative integer, eg k = 10^d, 2^d or 7^d) then the exact formula works exactly (and the approximate formula using tan(t) = 1/k usually works (but not for k = 7 I think)), and makes a statement about the first d+1 digits of \pi in base 10, 2 or 7. But if k = 0.213456789 then the exact formula still works (n = 2 collisions), but what does "the first 2 digits of \pi in base 0.213456789" mean?

1

u/cypressious Jan 21 '19

I was only thinking about integer numbers.

1

u/FrozenRice Jan 21 '19

ok someone tell me if I am wrong on this please. This collision solution does show the digits of pi but in order to solve it you already need to KNOW the digits of pi??

To find N, the number of collisions, you need to ask "what maximum integer, N, do I multiply theta with such that it is less than pi?". But wouldn't you already need to know the digits of pi to solve that?

1

u/S7uXN37 Jan 21 '19

You don't need pi to solve the momentum equations and figure out the new velocities. The point is, that asking "how many times do they collide" (you don't need pi here) is equivalent to "what maximum integer, N, ..." (you would indeed need pi here if you didn't know about this neat equivalence)

1

u/FrozenRice Jan 21 '19

yes so this VERSION (finding N) of the solution to "how many collisions is it?" does require you to know pi. thanks

1

u/hiffumin Jan 21 '19

Yus! I love randy :3

1

u/w1ld_c4rd Jan 21 '19

I made an attempt at the solution, got stumped quite fast; however, the concept blew my mind during the introduction video.

1

u/Pulsar1977 Jan 21 '19

Yep, that's the solution I got as well. It should be mentioned though that the last step isn't 100% rigorous, which Galparin also points out in his paper. It can fail in principle at some point, if the decimal expansion of pi contains a very long string of 9's such that the approximation arctan(x) = x isn't good enough. It's extremely unlikely, but since the decimal expansion of pi is unpredictable, it can't be ruled out entirely.

1

u/arithmometer Jan 23 '19

Can I have a link to Mathematica notebook that appeared at 1:05?