PHY6426: CLASSICAL MECHANICS HOMEWORK ASSIGNMENT #10: Solutions Canonical Trasformations and Hamilton-Jacobi Equation due by 9:35 a.m. Mon 11/05 Instructor: D. L. Maslov [email protected] 392-0513 Rm. 2114 Please help your instructor by doing your work neatly. 1. Goldstein, Problem 9.10 (note that x is used for q) (33 points) Solution a) αp x P = βx2 Q = is this a canonical transformation? A criterion for canonicity is ∂P ∂Q ∂P ∂Q − =1 ∂p ∂x ∂x ∂p α = −2αβ = 1 = −2βx x αβ = −1/2 [P, Q]p,x = b) Re-write the transformation as Qx α P = βx2 . p = Then it is a transformation from (x, Q) to (p, P ) . Hence the generating function is a function of x and Q (type I) Qx ∂F = ∂x α ∂F P = − = −βx2 . ∂Q p = Integrating the first equation, we get F = Qx2 . 2α Using the canonicity condition α = −1/2β, the equivalent form of F is F = −βQx2 Then the result for P is obtained as well. c) Harmonic oscillator H= 1 p2 + kx2 . 2m 2 Hamilton’s equations p ∂H = ∂p m ∂H ṗ = − = −kx ∂x ẋ = 2 or ẍ = ṗ k = − x = −ω 2 x m m Solution x (t) = x0 cos ωt + ẋ0 sin ωt, ω where x0 and ẋ0 are the initial coordinate and velocity, correspondingly. Choose ẋ0 = 0. Then x (t) = x0 cos ωt p (t) = mẋ = −mωx0 sin ωt New coord. and momentum p = −αmω0 tan ωt x P (t) = βx2 = x20 cos2 ωt. Q (t) = α 2. Goldstein, Problem 9.32 (33 points) Solution: H = q1 p1 − q2 p2 − aq12 + bq22 p1 − aq1 F1 = , F2 = q 1 q 2 . q2 Calculate the Poisson bracket for F1 [H, F1 ] = ∂H ∂F1 ∂H ∂F1 ∂H ∂F1 ∂H ∂F1 + − − . ∂p1 ∂q1 ∂p2 ∂q2 ∂q1 ∂p1 ∂q2 ∂p2 Calculate the derivatives ∂H ∂p1 ∂H ∂p2 ∂H ∂q1 ∂H ∂q2 ∂F1 ∂q1 ∂F1 ∂q2 ∂F1 ∂p1 ∂F1 ∂p2 = q1 = −q2 = p1 − 2aq1 = −p2 + 2bq2 a q2 p1 − aq1 = − q22 1 = q2 = − = 0. Substituting the derivatives into the Poisson brackets, we find p1 − aq1 1 a + (−q2 ) − − (p1 − 2aq1 ) =0 q1 − 2 q2 q2 q2 Jacobi identity (Goldstein, eq. 9.75e) states [u, [v, w]] + [v, [w, u]] + [w, [u, v]] = 0. 3 If u = H and v and w are the two constants of motion, so that [H, v] = [H, w] = 0 then the Poisson bracket of v and w is also a constant of motion. Check the Poisson bracket of F1 and F2 ∂F1 ∂F2 ∂F1 ∂F2 ∂F1 ∂F2 ∂F1 ∂F2 + − − ∂p1 ∂q1 ∂p2 ∂q2 ∂q1 ∂p1 ∂q2 ∂p2 1 = q2 = 1 q2 [F1 , F2 ] = This bracket is another constant of motion but it is a trivial constant, which is not an algebraic function of q, p.