Conference rusure::math

o.k. I'll solve half, and you solve the other half.  Fair deal ?  Doesn't
matter who takes which.  So, I'll arbitrarily take the half about no collisions
at all:

	If balls are bouncing back and forth from left rail to right rail, all
	balls running parallel to end rails, then there will be no collisions.

o.k. your turn to do the part about infinite collisions.

/Eric

    But you haven't finished YOUR half, Eric!

    OK Eric, I'll take this infinite half:
    
    Modelling this after your proof:
    
    	Take two balls bouncing off two rails perpendicularly that meet
    exactly misway between the rails and contact exactly head-on. 
    Obviously they will continue to rebound off the rails and each other
    forever (assuming no energy loss) with exactly the same motion vectors
    (in opposite directions).  Thus they have infinite collisions.
    
    OK?
    
    --  Barry

Hi Barry.  I don't think you answered the question.  The question isn't to
*demonstrate* that it's possible for balls to bounce forever.  The question is
to demonstrate that they either *must* bounce forever, or they must *never*
bounce at all.

Putting it another way, demonstrate that it's impossible to construct a ball
situation where they bounce a bit and then no more.

/Eric

	It would be enough to show the following:

	Given two balls that have just collided, show that (unless one of them
hits another ball in the mean time) they will collide again.   So there can
be no last collision.

Andrew.

    Let n=2 and assume the two balls are standard billard balls in diameter
    and mass.  Start the balls rolling toward each at the same velocity
    in opposite directions along the line between opposite corner pockets.
    
    When the two balls hit they will rebound along the same line and fall
    into the pockets never to hit again.
    
    Hypothisis fails;  proof by counter example.
    
    -Joe
    

>    Hypothisis fails;  proof by counter example.
    
    Flawed counterexample.
    
    Billiard tables do not have pockets.  "Pool" tables do.
    
      John

    I was just teasing you Eric...
    
    ;-)
    
    --  Barry

    
    I was disappointed :-) not to find the 2nd half of the solution, so 
    here goes...
    
    Assume there is at least one collision.  
      Let x and y denote the two balls which have last collided.  
      Let Tc be the time of the collision.

    Now x and y will travel paths Px and Py, oscillating with  periods 
    Tx and Ty respectively.  

    Since the two balls have a common starting point and are in periodic
    motion, there will be another collision between x and y at 
    Tc + LCM(Tx,Ty).
    
    The only thing that could prevent x and y from colliding again is if 
    one or both collide with some other ball(s) on the table. In this case,
    repeat the whole process with a new pair, say x and y'.
    
    Therefore if there is at least one collision, there must be an infinite 
    number of collisions.  :-)  
    
    /Al

        You need to explain why the paths oscillate, i.e., why you
        think they are periodic.
        
        Dan

    And even if the paths are periodic, the periods might not be in a
    rational ratio and hence would have no common multiples.  It would be
    necessary to show that no matter how small the balls are, the
    oscillations are eventually close enough that the balls touch.  There
    have been a couple of topics previously that touched on something like
    that.
    
    
    				-- edp

re .11
    
>It would be necessary to show that no matter how small the balls are, the
>oscillations are eventually close enough that the balls touch.
    
    I must admit I was pondering the same question. Are these balls point
    balls or real ones? If point balls, are you suggesting that the claim
    is false?

    Re .12:
    
    According to the problem, the balls have positive radii -- they are not
    points.  If they were points, simply build a table that is one meter by
    pi meters, and start two point-balls at the same place at the same
    moment moving at the same speed, one horizontally and one vertically. 
    They will never be in the same place at the same time again.
    
    (Of course, if you want the table to be constructable, you can't make
    it one meter by pi -- make it one meter by square root of two meters.)
    
    
    				-- edp

    Solution by the proposer and Bjorn Poonen, University of California at
    Berkeley, Berkeley, California.
    
    It suffices to show that if two billiard balls collide, there will
    ensue another collsion.
    
    Let the dimensions of the table be L by W.  Consider two balls of radii
    r[1] and r[2].
    
    To each ball associate a rectangular tiling of the plane by taking an
    L - 2*r[k] by W - 2*r[k] rectangle and reflecting it about its sides
    until the entire plane is filled.  This converts billiard paths of each
    ball into straight lines, assuming there are no collisions.
    
    Let E be the product of the two planes so formed.  Note that on E, the
    motion of both balls is modeled by a single straight line path.  Let D
    be the obvious lattice whose fundamental domain is a 2(L-2*r[1]) by
    2(W-2*r[1]) by 2(L-2*r[2]) by 2(W-2*r[2]) hyperblock in E.  The factors
    of 2 are designed so that points in E modulo D refer to the same point
    on the billiard table.
    
    Let C denote the set of points in E which represent impossible states;
    that is, states where the two balls strictly overlap.  Observe that C
    is an open subset of E.
    
    Now suppose our two balls have collided.  Let R(t) represent the
    ensuing straight line motion of the two balls in E, extended linearly
    in both directions, with t=0 at the time of collision.  If this
    straight line is not adhered to for positive t, it means a collision
    must take place.  Observe that R(0) is in dC [the original uses the
    partial derivative operator there -- the boundary of C?] and that
    R(-delta) is in C for some small delta > 0.  The first condition says
    that at the moment of collision, the two balls touch, and the second
    says that there was actually a collision.
    
    A ray starting at one point of a lattice must pass within epsilon of
    infinitely many other lattice points, so we can pick a time T > delta
    such that R(t)-R(0) is within epsilon of a vector in D.  Then
    R(t-delta) - R(delta) = R(t)-R(0) is within epsilon of a vector in D,
    so R(t-delta) is in C, if epsilon is small enough.  This implies that a
    collision takes place.