Homework 3. 1 Goldstein Part (a) Theoretical Dynamics September 24, The Hamiltonian is given by

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1 Theoretical Dynamics September 4, 010 Instructor: Dr. Thomas Cohen Homework 3 Submitted by: Vivek Saxena 1 Goldstein Part (a) The Hamiltonian is given by H(q i, p i, t) = p i q i L(q i, q i, t) (1) where all the q i s on the RHS are to be expressed in terms of q i, p i and t. Now, From (1), Comparing () and (3) we get dh = H dq i + H dp i + H p i t dh = p i d q i + q i dp i dl ( L = p i d q i + q i dp i dq i + L d q i + L q i = L ( dq i + q i dp i + p i L ) q i dt () ) t dt d q i L dt (3) t H = L = ṗ i (nd equality from Hamilton s equation) (4) q i = H (also Hamilton s equation) (5) p i L q i = 0 (H is not explicitly dependent on q i ) (6) L t From (4) and (6) we have d dt = H t ( L q i which are the Euler-Lagrange equations. ) L = 0, i = 1,,..., n (8) (7) 3-1

2 1. Part (b) L (p, ṗ, t) = ṗ i q i H(q, p, t) (9) = p i q i H(q, p, t) d dt (p iq i ) (10) = L(q, q, t) d dt (p iq i ) (11) = L(q, q, t) ṗ i q i p i q i (1) So, Comparing (1) and (13) we get dl = L dp i + L dṗ i + L p i ṗ i t dt (13) = q i dp i q i dṗ i + L dt (from (9)) (14) t q i = L p i (15) q i = L ṗ i (16) Thus the equations of motion are ( ) d L L dt ṗ i p i = 0, i = 1,,..., n (17) Goldstein 8.6 Hamilton s principle is or equivalently δ δ L dt = 0 (18) L dt = 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 q i and p i at the end points of the path, which vanish by assumption. Such a function is p i q i. So, the modified Hamilton s principle is ( δ L d ) dt (p iq i ) dt = 0 (0) 3 -

3 Using the Legendre transformation, this becomes δ (p i q i H p i q i ṗ i q i ) dt = 0 (1) = δ (H + ṗ i q i p i q i ) dt = 0 () now, ṗ i q i p i q i = [ q 1... q n p 1... p n ] So () becomes 1 n ( 0n n 1 n n 1 n n 0 n n ) n n q 1.. q n ṗ 1.. ṗ n (3) n 1 = η T J η (4) which is the required form of Hamilton s principle. 3 Goldstein 8.9 δ (H + η T J η ) dt = 0 (5) The constraints can be incorporated into the Lagrangian L by defining a constrained Lagrangian L c, as L c (q, q, t) = L(q, q, t) λ k ψ k (q, p, t) (6) k Applying Hamilton s principle, and using the Legendre transformation for L, we get ( δ p i q i H(q, p, t) ) λ k ψ k (q, p, t) dt = 0 (7) k By analogy with the constrained Lagrangian, we can define a constrained Hamiltonian H c as H c (q, p, t) = H(q, p, t) + λ k ψ k (q, p, t) (8) k Since both the terms are functions of q i, p i and t, this is a good Hamiltonian. Equation (7) can then be written as δ (p i q i H c (q, p, t)) dt = 0 (9) 3-3

4 This bears a resemblance to the usual variational principle in Hamiltonian mechanics, for a Hamiltonian H c. So the Hamilton equations are q i = H c p i ṗ i = H c which become q i = H + p i k ṗ i = H + k λ k ψ k p i (30) λ k ψ k (31) Time as a canonical variable If time t is treated as a canonical variable, we define q n+1 = t. By Hamilton s equations ṗ n+1 = H q n+1 (3) = H t = dh dt (33) (34) and H q n+1 = (35) p n+1 = 1 (since q n+1 = t) (36) As the Hamiltonian contains terms of the form p i q i for each coordinate and its canonical momentum, in order to incorporate the constraint imposed by the inclusion of time as the (n + 1) th canonical variable, we include a term of the form p n+1 q n+1 = p n+1 to the Hamiltonian to set up the constraint. Equivalently, the constraint can be obtained by integrating equation (34) above, and is given by H(q 1,..., q n, q n+1 ; p 1,..., p n ) + p n+1 = 0 (37) Hamilton s principle, can be written as δ δ (p i q i H)dt = 0 (38) (p i q i H)t dθ = 0 (39) 3-4

