Nonlinear Mechanics - Physics at Oregon State University

1.5. THE HAMILTONIAN FORMULATION 15
Now here’s what I mean that Legendre transformations are invertible:
First follow the steps from L → H. We start with L = L(q, ˙q). Equation
(1.39) gives p = p(q, ˙q). Invert this to find ˙q = ˙q(q, p). The Hamiltonian is
now
H(q, p) = ˙q(q, p)p − L[q, ˙q(q, p)]. (1.43)
Now suppose that we start from H = H(p, q). Use (1.40) to find ˙q = ˙q(q, p).
Invert to find p = p(q, ˙q). Finally
L(q, ˙q) = ˙qp(q, ˙q) − H[q, p(q, ˙q)] (1.44)
In both cases we were able to complete the transformation without knowing
ahead of time the functional relationship among q, ˙q, and p. To summarize:
Equations (1.37), (1.39), and (1.41) enable us to transform between the (q, ˙q)
(Lagrangian) prescription and the (q, p) (Hamiltonian) prescription; while
(1.40) and (1.41) are Hamilton’s equations of motion.
1.5.1 The Spherical Pendulum
A mass m hangs from a string of length R. The string makes an angle θ
with the vertical and can rotate about the vertical with an angle ϕ.
T = 1
2 mR2 ( ˙ θ 2 + sin 2 θ ˙ ϕ 2 ) (1.45)
V = mgR(1 − cos θ) (1.46)
The mgR constant doesn’t appear in the equations of motion, so we can
forget about it. The Lagrangian is L = T − V as usual.
pθ = ∂L
∂ ˙ θ = mR2 ˙ θ (1.47)
pϕ = ∂L
∂ ˙ ϕ = mR2 sin 2 θ ˙ ϕ ≡ lϕ
(1.48)
The angle ϕ is cyclic, so pϕ = lϕ is constant. At this point we are still in the
(q, ˙q) prescription. Invert (47) and (48) to obtain ˙ θ and ˙ ϕ as functions of pθ
and lϕ.
˙θ = pθ/mR 2
(1.49)
H = p2 θ +
2mR2 ˙ϕ = lϕ/mR 2 sin 2 θ (1.50)
l2 ϕ
2mR2 sin2 − mgR cos θ (1.51)
θ