 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-2/14d/b/solution

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

Choose ordered particle angles

q_{1}, q_{2}, q_{3}

with

α = q_{2} - q_{1}, β = q_{3} - q_{2}, γ = 2 π - α - β,

and put

Φ = (q_{1} + q_{2} + q_{3}) /3

. Solving for the individual angles gives

q_{1} = Φ - \frac{2 α + β}{3}, q_{2} = Φ + \frac{α - β}{3}, q_{3} = Φ + \frac{α + 2 β}{3} .

The kinetic energy is therefore

T = \frac{m r ^{2}}{2} \sum_{j = 1}^{3} \overset{q}{˙}_{j}^{2} = \frac{3 m r ^{2}}{2} \dot{Φ}^{2} + \frac{m r ^{2}}{3} (\overset{α}{˙}^{2} + \overset{α}{˙} \dot{β} + \dot{β}^{2}) .

The collective angle is an ignorable coordinate and decouples from the gaps, so its term may be omitted from the relative Lagrangian. Substituting

γ = 2 π - α - β

L_{rel} = \frac{m r ^{2}}{3} (\overset{α}{˙}^{2} + \overset{α}{˙} \dot{β} + \dot{β}^{2}) - V_{0} (e^{- 2 α} + e^{- 2 β} + e^{- 4 π + 2 α + 2 β}) .

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