Homework 4. Goldstein 9.7. Part (a) Theoretical Dynamics October 01, 2010 (1) P i = F 1. Q i. p i = F 1 (3) q i (5) P i (6)

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1 Theoretical Dynamics October 01, 2010 Instructor: Dr. Thomas Cohen Homework 4 Submitted by: Vivek Saxena Goldstein 9.7 Part (a) F 1 (q, Q, t) F 2 (q, P, t) P i F 1 Q i (1) F 2 (q, P, t) F 1 (q, Q, t) + P i Q i (2) F 1 (q, Q, t) F 3 (p, Q, t) p i F 1 q i (3) F 3 (p, Q, t) F 1 (q, Q, t) p i q i (4) F 1 (q, Q, t) F 4 (p, P, t) p i F 1 q i (5) P i F 1 Q i (6) F 4 (p, P, t) F 1 (q, Q, t) p i q i + P i Q i (7) F 2 (q, P, t) F 3 (p, Q, t) p i F 2 q i (8) Q i F 2 P i (9) F 3 (p, Q, t) F 2 (q, P, t) p i q i P i Q i (10) F 2 (q, P, t) F 4 (p, P, t) p i F 2 q i (11) F 4 (p, P, t) F 2 (q, P, t) p i q i (12) 4-1

2 F 3 (p, Q, t) F 4 (p, P, t) P i F 3 Q i (13) F 4 (p, P, t) F 3 (p, Q, t) + P i Q i (14) Part (b) For an identity transformation, F 2 function is q i P i and by equation (7), the type 4 generating F 4 (p, P, t) F 2 (q, P, t) p i q i (15) q i P i p i q i 0 as p i F 2 q i For an exchange transformation, F 1 q i Q i and by equation (4), the type 3 generating function is Part (c) P i F 3 (p, Q, t) F 1 (q, Q, t) p i q i (16) q i Q i p i q i 0 as p i F 1 q i Q i (17) Consider a type 2 generating function F 2 (q, P, t) of the old coordinates and the new momenta, of the form F 2 (q, P, t) f i (q 1,..., q n ; t)p i g(q 1,..., q n ; t) (18) where f i s are a set of independent functions, and g i s are differentiable functions of the old coordinates and time. The new coordinates Q i are given by In particular, the function Q i F 2 P i f i (q 1,..., q n ; t) (19) f i (q 1,..., q n ; t) R ij q j (20) where R ij is the (i, j)-th element of a N N orthogonal matrix, generates an orthogonal transformation of the coordinates. Now, p j F 2 q j This equation can be written in matrix form, as f i q j P i g q j R ij P i g q j (21) p f q P g q (22) 4-2

3 where p denotes the N 1 column vector (p 1,..., p N ) T, g/ q denotes the N 1 column vector ( g/ q 1,..., g/ q n ) T, and f q denotes the N N matrix with entries ( ) f f i R ij (23) q q j From (22), the new momenta are given by ij ( ) f 1 ( P p + g ) q q ( R 1 p + g ) q (24) (25) R 1 (p + q g) (26) As R is an orthogonal matrix, RR T R T R I, so R 1 R T is also an orthogonal transformation. This gives the required result: the new momenta are given by the orthogonal transformation (R 1 ) of an n-dimensional vector (p + q g), whose components are the old momenta (p) plus a gradient in configuration space ( q g). Goldstein 9.25 Part (a) The given Hamiltonian is The equation of motion for q is H 1 2 ( ) 1 q 2 + p2 q 4 (27) q H p pq4 (28) Part (b) Suppose we let Q 2 1/q 2 and P 2 p 2 q 4. Then, Q ±1/q and P ±pq 2. Now, {Q, P } {±1/q, ±pq 2 } {q 1, pq 2 } {q 1, p}q 2 + p{q 1, q 2 } ( ) q 1 p q p q 1 p q 2 + p 0 p q ( 1q ) 2 q

