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Unformatted text preview: rd Flow Super cial Question: What is the simplest possible billiard table? Super cial Answer: One with
no cushions. We rst consider this primordial case of billiards.
Here the table is the entire plane R R. Interpret K = 0; 2 ) as the circle of directions
(angles) , equipped with arclength measure \d ". The space of cueballs := K is thus 3dimensional, and has a natural productmeasure
:= area arclength ;
which simply measures 3dimensional volume. Given an arbitrary set S of cueballs, its crosssection in direction is
vol bSc := hv; i 2 S and so v2 vol(S) = Z K area bSc d; by Fubini's theorem. Cueball space also has a natural topology. Letting v denote the direction
of cueball v, a metric on is
dist v; w := dist v ; w + dist v ; w ;
where the righthand side uses the metrics on the plane and the \circle of directions", respectively.
Billiard ow . To write a formula for t(v), the location of cueball v after it has \rolled for
t seconds" at unit speed, interpret for a moment ~ as the unitvector in direction . Then the
billiard ow on the plane is the continuous map
t( v) := v + t~v ; v : Since area is translationinvariant, the ow leaves volume invariant:
vol t( S) =
= Notice also that Z K area b t Sc area bSc Z K d + t~ d = Z
K area bSc d = vol(S) : The set of cueballs which ever ow through any particular
point in the table is 2dimensional; hence it has zero volume.
Math. Intelligencer, vol.17 no.1, (1995), 8{16. (5) (6) 6
J.L. King
Billiards tables with cushions. Moving to a more advanced case of billiards, we now glance at tables where re ection is possible.
Suppose our table
R R is the closure of an open set and whose boundary, @ , is a
nice continuously di erentiable curve. When a cueball hits this cushion, it keeps its tangential
component of velocity but reverses its normal component. So cueball space essentially consists
of
K with an identi cation of cueball v1 with v2 if they have the same footpoint v 2 @ , the
same tangential component of velocity and opposite normal component.
In light of (6), one can freely permit the cushion to have nitelymany \corners" (e.g. the origin,
in gure 2) simply by deleting from the nullset of cueballs which ever roll into a corner. Thus
the boundary @ need only be piecewise continuously di erentiable.
Because of the presence of corners, for a xed t the \space map" v 7! t v can jump discontinuously as the trajectory of v is moved across a corner. On the other hand, the \time map" t 7! t v
is always continuous. The billiard ow leaves vol( ) invariant. In the case @ consists of a single straight line, the argument of (5) still applies, since a re ection of the plane does not change area. Together with (6),
this shows that when the table's cushion is a polygon, billiard measure is invariant. The lemma
we shall need is that for any cushion, volume is owinvariant. Billiard Lemma. Suppose the cushion of a billiard table is piecewise continuouslydifferentiable.
Then volumemeasure is invariant under the billiard ow. Sketch of proof of Billiard Lemma. Since we can partition both the cushion and into small
pieces, it su ces to check that vol tC = vol C when the cushion is the graph of a function
f : 0; 1] ! R which is continuously di erentiable, t is some xed time, and C is a set of cueballs
each of which hit f exactly once as time goes from 0 to t. Actually, we need but verify this
inequality:
vol t C vol C: (7) For then analogous reasoning gives the same inequality with \ t" replaced by t and \C" replaced
by t C. T...
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 Fall '09
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