Small Oscillations: Normal Modes and the Eigenvalue Problem for Coupled Oscillators

Intuition Beginner

Hang two identical pendulums side by side and connect them with a weak spring. Pull one pendulum to the side and release it. At first, only that pendulum swings. Gradually the second pendulum picks up the motion while the first slows down. After a while the first pendulum is nearly still and the second swings at full amplitude. Then the process reverses. Energy sloshes back and forth between the two pendulums in a slow beat pattern.

This is a coupled-oscillator system. The spring connects the two pendulums and lets them exchange energy. The key insight is that there are two special ways to start the system in which the energy does not slosh at all -- each pendulum swings at its own constant amplitude forever. These special patterns of motion are called normal modes. Every other motion is a mixture of the two normal modes, and the beating you observe is the interference between the two slightly different frequencies of the modes.

For any system of coupled oscillators near a stable equilibrium -- two pendulums, three masses on springs, or the atoms in a crystal lattice -- the motion can always be decomposed into independent normal modes. Each mode oscillates at its own characteristic frequency, and the number of modes equals the number of degrees of freedom. For two coupled pendulums there are two modes. For a molecule of $N$ atoms there are $3N-6$ vibrational modes (subtracting 3 translational and 3 rotational degrees of freedom).

The two normal modes of the coupled pendulums have a simple physical interpretation. In the symmetric mode, both pendulums swing together in the same direction at the same amplitude. The connecting spring does not stretch or compress, so it exerts no force. The frequency is just that of a single free pendulum: ${\omega }_1=\sqrt{g/L}$.

In the antisymmetric mode, the two pendulums swing in opposite directions. The spring stretches and compresses maximally, providing an extra restoring force. The frequency is higher: ${\omega }_2=\sqrt{g/L+2k{h}^2/(m{L}^2)}$, where $k$ is the spring constant, $h$ is the height of the spring attachment, $m$ is the mass, and $L$ is the pendulum length. The general motion is a superposition of these two modes, and the beat period is $T_{\text{beat}}=2\pi /({\omega }_2-{\omega }_1)$.

Visual Beginner

Figure: Two identical pendulums of length $L$ and mass $m$, connected by a horizontal spring of constant $k$ attached at height $h$ above the pivots. The upper panel shows the symmetric normal mode (both pendulums swing right together, spring relaxed). The lower panel shows the antisymmetric normal mode (left pendulum swings right while right pendulum swings left, spring stretched or compressed). A third panel shows a general motion -- the superposition of both modes -- producing the characteristic beating pattern where energy transfers periodically between the two pendulums.

Worked example Beginner

Two equal masses $m$ are connected by three springs of equal constant $k$: wall--spring--mass--spring--mass--spring--wall. Find the normal-mode frequencies and mode shapes.

Let $x_1$ and $x_2$ be the displacements from equilibrium. The spring forces are: on mass 1, the left wall spring pulls it back with force $-k{x}_1$, and the central spring pushes or pulls with force $-k(x_1-x_2)$, giving a total $-2k{x}_1+k{x}_2$. On mass 2 the same reasoning gives $-2k{x}_2+k{x}_1$. Newton's second law gives

$$m{\ddot{x}}_1=-2k{x}_1+k{x}_2,m{\ddot{x}}_2=k{x}_1-2k{x}_2.$$

Look for solutions where both masses oscillate at the same frequency: $x_1(t)=A_1\cos (\omega t)$ and $x_2(t)=A_2\cos (\omega t)$. Substituting:

$$-m{\omega }^2{A}_1=-2k{A}_1+k{A}_2,-m{\omega }^2{A}_2=k{A}_1-2k{A}_2.$$

In matrix form:

$$\left(\begin{array}{cc}2k& -k\\ -k& 2k\end{array}\right)\left(\begin{array}{c}{A}_1\\ A_2\end{array}\right)=m{\omega }^2\left(\begin{array}{c}{A}_1\\ A_2\end{array}\right).$$

Setting $\lambda =m{\omega }^2$: the characteristic equation is $(2k-\lambda {)}^2-k^2=0$, giving $\lambda =k$ or $\lambda =3k$.

