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Infrared spectroscopy measures absorption as the incident photon frequency is swept through molecular normal mode frequencies. In the electric-dipole approximation the interaction is $-\mathbf{p}\cdot \mathbf{E}$. Expanding the molecular electric dipole moment in a normal coordinate $Q$ givesThe fundamental one-quantum vibrational transition dipole moment is proportional to the derivative multiplying $Q$. Thusis the criterion for vibrational infrared activity; the derivative must also have a component along the incident polarization of an electromagnetic wave. A permanent molecular electric dipole moment is not necessary. Absorption occurs near $\hslash \omega =E_{v+1}-E_v$, with broadened lines reflecting finite lifetimes and the molecular environment.

General definitions. The Euler-Lagrange equation gives $m_i{\ddot{q}}_i+{\sum }_j{V}_{ij}{q}_j=0$. Write $M=\mathrm{diag}(m_i)$ and $K=(V_{ij})$. A normal mode has fixed displacement pattern $v$ and common time factor, $q(t)=vQ(t)$; its normal frequency satisfiesHere $\omega$ is angular frequency. A normal coordinate is the coefficient of one pattern in a complete modal expansion, so these coordinates decouple the equations. Since $M$ is positive definite, the change $z=M^{1/2}q$ turns the mode equation into the ordinary symmetric eigenvalue problem for $M^{-1/2}K{M}^{-1/2}$.

Three-mass equations. Let $x_1,x_2,x_3$ be horizontal displacements from the unstretched equilibrium, ordered left to right. The kinetic energy and potential energy areDifferentiating the potential energy givesThus $M=m\mathrm{diag}(1,2,3)$ and

Frequencies and patterns. Put $\lambda =m{\omega }^2/k$. Expansion of the generalized eigenvalue problem for small oscillations givesSolving at each root givesThe zero normal mode is uniform translation. In the lower nonzero mode, the central mass stays fixed, the two end masses move oppositely, and their unequal displacements balance both the end-mass momenta and the forces on the central mass. In the upper mode the end masses move together and the centre oppositely. Both vibrating modes leave the centre of mass fixed.

Orthogonality. Because $K$ is symmetric, two mode equations implyDistinct eigenvalues therefore give the mass-matrix orthogonality of normal modes:The Euclidean dot product need not vanish for unequal masses; for example $v_0^T{v}_1=-2$, while $v_0^{T}M{v}_1=m(-3+3)=0$. The modal squared norms are $6m,12m,12m$.

Normal coordinates. Expand $x=Q_0{v}_0+Q_1{v}_1+Q_3{v}_3$. Project using the mass-weighted inner product and divide by each modal norm to obtainFor example, the inverse equations are $x_1=Q_0-3Q_1+Q_3$, $x_2=Q_0-2Q_3$, $x_3=Q_0+Q_1+Q_3$. They reduce the energies toand henceThe first coordinate is the centre of mass and has free linear motion. Multiplying $Q_0,Q_1,Q_3$ by $\sqrt{6m},\sqrt{12m},\sqrt{12m}$ respectively gives mass-normalized normal coordinates if desired.

Order coordinates as $q=(x_1,x_2,x_3;y_1,y_2,y_3)$. For either uniform displacement, $t_x=(1,1,1;0,0,0)$ or $t_y=(0,0,0;1,1,1)$, every relative displacement and every extension vanishes. These normal modes have zero normal frequency and represent two independent planar translations.

For $r=(-1,-1,2;\sqrt{3},-\sqrt{3},0)$, substitution givesThus $Kr=0$ too. A small rigid rotation through angle $\alpha$ moves equilibrium position $(X_i,Y_i)$ by $\alpha (-Y_i,X_i)$. With the specified positions this displacement is $-\alpha r/2$, identifying the third vector as infinitesimal rigid rotation. HenceA zero-frequency normal coordinate obeys $\ddot{Q}=0$, so $Q=A+Bt$. This permits a constant displacement or a uniform velocity at the level of the linearized equations. In particular, the rotational pattern preserves lengths to first order; finite-angle rotation follows a curved path outside a single linear displacement pattern.

A normal mode is a fixed displacement pattern whose components share a common harmonic time dependence in the equations for small oscillations. Its normal frequency is the associated oscillation frequency; below, $\omega$ denotes the angular frequency, so the frequency in cycles per unit time is $\omega /(2\pi )$. A normal coordinate is the amplitude of one such pattern, chosen so that the linear equations decouple. For mass matrix $M$ and stiffness matrix $K$, the patterns solve $Kv={\omega }^{2}Mv$. Zero eigenvalues describe free rigid motions rather than periodic restoring oscillations.

For small oscillations with quadratic kinetic and potential energies, a normal mode has a fixed spatial amplitude pattern and one angular normal frequency. Its amplitude is a normal coordinate; choosing all modes diagonalizes the two quadratic forms and decouples their equations. A zero frequency describes a free symmetry motion.

Let $q_j=R[{\theta }_j-j\pi /2]$ be the small signed arc displacements, with cyclic indices $j=0,1,2,3$. The springs follow arcs, so their extensions are $q_{j+1}-q_j$. The Lagrangian of the normal modes of a four-bead spring ring isThe stiffness matrix divided by $k$ is the cycle graph Laplacian, with diagonal entries two, adjacent entries minus one, and zero opposite entries. An orthonormal basis of mode vectors and their frequencies isThe normal mode coordinates are $Q_{\alpha }=u_{\alpha }\cdot q$, with equations ${\ddot{Q}}_{\alpha }+{\omega }_{\alpha }^2{Q}_{\alpha }=0$. The zero mode rotates all beads together without stretching a spring and has $Q_0=A+Bt$. In each of the two degenerate middle modes, opposite beads have opposite signed arc displacements while the other pair remains stationary. Any fixed linear combination is another mode of that frequency. The last mode alternates the signed displacement of adjacent beads. The radius cancels from the frequencies because arc length supplies the same factor $R^2$ to the angular kinetic and spring energies.