PHY 5246: Theoretical Dynamics, Fall November 16 th, 2015 Assignment # 11, Solutions. p θ = L θ = mr2 θ, p φ = L θ = mr2 sin 2 θ φ.

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1 PHY 5246: Theoretical Dynamics, Fall 215 November 16 th, 215 Assignment # 11, Solutions 1 Graded problems Problem 1 1.a) The Lagrangian is L = 1 2 m(ṙ2 +r 2 θ2 +r 2 sin 2 θ φ 2 ) V(r), (1) and the conjugate momenta are By integrating these relations we get Thus the Hamiltonian is H = p i q i L = 1 2m Using our class discussion and Goldstein 8.1 we can find p r = L = mṙ, (2) ṙ p θ = L θ = mr2 θ, p φ = L θ = mr2 sin 2 θ φ. (3) ṙ = p r m, (4) θ = p θ mr, 2 p φ φ = mr 2 sin 2 θ. [ ] p 2 r + p2 θ r + p2 φ 2 r 2 sin 2 +V(r). (5) θ a =, 1 ˆT = m r 2 ˆT 1 = 1 1 r 2 sin 2 m θ H = 1 2 pt ˆT 1 p+v with p = p θ. p φ p r 1 r 2 1 r 2 sin 2 θ, (6)

2 Thus we see that H = T +V = E, as expected. Hamilton s equations of motion are ( ) r ṙ = p r = pr m p r ṗ r = = 1 r ( ) θ θ = p θ = p θ mr 2 p θ ṗ θ = ( ) φ φ = p φ = p φ ṗ φ = = φ mr 3 [ = p2 φ cosθ θ mr 2 sin 3 θ p φ mr 2 sin 3 θ p 2 θ + p2 φ sin 2 θ ] V (r) (7) 1.b) If p φ () = then p φ (t) = for all times (since ṗ φ = from the equations of motion). Then the equations of motion become ṙ = p r m θ = p θ ṗ r = p2 θ V mr φ = 2 (8) (r) ṗ mr 3 θ = ṗ φ = Thus if φ() = φ(t) = at all times, and the motion is planar (in the φ = plane) as we would expect. Given the initial conditions, it will be in the φ = plane. The (r,p r ) and (θ,p θ ) sets of equations reduce to the usual equations for central-force motion: θ = p θ p mr 2 θ = mr 2 θ, conserved. (9) ṗ θ = ṗ θ = mr(r θ+2ṙ θ) = = ma θ =. So we have that p θ = l is the magnitude of the angular momentum, and from ṗ θ = we get Newton s second law in the ˆθ direction. For the radial part: ṙ = p r m (1) ṗ r = p2 θ V (r) = l2 V (r). mr 3 mr 3 Taking a derivative of the first equation and plugging it into the second equation we find Thus we can write r = ṗr m = l2 m 2 r 3 V (r) m. (11) ṗ r = m r = l2 mr 3 V (r) = mr θ 2 V (r), (12) = m( r r θ 2 ) = ma r = V (r) = F(r). (13) So we find this set of equations gives us Newton s 2nd law in the radial direction.

3 1.c) y r F m For the specific case of F(r) = k r 2ˆr, since this is a conservative force with no constraint we already know that H = T +U = E, for constant E. From the diagram we see the coordinates are x = rcosθ y = rsinθ = v2 = ṙ 2 +r 2 θ2. (14) The kinetic and potential energy, and the Lagrangian are O x T = 1 2 m(ṙ2 +r 2 θ2 ), (15) U = ( kr ) dr = k 2 r, L = T U = 1 2 m(ṙ2 +r 2 θ2 )+ k r. For the conjugate momenta we have p r = L = mṙ ṙ = pr ṙ m p θ = L θ = mr2 θ θ = p θ (16) mr 2 Our Hamiltonian is H = ṙp r + θp θ L = p2 r 2m + p θ 2mr 2 k r = T +U, (17) and E is conserved since = E = constant. (18) t The Hamiltonian (canonical) equations of motion are From the equations of motion for (θ,p θ ) we get; = θ = p θ mr 2 p θ p r = ṙ p θ = θ (19) = ṗ r r = ṗ θ θ θ = ṗ θ = p θ = mr 2 θ = constant, (2) which shows the angular momentum is conserved. From the set of equations for (r,p r ), we find p r r = ṙ = p r m, (21) = ṗ r = p2 θ mr 3 + k r 2 ṗ r = m r = p2 θ mr 3 k r 2 = mr θ 2 k r. (22)

