Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 14D (Classical Dynamics)

Because the three positive gaps make one circuit,They describe only relative positions. The missing degree of freedom is the collective rigid rotation of all three particles around the circle, represented by the mean angular coordinate $\Phi$ in collective rotation and gap coordinates on a circle.

Choose ordered particle angles $q_1,q_2,q_3$ withand put $\Phi =(q_1+q_2+q_3)/3$. Solving for the individual angles givesThe kinetic energy is thereforeThe collective angle is an ignorable coordinate and decouples from the gaps, so its term may be omitted from the relative Lagrangian. Substituting $\gamma =2\pi -\alpha -\beta$ into the potential energy gives

The function $U(s)=V_0{e}^{-2s}$ is strictly convex. Subject to $\alpha +\beta +\gamma =2\pi$, the Jensen inequality giveswith equality only atThis 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 putThe gap perturbations arewhose sum vanishes. A Taylor expansion gives the quadratic energiesThus the linearized equations areThe 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 thereforeThe 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.

Permuting the gaps, or equivalently relabelling the particles, leaves both quadratic forms invariant. The three-dimensional permutation representation splits as the rigid-rotation line spanned by $(1,1,1)$ and the two-dimensional sum-zero standard representation of the symmetric group.

The two displayed vibrational vectors merely choose a basis of that sum-zero plane. A permutation generally mixes them, but their common frequency leaves the whole plane invariant. This is precisely the degenerate normal modes from permutation symmetry required by the physical symmetry.