Mode 1 (symmetric): ${\omega }_1=\sqrt{k/m}$. The eigenvector gives $A_1=A_2$. Both masses move in the same direction by the same amount -- the central spring is neither stretched nor compressed.

Mode 2 (antisymmetric): ${\omega }_2=\sqrt{3k/m}$. The eigenvector gives $A_1=-A_2$. The masses move in opposite directions by the same amount -- the central spring is maximally stretched or compressed.

For $m=1$ kg and $k=4$ N/m: ${\omega }_1=2$ rad/s and ${\omega }_2=2\sqrt{3}\approx 3.46$ rad/s. If you displace only mass 1 by distance $d$ and release, the motion is $x_1=\frac{d}{2}[\cos ({\omega }_{1}t)+\cos ({\omega }_{2}t)]$ and $x_2=\frac{d}{2}[\cos ({\omega }_{1}t)-\cos ({\omega }_{2}t)]$. Using the cosine product identity this becomes $x_1=d\cos (\frac{{\omega }_2-{\omega }_1}{2}t)\cos (\frac{{\omega }_2+{\omega }_1}{2}t)$ -- a rapid oscillation modulated by a slow beat envelope.

Check your understanding Beginner

Formal definition Intermediate+

Taylor expansion of the Lagrangian about equilibrium

Consider a mechanical system with $n$ generalised coordinates $q^1,\dots ,q^n$ and a stable equilibrium at $q^i=0$. The Lagrangian is $L=T-V$. The kinetic energy is a quadratic form in the velocities:

$$T=\frac{1}{2}{M}_{ij}{\dot{q}}^i{\dot{q}}^j,$$

where $M_{ij}=M_{ji}$ is the mass matrix (or inertia matrix), evaluated at equilibrium. It is symmetric and positive-definite. The potential energy expanded to second order about the equilibrium is

$$V=V_0+\frac{1}{2}{K}_{ij}{q}^i{q}^j+\cdots ,$$

where $K_{ij}=K_{ji}={\frac{{\partial }^{2}V}{\partial q^i\partial q^j}|}_{q=0}$ is the stiffness matrix (or force-constant matrix). The linear term vanishes because the equilibrium condition requires $\partial V/\partial q^i=0$. Stability requires $K$ to be positive-semidefinite (all eigenvalues non-negative). Dropping the irrelevant constant $V_0$, the quadratic Lagrangian is

$$L=\frac{1}{2}{M}_{ij}{\dot{q}}^i{\dot{q}}^j-\frac{1}{2}{K}_{ij}{q}^i{q}^j.$$

The Euler-Lagrange equations yield the equations of motion

$$M_{ij}{\ddot{q}}^j+K_{ij}{q}^j=0,$$

or in matrix form $M\ddot{\mathbf{q}}+K\mathbf{q}=0$.

The generalised eigenvalue problem

Seek harmonic solutions $\mathbf{q}(t)=\mathbf{a}\cos (\omega t)$, where $\mathbf{a}$ is a constant amplitude vector. Substituting:

$$(-{\omega }^{2}M+K)\mathbf{a}=0.$$

This is a generalised eigenvalue problem:

$$\boxed{K\mathbf{a}={\omega }^{2}M\mathbf{a}.}$$

Nonzero solutions exist when

$$\det (K-{\omega }^{2}M)=0,$$

the secular equation (also called the characteristic or frequency equation). It is a polynomial of degree $n$ in ${\omega }^2$.

