OurBigBook.comSite Source code

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 14D / c / Solution

The function

U (s) = V_{0} e^{- 2 s}

is strictly convex. Subject to

α + β + γ = 2 π

, the Jensen inequality gives

U (α) + U (β) + U (γ) \geq 3 U (\frac{α + β + γ}{3}),

with equality only at

α = β = γ = \frac{2 π}{3} .

This is the minimum of the relative potential, so it is a stable relative equilibrium.

Let

q_{j} = q_{j}^{(0)} + x_{j}

near an equally spaced configuration, and put

A = V_{0} e^{- 4 π /3} .

The gap perturbations are

δ α = x_{2} - x_{1}, δ β = x_{3} - x_{2}, δ γ = x_{1} - x_{3},

whose sum vanishes. A Taylor expansion gives the quadratic energies

T_{2} = \frac{m r ^{2}}{2} \sum_{j} \overset{x}{˙}_{j}^{2}, V_{2} = 2 A [(x_{2} - x_{1})^{2} + (x_{3} - x_{2})^{2} + (x_{1} - x_{3})^{2}] .

Thus the linearized equations are

m r^{2} \overset{x}{¨}_{i} + 4 A (2 x_{i} - x_{j} - x_{k}) = 0, {i, j, k} = {1, 2, 3} .

The matrix in parentheses is the graph Laplacian of

K_{3}

. Its constant eigenvector has eigenvalue zero, while every vector whose components sum to zero has eigenvalue three. One orthonormal set of three normal modes and their angular frequencies is therefore

mode \frac{1}{3} (1, 1, 1) \frac{1}{2} (1, - 1, 0) \frac{1}{6} (1, 1, - 2) ω 0 \frac{12 V _{0} e ^{- 4 π /3}}{m r ^{2}} \frac{12 V _{0} e ^{- 4 π /3}}{m r ^{2}}

The zero mode is rigid rotation. The positive and equal squared frequencies on the two-dimensional relative subspace prove stability and agree with the general normal modes of three equal masses with a symmetric gap potential.

Solved by gpt-5.6-sol high.

 Ancestors (11)