Unformatted text preview: nservative if every set S
is Poincarerecurrent. De ne conservativity for
a transformation analogously.
Motivated by his study of the 3body problem, Henri Poincare made this simple, but tremendously useful, observation.
Poincarerecurrence theorem. If is a measurepreserving ow on a nite z measure space,
then is conservative. A measurepreserving transformation T on a nite measure space is likewise
conservative.
Proof. Fix time > 0 and let B S consist of those points which never recur to S after time .
Thus t (B) is disjoint from B for all t . Consequently these sets B; (B ); 2 (B ); 3 (B ); : : : are mutually disjoint. Since they all have the same mass, B must have been a nullset.
On nitearea table, that almostevery cueball is recurrent is a consequence of the following
elementary fact, which is left as an exercise.
Lemma 9. Suppose is a separable metric space and is a ( nite or in nite) measure. Then
under any conservative measurepreserving ow transformation on ( ; ), almostevery point is
topologically recurrent.
As a consequence, since table f (x) = 3 + sin(px) (x + 1)2 of question (4a) has nite area, its
escapeset is null. I do not know whether it is empty. One can certainly manufacture a nitearea
nonconvex uppercushion f which coaxes one particular cueball v monotonically out to in nity;
simply draw the desired orbit of v rst, then draw f to match the desired slope at the re ection
points of the orbit. With a bit of extra e ort, one can even arrange for f to have negative slope
everywhere.
Weiss's proof of empty escapeset. Sometimes an \everywhere" rabbit can be pulled out of
an \almosteverywhere" hat. A case in point is the neat proof by my friend Benjamin Weiss that
under an eventuallyconvex f of nite area the escapeset is indeed empty . yConsider a table bounded by a noncircular ellipse. A cueball hit along the major axis has a periodic orbit.
Conversely, a cueball v pointed at a focus {but with footpoint not on the major axis{ has an orbit which converges
to the periodic orbit along the major axis. So v is not a recurrent point. Notice, though, that this example exhibits
only a nullset of nonrecurrent points, since the set of cueballs pointing at a focus is but 2dimensional.
zA ow on an in nite measure space need not be conservative; witness t(x) := x + t on the real line equipped
with Lebesgue measure.
Printed: February 14, 1995 Filename is Article/18Billiards/billiards.ams.tex 9 Billiards inside a cusp The strategy is to show that if even one cueball, v, escapes, then an entire open set of cueballs
escape. We may assume that the orbit of this v = hv; v i is already in the convex part of the cusp
where the slope of f is always negative (as shown in gure 2), and that 0 < v < =2.
Consider any cueball e = he; i with > 0 and having a shallower angle than v; by \shallower"
we mean that 0 < j j < j v j. Moreover, we ask that e lie \further right" than v in the sense that its
footpoint, e, lie on the southeast side of the line through v in direction v . Compare the orbits of e
and v: When they rst hit the uppercushion, e hits to the right of v and consequently bounces o
with a shallower angle than does v. Thus, after they bounce o the oor, cueball e is still further
right, and is shallower than, v. Iterating shows that e escapes.
Were such a v to exist, this reasoning would hold for the above open set of cueballs e, which
perforce has positive volume. The inescapable conclusion is that E is empty. x3 P...
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 Carom billiards, measure, Lebesgue measure, billiard ow, cueballs, cueball

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