4 Problem 2 2.a) Our particle moves in 1D, so we use one generalized coordinate x. The potential is given by integrating the force: F(x,t) = k U(x,t) = F(x,t)dx = k x 2e t/τ x et/τ +C, (23) where we assume as x that U so we take C =. The kinetic energy and the Lagrangian are therefore T = 1 2 mẋ2, (24) The conjugate momentum of x is = L = T U = 1 2 mẋ2 k x e t/τ. (25) so the Hamiltonian is p x = L ẋ = mẋ ẋ = p x m. (26) H = ẋp x L = 1 p 2 x 2m + k x e t/τ = T +U = E. (27) 2.b) From above we see H = E, since there are no constraints and U is not a function of ẋ. However, since U = U(x,t) (explicitly depends on time!), the energy of the system is not conserved: Problem 3 3.a) de dt = dh dt = t. (28) The magnetic field is given to us as B( r) = B ẑ, and we can verify that the vector potential A( r) = 1 2 B r satisfies B = A in the following way: ( A) i = ǫ ijk j A k = ǫ ijk j ( 1 2 ǫ klmb l x m ) (29) = 1 2 ǫ ijkǫ klm B l δ lm = 1 2 ǫ ijkǫ klj B l = 1 2 2δ ilb l = B i. This implies exactly that A = 1 2 B ẑ r = 1 2 B (xŷ yˆx). (3)

5 3.b) The Lagrangian is The conjugate momenta are L( r, r) = 1 2 m r 2 + q c r A( r) (31) = 1 2 m(ẋ2 +ẏ 2 +ż 2 )+ q c (ẋa x +ẏa y +ża z ) = 1 2 m(ẋ2 +ẏ 2 +ż 2 )+ qb 2c ( yẋ+xẏ). p x = L ẋ = mẋ qb 2c y ẋ = 1 ( ) qb m 2c y +p x ( qb ) 2c +p y p y = L ẏ = mẏ + qb 2c x ẏ = 1 m p z = L ż = mż ż = p z m. Thus the Hamiltonian is 3.c) H = p x ẋ+p y ẏ +p z ż L = qb 2 p xy + p2 x m qb 2 p yx+ p2 y m + p2 y = 1 2m (p2 x +p 2 y +p 2 z) qb 2 = 1 ( p x + qb ) 2 2m 2c y + 1 2m The mechanical momenta are ( qb (32) (33) m 1 2m (p2 x +p2 y +p2 z ) qb ( qb 2 4c y2 qb ) 4c x2 4c y2 p x y qb ) 4c x2 +p y x ) m p2 z. ( p y qb 2c x πx = mẋ = p x + qb y 2c x 2c π y = mẏ = p y qb π z = mż So we have (using Landau s definition of Poisson s bracket, as also given in this problem), π x,π y } = p x + qb 2c y,p y qb } 2c x = p x,p y }+ qb 2c y,p y} qb ( ) 2 2c p qb x,x} y,x} 2c ( ) qb = 2 = qb, 2c c where we have used that p x,p y } = and y,x} =. Similarly we have π y,π z } = p y qb } 2cx,p z =, (36) π z,π x } = p z,p x + qb } 2c y =. (34) (35)

6 (3.d) In terms of the mechanical momenta: Now using we get H = π2 x 2m + π2 y 2m + π2 z 2m. (37) d π dt = H, π}, (38) πx = H,π x } = 1 2m π2 y,π x } = qb π y, π y = H,π y } = 1 2m π2 x,π y } = qb π x, π z = H,π z } =. The last expression implies π z (t) = constant = π z (). Taking a derivative of the first expression and plugging in the second gives (39) π x = qb ( ) 2 π qb y = π x = ω 2 π x = π x +ω 2 π x. (4) The solution to this is while for π y we have π x (t) = Acos(ωt)+Bsin(ωt), (41) With the initial conditions we can write π y (t) = 1 ( Aωsin(ωt)+Bωcos(ωt)) = Asin(ωt)+Bcos(ωt). (42) ω From Newton s 2nd law we would get π x () = A, π y () = B, (43) πx (t) = π x ()cos(ωt)+π y ()sin(ωt) π y (t) = π x ()sin(ωt)+π y ()cos(ωt) π z (t) = π z (). (44) m a = m d v dt = d π dt = q c v B = qb π ẑ. (45) This means that π moves precessing about the z-axis with frequency ω = qb, as we have found in the explicit expression for π x, π y, and π z above. Also note that, since we have also found (38). H, π} = qb π ẑ, (46)