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Direct product of integrable Hamiltonian systems 2026-09-29

The direct product of integrable Hamiltonian systems is integrable. Pull each family's commuting integrals of motion back to the product phase space. Integrals from different factors have zero Poisson bracket because they use disjoint canonical coordinates, and their differentials are independent wherever the differentials within each factor are independent.

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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 32C / a / Solution Created 2026-09-24 Updated 2026-09-29

In canonical coordinates, Hamilton's equations are

\overset{q}{˙}_{i} = \frac{\partial H}{\partial p _{i}}, \overset{p}{˙}_{i} = - \frac{\partial H}{\partial q _{i}} (1 \leq i \leq n) .

F (q, p)

is an integral of motion if it is constant along every Hamiltonian flow, equivalently

\dot{F} = {F, H} = 0

for a time-independent

F

. The system is an integrable Hamiltonian system when it has

n

first integrals

F_{1}, \dots, F_{n}

whose differentials are linearly independent almost everywhere and which are in mutual Poisson involution:

{F_{i}, F_{j}} = 0

for all

i, j

.

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Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 3 / 9A / b / Solution Created 2026-09-24 Updated 2026-09-25

Differentiating the spherical coordinate system

x = (r sin θ cos ϕ, r sin θ sin ϕ, r cos θ)

gives the orthogonal decomposition

d x = e_{r} d r + r e_{θ} d θ + r sin θ e_{ϕ} d ϕ,

where

e_{r} e_{θ} e_{ϕ} = (sin θ cos ϕ, sin θ sin ϕ, cos θ), = (cos θ cos ϕ, cos θ sin ϕ, - sin θ), = (- sin ϕ, cos ϕ, 0) .

Thus the scale factors of orthogonal coordinates are

h_{r} = 1, h_{θ} = r, h_{ϕ} = r sin θ .

Since the total differential satisfies

df = f_{r} d r + f_{θ} d θ + f_{ϕ} d ϕ = d x \cdot \nabla f,

comparison of coefficients gives the gradient

\nabla f = e_{r} \frac{\partial f}{\partial r} + e_{θ} \frac{1}{r} \frac{\partial f}{\partial θ} + e_{ϕ} \frac{1}{r sin θ} \frac{\partial f}{\partial ϕ} .

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