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Let $q_i=z_i-z_i^{(0)}$ denote a small longitudinal displacement from equilibrium, and let every atomic mass be $m$. The constant equilibrium lengths disappear from the second derivatives. The force constant matrix isThe mass-weighted dynamical matrix is $D=K/m$, with equations $\ddot{\mathbf{q}}=-D\mathbf{q}$. If the phrase dynamical matrix denotes the unweighted Hessian, that object is $K$ instead; the normal-mode equation is $K\mathbf{v}=m{\omega }^2\mathbf{v}$ in either notation. Every row sums to zero, as required for a freely translating chain.

In the totally symmetric $(r,s)$ basis, the force constant matrix reduces toThe unperturbed eigenvalues are distinct within this symmetry block, so nondegenerate perturbation theory applies even though the full unperturbed matrix has degeneracies in different symmetry sectors. The first-order shifts are both $\gamma$. At second order the off-diagonal matrix element gives $-{\gamma }^2/(2k)$ for the lower mode and $+{\gamma }^2/(2k)$ for the upper one. ThereforeIn frequency form,The corresponding eigenvectors are $r+(\gamma /2k)s+\cdots$ and $s-(\gamma /2k)r+\cdots$. As a check, the exact squared-frequency eigenvalues arewhose expansions reproduce the perturbation theory result. The newly lifted zero frequency scales as $\sqrt{\gamma }$, so expanding ${\omega }_{-}$ as an ordinary Taylor series in $\gamma$ would be inappropriate.