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Typology: Exercises

2017/2018

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Download Selected Solutions to Chapter 8 for Classical Mechanics by Goldstein and more Exercises Physics in PDF only on Docsity! Theoretical Dynamics September 24, 2010 Homework 3 Instructor: Dr. Thomas Cohen Submitted by: Vivek Saxena 1 Goldstein 8.1 1.1 Part (a) The Hamiltonian is given by H(qi, pi, t) = piq̇i − L(qi, q̇i, t) (1) where all the q̇i’s on the RHS are to be expressed in terms of qi, pi and t. Now, dH = ∂H ∂qi dqi + ∂H ∂pi dpi + ∂H ∂t dt (2) From (1), dH = pidq̇i + q̇idpi − dL = pidq̇i + q̇idpi − ( ∂L ∂qi dqi + ∂L ∂q̇i dq̇i + ∂L ∂t dt ) = −∂L ∂qi dqi + q̇idpi + ( pi − ∂L ∂q̇i ) dq̇i − ∂L ∂t dt (3) Comparing (2) and (3) we get ∂H ∂qi = −∂L ∂qi = −ṗi (2nd equality from Hamilton’s equation) (4) q̇i = ∂H ∂qi (also Hamilton’s equation) (5) pi − ∂L ∂q̇i = 0 (H is not explicitly dependent on q̇i) (6) −∂L ∂t = ∂H ∂t (7) From (4) and (6) we have d dt ( ∂L ∂q̇i ) − ∂L ∂qi = 0, i = 1, 2, . . . , n (8) which are the Euler-Lagrange equations. 3 - 1 1.2 Part (b) L′(p, ṗ, t) = −ṗiqi −H(q, p, t) (9) = piq̇i −H(q, p, t)− d dt (piqi) (10) = L(q, q̇, t)− d dt (piqi) (11) = L(q, q̇, t)− ṗiqi − piq̇i (12) So, dL′ = ∂L′ ∂pi dpi + ∂L′ ∂ṗi dṗi + ∂L′ ∂t dt (13) = −q̇idpi − qidṗi + ∂L ∂t dt (from (9)) (14) Comparing (12) and (13) we get q̇i = − ∂L′ ∂pi (15) qi = − ∂L′ ∂ṗi (16) Thus the equations of motion are d dt ( ∂L′ ∂ṗi ) − ∂L ′ ∂pi = 0, i = 1, 2, . . . , n (17) 2 Goldstein 8.6 Hamilton’s principle is δ ∫ Ldt = 0 (18) or equivalently δ ∫ 2Ldt = 0 (19) We can subtract the total time derivative of a function whose variation vanishes at the end points of the path, from the integrand, without invalidating the variational principle. This is because such a function will only contribute to boundary terms involving the variation of qi and pi at the end points of the path, which vanish by assumption. Such a function is piqi. So, the ‘modified’ Hamilton’s principle is δ ∫ ( 2L− d dt (piqi) ) dt = 0 (20) 3 - 2 where t′ = dt/dθ and θ is some parameter. Using the constrained form of Hamilton’s equations we get q̇i = (1 + λ) ∂H ∂pi , i = 1, 2, . . . n (40) ṗi = −(1 + λ) ∂H ∂qi , i = 1, 2, . . . n (41) q̇n+1 = λ (42) ṗn+1 = −(1 + λ) ∂H ∂t = −∂L ∂t (43) By regarding H ′ = (1+λ)H as an equivalent Hamiltonian, these equations are the required (2n+ 2) equations of motion. Also, λ = q̇n+1 = dt/dθ. 4 Goldstein 8.26 4.1 Part (a) In the given configuration, both springs elongate or compress by the same magnitude. Suppose q denotes the position of the mass m from the left end. At t = 0, q(0) = a/2, but the unstretched lengths of both springs are given to be zero. Therefore, the elongation (compression) of spring k1 is q and the compression (elongation) of spring k2 is q. The potential energy is V = 1 2 k1q 2 + 1 2 k2q 2 = 1 2 (k1 + k2)q 2 (44) The kinetic energy is T = 1 2 mq̇2 (45) The Lagrangian is L = T − V = 1 2 mq̇2 − 1 2 (k1 + k2)q 2 (46) The momentum canonically conjugate to the coordinate q is pq = ∂L ∂q̇ = mq̇ (47) So the Hamiltonian is H = pq q̇ − L = 1 2 mq̇2 + 1 2 (k1 + k2)q 2 (48) that is, H(q, pq, t) = p2q 2m + 1 2 (k1 + k2)q 2 (49) Clearly, the Hamiltonian equals the total energy E. The energy is conserved since, dE dt = mq̇q̈ + (k1 + k2)qq̇ = q̇(−(k1 + k2)q) + (k1 + k2)qq̇ = 0 (50) where we have used the equation of motion1. In this case, the Hamiltonian is also conserved. 1 d dt ( ∂L ∂q̇ ) − ∂L ∂q = 0 =⇒ mq̈ + (k1 + k2)q = 0 3 - 5 4.2 Part (b) Substituting q = Q+b sin(ωt) and q̇ = Q+bω cos(ωt) into the expression for the Lagrangian, we get L(Q, Q̇, t) = 1 2 m(Q̇+ bω cos(ωt))2 − 1 2 (k1 + k2)(Q+ b sin(ωt)) 2 (51) and the momentum canonically conjugate to the coordinate Q is given by pQ = ∂L ∂Q̇ = m(Q̇+ bω cos(ωt)) (52) So the Hamiltonian becomes H(Q, pQ, t) = pQQ̇− L(Q, Q̇, t) (53) = m(Q̇+ bω cos(ωt))Q̇− 1 2 m(Q̇+ bω cos(ωt))2 + 1 2 (k1 + k2)(Q+ b sin(ωt)) 2 = mQ̇2 2 − mb 2ω2 2 cos2(ωt) + 1 2 (k1 + k2)(Q+ b sin(ωt)) 2 = p2Q 2m − pQbω cos(ωt) + 1 2 (k1 + k2)(Q+ b sin(ωt)) 2 (54) The Hamiltonian is now explicitly dependent on time, and hence is not conserved, as is confirmed by the fact that dH/dt 6= 0. The energy is given by E = T + V = 1 2 (Q̇+ bω cos(ωt))2 + 1 2 (k1 + k2)(Q+ bω sin(ωt)) 2 (55) So, dE dt = m(Q̇+ bω cos(ωt))(Q̈− bω2 sin(ωt)) + (k1 + k2)(Q+ b sin(ωt))(Q̇+ bω cos(ωt)) = (Q̇+ bω cos(ωt))(m(Q̈−Bω2 sin(ωt)) + (k1 + k2)(Q+ b sin(ωt))) = (Q̇+ bω cos(ωt))(mq̈ + (k1 + k2)q) (56) = 0 (c.f. footnote on prev. page) (57) Therefore, the energy is conserved, as expected (the energy is still given by T + V , but the Hamiltonian is not T+V anymore, as the relationship connecting the generalized coordinate to the cartesian coordinate is now explicitly dependent on time). 5 Goldstein 8.23 5.1 Part (a) The Lagrangian for the system is L = 1 2 m(v · v) + eA(r) · v − eV (r) (58) The canonical momentum is p = ∂L ∂v = mv + eA (59) 3 - 6 So the Hamiltonian is H = p · v − L (60) = (mv + eA) · v − ( 1 2 m(v · v) + eA(r) · v − eV (r) ) = m 2 v · v + eV (r) = (p− eA) · (p− eA) 2m + eV (r) (61) = 1 2m (p2 − 2ep ·A+ e2A2) + eV (r) (62) Now, p ·A = p · 1 2 B × r = 1 2 B · (r × p) (63) = 1 2 B · J (64) where J = r × p denotes the angular momentum. Also, A2 = 1 4 (B × r) · (B × r) = 1 4 B2r2 (as B is perpendicular to r) (65) So the Hamiltonian of equation (58) becomes H = p2 2m − e 2m B · J + e 2 8m B2r2 + eV (r) (66) 5.2 Part (b) Let vlab = (ẋ, ẏ) denote the velocity of the particle in