Since $M$ is positive-definite and $K$ is symmetric, the eigenvalues ${\omega }_1^2,\dots ,{\omega }_n^2$ are all real. If $K$ is positive-definite, all ${\omega }_{\alpha }^2>0$ and the frequencies are real. If $K$ is positive-semidefinite with zero eigenvalues, the corresponding modes have $\omega =0$ -- these are zero-frequency modes representing free translations or rotations.

Orthogonality and normal coordinates

The eigenvectors ${\mathbf{a}}^{(\alpha )}$ satisfy the orthogonality relations

$${\mathbf{a}}^{(\alpha )\mathsf{T}}M{\mathbf{a}}^{(\beta )}={\delta }_{\alpha \beta },{\mathbf{a}}^{(\alpha )\mathsf{T}}K{\mathbf{a}}^{(\beta )}={\omega }_{\alpha }^2{\delta }_{\alpha \beta }.$$

Define the modal matrix $A=[{\mathbf{a}}^{(1)}\mid \cdots \mid {\mathbf{a}}^{(n)}]$. Then $A^{\mathsf{T}}MA=I$ and $A^{\mathsf{T}}KA={\Omega }^2=\mathrm{diag}({\omega }_1^2,\dots ,{\omega }_n^2)$.

The normal coordinates ${\eta }_{\alpha }$ are defined by $\mathbf{q}=A\mathbf{\eta }$, or equivalently ${\eta }_{\alpha }={\mathbf{a}}^{(\alpha )\mathsf{T}}M\mathbf{q}$. In normal coordinates the Lagrangian decouples:

$$\boxed{L=\frac{1}{2}\sum _{\alpha =1}^n\left({\dot{\eta }}_{\alpha }^2-{\omega }_{\alpha }^2{\eta }_{\alpha }^2\right).}$$

Each normal coordinate satisfies an independent harmonic-oscillator equation ${\ddot{\eta }}_{\alpha }+{\omega }_{\alpha }^2{\eta }_{\alpha }=0$. The general solution is ${\eta }_{\alpha }(t)=C_{\alpha }\cos ({\omega }_{\alpha }t)+D_{\alpha }\sin ({\omega }_{\alpha }t)$, and the physical coordinates are reconstructed by $\mathbf{q}(t)={\sum }_{\alpha }{\mathbf{a}}^{(\alpha )}{\eta }_{\alpha }(t)$.

Degeneracy

When two or more normal-mode frequencies coincide, the system has a degeneracy. Degeneracy almost always arises from a symmetry of the system. For a degenerate eigenvalue ${\omega }^2$ with multiplicity $d$, any linear combination of the $d$ independent eigenvectors is also an eigenvector with the same eigenvalue. The eigenvectors spanning the degenerate subspace are not unique; any $M$-orthonormal basis of the subspace serves, and this freedom reflects the underlying symmetry group.

Worked example: the double pendulum

A double pendulum has mass $m_1$ on a rod of length $l_1$ pivoted at the ceiling, and mass $m_2$ on a rod of length $l_2$ pivoted at $m_1$. Take $m_1=m_2=m$ and $l_1=l_2=l$. The coordinates are the small angles ${\theta }_1$ and ${\theta }_2$.

Position of $m_1$: $(l\sin {\theta }_1,-l\cos {\theta }_1)$. Position of $m_2$: $(l\sin {\theta }_1+l\sin {\theta }_2,-l\cos {\theta }_1-l\cos {\theta }_2)$. For small angles ($\sin \theta \approx \theta$, $\cos \theta \approx 1-{\theta }^2/2$):

$$T\approx \frac{1}{2}m{l}^2(2{\dot{\theta }}_1^2+{\dot{\theta }}_2^2+2{\dot{\theta }}_1{\dot{\theta }}_2),$$

so $M=m{l}^2\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$.

$$V=-mgl\cos {\theta }_1-mg(l\cos {\theta }_1+l\cos {\theta }_2)\approx -2mgl+mgl{\theta }_1^2+\frac{1}{2}mgl{\theta }_2^2,$$

so $K=mgl\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$ (dropping the constant).

