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17. Action-Angle Coordinates Gerhard Müller University of Rhode Island, Gmuller@Uri.Edu Creative Commons License University of Rhode Island DigitalCommons@URI Classical Dynamics Physics Course Materials 2015 17. Action-Angle Coordinates Gerhard Müller University of Rhode Island, [email protected] Creative Commons License This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 4.0 License. Follow this and additional works at: http://digitalcommons.uri.edu/classical_dynamics Abstract Part seventeen of course materials for Classical Dynamics (Physics 520), taught by Gerhard Müller at the University of Rhode Island. Documents will be updated periodically as more entries become presentable. Recommended Citation Müller, Gerhard, "17. Action-Angle Coordinates" (2015). Classical Dynamics. Paper 5. http://digitalcommons.uri.edu/classical_dynamics/5 This Course Material is brought to you for free and open access by the Physics Course Materials at DigitalCommons@URI. It has been accepted for inclusion in Classical Dynamics by an authorized administrator of DigitalCommons@URI. For more information, please contact [email protected]. Contents of this Document [mtc17] 17. Action-Angle Coordinates • Action-angle coordinates [mln92] • Actions and angles for librations [mln93] • Actions and angles for rotations [mln94] • Action-angle coordinates of the harmonic oscillator [mex91] • Action-angle coordinates of an anharmonic oscillator [mex92] • Unbounded motion in piecewise constant periodic potential [mex96] • Unbounded motion in piecewise linear periodic potential [mex93] • Bounded motion in piecewise constant periodic potential [mex95] • Poisson brackets [msl30] • Specifications of Hamiltonian system [mln95] • Poisson’s theorem [mex191] • Poisson brackets of angular momentum variables [mex192] • Action-angle coordinates of plane pendulum: librations [mex200] • Hamiltonian system specified by noncanonical variables [mex94] • Generating a pure Galilei transformation [mex197] • Exponential potential [mex199] Action-Angle Coordinates [mln92] An elegant way of using Hamiltonian mechanics to solve a dynamical problem is to search for a canonical transformation to action-angle coordinates, (q1, . , qn; p1, . , pn) → (θ1, . , θn; J1,...,Jn), | {z } | {z } angles actions such that the Hamiltonian turns into a function of the actions alone: H(q1, . , qn; p1, . , pn) → K(J1,...,Jn). If such a transformation exists and can be found then the solution of the canonical equations is simple: ˙ ∂K Ji = − = 0 ⇒ Ji = const. ∂θi ˙ ∂K . (0) θi = = ωi(J1,...,Jn) = const. ⇒ θi(t) = ωit + θi . ∂Ji The inverse canonical transformation then yields qi(t) and pi(t). Two or more degrees of freedom: The existence of a transformation to action-angle coordinates is exceptional. Such systems are named integrable. Nonintegrable systems exhibit symptoms of Hamiltonian chaos (to be discussed later). One degree of freedom: Integrability is guaranteed. There exists a general prescription for finding the canonical transformation to action-angle coordinates. The prescription for two modes of bounded motion is discussed in detail: • libration (oscillation) [mln93], • rotation [mln94]. The two modes are realized, for example, in the plane pendulum. The ro- tational motion can also be interpreted as unbounded motion in a periodic potential. Actions and Angles for Librations [mln93] p2 Hamiltonian: H(q; p) = + V (q) = E = const: 2m Canonical momentum: p(q; E) = ±p2m [E − V (q)]: Action J and Hamiltonian K(J) from area A inside trajectory: I Z q2 Z 2π A = dq p(q; E) = 2 dqp2m [E − V (q)] = dθ J = 2πJ: q1 0 1 Z q2 ) J(E) = dqp2m [E − V (q)] ) E = K(J) = H(p; q): π q1 Angle variable θ(q; J) from area dA between nearby trajectories: Z q Z q @ dA = dq [p(q; J + dJ) − p(q; J)] = dJ dq p(q; J) = dJθ(q; J): 0 0 @J @ Z q @ Z q ) θ(q; J) = dq p(q; J) = dq p2m [K(J) − V (q)]: @J 0 @J 0 dK Time evolution: J = const:; θ(t) = !(J)t + θ ;!(J) = : 0 dJ ) q(θ; J) = q(t) ) p(q; J) = p(t): p J dA dA dJ A q q A θ θ 2π Generating function of the canonical transformation (q; p) ! (θ; J): Z q F2(q; J) = dq p(q; J): 0 Actions and Angles for Rotations [mln94] p2 Hamiltonian: H(q; p) = + V (q) = E = const: with V (q + Q ) = V (q). 