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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/8b/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 8B / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

Put

ϵ = ∣ E ∣ > 0

. On the energy surface,

\frac{p ^{2}}{2 m} = \frac{1}{∣ q ∣} - ϵ \geq 0,

so

∣ q ∣ \leq q_{m a x} = \frac{1}{ϵ} .

Equality occurs at a turning point, so this is the smallest possible bound.

Using the natural collision continuation through the singular point, a full orbit runs between both turning points. Hence

I = \frac{1}{2 π} \oint p d q = \frac{2}{π} \int_{0}^{1/ ϵ} 2 m (\frac{1}{q} - ϵ) d q .

q = x^{2} / ϵ

. Then

\int_{0}^{1/ ϵ} 2 m (\frac{1}{q} - ϵ) d q = 2 \frac{2 m}{ϵ} \int_{0}^{1} 1 - x^{2} d x = \frac{π}{2} \frac{2 m}{ϵ} .

Therefore

I = \frac{2 m}{∣ E ∣} .

If instead one identifies the singularity as a reflecting endpoint and regards one half-line as the orbit, the action is half this value; the following scaling is identical.

Adiabatic invariance keeps

m /∣ E ∣

constant. Thus doubling

m

doubles

∣ E ∣

:

∣ E ∣_{final} = 2∣ E ∣_{initial}, E_{final} = 2 E_{initial} .

The energy becomes twice as negative.

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