NIU PHYS 500, Fall 2006 Classical Mechanics Solutions for HW6. Solutions

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1 NIU PHYS 500, Fall 006 Classical Mechanics Solutions for HW6 Assignment: HW6 [40 points] Assigned: 006/11/10 Due: 006/11/17 Solutions P6.1 [ = 10 points] Consider a particle of mass m moving in two dimensions in a potential well. Let us choose the origin of our coordinate system at the minimum of this well. The well would be termed isotropic if the potential did not depend on the polar angle. (a) First, consider the anisotropic potential in a given Cartesian coordinate system: V (x 1, x ) = k (x 1 + x ) + k x 1 x ; k > k > 0. (1) Find the eigenfrequencies normal modes, preferably by reasoning rather than brute-force matrix diagonalization. Give a physical interpretation of the normal modes. (b) Use a qualitative physics-based argument to write down two independent constants of the motion. Verify your choice using the Poisson bracket equation u = {u, H} PB + u t, () where u = u(q, p, t) H is the Hamiltonian. (c) The oscillator becomes isotropic if k = 0. Again use a qualitative physics-based argument to write down an additional independent constant of motion if k = 0, verify your choice with the PB equation above. S6.1 (a) The potential funcion describes an ellipsoid (i.e. the equipotential contours are ellipses in the (x 1, x ) plane). The x 1 x term in V indicates that oscillations along x 1 x are coupled: those are not normal coordinates. The symmetrical dependence of V on x 1, x suggests, however, that a π rotation is a sensible guess for a set of 4 normal Cartesian coordinates: q 1 = 1 (x 1 + x ), q = 1 (x 1 x ), (3) whence, x 1 = 1 (q 1 + q ), x = 1 (q 1 q ). (4) In terms of the new coordinates velocities, the kinetic energy is T = 1 m(ẋ 1 + ẋ ) = 1 4 m(( q 1 + q ) + ( q 1 q ) ) (5) = 1 m( q 1 + q ), 1

2 NIU PHYS 500, Fall 006 Classical Mechanics Solutions for HW6 the potential energy is V = 1 k(x 1 + x ) + k x 1 x = 1 4 k((q 1 + q ) + (q 1 q ) ) + 1 k (q 1 q ) = 1 k(q 1 + q ) + 1 k (q 1 q ) (6) = 1 ((k + k )q 1 + (k k )q ). So, the kinetic energy tensor is diagonal in q 1, q potential energy tensor is diagonal in q 1, q. Thus, the system executes independent simple harmonic motions in each normal coordinate q 1 q with the respective angular frequencies ω 1 = k + k m ω = k k m. (7) The first normal mode represents a SHM along the steepest slope inside the potential well (the minor axis of the ellipse), while the second represents that along the least steep slope, which is perpendicular to the first (the major axis of the ellipse). (b) The total energy E = T + V of any simple harmonic oscillator is a constant of the motion. In this example, the energy associated with each of the two normal modes, E 1 = m q 1 + k + k q1 E = m q + k k q (8) is conserved separately. Since the potential is conservative, H = E 1 + E. Denoting the momentum conjugate to q i by p i = L = (T V ) = m q i, we have q i q i for the first normal mode, Ė 1 = {E 1, H} + E 1 t = {E 1, E 1 + E } = {E 1, E 1 } + {E 1, E } = {E 1, E } = E 1 E E 1 E = E 1 q 1 E p 1 + E 1 q E p E 1 p 1 E q 1 E 1 p E q = 0. (9) Similarly, Ė = 0. (c) If k = 0, then the potential is isotropic in the x 1, x plane. The rotational symmetry implies that the polar angle is cyclic. Consequently, the angular momentum l 3 = m(q 1 p q p 1 ) is a constant of

