 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-4/15b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

p_{i} = \frac{\partial L}{\partial q ˙ _{i}} .

When the velocity Hessian is nonsingular, these equations can be inverted for

\overset{q}{˙}_{i} (q, p, t)

H (q, p, t) = i \sum p_{i} \overset{q}{˙}_{i} - L (q, \overset{q}{˙}, t) .

Its differential is

d H = \sum_{i} \overset{q}{˙}_{i} d p_{i} - \sum_{i} \frac{\partial L}{\partial q _{i}} d q_{i} - \frac{\partial L}{\partial t} d t,

because the

p_{i} d \overset{q}{˙}_{i}

terms cancel. The Euler-Lagrange equation identifies

\partial L / \partial q_{i} = \overset{p}{˙}_{i}

\overset{q}{˙}_{i} = \frac{\partial H}{\partial p _{i}}, \overset{p}{˙}_{i} = - \frac{\partial H}{\partial q _{i}} .

For the symmetric top, the momenta are

p_{θ} = A \dot{θ}, p_{ψ} = B (\dot{ψ} + \dot{ϕ} cos θ), p_{ϕ} = A \dot{ϕ} sin^{2} θ + p_{ψ} cos θ .

Thus

\dot{θ} = \frac{p _{θ}}{A}, \dot{ϕ} = \frac{p _{ϕ} - p _{ψ} c o s θ}{A s i n ^{2} θ}, \dot{ψ} = \frac{p _{ψ}}{B} - \dot{ϕ} cos θ .

The Legendre transform gives

H = \frac{p _{θ}^{2}}{2 A} + \frac{( p _{ϕ} - p _{ψ} cos θ ) ^{2}}{2 A sin ^{2} θ} + \frac{p _{ψ}^{2}}{2 B} + M g l cos θ .

Both

ϕ

and

ψ

p_{ϕ}

and

p_{ψ}

are conserved. The Hamiltonian has no explicit time dependence and is also conserved. Since

H

is independent of

ϕ, ψ

{H, p_{ϕ}} = {H, p_{ψ}} = {p_{ϕ}, p_{ψ}} = 0

H, p_{ϕ}, p_{ψ} .

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