This preview shows page 1. Sign up to view the full content.
Unformatted text preview: his replacement gives vol t t C vol t C , forcing equality in (7).
Furthermore, we may assume that C is a cube, since the cubes generate the Borel eld on . A
\cube" is of the form C = C I , where I K is an interval of directions and C is a square in the
plane. We must therefore prove that
vol S vol C ; whenever C = t(S) is a cube (70 ) as in gure 8 below.
Suppose g : 0; 1] ! R is a piecewise linear approximation of f , and let gt denote the transformation of owing for time t but bouncing o the graph of g rather than f .
Printed: February 14, 1995 Filename is Article/18Billiards/billiards.ams.tex 7 Billiards inside a cusp
f
g
t Φ ( v)
t Φg( v) S C C v
I Figure 8 A small set S of cueballs ows for time t. It bounces o the cushion, f , exactly once to
form a small cube C = C I , whose interval I of directions points to the southeast. Polygonal
approximation g is su ciently close to f that every cueball of S hits g exactly once. The solid line shows the true trajectory of v. The dotted segment shows the altered trajectory of v when
it bounces o of g rather than f .
Since f is continuously di erentiable, given " we may take g uniformly close to f in both position
and slope so as to arrange, for any v 2 S, that dist gt (v); t (v) ". (The nullset of cueballs v which
hit a vertex of g is immaterial.) This implies that gt (S) is a subset of the set Ball" := Ball" (C) of
cueballs which are within distance " of some cueball in C. But billiard measure is preserved when
bouncing o the polygonal cushion g and so S = vol t
g( S) Ball" :
And the volume of Ball" tends to the volume of the cube C, as " & 0.
vol vol x2 A second shot at Billiards
The seriesdivergence proof of the Introduction showed that if the (piecewise smooth) uppercushion f : 0; 1) ! R+ of gure 2 is eventuallyconvex, then the set of cueballs which escape is
empty. For a more general f this escapeset E = E( ), the set of cueballs e such that liminf xcoord
t!1 t e = +1 ; might not be empty but may nonetheless be small in another sense. A nitearea cusp has a null escapeset: The Squeeze Play. Replacing the convexity of f
R
with a nite area requirement, 11 f (x) dx < 1, allows the weaker conclusion that E is a nullset.
\A gallon of water won't t inside a pintsized cusp" is the proof: Pick x0 su ciently large that
area(S ) is pintsized, where S consists of those points (x; y ) 2 with x x0 ; indeed, area(S ) is to
be taken so small that
area(S ) 2 < vol(E):
But vol(S K ) = area(S ) 2 . Hence S K has strictly less volume than E. But this contradicts
that
vol t (E) \ (S K ) ! vol(E) ;
as t ! 1,
which follows from the de nition of the escapeset. So no such x0 exists and thus vol(E) = 0.
Math. Intelligencer, vol.17 no.1, (1995), 8{16. 8
J.L. King
A second proof that vol(E) is zero: Recurrence. For a continuous ow on a metric space ,
a point ! is (topologically) recurrent if t (!) ! ! along some sequence of times ti ! 1.
Showing that almostevery cueball is recurrent would emphatically prove that the escapeset is
null. We will not, however, be able to prove that E is empty , since a table of nite area {even a
i bounded table{ need not have all its cueballs recurrenty.
The key to showing that a.e. cueball is recurrent is to de ne a measuretheoretic notion of
recurrence. Consider a measurepreserving ow on measurespace ( ; ). A point ! 2 S \recurs
to S " if t! 2 S for arbitrarily large times t. A set S is Poincarerecurrent if a.e. ! 2 S recurs
to S . Flow is co...
View
Full
Document
 Fall '09
 Staff

Click to edit the document details