Classical Mechanics - Homework Assignment 9Alejandro G´mez Espinosa∗ o November 29, 2012 Goldstein, Ch.9, 11 Determine whether the transformation is canonical Q1 = q1 q2 Q2 = q1 + q2 p1 − p2 +1 q2 − q1 q2 p2 − q1 p1 P2 = + (q2 + q1 ) q2 − q1 P1 = To determine if this transformation is canonical, let us use the Poisson brackets: [u, v]q,p = Then, [Q1 , P1 ]q,p = ∂Q1 ∂P1 ∂Q1 ∂P1 ∂Q1 ∂P1 ∂Q1 ∂P1 − + − ∂q1 ∂p1 ∂p1 ∂q1 ∂q2 ∂p2 ∂p2 ∂q2 1 1 = q2 − q2 q2 − q1 q2 − q1 = 0 ∂u ∂v ∂u ∂v − ∂qi ∂pi ∂pi ∂qi [Q1 , P2 ]q,p = ∂Q1 ∂P2 ∂Q1 ∂P2 ∂Q1 ∂P2 ∂Q1 ∂P2 − + − ∂q1 ∂p1 ∂p1 ∂q1 ∂q2 ∂p2 ∂p2 ∂q2 q1 q2 = −q2 + q1 q2 − q1 q2 − q1 = 0 [Q2 , P2 ]q,p = ∂Q2 ∂P2 ∂Q2 ∂P2 ∂Q2 ∂P2 ∂Q2 ∂P2 − + − ∂q1 ∂p1 ∂p1 ∂q1 ∂q2 ∂p2 ∂p2 ∂q2 q1 q1 = − + q2 − q1 q2 − q1 = 0 ∗
[email protected] 1 [Q2 , P1 ]q,p = ∂Q2 ∂P1 ∂Q2 ∂P1 ∂Q2 ∂P1 ∂Q2 ∂P1 − + − ∂q1 ∂p1 ∂p1 ∂q1 ∂q2 ∂p2 ∂p2 ∂q2 1 1 − = q2 − q1 q2 − q1 = 0 [P1 , P2 ]q,p = ∂P1 ∂P2 ∂P1 ∂P2 ∂P1 ∂P2 ∂P1 ∂P2 − + − ∂q1 ∂p1 ∂p1 ∂q1 ∂q2 ∂p2 ∂p2 ∂q2 p1 − p2 q1 1 p1 = − − − −1 2 (q2 − q1 ) q2 − q1 q2 − q1 (q2 − q1 )2 p1 − p2 q2 1 p2 + − − −1 (q2 − q1 )2 q2 − q1 q2 − q1 (q2 − q1 )2 = 0 [Q1 , Q2 ]q,p = ∂Q1 ∂Q2 ∂Q1 ∂Q2 ∂Q1 ∂Q2 ∂Q1 ∂Q2 − + − ∂q1 ∂p1 ∂p1 ∂q1 ∂q2 ∂p2 ∂p2 ∂q2 = 0 Since [Q1 , P1 ] = [Q1 , P2 ] = [Q2 , P1 ] = [Q2 , P2 ] = [Q1 , P2 ] = [P1 , P2 ] = 0, therefore this transformation is canonical. Goldstein, Ch.9, 17 Show that the Jacobi identity is satisfied if the Poisson bracket sign stands for the commutator of two square matrices: [A, B] = AB − BA Show also that for the same representation of the Poisson bracket that [A, BC] = [A, B]C + B[A, C] The Jacobi identity is given by: [A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0 then, if (1) is satisfied: [A, [B, C]] = [A, BC − CB] = A(BC − CB) − (BC − CB)A = ABC − ACB − BCA + CBA [B, [C, A]] = [B, CA − AC] = B(CA − AC) − (CA − AC)B = BCA − BAC − CAB + ACB 2 (2) (1) [C, [A, B]] = [C, AB − BA] = C(AB − BA) − (AB − BA)C = CAB − CBA − ABC + BAC where is easy to see that all the terms will vanish. For (2): [A, B]C + B[A, C] = ABC − BAC + BAC − BCA = ABC − BCA = [A, BC] Goldstein, Ch.9, 22 For the point transformation in a system of two degrees of freedom, 2 Q1 = q1 , Q2 = q1 + q2 find the most general transformation equations for P1 and P2 consistent with the overall transformation being canonical. Show that with a particular choise for P1 and P2 the Hamiltonian H= p1 − p2 2q1 2 + p2 + (q1 + q2 )2 can be transformed to one in which both Q1 and Q2 are ignorable. By this means solve the problem and obtain expressions for q1 , q2 , p1 , and p2 as functions of time and their initial values. Using the relations for a point transformation: Q1 = Q2 = ∂F2 2 = q1 ∂P1 ∂F2 = q1 + q2 ∂P2 Then, the generating function must be: 2 F2 = q1 P1 + (q1 + q2 )P2 and the momentum coordinates are: p1 = p2 = ∂F2 = 2q1 P1 + P2 ∂q1 ∂F2 = P2 ∂q2 Solving for P1 and P2 , we found the most general transformations: P2 = p2 p1 − p2 P1 = 2q1 Therefore, the Hamiltonian is given by: 2 H = P1 + P2 + Q2 