4 the signs on both Q and P cannot be identical. We take Q 1 q (29) P pq 2 (30) which is a valid canonical transformation. This gives the Hamiltonian, The equations of motion are H(Q, P ) 1 2 (P 2 + Q 2 ) (31) Q H P P (32) P H Q (33) Q Q + Q 0, the solution to which is of the form Q A cos t + B sin t. P Q B cos t A sin t. Now, This gives q 1 Q (A cos t + B sin t) 1 (34) p P Q 2 (B cos t A sin t)(a cos t + B sin t) 2 (35) so, and hence q (A cos t + B sin t) 2 ( A sin t + B cos t) (36) pq 4 (B cos t A sin t)(a cos t+b sin t) 2 (A cos t+b sin t) 4 (B cos t A sin t)(a cos t+b sin t) 2 q (37) the solution to the transformed equation for Q satisfies the original equation of motion for q. Problem 1 Part (a) {X, P x } {x + ɛ, p x } {x, p x } 1 (38) {Y, P y } {y, p x } 1 (39) 4-4

5 {Z, P z } {z, p z } 1 (40) {X, P y } {X, P z } {Y, P x } {Y, P z } {Z, P x } {Z, P y } 0 (41) {X, X} {Y, Y } {Z, Z} {P x, P x } {P y, P z } {P z, P z } {P x, P y } {P y, P z } {P z, P x } 0 (42) this is a canonical transformation. It corresponds to a translated canonical coordinate system (translation along the x-direction in phase space). So P x is the generator of the canonical transformation. Part (b) dx dɛ [X, P x] 1 (43) {X, P x } {x cos ɛ + y sin ɛ, p x cos ɛ + p y sin ɛ} cos 2 ɛ{x, p x } + sin 2 {y, p y } 1 (44) {Y, P y } { x sin ɛ + y cos ɛ, p x sin ɛ + p y cos ɛ} sin 2 ɛ{x, p x } + cos 2 ɛ{y, p y } 1 (45) {Z, P z } {z, p z } Using properties of the Poisson Bracket, we also have 1 (46) {X, P y } {X, P z } {Y, P x } {Y, P z } {Z, P x } {Z, P y } 0 (47) {X, X} {Y, Y } {Z, Z} {P x, P x } {P y, P z } {P z, P z } {P x, P y } {P y, P z } {P z, P x } 0 (48) this is a canonical transformation. It corresponds to a rotation about the z-axis in phase space. dx x sin ɛ + y cos ɛ (49) dɛ whereas {X, L z } {x cos ɛ + y sin ɛ, xp y yp x } x sin ɛ y cos ɛ (50) dx dɛ {X, L z} (51) Therefore, L z is the generator of the canonical transformation. 4-5

6 Part (c) {X, P x } {x, p x + ɛ} 1 (52) {Y, P y } {y, p y } 1 (53) {Z, P z } {z, p z } Using properties of the Poisson Bracket, we also have 1 (54) {X, P y } {X, P z } {Y, P x } {Y, P z } {Z, P x } {Z, P y } 0 (55) {X, X} {Y, Y } {Z, Z} {P x, P x } {P y, P z } {P z, P z } {P x, P y } {P y, P z } {P z, P x } 0 (56) this is a canonical transformation. It corresponds to a translation along the p x direction in phase space. Now, ( Px X {P x, X} P ) x X 1 dp x (57) q i p i p i q i dɛ Therefore, X is the generator of the canonical transformation. Part (d) {X, P x } {(1 + ɛ)x, (1 + ɛ) 1 p x } 1 (58) {Y, P y } {(1 + ɛ)y, (1 + ɛ) 1 p y } 1 (59) {Z, P z } {(1 + ɛ)z, (1 + ɛ) 1 p z } Using properties of the Poisson Bracket, we also have 1 (60) {X, P y } {X, P z } {Y, P x } {Y, P z } {Z, P x } {Z, P y } 0 (61) {X, X} {Y, Y } {Z, Z} {P x, P x } {P y, P z } {P z, P z } {P x, P y } {P y, P z } {P z, P x } 0 (62) 4-6

