Lecture 11 : Overview

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1 Lecture 11 : Overview Error in Assignment 3 : In Eq. 1, Hamiltonian should be H = p2 r 2m + p2 ϕ 2mr + (p z ea z ) 2 2 2m + eφ (1) Error in lecture 10, slide 7, Eq. (21). Should be S(q, α, t) m Q = β = = dq/ 1 mω2 q 2 α 2α 2α t (2) Today s lecture: Revise : Hamilton-Jacobi equations for Hamiltons principle and characteristic functions. Example : Simple harmonic oscillator Action angle variables Invariant tori vs. chaos

2 Hamilton-Jacobi Eqn for Hamilton s Principle Function Aim : Seek a canonical transformation such that Q = 0 and q(t) Q, p(t) P, H(t) K, (3) Ṗ = 0. New equations of motion : K P = Q = 0, K Q = Ṗ = 0 (4) Both Q = 0 and Ṗ = 0 are guarenteed if K = 0. If such a transformation found, then Eq. (3) yields solutions for q(t) and p(t). Method : Use a generating function. Why? Want to convert expression for K to a mixed PDE in t, p or t, q. Any generating function will do, but lead to different free choices for Q or P. The choice F 2 (q, P, t) leads to the Hamilton-Jacobi equation. ( K = H q, F ) 2 q, t + F 2 t = 0 (5) The solution of F 2 is often referred to as Hamilton s principal function, and labeled S.

3 Suppose there exists a solution of form S = S(q 1,..., q n, α 1,..., α n, t) (6) where α 1,..., α n are independent constants of integration of Eq. (5). Choose P = α. The F 2 transformation equations give : p i S(q, α, t) = q i (7) Q i S(q, α, t) = β i = α i (8) The β i values are labels for the new generalized coordinates. They are constant (time-independant) becuase K = 0. Eq. (24) can be inverted to give q in terms of α, β and t. The RHS of Eq. (7) can then be written in terms of α, β and t. That is q = q(α, β, t), p = p(α, β, t) (9) When solving the Hamilton-Jacobi equations, we are at the same time obtaining a solution to the mechanical problem.

4 Suppose the Hamiltonian is time independant. Then ( H q, F ) 2 + F 2 q t = 0 (10) Use the transformation S(q, α, t) = W (q, α) α 1 t (11) where W (q, α) is Hamilton s characteristic function. The transformed Hamiltonian is now ( K = H q, W ) q, t + W t = 0 (12) and so H ( q, W ) q = α 1 (13)

5 Hamilton-Jacobi Eqn for Hamilton s Characteristic Fn Consider a new transformation q(t) Q(t), p(t) P, H(t) K, (14) such that Ṗ = 0. Choose P i = α i, where α i are the constants of the motion, and α 1 = H. Using generating function W (q, P ), then p i = W q i, New equations of motion: with solutions Q i = W α i, K = H + W t = α 1 (15) Q i = K P i, Ṗ i = K Q i = 0 (16) Q 1 = t + β 1, Q i = β i, i 1 (17) P i = α i (18) The new Hamilton-Jacbi PDE is ( H q, W ) q, t + W t = 0 (19) First Prev Next Last Go Back Full Screen Close Quit

6 1. 1D harmonic oscillator - revisited Aim : Recall H = p2 2m + mω2 q 2 /2 E (20) Solution : We seek a transformation q(t) Q(t), p(t) P, H(t) K, (21) Choose P = α = E. New Hamiltonian is [ ( W K = H(P ) = 1 ) ] 2 + m 2 ω 2 q 2 2m q = α (22) What is Q? Hamilton s equations of motion + generating function transformation : Q = K P = 1, Q = 1 + β (23) plus generating function transformation S(q, α, t) Q = = 2mα dq/ 1 mω2 q 2 α 2α t (24) First Prev Next Last Go Back Full Screen Close Quit

7 1D harmonic oscillator - continued Hence, Q = t + β = 1 mω ω arcsin q 2 2α Therefore, same dynamics, but different transformed variables S(q, p, t) : Q = β initial phase angle, P = α system energy E (25) W (q, p) : Q = t + β phase angle, P = α system energy E

8 2. Action-Angle Variables Systems in physics that are periodic, with Hamiltonian independant of time are common. An elegant and powerful method is a variation on Hamilton-Jacobi theory. Instead of using integration constants to define new momenta, choose P = J, with J i = p i dq i (26) where the integration is carried out over one complete period of libration, or rotation. The new Hamiltonian is K = H(J), with Hamilton s characteristic function W = W (q, J). The new coordinates are action-angles, defined by the transformation θ = W/ J (27) with equation of motion ω(j) θ = K(J)/ J (28) First Prev Next Last Go Back Full Screen Close Quit

9 Action-Angle variables (cont.) and solution θ = θ 0 + ωt (29) Equation of motion for J is and so J retains it s initial value. J = K θ = 0 (30) Such a set is an n-torus (embedded in the 2n-dimensional phase space) because specifying θ uniquely determines where we are on the invariant set it is an invariant torus. The phase space is filled (foliated) with these invariant tori.

