Solutions to Problems in Goldstein, Classical Mechanics, Second Edition Homer Reid June 17, 2002 Chapter 8 Problem 8.4 The Lagrangian for a system can be written as L = ax˙ 2 + b p y˙ + cx˙ y˙ + f y 2 x˙ z˙ + g y˙ − k x2 + y 2 , x where a, b, c, f, g, and k are constants. What is the Hamiltonian? What quantities are conserved? Problem 8.5 A dynamical system has the Lagrangian L = q˙12 + q˙22 + k1 q12 + k2 q˙1 q˙2 , a + bq12 where a, b, k1 and k2 are constants. Find the equations of motion in the Hamiltonian formulation. Rewriting the Lagrangian in the form of Goldstein’s (8-16), we have L = k1 q12 + 1 2 q˙1 q˙2 1 2 k2 k2 2 a+bq12 q˙1 q˙2 Homer Reid’s Solutions to Goldstein Problems: Chapter 8 2 From this we can immediately identify the T matrix and its inverse: 2 2 k2 −k2 a + bq12 2 −1 a+bq 1 T= T = 2 k2 a+bq 2 4 − k22 (a + bq12 ) −k2 2 1 Then the Hamiltonian is 2 −k2 1 a + bq12 p1 2 a+bq 1 H= p1 p2 − k1 q12 p2 2 4 − k22 (a + bq12 ) −k2 2 p21 a + bq12 2 = − k2 p1 p2 + p2 − k1 q12 . 4 − k22 (a + bq12 ) a + bq12 Then the equations of motion are ∂H 2p1 a + bq12 q˙1 = = − k 2 p2 ∂p1 4 − k22 (a + bq12 ) a + bq12 ∂H a + bq12 {−kp1 − 2p2 } q˙2 = = ∂p2 4 − k22 (a + bq12 ) ∂H = something ugly p˙1 = − ∂q1 ∂H p˙2 = − =0 ∂q2 So in the Hamiltonian formulation there is one cylic variable. . but I still think this is much harder than the Lagrangian formulation for this problem. (b) Find an equivalent Lagrangian that is not explicitly dependent on time. (c) What is the Hamiltonian corresponding to this second Lagrangian. ∂p a (1) Then. and k are constants. which doesn’t make sense dimensionally in light of the rest of the terms in the Hamiltonian. α. using a reverse Legendre transformation. and what is the relationship between the two Hamiltonians? (a) From the Hamilton equations of motion. Note: I think there must be a misprint in the book. L = pq˙ − H 2 p2 p ba kq 2 − bqpe−αt − − bqpe−αt + q 2 e−αt (α + be−αt ) + a 2a 2 2 2 2 ba kq p − q 2 e−αt (α + be−αt ) − . the coefficient of p2 in the first term is printed there as 1/2α. = 2a 2 2 = (2) We would now like to eliminate p from this equation in favor of q. (a) Find a Lagrangian corresponding to this Hamiltonian.Homer Reid’s Solutions to Goldstein Problems: Chapter 8 3 Problem 8.6 A Hamiltonian of one degree of freedom has the form H= ba kq 2 p2 − bqpe−αt + q 2 e−αt (α + be−αt ) + . (3) q˙ − αq − 2 2 2 L= . ˙ From (1) we have p = aq˙ + bqae−αt p2 = a2 q˙2 + 2bq qa ˙ 2 e−αt + b2 q 2 a2 e−2αt so (2) becomes aq˙2 1 baαq 2 −αt b2 aq 2 −2αt kq 2 + bq qae ˙ −αt + b2 q 2 ae−2αt − e − e − 2 2 2 2 2 aq˙2 kq 2 1 −αt = + bqae . as the units do work out correctly if we put 1/2a for the coefficient of that term. b. 2a 2 2 where a. q˙ = p ∂H = − bqe−αt . It seems reasonable to assume that someone got their Greek and Roman letters mixed up. ax2 ). (5) . (c) From (4). ˙ t) to the Lagrangian without changing the resulting equations of motion. We’ll denote the coordinates of the suspension point as (x. zm ) = (x + L sin θ. q. ax2 − L cos θ). Obtain the Hamilton’s equations of motion. if θ is the angle the pendulum makes with the vertical (θ = 0 when the mass point is precisely at 6:00.9 The point of suspension of a simple pendulum of length l and mass m is constrained to move on a parabola z = ax2 in the vertical plane. Then. we consider d ab 2 −αt 0 . q e L =L− dt 2 The derivative term just cancels the second term in (3). 2a 2 Problem 8. z − L cos θ) = (x + L sin θ. Derive a Hamiltonian governing the motion of the pendulum and its point of suspension. The potential energy of the system is L = mgz = mg(ax2 − L cos θ). the new canonical momentum is p= ∂L0 = aq˙ ∂ q˙ Then the Legendre transformation defining the Hamiltonian reads H = pq˙ − L = = kq 2 aq˙2 + 2 2 p2 kq 2 + . leaving L0 = aq˙2 kq 2 − 2 2 (4) which is just the Lagrangian of a one-dimensional harmonic oscillator. and grows in the positive direction as the mass point moves counter-clockwise) then the coordinates of the mass point are (xm .4 Homer Reid’s Solutions to Goldstein Problems: Chapter 8 (b) Since we can the total time derivative of any function f (q. z) = (x. L=T −L o mn (1 + 4a2 x2 )x˙ 2 + L2 θ˙2 + 2Lθ˙x˙ [cos θ + 2ax sin θ] − mgax2 + mgL cos θ. Then from Goldstein’s (??) we can write 1 1 × H= 2m L2 (sin θ − 2ax cos θ)2 px L2 −L[cos θ + 2ax sin θ] − L0 (x. θ) + x˙ θ˙ L[cos θ + 2ax sin θ] L2 θ˙ 2 (7) where L0 (x.Homer Reid’s Solutions to Goldstein Problems: Chapter 8 5 The kinetic energy is m 2 2 (x˙ + z˙m ) 2 m m (x˙ + Lθ˙ cos θ)2 + (2axx˙ + Lθ˙ sin θ)2 = 2 o mn (1 + 4a2 x2 )x˙ 2 + L2 θ˙2 + 2Lθ˙ x˙ [cos θ + 2ax sin θ] . = 2 For convenience in converting to the Hamiltonian. θ) (px pθ ) pθ −L[cos θ + 2ax sin θ] (1 + 4a2 x2 ) 1 1 = × 2 2m L (sin θ − 2ax cos θ)2 2 2 2 2 2 L px − 2L[cos θ + 2ax sin θ]px pθ + (1 + 4a x )pθ − L0 (x. θ) . we may write this in the language of Goldstein’s (8-16): m x˙ (1 + 4a2 x2 ) L[cos θ + 2ax sin θ] L = L0 (x. from (7) and (6). θ) = −mgax2 + mgL cos θ. = 2 T = (6) Then the Lagrangian for the system is.