7 this is a canonical transformation. It is a scaling transformation, which preserves the volume element in phase space. Suppose g is the generator of the scaling transformation. Then, X x [X, g] [(1 + ɛ)x, g] (63) ɛ which implies that is, the solution to which is x 1 + ɛ x g [x, g] x g x p x p x x (64) x 1 + ɛ g (65) p x g xp x 1 + ɛ + f(y, z, p y, p z ) (66) As dy/dɛ y [(1 + ɛ)y, g] and dz/dɛ z [(1 + ɛ)z, g], following a similar argument for Y and Z (or by symmetry) we get as the generator of the scaling transformation. Problem 2 As η is a canonical transformation, we have g xp x 1 + ɛ + yp y 1 + ɛ + zp z + constant (67) 1 + ɛ H ɛ H ɛ H {η i, g} H ξ j J jk g ξ k H η j J jk g η k H δ ij J jk g η k H J ij g η j {H, g} ġ ɛ {η i, g} (68) (as ξ is a canonical transformation) (as Poisson Brackets are invariant under canonical transformations) (69) But since H is conserved, H ɛ 0 and hence ġ 0. Therefore, g is conserved. 4-7

8 Problem 3 The quantity, which was found to be an invariant of the system, can be expressed in terms of the canonical coordinates x, y, p x, p y as (x, y, p x, p y ) 1 (p2 x p 2 y) mω2 (x 2 αy 2 ) (70) As is the conserved generator of a family of canonical transformations parametrized by an infinitesimal parameter ɛ, we must have We consider each condition separately below. δx ɛ{x, } (71) δy ɛ{y, } (72) δp x ɛ{p x, } (73) δp y ɛ{p x, } (74) δx ɛ{x, } 1 ɛ{x, (p2 x p 2 y) mω2 (x 2 αy 2 )} ɛ{x, p2 x } ɛp x m (75) (76) δy ɛ{y, } 1 ɛ{y, (p2 x p 2 y) mω2 (x 2 αy 2 )} ɛ{y, p2 y } ɛp y m (77) δp x ɛ{p x, } 1 ɛ{p x, (p2 x p 2 y) mω2 (x 2 αy 2 )} ɛ{p x, 1 2 mω2 x 2 } ɛmω 2 x (78) δp y ɛ{p y, } 1 ɛ{p y, (p2 x p 2 y) mω2 (x 2 αy 2 )} ɛ{p y, 1 2 mω2 αy 2 } ɛmω 2 αy (79) 4-8

9 Now, let ɛ δθ where θ is a parameter. Then, the above equations become dx p x dθ m (80) dy p y dθ m (81) dp x dθ mω 2 x (82) dp y dθ mω 2 αy (83) The solutions to which are and correspondingly d 2 x dθ 2 + ω2 x 0 (84) d 2 y dθ 2 + ω2 αy 0 (85) x A cos(ωθ) + B sin(ωθ) (86) y C cos(ω αθ) + D sin(ω αθ) (87) p x mωa sin(ωθ) + mωb cos(ωθ) (88) p y mω αc sin(ω αθ) mω αd cos(ω αθ) (89) Using the subscript 0 to denote the initial coordinates and momenta, we have x 0 A (90) y 0 C (91) p x0 mωb (92) p y0 mω αd (93) x x 0 cos(ωθ) + p x0 mω sin(ωθ) y y 0 cos(ω θ) p y 0 mω α sin(ω αθ) p x mωx 0 sin(ωθ) + p x0 cos(ωθ) p y mω αy 0 sin(ω αθ) + p y0 cos(ω αθ) Reverting to the notation in which x 0, p x0, y 0, p y0 denote the original coordinates and X, Y, P x, P y denote the canonically transformed coordinates, we get the form of the canonical transformation as X x cos(ωθ) + p x sin(ωθ) mω (94) P x p x cos(ωθ) mωx sin(ωθ) (95) Y y cos(ω αθ) p y mω α sin(ω αθ) (96) P y p y cos(ω αθ) + mω αy sin(ω αθ) (97) 4-9

10 where θ is an arbitrary parameter, such that θ 0 corresponds to the untransformed coordinates. This canonical transformation is composed of two rotations in the 4-dimensional phase space (one involving X and P x and the other involving Y and P y ), and its generator is the conserved quantity. Problem 4 Part (a) Now, F 2 (q, P, t) (q + 12 gt2 ) (P mgt) P 2 t (98) p F 2 q the canonical transformation is Part (b) P mgt (99) Q F 2 P q gt2 P t m q gt2 pt m gt2 (100) P p + mgt (101) Q q pt m 1 2 gt2 (102) {Q, Q} {q pt m 1 2 gt2, q pt m 1 2 gt2 } 0 (103) {P, P } {p + mgt, p + mgt} 0 (104) {Q, P } {q pt m 1 2 gt2, p + mgt} Q P q p Q P p q ( (1)(1) t ) (0) m 1 (105) the transformation satisfies the canonical Poisson Bracket relations. Part (c) The Lagrangian is L(q, q) 1 2 m q2 mgq (106) 4-10

