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A645: Exercise #1 Dynamical Billiards A645: Exercise #1 Dynamical Billiards An exploration of regular and irregular orbits on the billiards table. Due: 2014 Jan 30 As many of you are aware, Newton mechanics experienced a resurgence several decades ago. This was driven by the rediscovery that classical dy- namical systems exhibit exponential sensitivity to initial conditions, called chaos. The surprising result that a deterministic system can yield unpre- dictable results has lead to new insights into meaning of F = ma. This problem “jumps the gun”: the deeper meaning of the findings here will be- come clear as the semester moves on. We will start off with a simple example, the billiard problem, which is quite easy to implement numerically but has much of the behavior of the general case. For the dynamicist, the billiard problem considers the trajec- tory of a freely moving point particle in the region of a plane bounded by some closed curve B (called a billiard). The particle reflects elastically at the billiard conserving the tangential component of momentum (so that the angle of incidence equals the angle of reflection). The state of the particle is fully specified by position and momentum. Because the collisions are elastic, the energy is conserved and the position and direction suffice to label a state. Between impacts with the wall, the particle moves on straight lines and therefore a particular orbit is fully spec- ified by its position and direction immediately following each impact. The position may be labeled by the arc length around B at the point of impact. The direction of the orbit φ after each impact can be labeled by angle with respect to the tangent to B or the tangential momentum p ≡ cosφ (choosing unit speed). The orbit is then fully specified by the sequence of pairs (s j, p j). This discrete dynamics is an example of a mapping M of phase space. Given an initial condition (s0, p0), the orbit is then obtain from the mapping as 1 follows: s s j+1 = M j . p j+1 p j The map M is not in general a simple matrix (it may be non-linear, for example) but one can show after a fair bit of work that it is area preserving and the system is Hamiltonian. There are three possible outcomes of this mapping, each of which de- scribes a behavior which is typical of continuous Hamiltonian problems: 1. The orbit may be periodic. Specifically, the sequence of phase space points may repeat after N iterations: s s s j+N = MN j = j . p j+N p j p j 2. The orbit may fill a curve in phase space. This indicates the exis- tence of some invariant or conserved quantity such that G(s j, p j)= G(s0, p0). 3. The orbit may fill all of phase space. The orbit, therefore, has no other conserved quantities other than energy and is said to be ergodic. There are many interesting properties of the billiard problem and mono- graphs on the subject can be found in the library (e.g. “Billiards : a ge- netic introduction to the dynamics of systems with impacts” by Kozlov and Treshchev).¨ In the rest of this problem set, we will take an experimental point of view. 1 Billiards simulation code To investigate this problem, we will write a very simple particle simu- lation code. The steps are: (a) Propagate the particle forward along its direction (momentum). Let (xn,yn) be the current position of the particle and p = (px, py) be the direction vector (|p| = 1). Then (xn+1,yn+1) = (xn,yn)+ ∆p. Choose ∆ to be some small step (e.g. ∆ = 0.001). (b) After every step, check to see if the particle has hit the billiard. Assume that the billiard curve can be written as F(x,y)= 0 where F(x,y) > 0 inside the curve and F(x,y) < 0 outside. For example, for a circular billiard F(x,y)= 1 − x2 + y2. p 2014 Jan 20/MDW 2 After each iteration, check to see if value F(xn+1,yn+1) < 0. Let δ be the value (x,y) = (xn,yn)+ ∆p such that F(x,y)= 0. This is easily obtained by some root finder (e.g. Brent’s method). (c) At the contact point, the projection of the momentum p parallel to the wall is conserved and the wall delivers and impulse equal to twice that of the projection perpendicular to the wall. This is the standard elastic bounce boundary condition. These vectors are easily computed from F(x,y). The location along the billiard s j+1 and the new direction p j+1 is computed and saved. Goto Step (a). There is nothing tricky