5 where t = dt/dθ and θ is some parameter. Using the constrained form of Hamilton s equations we get q i = (1 + λ) H p i, i = 1,,... n (40) ṗ i = (1 + λ) H, i = 1,,... n (41) q n+1 = λ (4) ṗ n+1 = (1 + λ) H t = L t (43) By regarding H = (1 + λ)h as an equivalent Hamiltonian, these equations are the required (n + ) equations of motion. Also, λ = q n+1 = dt/dθ. 4 Goldstein 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/, but the unstretched lengths of both springs are given to be zero. Therefore, the elongation (compression) of spring k 1 is q and the compression (elongation) of spring k is q. The potential energy is The kinetic energy is The Lagrangian is V = 1 k 1q + 1 k q = 1 (k 1 + k )q (44) T = 1 m q (45) L = T V = 1 m q 1 (k 1 + k )q (46) The momentum canonically conjugate to the coordinate q is So the Hamiltonian is that is, p q = L q = m q (47) H = p q q L = 1 m q + 1 (k 1 + k )q (48) H(q, p q, t) = p q m + 1 (k 1 + k )q (49) Clearly, the Hamiltonian equals the total energy E. The energy is conserved since, de dt = m q q + (k 1 + k )q q = q( (k 1 + k )q) + (k 1 + k )q q = 0 (50) where we have used the equation of motion 1. In this case, the Hamiltonian is also conserved. ( ) 1 d L L = 0 = m q + (k1 + k)q = 0 dt q q 3-5

6 4. Part (b) Substituting q = Q+b sin(ωt) and q = Q+bω cos(ωt) into the expression for the Lagrangian, we get L(Q, Q, t) = 1 m( Q + bω cos(ωt)) 1 (k 1 + k )(Q + b sin(ωt)) (51) and the momentum canonically conjugate to the coordinate Q is given by So the Hamiltonian becomes p Q = L Q = m( Q + bω cos(ωt)) (5) H(Q, p Q, t) = p Q Q L(Q, Q, t) (53) = m( Q + bω cos(ωt)) Q 1 m( Q + bω cos(ωt)) + 1 (k 1 + k )(Q + b sin(ωt)) = m Q mb ω cos (ωt) + 1 (k 1 + k )(Q + b sin(ωt)) = p Q m p Qbω cos(ωt) + 1 (k 1 + k )(Q + b sin(ωt)) (54) The Hamiltonian is now explicitly dependent on time, and hence is not conserved, as is confirmed by the fact that dh/dt 0. The energy is given by So, E = T + V = 1 ( Q + bω cos(ωt)) + 1 (k 1 + k )(Q + bω sin(ωt)) (55) de dt = m( Q + bω cos(ωt))( Q bω sin(ωt)) + (k 1 + k )(Q + b sin(ωt))( Q + bω cos(ωt)) = ( Q + bω cos(ωt))(m( Q Bω sin(ωt)) + (k 1 + k )(Q + b sin(ωt))) = ( Q + bω cos(ωt))(m q + (k 1 + k )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 Part (a) The Lagrangian for the system is L = 1 m(v v) + ea(r) v ev (r) (58) The canonical momentum is p = L v = mv + ea (59) 3-6

7 So the Hamiltonian is Now, H = p v L (60) ( ) 1 = (mv + ea) v m(v v) + ea(r) v ev (r) = m v v + ev (r) = (p ea) (p ea) + ev (r) m (61) = 1 m (p ep A + e A ) + ev (r) (6) p A = p 1 B r where J = r p denotes the angular momentum. Also, A = 1 (B r) (B r) 4 So the Hamiltonian of equation (58) becomes 5. Part (b) = 1 B (r p) (63) = 1 B J (64) = 1 4 B r (as B is perpendicular to r) (65) H = p m e m B J + e 8m B r + ev (r) (66) Let v lab = (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 counterclockwise 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

8 Therefore The Lagrangian in the lab frame is v lab = ẋ + ẏ = ẋ + ẏ + ω(xẏ ẋy) + ω r (71) L = 1 mv lab ev (r) (7) = 1 m(ẋ + ẏ ) + mω(x ẏ ẋ y ) + 1 mω r ev (r) (73) The momenta canonically conjugate to x and y in the rotating system are So the Hamiltonian in the rotating frame is p x = L ẋ = m(ẋ ωy ) (74) p y = L ẏ = m(ẏ + ωx ) (75) H = p x ẋ + p y ẏ L (76) = p x + p y m + ω(y p x x p y ) + ev (r) (77) = p x + p y m J zω + ev (r) (78) where J z denotes the angular momentum in the z-direction (direction of ω) as measured in the rotating frame. This means that for counterclockwise rotation along the z-axis, H rotating frame = H lab frame ωj z (79) This is the general relationship between Hamiltonians in the lab frame and rotating frame. For this problem, from equation (6) above, we have H lab frame = p m eb m J + e 8m B r + ev (r) (80) as B = Bẑ and J = J z ẑ = Jẑ. So, the Hamiltonian in the rotating frame is ( H rotating frame = p m ω + eb ) J z + e m 8m B r + ev (r) (81) It is interesting to note that if ω = ω c = eb m, then the term linear in the magnetic field vanishes. In this problem, it is given that ω = eb m (8) which is twice the frequency ω c. So, in this case, the Hamiltonian becomes H rotating frame = p m + eb m J z + e 8m B r + ev (r) (83) 3-8