the lab frame, and v ′ = (ẋ′, ẏ′) denote the velocity in the rotating frame. Without loss of generality, we may assume that motion is confined to the xy-plane. We first derive a relationship between the Hamiltonian in a rotating frame with that in a non-rotating frame (in this case, the lab frame). The coordinates are related by x = x′ cos(ωt)− y′ sin(ωt) (67) y = x′ sin(ωt) + y′ cos(ωt) (68) Here, it has been assumed that the rotation is counterclockwise, i.e. ω > 0 for counter- clockwise rotation. So the velocity components are related by ẋ = ẋ′ cos(ωt)− ẏ′ sin(ωt)− ω(x′ sin(ωt) + y′ cos(ωt)) (69) y = ẋ′ sin(ωt) + ẏ′ cos(ωt)− ω(x′ cos(ωt)− y′ sin(ωt)) (70) 3 - 7 So, [Li, f(r)]PB = − ∂Li ∂pα ∂f(r) ∂xα = −xα r ∂(ijkxjpk) ∂pα ∂f(r) ∂r = − ijkxαxjδα,k r ∂f ∂r = − ijkxjxk r ∂f ∂r = −(r × r)i r ∂f ∂r = 0 (88) 7 Problem 2 7.1 Part (a) [ξi, ξj ] = ∂ξi ∂ηα Jαβ ∂ξj ∂ηβ (89) So, d d [ξi, ξj ] = d d ( ∂ξi ∂ηα Jαβ ∂ξj ∂ηβ ) = d d ( ∂ξi ∂ηα ) Jαβ ∂ξj ∂ηβ + ∂ξi ∂ηα Jαβ d d ( ∂ξj ∂ηβ ) = ∂ ∂ηα ( dξi d ) Jαβ ∂ξj ∂ηβ + ∂ξi ∂ηα Jαβ ∂ ∂ηβ ( dξj d ) = ∂ ∂ηα ([ξi, g]) Jαβ ∂ξj ∂ηβ + ∂ξi ∂ηα Jαβ ∂ ∂ηβ ([ξj , g]) = ∂ ∂ηα ( ∂ξi ∂ηγ Jγδ ∂g ∂ηδ ) Jαβ ∂ξj ∂ηβ + ∂ξi ∂ηα Jαβ ∂ ∂ηβ ( ∂ξj ∂ηω Jωθ ∂g ∂ηθ ) = ( ∂2ξi ∂ηα∂ηγ Jγδ ∂g ∂ηδ + ∂ξi ∂ηγ Jγδ ∂2g ∂ηαηδ ) Jαβ ∂ξj ∂ηβ + ∂ξi ∂ηα Jαβ ( ∂2ξj ∂ηβ∂ηω Jωθ ∂g ∂ηθ + ∂ξj ∂ηω Jωθ ∂2g ∂ηβ∂ηθ ) (90) Now, for = 0, ξ = η, so the second order terms ∂2ξi ∂ηα∂ηγ ∣∣∣∣ =0 , ∂2ξj ∂ηβ∂ηω ∣∣∣∣ =0 3 - 10 equal zero. So, d d [ξi, ξj ] ∣∣∣∣ =0 = ∂ξi ∂ηγ Jγδ ∂2g ∂ηα∂ηδ Jαβ ∂ξj ∂ηβ + ∂ξi ∂ηα Jαβ ∂ξj ∂ηω Jωθ ∂2g ∂ηβ∂ηθ (91) = ∂ξi ∂ηα Jαβ ∂2g ∂ηγ∂ηβ Jγδ ∂ξj ∂ηδ + ∂ξi ∂ηα Jαβ ∂2g ∂ηγ∂ηβ Jδγ ∂ξj ∂ηδ (92) which is obtained by switching γ ↔ α, δ ↔ β in the first term and θ ↔ γ, ω ↔ δ in the second term. As J is antisymmetric, Jδγ = −Jγδ. So, d d [ξi, ξj ] ∣∣∣∣ =0 = ∂ξi ∂ηα Jαβ ∂2g ∂ηγ∂ηβ Jγδ ∂ξj ∂ηδ − ∂ξi ∂ηα Jαβ ∂2g ∂ηγ∂ηβ Jγδ ∂ξj ∂ηδ (93) = 0 (94) So upto O(2), we have d[ξi, ξj ]/d = 0 and so [ξi, ξj ] is constant up to O( 2). As [ξi, ξj ]|=0 = [ηi, ηj ] = Jij , we have [ξi, ξj ] = Jij upto O( 2). So, ξ is a canonical transformation upto O(2). 7.2 Part (b) First of all, η(ξ, 0) = ξ (95) because at t = 0, the canonical coordinates overlap in phase space. So, using the result of part (a), η(ξ, t) can be treated as a one-parameter family of canonical transformations (parametrized by t), provided there exists some function g satisfying dξi dt = [ξi, g]PB (96) This condition is seen to be satisfied if g is taken as the Hamiltonian H, for in this case, [ξi, H]PB = ∂ξi ∂ξk Jkj ∂H ∂ξj (97) = δikJkj ∂H ∂ξj (98) = Jij ∂H ∂ξj (99) = ξ̇i (100) where the last equality is obtained via Hamilton’s equations. So, taking g = H satisfies the Poisson bracket relation with H acting as the generator of time translations. 3 - 11