10 Example : 2-torus. Locii θ 1 and θ 2 trace out a torus. First Prev Next Last Go Back Full Screen Close Quit

11 3. Invariant tori vs. chaos Survival of invariant tori KAM Theorem Suppose we perturb an integrable system with Hamiltonian H 0 : H = H 0 + ɛh 1. Then we can try to find new action-angle coordinates, but in general this procedure breaks down on resonant tori, where there exists an integer array m such that m ω = 0. That is, on resonant torii there are no solutions. If however we choose to study invariant tori that are sufficiently far from resonance, then Kolmogorov, Arnol d and Moser (KAM) showed that some invariant tori survive, even in nonintegrable systems, provided ɛ is sufficiently small. Destruction of invariant tori chaos When ɛ exceeds a critical value, the KAM torus breaks up and chaotic orbits can diffuse through the remnant invariant set (a Cantor set). The chaotic orbits fill an invariant volume and separate exponentially fast, so memory of initial conditions is rapidly lost.

12 4. Area-preserving map: kicked rotor Consider the period map of a periodically forced 1-degree-of-freedom system (sometimes called a degree of freedom Hamiltonian system). Then the phase-space is only 2-dimensional and the conserved volume is in fact an area. I.e. the period map is area preserving. Consider a pendulum (rotor): in which gravity is applied purely impulsively in a periodic fashion: z g(t) = g 0 t n= δ(t t n ) (31) where t n n t, with n an integer, and δ(t) is the Dirac δ function, so that g 0 is the time average of g(t): θ ( 1 cosθ ) l 0 l mg

13 5. Trial function The Hamiltonian is H(θ, p, t) = p2 + mg(t)l(1 cos θ). (32) 2ml2 Between the kicks the motion is one of uniform rotation with constant angular velocity p/ml. That is, θ is a continuous, piecewise linear function of t and p a piecewise constant step function with only the values θ n at, and p n immediately before, t = t n as yet unknown: p n θ n p(t) g(t) θ(t) t ε t n 1 t n t n+1 θ(t) = 1 t [(t n+1 t)θ n + (t t n )θ n+1 ], p(t) = p n+1 (33) for t in each range t n n t < t < t n+1 (n + 1) t. t

14 6. Variational solution Inserting this trial function in the phase-space action integral over the interval t N ε t t N + ε, N 1, ε 0+, we get S ph = N 1 n = N [ ] p n +1 (θ n +1 θ n ) p2 n +1 t mg 2ml 2 0 l t (1 cos θ n +1) (34) The conditions for S ph to be stationary under variations of θ n and p n+1, N < n < N, (noting that θ n occurs in the terms of the sum for both n = n and n = n 1) are S ph p n+1 = θ n+1 θ n p n+1 t ml 2 = 0, (35) and S ph = p n p n+1 mg 0 l t sin θ n = 0. (36) θ n These equations define an area-preserving map known as the Standard Map:.

15 7. Standard Map Commonly the map is re-expressed in terms of an angle coordinate x with period normalized to unity and a nondimensionalized momentum y, which is p expressed in units of 2πml 2 / t, Then the map becomes θ = 2πx, p = 2πml2 y. (37) t y n+1 = y n k 2π sin 2πx n, x n+1 = x n + y n+1, (38) where the chaos parameter k is defined by k g 0( t) 2 l. (39)

16 8. Poincaré plot Some typical iterated orbits of the map, showing both chaotic and regular regions. The remnants of the librating pendulum orbits around the O points are seen as large islands around (x, y) = (0, 0) and (0, 1) [also (1, 0) and (1, 1) because the phase space is periodic in y as well as x], but the region around the pendulum X point is highly chaotic. There are also new islands due to resonances not present in the physical pendulum problem (in fact an infinite number of them), each with chaotic separatrices: 1 y 0 0 x 1

17 9. Green s critical k The value of k used in the Poincaré is k c = , which is the value at which the last KAM invariant curves with the topology of a rotating pendulum orbit become unstable and break up into invariant Cantor sets. Two of these KAM curves are shown, being the only curves completely crossing the figure from left boundary to right. For k < k c the motion is bounded in y, while for k greater than this value a phase-space point can diffuse without bound in the positive or negative y-direction. From eq. (39) we see that the integrable limit t 0 corresponds to k 0 [noting that eq. (37) shows that the width of the physical pendulum s separatrix shrinks to zero as k 0 when represented in terms of y].

18 10. A 4-D Symplectic Map Froeschle map (two coupled Standard Maps): y 1 = y 1 + a 2π sin 2πx 1 + c 2π sin 2π(x 1 + x 2 ) y 2 = y 2 + b 2π sin 2πx 2 + c 2π sin 2π(x 1 + x 2 ) x 1 = x 1 + y 1 x 2 = x 2 + y 2 (40)

19 11. Lecture 11 : Summary Revised Hamilton-Jacobi equations for Hamilton s principle and characteristic functions. Discussed action-angle variables and coordinates Dynamics for set of constant actions J describes an invariant n torus, embdedd in 2n space. Introduced KAM theory Discussed the transition to chaos for the kicked rotor. Last Lecture Tommorrow: Revision.

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