11 The canonical momentum is p L q m q (107) So the Hamiltonian is Now, Q q pt m 1 2 gt2, so {Q, H} H p q L p2 + mgq (108) { q pt {q, p2 m 1 2 gt2, p2 } } + mgq } { pt m, mgq 1 {q, p2 } gt{p, q} p + gt (109) m Also Also, P p + mgt, so Q p gt (110) m dq dt Q + {Q, H} 0 (111) {P, H} {p + mgt, p2 + mgq} mg{p, q} mg (112) and P dp dt P mg (113) + {P, H} 0 (114) 4-11

12 Part (d) F 2 the Hamiltonian associated with Q, P is gt(p mgt) + (q + 12 ) gt2 ( mg) P 2 P gt 3 2 mg2 t 2 mgq P 2 (p + mgt)gt 3 2 mg2 t 2 (p + mgt)2 mgq p2 mgq mg2 t 2 (115) K H + F 2 p2 p2 + mgq mgq mg2 t 2 mg 2 t 2 (116) the Hamiltonian K is zero up to time-dependent constant term mg 2 t 2, but it is not a function of P and Q (which are constant with time, since {Q, H} {P, H} 0 as shown above). Part (e) Q and P are conserved quantities, that equal the initial position and the initial momentum respectively. They are constant with time, as q and p vary: Part (f) q(t 0) Q p(t 0) P F 2 q F 2 P mgt p (117) P gt mg2 t 2 mgq P 2 2 (118) H p2 + mgq 1 ( ) 2 F2 + mgq (119) q K H + F 2 mg 2 t 2 (as shown in part d) 4-12

13 implies H ( q, F ) 2 + F 2 q mg 2 t 2 (120) the Hamilton-Jacobi equation is satisfied, except for a time-dependent constant term appearing on the right hand side. Part (g) f(q, P, t) F 2 (q(q, P, t), P, t) (q + 12 ) gt2 (P mgt) P 2 t ( Q + pt ) m + gt2 (P mgt) P 2 t ( ) (P mgt)t Q + + gt 2 (P mgt) P 2 t m Also, p m q P mgt. ( Q + P t m L(q, q) p2 mgq ) (P mgt) P 2 t QP + P 2 t Qmgt gp t2 (121) f P 2 mgq 2P gt (122) 1 (P mgt)2 mg ( Q + P t m 1 ) 2 gt2 1 (P 2 + m 2 g 2 t 2 2P mgt) mgq P gt mg2 t 2 P 2 2P gt mgq + mg2 t 2 f(q, P, t) + mg 2 t 2 (123) L(q(Q, P, t), q(q, P, t)) f(q,p,t) up to a time-dependent term mg 2 t 2. Problem 5 The Hamilton-Jacobi equation, as expressed in the form H(q, S(q, P )) + S(q, P ) 0 (124) 4-13

14 was obtained by constructing a generating function of the form F F 2 (q, P, t) Q i P i where F 2 denotes a generic type-2 generating function. For such a choice of F, the Hamiltonian K H + F is zero. Now, consider a type-3 generating function F 3 of the old momenta and the new coordinates, such that the Hamiltonian K is zero. Therefore, Now, so, Q i K P i 0 (125) P i K 0 (126) Q i q i F 3 p i ( p F 3 ) i (127) H(q(Q, p), p, t) + F 3 (Q, p, t) 0 (128) where the old coordinates q have been expressed in terms of the old momenta and the new coordinates using equation (127). This is a PDE in (n + 1) variables p 1,..., p n, t. Let S denote the solution of this PDE. Then, a solution of the form, F 3 S S(p 1,..., p n ; α 1,..., α n+1 ; t) (129) where Q i α i are the constants of motion (for i 1,..., n), is consistent with equation (125). Here the constant α n+1 must be a constant of integration, so the physically meaningful solution is of the form in terms of S, equation (128) can be written as which is of the desired form. S S(p 1,..., p n ; α 1,... α n ; t) (130) H( p S, S p, t) + (Q, p, t) 0 (131) 4-14