Nonlinear Mechanics - Physics at Oregon State University

4.1. ONE-DIMENSIONAL SYSTEMS 49
and
∂
∂ψ F1(ψ, J) = 1 [
H1 − H1(ψ, J)
ω0(J)
] ≡ − ˜ H1(ψ, J)
ω0
(4.16)
˜H1 is the periodic part of H1. We are left with a differential equation that
is easy to integrate.
F1(ψ, J) = − 1
∫
dψ
ω0(J)
˜ H1(ψ, J) (4.17)
4.1.1 Summary
I will summarize all these technical details in the form of an algorithm for
doing first order perturbation theory. Remember that the object is to find
equations of motion in the form q = q(t) and p = p(t). We do this in three
steps: (1) Find q = q(I, ψ) and p = p(I, ψ). (2) Find I = I(J, φ) and
ψ = ψ(J, φ). (3) J and ˙φ are constant, so φ = ˙φt + φ0.
1. Identify the H0 part of the Hamiltonian. Find the transformation
equations q = q(I, ψ) and p = p(I, ψ) using the Hamiltonian-Jacobi
equation as described in the previous section. Use (??) to get ω0.
2. Equation (4.8) can be used to find J in terms of the total energy
E. The integral presents no difficulties in principle, especially if the
Hamiltonian is separable. In fact, textbooks never bother to do this.
It seems sufficient to display the results in terms of J, the assumption
being that we could find J = J(E) if we really had to.
3. The first order correction to the energy is obtained from the integral in
(4.15). Get the first order correction to the frequency by differentiating
it with respect to J.
4. The generating function F1 is calculated from (4.17). It is then substituted
into (4.4) and (4.5). These give implicit equations for ψ =
ψ(J, φ) and I = I(J, φ). Unfortunately, it is usually impossible to
invert them to obtain these formula explicitly.
4.1.2 The simple pendulum
The pendulum makes a nice example
H = l2
+ mgR(1 − cos θ)
2mR2