Section 1. Classical Dynamics Section 1.10 Chaos in Billards. Consider a particle moving freely in a region of the plane bounded by a closed curve B. Assume that the particle moves without friction and is reflected elastically when it hits B. In this case the angle of reflection equals the angle of incidence. Billard α α Suppose as a simple example B is a circle. Then there are certain orbits which close. These orbits trace out regular ploygons such as pentagons or star shapes. But there are other orbits which never close so that they cover the whole boundary. Circular Billard Β S S Periodic orbits Orbit covers the whole of B S Any orbit can be specified by the arc position s i on the boundary and the angle it s trajectory makes with the tangent to B at s. So one can think of the orbit as a map. s 0, α 0 s 1, α 1 s, α... However it turns out to be more convienient to think interms of the tangential momentum sn1 sn p cos α M on S R p n1 where M depends on the boundary B. If B is a circle α 0 α 1 α i, so that p is a constant of the motion. If α πk the orbit closes after N bounces. For instance if α π p 0 it closes N after two bounces. If α 3π p 5 cos3π closes after 5 bounces. 5 p n Circle Billard is just a twist map. p p is a constant. O pcos3π/5 1 pcosα απ irrational S α πk orbit closes after N bounces. N α π p 0 it closes after two bounces. α 3π p 5 cos3π 5 orbit closes after 5 bounces.
Any elastic billard mapping is area preserving Proving this result is tricky. The canonical variables are s, p as described. So what we need to prove is that, p 1 s 0, p 0 det s1 Consider the small change α 0 α 0 dα 0. Define ψ as the angle the forward tangent to B at s 0 makes with the horiziontal, then ψ also undergoes a small change: ψ 0 ψ 0 dψ 0. Billards with boundary B 1 Condition 1 ψ1 α1 ψ0 α0 B Condition α0 d α0 α0 s0 ψ0 ψ0 α0 δs0 sinα0 δs1 sinα1 ρ01δα0 δψ0 ψ1 d ψ1 α1 d α1 ψ1 s1 α1 ψ0 d ψ0 ψ1 α1 α1 δs1 sinα1 δs0 sinα0 δs0 sinα0 ρ01 α0 δα0 δψ0 s0 δs0 Then from the diagram ψ 1 α 1 ψ 0 α 0 and if ds 0, ds 1, dα 1, dψ 0 are small δs 0 sin α 0 δs 1 δα 0 δψ 0 Since ψ i s i only and s 0 and α 0 are independent sin α 0 ψ 0 and α 0
Now the radius of curvature is Rψ ds dψ sψ ψ so that since s 0 and α 0 are independent α 0 0 And since p cosα π dψ Rψ ds 0 dψ 0 Rψ 0 and ds 1 dψ 1 Rψ 1 sin α 0 ρ 10 dψ 0 ds 0 sin α 0 1 sin α 0 α 0 α 1 But ψ 1 ψ 1 sin α 0 ρ 01 Rψ 0 sinα 1 so that Lastly ρ 10 Rψ 0 sin α 0 ψ1 ψ 0 sin α 0 Rψ 1 Rψ 0 Rψ 0 Rψ 1 sin α 0 α 1 α 0 Where since ψ 0 is not a function of α 0 then α 1 α 0 1 ψ 1 α 1 1 Rψ 1 sin α 0 sin α 0 Rψ 1 sinα 1 This means that s1 sin α 0 sinα 1 sinα 0 Rψ 1 Rψ 0 Rψ 0 Rψ 0 Rψ 1 sinα 0 sinα 1 sinα 0 Rψ 1 sinα 1 sinα 0 The determinant is then sin α 0 ρ 10 sin α 0 Rψ 0 sin α0 Rψ 1 Rψ 0 sin α 1 Rψ 1 sin α 0 sin α 0 Rψ 0 Rψ 1 1
Calculating the Map Given the curvature at any point on the boundary B you can obtain the map ψ 1 ψ 0, α 0, α 1 ψ 0, α 0 numerically. From this you can deduce the map on s, p through sψ ψ π Rψ dψ pα cosα. Consider the slope of the arc between s 0 and s 1. As s changes so does ψ. Let x, y be a point on B. Then x dx cosψ s ds cosψ dy ds sin ψ so that B s0 B s ψ x y s0 Similarily yψ 1 yψ 0 α0 ψ0 xψ1 xψ0 s1 xψ 1 xψ 0 ψ1 yψ1 yψ0 yψ1 yψ0 tan ψ0α0 xψ1 xψ0 cosψds ψ1 ψ 0 Rψ cosψdψ ψ 0 Rψ sin ψdψ But the slope of the arc between s 0 and s 1 is ψ 0 α 0 yψ 1 yψ 0 xψ 1 xψ 0 tanψ 0 α 0 so that the equation tanψ 0 α 0 ψ1 ψ 0 Rψ sin ψdψ ψ1 ψ 0 Rψ cosψdψ defines ψ 1 ψ 0, α 0. Then we can use the fact that ψ 1 α 1 ψ 0 α 0 to give α 1 ψ 0, α 0 ψ 1 ψ 0, α 0 ψ 0 α 0 Integrability Like all maps the map is integrable if there is a constant of the motion. Fs 1, p 1 Fs 0, p 0 The simplest billard, the circular billiard is integrable because α 1 α 0 p 1 p 0 so that p is a constant. Also since, ψ 1 α 1 ψ 0 α 0 ψ 1 ψ 0 α 0. So using the fact that the curvature is the radius R we have that sψ ψ π Rdψ Rψ π s 1 s 0 R cos 1 p 0 and p 1 p 0