here, but be methodical! Note that for every billiard curve, one needs to evaluate F(x,y), the arc length s given (x,y), and the slope at any position s (to get the tangen- tial and perpendicular directions to the billiard). It is also convenient to be able to get (x,y) for a given s to specify initial conditions. Although it is numerically convenient to work numerically in coordinates (x,y), coordinates (s, p) are nice canonical variables. You should write your code with enough modularity that you can eas- ily change the billiard curve with no changes to your main simulation code. The easiest is source-code based subroutines or functions that one compiles with the main routine. In Java, F90 or C++, one might use classes. 2014 Jan 20/MDW 3 2 Circular billiards This is the simplest, most ideal case. You can easily convince yourself from the proper- ties of trigonometry that p ≡ cosα = constant α α (see Figure 1). αα There are two qualitatively different cases here α α depending on whether α is a rational or ir- α α rational fraction of π (A rational fraction of π would be α = mπ/n for relatively integers Figure 1: Geometry for tra- m and n.). Describe the orbits for rational jectories in a circular bil- and irrational orbits. Run your simulation for liard. Simple trigonometry both cases and describe the difference in the shows that all angles α are identical. phase space plots for the sequence (sn, pn). As a test of your code, make sure that degenerate cases such m = 1,n = 2 behave as expected. How about α → 0? 3 Stadium billiards Wall This is a non-trivial one. A stadium con- α sists of ends which are semicircles connected s by straight lines. The let semicircles have l=1 unit radius and the length of the wall η then uniquely describes the problem (see Figure 2). η Write a set of subroutines or functions (or Figure 2: Geometry for the class) to add the stadium billiard to your stadium billiard. The wall simulation. One may construct F(x,y) by length η and a phase-space using using the circle for the ends and con- state (s, p = cosα) are shown. necting the contours of by horizontal lines parallel to the walls. That is: 1 − (x − η/2)2 + y2 if x > η/2 F(x,y)= 1 − p(x + η/2)2 + y2 if x < −η/2 . 1 − |py| otherwise The arc length and other conversions and derivatives also may be straight- forwardly implemented as a set of cases. 2014 Jan 20/MDW 4 (a) Test this implementation on degenerate cases. For example the initial s = 0,α = π/2 should bounce from stadium end to stadium end. Similarly s = π/2+η/2,α = π/2 should bounce between the walls (at least for a some small number of orbits). Also, you should recover the circle case for η = 0. (b) Now, one might think that for very small η that the the stadium will behave like the circular billiard. Try choosing some small value of η, η = 0.01 for example, run your simulation for s = 1, p = 1/e (that is, α = 64.8◦. Plot 1000 steps and compare with the phase space for the η = 0. (c) Are you surprised? The results are quite different. One can prove that the stadium is an ergodic billiard1; in other words, that as n → ∞ sn, pn will come arbitrarily close to any point is phase space. To see this, make a series of phase space plots for η = 0.001, η = 0.01, η = 0.02, η = 0.05, η = 0.1, and finally η = 1. Notice, that for a finite number of points that the phase space is not uniformly covered (for 1000 points). It is still ergodic; any holes will shrink as n increases (try it). (d) Let us concentrate on the η = 1 case. Run your simulation and plot the results for n = 5000. Note that there are two holes for p = 0,s = (π + 1)/2 and p = 0,s = 3(π + 1)/2. Try to find an explanation for why these occur. (Hint: think about the trajectories at these points.) 4 Sensitivity to initial conditions Here, we will choose a few different initial conditions leading to regu- lar and chaotic orbits and explore what happens when the sense of time is reversed. You may reverse time by changing the sign of the velocity components. Iterate your system forward for n = 5,10,20,30,50,100,1000 bounces. Reverse time and iterate backwards for the same n number of bounces. Compare the resulting values of s and p with the initial conditions. Some suggestions: (a) For a given initial condition, plot the difference between the initial condition and final state as a function of n.
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