9 6 Problem Part (a) The subscript P B is suppressed for clarity. [L i, L j ] P B = [ɛ iαβ x α p β, ɛ jγδ x γ p δ ] = ɛ iαβ ɛ jγδ [x α p β, x γ p δ ] = ɛ iαβ ɛ jγδ (x α [p β, x γ p δ ] + [x α, x γ p δ ]p β ) (as [A, BC] P B = B[A, C] P B + [A, B] P B C) = ɛ iαβ ɛ jγδ (x α [p β, x γ ]p δ + x α x γ [p β, p δ ] + [x α, x γ ]p δ p β + x γ [x α, p δ ]p β ) = ɛ iαβ ɛ jγδ ( δ βγ x α p δ + δ αδ x γ p β ) = ɛ iαβ ɛ jβδ ( x α p δ ) + ɛ iαβ ɛ jγα x γ p β = ɛ iαβ ɛ jδβ x α p δ ɛ iβα ɛ jγα x α p β = (δ ij δ αδ δ iδ δ jα )x α p δ (δ ij δ βγ δ iγ δ jβ )x γ p β = (δ ij x α p α x j p i ) (δ ij x β p β x i p j ) = x i p j x j p i Now, [L 1, L ] P B = x 1 p x p 1 = L 3, [L 1, L 3 ] P B = x 1 p 3 x 3 p 1 = L, [L 3, L ] P B = x 3 p x p 3 = ɛ 31 L 1, etc. So, x i p j x j p i = ɛ ijk L k. Hence, [L i, L j ] P B = ɛ ijk L k. 6. Part (b) For each i = 1,, 3, [L i, L ] P B = [L i, L j L j ] (sum over j) = L j [L i, L j ] + [L i, L j ]L j = L j (ɛ ijk L k ) + (ɛ ijk L k )L j So, = ɛ ijk L j L k = 0 (as ɛ ijk is antisymmetric under j k, while L j L k is symmetric.) [L, L ] P B = ê i [L i, L ] P B = 0 (84) 6.3 Part (c) For each i = 1,, 3, Now, r = x i x i and L i = ɛ ijk x j p k, so [L i, f(r)] P B = L i f(r) L i f(r) (85) x α p α p α x α f = r f x α x α r = x α f (86) r r f = 0 (87) p α 3-9

10 So, [L i, f(r)] P B = L i f(r) p α x α = x α r (ɛ ijk x j p k ) f(r) p α r f r = ɛ ijkx α x j δ α,k r = ɛ ijkx j x k f r r = (r r) i f r r = 0 (88) 7 Problem 7.1 Part (a) So, [ξ i, ξ j ] = ξ i η α η β (89) d dɛ [ξ i, ξ j ] = d ( ) ξi dɛ η α η β = d ( ) ξi + ξ ( ) i d ξj dɛ η α η β η α dɛ η β = ( ) dξi + ξ ( ) i dξj η α dɛ η β η α η β dɛ = ([ξ i, g]) + ξ i ([ξ j, g]) η α η β η α η β = ( ) ξi g J γδ + ξ i η α η γ η δ η β η α η β = ( ξ i η α η γ J γδ g + ξ i η α + ξ i g J γδ η δ η γ η α η δ ( ξ j g J ωθ + η β η ω η θ ) η β ) g J ωθ η ω η β η θ ( ) ξj g J ωθ η ω η θ (90) Now, for ɛ = 0, ξ = η, so the second order terms ξ i ξ j η α η γ, ɛ=0 η β η ω ɛ=0 3-10

11 equal zero. So, d dɛ [ξ i, ξ j ] = ξ i g J γδ + ξ i g J ωθ (91) ɛ=0 η γ η α η δ η β η α η ω η β η θ = ξ i g J γδ + ξ i g J δγ (9) η α η γ η β η δ η α η γ η β η δ 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, J δγ = J γδ. So, d dɛ [ξ i, ξ j ] = ξ i g J γδ ξ i g J γδ (93) ɛ=0 η α η γ η β η δ η α η γ η β η δ = 0 (94) So upto O(ɛ ), we have d[ξ i, ξ j ]/dɛ = 0 and so [ξ i, ξ j ] is constant up to O(ɛ ). As [ξ i, ξ j ] ɛ=0 = [η i, η j ] = J ij, we have [ξ i, ξ j ] = J ij upto O(ɛ ). So, ξ is a canonical transformation upto O(ɛ ). 7. 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 dξ i dt = [ξ i, g] P B (96) This condition is seen to be satisfied if g is taken as the Hamiltonian H, for in this case, [ξ i, H] P B = ξ i ξ k J kj H (97) = δ ik J kj H (98) = J ij H (99) = ξ i (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