Nonlinear Mechanics - Physics at Oregon State University

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30 CHAPTER 2. CANONICAL TRANSFORMATIONS however, as the following argument shows. The Hamilton-Jacobi equation is [ (dW ) 2 1 + m 2m dq 2 ω 2 q 2 ] = E ∫ √2mIω W = − m2ω2q 2 dq ∂W ∂I ∫ = mω dq √ 2mIω − m 2 ω 2 q 2 = sin −1 ( √ ) mω2 q − ψ0 = ψ 2I √ 2E q = sin(ψ − ψ0) mω2 In the last equation ψ0 appears as an integration constant. Evidentally, ψ is the angle variable conjugate to I. In summary, to use action-angle variables for problems with one degree of freedom: 1. Find p as a function of E = α and q. 2. Calculate I(E) using (2.46). 3. Solve the Hamilton-Jacobi equation to find W = W (q, I). 4. Find ψ = ψ(q, I) using (2.49). 5. Invert this equation to get q = q(I, ψ). 6. Use (2.47) to get ω(I). 7. Calculate p = p(I, q) from (2.49). One attractive feature of this scheme is that you can find the frequency without using the characteristic function and without finding the equations of motion. The phase space plot is particularly important. Use polar coordinates (what else) for (I, ψ). Every trajectory, whatever the system, is a circle! Our derivation was based on the following assumptions: (1) The system had one degree of freedom. (2) Energy was conserved and the Hamiltonian had no explicit time dependence. (3) The motion was periodic. Every such system is at heart, a harmonic oscillator. Phase space trajectories are circles.
2.5. ACTION-ANGLE VARIABLES 31 The frequency can be found with a few deft moves. From a philosophical point of view, (and we will be getting deeper and deeper into philosophy as these lectures proceed) problems in this category are “as good as solved,” nothing more needs to be said about them. The same is definitely not true true with more than one degree of freedom. I will take a paragraph to generalize before going on to some more abstract developments. We must assume that the system is separable, so W (q1, . . . , qn, α1, . . . , αn) = ∑ Wk(qk, α1, . . . , αn) (2.51) k pk = ∂ Wk(qk, α1, . . . , αn) (2.52) ∂qk Ik = 1 pk(qk, α1, . . . , αn) (2.53) 2π Next find all the q’s as function of the I’s and substitute into W . Finally ψk = ∂W ∂Ik W = W (q1, . . . , qn; I1, . . . , In) ˙ Ik = 0 ˙ ψk = ∂H ∂Ik = ωk (2.54)
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Nonlinear Mechanics - Physics at Oregon State University

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