Fixed Points and Stability There are no period-1 orbits, but there are two bounce orbits which are often easy to identify. In the elliptical billiard they exist along the major and minor axes. To work out their stability we need to work out the linearized matrix. For any two bounce orbit α 0 α 1 π and Rψ 0 Rψ 1 Rso that, p 1 s 0, p 0 sin α 0 sinα 1 sinα 0 Rψ 1 Rψ 0 Rψ 0 s1, p 1 s 0, p 0 Rψ 0 Rψ 1 R sinα 0 sinα 1 sinα 1 sinα 0 Rψ 1 sinα 0 ρ R 1 1 ρ 1 ρ R ρ 1 ρ ρ R R R 1 1 ρ 1 ρ R ρ ρ 1 R R R ρ ρ This has trace 4 R 1. So the orbit is stable if < 4 R 1 <. That is ρ R 1 ρ < and if ρ > R that the orbit is stable if R < Period- orbits two bounce orbits circular billard two bounce orbits pass through the center elliptical billard just two bounce orbits Period- points for the Circular billiard. Since for every period- orbit in a circular billiard ρ R so that ρ R bounce orbits are neutrally stable. Infact for the circular billiard s1 s Df 0 1 R sin α 0 1 Df n 1 R sin α 0 1 n 1 Rn sin α 0 1 which has trace so all periodic orbits are neutrally stable. Period- points for the Elliptical billard. If we parametrise the ellipse by λ. All two x a cosλcoshm y a sin λ sin hm the ellipse is which has eccentricity e 1 cos h M. x y 1 a cos hm a sin hm
Since Now the curvature is That is Rψ ds dψ dx dλ dx dλ dy dλ dλ dψ dy dy dx tan ψ dλ tan hm cotλ tan ψ tanhm cot λ dψ dλ which you can then use to evaluate Rψ Exercise show that Rψ tanhm sin λ sec ψ a sin hm cos hm cosh M sin ψ sinh M cos ψ 3 Now consider the two bounces on the major and minor axes. ψ π or π. On the major axis R π a sin hm cosh M ρ a coshm ρ R cosh3 M sin hm > unstable On the minor axis Rπ a coshm sin h M ρ a sin hm ρ R sin h3 M coshm < Infact the elliptical billiard is also integrable. stable Fs, p p e cos ψs 1 e cos ψs is a constant of the map. Proving this involves alot of algebra.
The stadium considtes of two semicircles joined by straight lines. In the semi circles Rψ R, but for the straight lines R. The two bounce orbit, lengthways is unstable as ρ Rη >. But the family of R r nonisolated two bounce orbits across the stadium are neutrally stable. Period- orbits for the stadium. η R However the stadium is not integrable, infact there are no invariant lines and it has been shown that it is an ergodic billiard. That means that for almost every initial condition s 0, p 0 the iterates will come arbitraily close to every point in the phase space as n.
Oval billiards, not elliptical billiards provide us with a near integrable system. Let Rψ a1 δ cos ψ Then ds dψ Rψ and dx ds cos ψ implies that dx dψ a cosψ1 δ cos ψ x a1 δ sin ψ δ sin 3ψ 6 Similarily and y a 1 δ cosψ δ cos 3ψ 6 sψ aψ π δ sin ψ There two bounce orbits at ψ π, 3π and at ψ 0π. Like the ellipse the long one is unstable. R π a1 δ and ρ a1 δ 3 ρ R 1 δ 3 1 δ > Unlike the ellipse the oval is not integrable. As δ is increased from zero the circular billiard it appears to mimick the elliptical billiard but infact the separatrix is chaotic. Infact thought of as a perturbation from the elliptical billiard it exhibits all the usual resonance behaviour we expect from near integrable systems. The separatrix is chaotic and increasingly so as δ is increased from zero. Interesting four bounce orbits or period-4 orbits are seen for larger values of δ.