2m 0 Canonical momentum: p(q; E) = p2m [E − V (q)]: Action J and Hamiltonian K(J) from area A under trajectory: Z Q0 Z Q0 Z 2π A = dq p(q; E) = dqp2m [E − V (q)] = dθ J = 2πJ: 0 0 0 1 Z Q0 ) J(E) = dqp2m [E − V (q)] ) E = K(J) = H(p; q): 2π 0 Angle variable θ(q; J) from area dA between nearby trajectories: Z q Z q @ dA = dq [p(q; J + dJ) − p(q; J)] = dJ dq p(q; J) = dJθ(q; J): 0 0 @J @ Z q @ Z q ) θ(q; J) = dq p(q; J) = dq p2m [K(J) − V (q)]: @J 0 @J 0 dK Time evolution: J = const:; θ(t) = !(J)t + θ ;!(J) = : 0 dJ ) q(θ; J) = q(t) ) p(q; J) = p(t): p J dA dA dJ A A q q Q0 θ θ 2π In the case of rotations there is no natural boundary for J. Here J is only determined up to a constant. [mex91] Action-angle coordinates of the harmonic oscillator Determine the canonical transformation (q; p) ! (θ; J) which produces the action-angle coordinates for the harmonic oscillator: p2 1 H(q; p) = + m!2q2 ! K(J): 2m 2 0 (a) Find the transformed Hamiltonian K(J). (b) Find the transformation relations q(θ; J); p(θ; J). (c) Reconstruct the generating function F2(q; J). (d) Determine from F2 the generating function F1(q; θ) and verify that it is equal to function F1(q; Q) used in [mex86]. Solution: [mex92] Action-angle coordinates of an anharmonic oscillator Determine the canonical transformation (q; p) ! (θ; J) which produces the action-angle coordinates for the anharmonic oscillator: p2 H(q; p) = + U tan2(αq) ! K(J): 2m (a) Find the transformed Hamiltonian K(J) and determine the angular frequency !(J) which determines the linear time evolution θ(t) = !(J)t + θ0 of the angle coordinates. (b) Find the transformation relations q(θ; J); p(θ; J), which amount to a solution of the dynamical problem. Solution: [mex96] Unbounded motion in piecewise constant periodic potential Consider a particle of mass m moving in the potential V (q) = 0 for 0 < jqj < D=2 and V (q) = U for D=2 < jqj < D with periodicity V (q + 2D) = V (q). For energies E > U the motion is unbounded and can be reinterpreted as a rotational mode of bounded motion. Solve this dynamical problem via transformation (q; p) ! (θ; J) to action-angle coordinates for motion with initial conditions q(0) = 0; p(0) > 0: (a) Find the function J(E), which expresses the action as a function of the energy. (b) Find the period T ≡ 2π=!(E) of the rotational motion. (c) Find the function θ(q; E) for 0 < q < 2D. (c) Plot in one diagram the functions J = const and p(t) for 0 < t < T . (d) Plot in a second diagram the functions q(t) and θ(t) for 0 < t < T . V(q) U q −D −D/2 0 D/2 D Solution: [mex93] Unbounded motion in piecewise linear periodic potential Consider a particle of mass m moving in a periodic potential V (q) = (U/π)jqj for −π ≤ q ≤ π and V (q + 2π) = V (q). For energies E > U > 0, the motion is unbounded and can be reinterpreted as a rotational mode of bounded motion. Solve this dynamical problem via transformation (q; p) ! (θ; J) to action-angle coordinates by establishing the following relations: p 2 2m h i p(q; E) = p2m [E − (U/π)jqj];J(E) = E3=2 − (E − U)3=2 ; p 3U 1 hp p i E − pE − (U/π)jqj !(E) = p E + E − U ; θ(q; E) = ±π p p ; 0 ≤ ±q ≤ π: 2m E − E − U V(q) U −π π q Solution: [mex95] Bounded motion in piecewise constant periodic potential Consider a particle of mass m moving in the potential V (q) = 0 for 0 < jqj < D=2 and V (q) = U for D=2 < jqj < D with periodicity V (q + 2D) = V (q). For energies E < U the motion is bounded. Solve this dynamical problem via transformation (q; p) ! (θ; J) to action-angle coordinates for motion with initial conditions q(0) = 0; p(0) > 0. (a) Find the function K(J), which expresses the Hamiltonian as a function of the action coordinate. (b) Find the period T ≡ 2π=!(J) of the librational motion. (c) Find the function q(θ; J) for 0 < θ < 2π. (d) Plot in one diagram the functions J = const and p(t) for 0 < t < T . (e) Plot in a second diagram the functions q(t) and θ(t) for 0 < t < T . V(q) U q −D −D/2 0 D/2 D Solution: Poisson Brackets [msl30] n X ∂f ∂g ∂g ∂f Definition: {f, g} = − , ∂q ∂p ∂q ∂p j=1 j j j j where f(q1, . , qn, p1, . , pn), g(q1, . , qn, p1, . , pn) are arbitrary dynam- ical variable expressed as functions of canonical coordinates. Algebraic properties: •{f, g} = −{g, f}, •{f, c} = 0 if c = const., •{f1 + f2, g} = {f1, g} + {f2, g}, •{f1f2, g} = f1{f2, g} + f2{f1, g}, ∂ ∂f ∂g • {f, g} = , g + f, , ∂t ∂t ∂t ∂f ∂f •{qj, f} = , {pj, f} = − . ∂pj ∂qj Fundamental Poisson brackets: {qi, qj} = 0, {pi, pj} = 0, {qi, pj} = δij Invariance under canonical transformations: Qj = Qj(q1, . , qn, p1, . , pn),Pj = Pj(q1, . , qn, p1, . , pn) ⇒ {Qi,Qj}q,p = 0, {Pi,Pj}q,p = 0, {Qi,Pj}q,p = δij ∂H ∂H Canonical equations:q ˙i = = {qi,H}, p˙i = − = {pi,H}.
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