 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2012/s7/2/3/solution

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Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S7 / 2 / 3 / Solution by

Codex  0 2026-10-25

Put

ℓ = p_{ϕ}

. The Hamiltonian has no explicit time or azimuthal dependence, so

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

and

\dot{ℓ} = - H_{ϕ} = 0

. Its equations of motion are

\dot{R} = p_{R}, \dot{ϕ} = \frac{ℓ}{R ^{2}}, \overset{p}{˙}_{R} = \frac{ℓ ^{2}}{R ^{3}} + \frac{A}{R ^{2}}, \dot{ℓ} = 0.

Write

C = p_{R} ℓ sin ϕ + (ℓ^{2} / R + A) cos ϕ

. Differentiating along these equations gives

\dot{C} = (\frac{ℓ ^{3}}{R ^{3}} + \frac{A ℓ}{R ^{2}}) sin ϕ + \frac{p _{R} ℓ ^{2}}{R ^{2}} cos ϕ - \frac{p _{R} ℓ ^{2}}{R ^{2}} cos ϕ - (\frac{ℓ ^{3}}{R ^{3}} + \frac{A ℓ}{R ^{2}}) sin ϕ = 0.

Equivalently, the same cancellation evaluates

{C, H}

. Hence

H, p_{ϕ}, C are all constants of motion .

The extra conserved component is associated with the Runge-Lenz symmetry of this inverse-distance central potential, rather than with conservation of radial momentum.

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