3 NIU PHYS 500, Fall 006 Classical Mechanics Solutions for HW6 the motion. 1 With the Hamiltonian simplified to H = 1 ( p 1 + p ) m + k(q 1 + q), (10) we explicitly verify the conservation of the angular momentum: l 3 = {l 3, H} + l 3 t = {l 3, H} = l 3 H l 3 H = p p 1 p 1 p + mkq q 1 mkq 1 q = 0. (11) P6. [ = 8 points] (a) Verify the Poisson bracket equation {L i, L j } = ɛ ijk L k (1) among the Cartesian components of angular momentum of a spherical pendulum of mass m in a gravitaional field of acceleration g pointing opposite to the pole. ɛ ijk represents the Levi-Civita tensor. Hint: Start with expressing the Lagrangian in spherical coordinates: L = L(θ, φ, θ, φ). (b) Likewise, verify for the spherical pendulum. {p θ, p φ } = 0 (13) (c) The mathematical machinery of Poisson brackets evidently tells us that some perpendicular momentum components are valid canonical momenta (e.g., p θ p φ ), while others are not (e.g., the Cartesian components of angular momentum above). Explain the physics behind this. S6. Let the distance between the bob the fixed support point be l. (a) Then, the Lagrangian is L(θ, φ, θ, φ) = T V = m l ( θ + sin θ φ ) + mgl cos θ. (14) This gives the conjugate momenta p θ = L θ = ml θ pφ = L φ = ml sin θ φ, (15) 1 The subscript 3 after l merely states that the angular momentum is perpendicular to the plane of motion. In 3 dimensions, the (antisymmetric) Levi-Civita tensor is defined as ɛ 13 = ɛ 31 = ɛ 31 = 1, ɛ 13 = ɛ 13 = ɛ 31 = 1, all other ɛ ijk = 0. In n dimensions ɛ 13...n its even permutations (i.e., even number of swapping of adjacent indices) are 1, odd permutations 1, all others 0. 3

4 NIU PHYS 500, Fall 006 Classical Mechanics Solutions for HW6 the Hamiltonian H(θ, φ, p θ, p φ ) = p θ θ + pφ φ L = 1 ml ( p θ + p φ sin θ ) mgl cos θ. (16) Now, the Cartesian coordinates can be expressed in terms of the sphericals: x = l sin θ cos φ, y = l sin θ sin φ, z = l cos θ. (17) And the Cartesian momenta are p x = mẋ = ml( θ cos θ cos φ φ sin θ sin φ) = p θ l cos θ cos φ p φ sin φ l sin θ, p y = mẏ = ml( θ cos θ sin φ φ sin θ cos φ) = p θ l cos θ sin φ + p φ cos φ l sin θ, p z = mż = ml θ sin θ = p θ l sin θ. (18) Now we can compute the Cartesian angular momentum components: L x = yp z zp y Similarly, = l sin θ sin φ ( p ) ( θ l sin θ pθ l cos θ l cos θ sin φ + p ) φ cos φ l sin θ = p θ sin φ p φ cot θ cos φ. (19) L y = zp x xp z = p θ cos φ p φ cot θ sin φ, (0) So, L z = xp y yp x = p φ. (1) {L y, L z } = L y θ L z + L y L z L y L z p θ φ p φ p θ θ L y L z p φ φ = 0 + L y φ = p θ sin φ p φ cot θ cos φ = L x. () Similarly it can be shown that {L z, L x } = L y {L x, L y } = L z. These results are compactified in {L i, L j } = ɛ ijk L k. (3) (b) {p θ, p φ } = p θ θ p φ + p θ p φ p θ p φ p θ φ p φ p θ θ p θ p φ = 0. (4) p φ φ 4

5 NIU PHYS 500, Fall 006 Classical Mechanics Solutions for HW6 (c) The orientation of a rigid body in space is fully specified by two independent coordinates, e.g. θ φ. It is therefore impossible to find 3 independent components of angular momentum. Consequently, the Poisson brackets {L i, L j } do not vanish. P6.3 [ = 10 points] Consider a system with a time-dependent Hamiltonian H(q, p, t) = H 0 (q, p) ɛq sin (ωt), (5) where ɛ ω are known constants H 0 t = 0. (a) Derive Hamilton s canonical equations of motion for the system. (b) Use a canonical transformation generating function G(q, P, t) to find a new Hamiltonian H new canonical variables Q, P such that H (Q, P ) = H 0 (q, p). Hint: The partial differential equations do not tell us how q P are related in the generating function. We can take an educated guess though. G = qp ɛq cos (ωt) works. ω (c) Verify that Hamilton s canonical equations of motion are invariant under the transformation. (d) Suggest a possible physical interpretation of the time-dependent term in H. S6.3 (a) Hamilton s canonical equations of motion are q = H p = H 0 p ṗ = H q = H 0 q + ɛ sin(ωt) (b) The generating function has to satisfy the following conditions: p = G q, (6) Q = G P, H = H + G t. (7) The last one means G t = H H = ɛq sin(ωt). (8) While G cannot be fully determined from this, the first two relations in Eq. 7 suggest that the identity transformation q i P i be added to the time integral of ɛq sin(ωt). So, G = qp ɛq ω cos (ωt) (9) is a sensible trial function. This leads to P = p + ɛ cos (ωt), ω Q = q, H (Q, P, t) = H(q, p, t) + ɛq sin(ωt) = H 0 ɛq sin(ωt) + ɛq sin(ωt) = H 0, (30) 5

