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# Furthermore we may assume that c is a cube since the

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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 series-divergence proof of the Introduction showed that if the (piecewise smooth) uppercushion f : 0; 1) ! R+ of gure 2 is eventually-convex, then the set of cueballs which escape is empty. For a more general f this escape-set E = E( ), the set of cueballs e such that liminf x-coord t!1 t e = +1 ; might not be empty but may nonetheless be small in another sense. A nite-area cusp has a null escape-set: 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 pint-sized cusp" is the proof: Pick x0 su ciently large that area(S ) is pint-sized, 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 escape-set. 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 almost-every cueball is recurrent would emphatically prove that the escape-set 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 measure-theoretic notion of recurrence. Consider a measure-preserving ow on measure-space ( ; ). A point ! 2 S \recurs to S " if t! 2 S for arbitrarily large times t. A set S is Poincare-recurrent if a.e. ! 2 S recurs to S . Flow is co...
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