2 3 but, whether we choose P2 = p2 + (q1 + q2 )2 : 2 H = P1 + P2 the Hamiltonian does not depend upon Q1 and Q2 . Now, solving this Hamiltonian: ∂H ˙ P1 = − =0 ∂Q1 ∂H ˙ P2 = − =0 ∂Q2 ∂H ˙ Q1 = = 2P1 ∂P1 ∂H ˙ =1 Q2 = ∂P2 ⇒ P1 = a ⇒ P2 = b ⇒ Q1 = 2P1 t + c ⇒ Q2 = t + d where a, b, c, d are constant, i.e., the initial values. Replacing with the old coordinates: p1 + p2 = a 2q1 p2 + (q1 + q2 )2 = b p1 − p2 2 t+c q1 = q1 q1 + q2 = t + d Solving this equations, we found: √ p1 = 2a 2at + c + b(t + d)2 p2 = b + (t + d)2 √ q1 = 2at + c √ q2 = t − 2at + c + d Goldstein, Ch.9, 28 A charged particle moves in space with a constant magnetic field B such that the vector potential, A, is 1 A = (B × r) 2 (a) If vj are the Cartesian components of the velocity of the particle, evaluate the Poisson brackets [vi , vj ], i = j = 1, 2, 3 We know that the mometum of a charged particle in an electric field is given by pi = mvi + qAi then, [vi , vj ] = 1 [pi − qAi , pj − qAj ] m2 1 = ([pi , pj ] − [pi , qAj ] − [qAi , pj ] + q[Ai , Aj ]) m2 q = − 2 ([pi , Aj ] + [Ai , pj ]) m q = ([pj , Ai ] − [pi , Aj ]) m2 4 ⇒ vi = pi − qAi m But, the vector potential in terms of the Levi-Civita symbols are: Ai = iab Ba xb Calculate the first term in the previous relation: [pj , Ai ] = = = = Consequently, the second term: [pi , Aj ] = Replacing in the relation: [vi , vj ] = = = = = q ([pj , Ai ] − [pi , Aj ]) m2 1 q 1 iaj Ba − jai Ba 2 m 2 2 qBa ( iaj − jai ) 2m2 qBa 2 iaj 2m2 qBa iaj m2 1 2 jai Ba 1 [pj , iab Ba xb ] 2 1 iab Ba [pj , xb ] 2 1 iab Ba δjb 2 1 iaj Ba 2 (b) If pi is the canonical momentum conjugate to xi , also evaluate the Poisson backets [xi , vj ], [pi , vj ], [x1 , pj ], ˙ [pi , pj ], ˙ [xi , vj ] = = = = = [Pi , vj ] = 1 [xi , pj − qAj ] m 1 ([xi , pj ] − q[xi , Aj ]) m 1 q δij − [xi , jab Ba xb ] m 2 1 q δij − jab Ba [xi , xb ] m 2 δij m 1 [pi , pj − qAj ] m 1 = ([pi , pj ] − q[pi , Aj ]) m q = − jab Ba [pi , xb ] m q = − jai Ba m 5 For pi , we know that the Hamiltonian for a charge particle moving in a magnetic field is given ˙ by: 1 H = (pi − qAi )2 m then, pi is: ˙ pi = − ˙ = = = = = ∂H ∂xi ∂Ai 1 − (pi − qAi ) m ∂xi 1 1 ∂xk − (pi − qAi ) ijk Bj m 2 ∂xi 1 − (pi − qAi ) ijk Bj δki 2m 1 (pi − qAi ) iji Bj − 2m 0 thus, [xi , pj ] = 0 ˙ [pi , pj ] = 0 ˙ Goldstein, Ch.9, 34 Obtain the motion in time of a linear harmonic oscillator by means of the formal solution for the Poisson bracket version of the equation of motion as derived from Eq.(9.116). Assume that at time t = 0 the initial values are x0 and p0 . The derivation of eq. (9.116) ends up in this relation: u(t) = u0 + t[u, H]0 + t2 t3 [[u, H], H]0 + [[[u, H], H], H]0 + ... 2! 3! Knowing the Hamiltonian of the linear harmonic oscillator: H= p2 mw2 x + 2m 2 we can use the previous relation to find the motion in time: [x, H]0 = [[x, H], H]0 = Pluging them in the initial relation: x(t) = x0 + p0 t w2 t2 − m 4 ∂x ∂H ∂x ∂H p − = ∂x ∂p ∂p ∂x m 1 1 ∂p ∂H ∂p ∂H [p, H] = − m m ∂x ∂p ∂p ∂x =− w2 2 6