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        riddles >> easy >> bouncing billiard ball
        
(Message started by: towr on Apr 26^th, 2004, 2:51am)

Title: bouncing billiard ball
Post by towr on Apr 26^th, 2004, 2:51am

You have a one by one meter billiard table, and a billiard ball. Show there are both periodic and non-periodic paths the ball can follow when bouncing off the sides (assuming elastic collisions with the sides and no friction)
What can you say about the paths on a 1 by [sqrt]2 meter table?

Title: Re: bouncing billiard ball
Post by Barukh on Apr 26^th, 2004, 5:25am

What happens when the ball hits the corner of the table? :-/

Title: Re: bouncing billiard ball
Post by towr on Apr 26^th, 2004, 5:55am

The probability of it hitting the corner is 0 :P
Feel free to assume it allways hits one wall before the other

Title: Re: bouncing billiard ball
Post by THUDandBLUNDER on Apr 26^th, 2004, 12:13pm

on 04/26/04 at 02:51:00, towr wrote:

What can you say about the paths on a 1 by [sqrt]2 meter table?

:
[hide]They are never periodic.[/hide]?

Title: Re: bouncing billiard ball
Post by towr on Apr 26^th, 2004, 1:22pm

There's two reasons I'd disagree with that
::[hide]1) trivial case, a path parallel to any of the sides. But those are really to trivial to be interesting.
2) If I'm not mistaken (which I don't think I am), the length of the sides doesn't matter. I'd show why I think that but it'd also solve the first part, so I'll wait for someone else to tackle that one first. Supposedly it's easy ;) [/hide]::

Title: Re: bouncing billiard ball
Post by Icarus on Apr 26^th, 2004, 3:44pm

I would say it is easy. I'm not going to provide the solution, as I have produced proofs of essentially the same thing before (periodic vs non-periodic geodesics on the torus - and i haven't done so in these forums), so I'll leave it to someone else.

Also, for elastic collisions with an immobile wall, the component of velocity orthogonal to the wall is reversed. So if you hit a corner, you hit both walls at the same time, and therefore both orthogonal directions of the velocity vector are reversed. I.e. the ball reverses course. If the path would have been periodic were it shifted so that the ball does not hit the corner, then the path will still be periodic, as the ball will somewhere in the opposite direction once again hit a corner, and so will repeatedly cover the path between the two corners (or possibly the same corner twice), reversing at each end. On the other hand, if the shifted path would not be periodic, then neither will the original: The path will hit only the one corner and never strike a corner nor ever repeat itself.

Lastly, once you answer the first question, a simple transformation shows that the same is true for any rectangle, so it does not matter what the ratio of the sides is.

More generally, the same will be true for any bounded region, regardless of the shape.

Title: Re: bouncing billiard ball
Post by towr on Apr 26^th, 2004, 11:49pm

on 04/26/04 at 15:44:34, Icarus wrote:

More generally, the same will be true for any bounded region, regardless of the shape.

That's interesting. I don't suppose that's as easy to show though.

Title: Re: bouncing billiard ball
Post by Icarus on Apr 27^th, 2004, 3:18pm

Yes it is harder, but mostly in abstraction. Continuity demands it.

Title: Re: bouncing billiard ball
Post by grimbal on May 18^th, 2004, 8:05pm

Am I missing some condition?

It seems to me that you can do exactly the same on a 1 by v2 table as on a 1 by 1. Just expand one dimension.

For instance, if the ball goes from the middle of one side to the middle of an adjacent side, it will make a nice lozenge. (or a big viagra tablet :))

Title: Re: bouncing billiard ball
Post by towr on May 18^th, 2004, 11:52pm

on 05/18/04 at 20:05:01, grimbal wrote:

Am I missing some condition?

It seems to me that you can do exactly the same on a 1 by v2 table as on a 1 by 1. Just expand one dimension.

Nope you're not missing anything, well except the solution to the first part ;)
1 by 1, 1 by [sqrt]1 and x > 0 by y > 0 are all equivalent. If you prove the first part in the way I did when I came across the problem, this follows from it naturally.

Title: Re: bouncing billiard ball
Post by SWF on May 19^th, 2004, 8:31pm

Is this the approach you are thinking of?
:[hide]Tile a plane with a grid of copies of the table, and consider how a straight line on that grid relates to the path of a bouncing ball[/hide].

Title: Re: bouncing billiard ball
Post by towr on May 20^th, 2004, 1:11am

pretty much, ::[hide]Each time the ball hits the wall the path get's reflected, so you could instead continue the path straight through into the mirrorimage of the table. Every mirror image an even number of steps in any direction is an exact copy of the original. So if you target those it's easy to put a line that goes through the same point over and over, or to draw a line that can never hit the same point again[/hide]::

Title: Re: bouncing billiard ball
Post by PermanentInk on May 27^th, 2004, 7:50pm

In other words, a line with either [hide] rational or irrational slope[/hide], respectively.

Title: Re: bouncing billiard ball
Post by towr on May 28^th, 2004, 1:00am

If you've normalized the sides of the table to one, yes. Otherwise you have to take the length of the sides into account as well ::[hide]so the slope * width/height is rational, resp. irrational[/hide]::

Title: Re: bouncing billiard ball
Post by grimbal on May 28^th, 2004, 3:38am

My idea is that [hide]If you draw the path of a ball on a axb table and you stretch the picture to match a cxd table, a valid path on the axb table is also a valid path on the cxd table and vice versa. So, it doesn't matter the proportion. A periodic path on an axb table is periodic iff it starts with a velocity vx,vy such that (vx/a)/(vy/b) is rational.[/hide]