Nonlinear Mechanics - Physics at Oregon State University

4.1. ONE-DIMENSIONAL SYSTEMS 47
φ = ∂F
∂J
= ψ + ϵ∂F1 (ψ, J) + · · · (4.5)
∂J
Before going on there are some technical points about ψ and J that need
to be discussed. When ϵ = 0, ψ is the exact angle variable for the system.
This means that we can find p and q as functions of ψ such that p and q
return to their original values when ∆ψ = 2π. We can in principle invert
this transformation to find ψ as a function of p and q.
ψ = ψ(q, p) (4.6)
When p and q run through a complete cycle, ψ advances by 2π. When ϵ ̸= 0
the orbit will be different from the unperturbed case, but the functional
relationship doesn’t change, so when p and q run through a complete cycle,
we must still have ∆ψ = 2π. Of course, the exact angle variable will also
advance 2π. In summary
∆ψ = ∆φ = 2π (4.7)
for one complete cycle.
The following integrals are all equal because canonical transformations
preserve phase space volume.
J = 1
2π
p dq = 1
2π
Now integrate (4.4) around one orbit:
that is
1
2π
I dψ = 1
2π
J dφ = 1
2π
J dψ + 1
2π ϵ
∂F1
∂ψ
J = J + 1
2π ϵ
∂F1
∂ψ
dψ + · · · ;
I dψ (4.8)
dψ + · · · ;
We have just seen that ∆ψ = 2π around one cycle. Consequently
∂F1
dψ = 0 (4.9)
∂ψ
implies that the derivative of F1 is purely oscillatory with a fundamental
period of 2π in ψ. (The same is true of the higher order terms as well.)
The Hamiltonian is transformed using (??) with the new variables.
K(φ, J) = H(ψ(φ, J), I(φ, J)) + ∂
∂t F2(ψ(φ, J), J, t) (4.10)