6 NIU PHYS 500, Fall 006 Classical Mechanics Solutions for HW6 (c) where the transformed Hamiltonian can be written as H (Q, P ) = H 0 (q, p) = H 0 (Q, P ɛ cos (ωt)). (31) ω So, our trial function passes the requirements. H P = H 0 p p P + H 0 q q P = H 0 p + 0 = q = Q, (3) H Q = H 0 q q Q H 0 p p Q = H 0 q 0 = ṗ ɛ sin(ωt) = P. (33) Therefore, the transformation is canonical Hamilton s equations of motion remain invariant. (d) The difference between the old the new Hamiltonians, ɛq sin(ωt), must have the dimensions of energy, ɛ sin(ωt) has the form of a generalized force. Thus, in the transformed system, the particle experiences a sinusoidal force that has an angular frequency ω. The variation of the force in phase space is undetermined. P6.4 [4 points] Show that canonical transformations leave the physical dimension of the product p i q i unchanged, i.e., [P i Q i ] = [p i q i ]. Let Φ be the generating functin for a canonical transformation. Show that where H is the Hamiltonian t the time. [P i Q i ] = [p i q i ] = [Φ] = [Ht], (34) S6.4 Set w α = (q 1,..., q n, p 1..., p n ), w β = (Q 1,..., Q n, P 1..., P n ), define M αβ w β w α, (35) ɛ = 0 n n 1 n n 1 n n 0 n n. (36) Then the equation that relates Q k to P k, q j P l to Q l p j etc. is given by M T ɛm = ɛ. (37) From this it follows that [P i Q i ] = [p i q i ] = [Φ]. If Φ(w α, w β ) is the generating function of the canonical transformation, as H = H + Φ t, the function Φ has thte dimension of the product Ht. The last part of the assertion then follows from the canonical equations. 6

7 NIU PHYS 500, Fall 006 Classical Mechanics Solutions for HW6 P6.5 [4 + 4 = 8 points] The Hamiltonian H = p m + mω q describes a simple harmonic oscillator of mass m frequency ω. Introducing the transformation x 1 ω mq, x p m, τ ωt, (38) we obtain H = 1 (x 1 + x ). (a) What is the generating function ˆΦ 1 (x 1, y 1 ) for the canonical transformation {x 1, x } {y 1, y } that corresponds to the function Φ(q, Q) = mωq cot Q? (b) Calculate the matrix M ij x i confirm that det M = 1 y j M T ɛm = ɛ (ɛ is the antisymmetric matrix used in the lectures to put the coordinates q i momenta p i in a single array w µ ). Hint: y 1 = Q, y = ωp, where Q P are the new generalized coordinates momenta, respectively. S6.5 (a) We can wrirte H = H + Φ τ, (39) such that [Φ] = [H] = [x 1 x ] = [ωpq]. The new generalized coordinate y 1 = Q is dimensionless. As y 1 y has the same dimension as x 1 x, y must have the dimension of H. So, y = ωp. Therefore, ˆΦ 1 (x 1, y 1 ) = 1 x 1 cot y 1. (40) (b) or, Thus, x = ˆΦ x 1 = x 1 cot y 1, y = ˆΦ x 1 = y 1 sin, (41) y 1 x 1 = y sin y 1, x = y cos y 1. (4) M αβ = x α = y β (y ) 1 cos y 1 (y ) 1 sin y 1 (y ) 1 sin y 1 (y ) 1 cos y 1 (43) It is easy to verify that det M = 1 M T ɛm = ɛ. 7