The secular equation $\det (K-{\omega }^{2}M)=0$ gives

$$(2mgl-2m{l}^2{\omega }^2)(mgl-m{l}^2{\omega }^2)-m^2{l}^4{\omega }^4=0.$$

Dividing by $m^2{l}^2$: $(2g-2l{\omega }^2)(g-l{\omega }^2)-l^2{\omega }^4=0$, which simplifies to ${\omega }^4-4(g/l){\omega }^2+2(g/l{)}^2=0$. The solutions are

$${\omega }_1^2=\frac{g}{l}(2-\sqrt{2})\approx 0.586\frac{g}{l},{\omega }_2^2=\frac{g}{l}(2+\sqrt{2})\approx 3.414\frac{g}{l}.$$

The low-frequency mode (${\omega }_1$) has eigenvector with $a_2/a_1=\sqrt{2}$: both arms swing roughly together. The high-frequency mode (${\omega }_2$) has $a_2/a_1=-\sqrt{2}$: the arms swing in opposite directions.

Worked example: the triple coupled oscillator

Three equal masses $m$ connected in a line by four equal springs $k$: wall--spring--mass--spring--mass--spring--mass--spring--wall. The mass matrix is $M=m{I}_3$. The stiffness matrix is the tridiagonal Toeplitz matrix

$$K=k\left(\begin{array}{ccc}2& -1& 0\\ -1& 2& -1\\ 0& -1& 2\end{array}\right).$$

The secular equation gives three eigenvalues ${\omega }_j^2=(2k/m)(1-\cos (j\pi /4))$ for $j=1,2,3$:

• ${\omega }_1^2=(2-\sqrt{2})k/m$: the slowest mode, with all masses moving in the same direction, amplitude largest in the centre.
• ${\omega }_2^2=2k/m$: the middle mode, with the centre mass stationary and the outer masses moving in opposite directions.
• ${\omega }_3^2=(2+\sqrt{2})k/m$: the fastest mode, with adjacent masses moving in opposite directions.

The eigenvectors have the form $a_n^{(j)}=\sin (jn\pi /4)$, reflecting the standing-wave pattern on a lattice with fixed endpoints.

Key theorem with proof Intermediate+

Theorem (Normal-mode decomposition). Let $M$ be a real symmetric positive-definite $n\times n$ matrix and $K$ a real symmetric positive-semidefinite $n\times n$ matrix. Then there exists an invertible matrix $A$ such that

$$A^{\mathsf{T}}MA=I,A^{\mathsf{T}}KA=\mathrm{diag}({\omega }_1^2,\dots ,{\omega }_n^2),$$

where ${\omega }_{\alpha }^2\ge 0$ are the generalised eigenvalues of $K\mathbf{a}={\omega }^{2}M\mathbf{a}$. The columns of $A$ are the eigenvectors (normal modes), and the transformation $\mathbf{q}=A\mathbf{\eta }$ decouples the system into $n$ independent harmonic oscillators.

Proof. Since $M$ is positive-definite, it has a unique positive-definite square root $M^{1/2}$. Define $B=M^{-1/2}K{M}^{-1/2}$, which is real and symmetric (because $K$ is symmetric). By the spectral theorem, there exists an orthogonal matrix $Q$ such that $Q^{\mathsf{T}}BQ={\Omega }^2=\mathrm{diag}({\omega }_1^2,\dots ,{\omega }_n^2)$. Set $A=M^{-1/2}Q$. Then:

$$A^{\mathsf{T}}MA=Q^{\mathsf{T}}{M}^{-1/2}M{M}^{-1/2}Q=Q^{\mathsf{T}}Q=I,$$ $$A^{\mathsf{T}}KA=Q^{\mathsf{T}}{M}^{-1/2}K{M}^{-1/2}Q=Q^{\mathsf{T}}BQ={\Omega }^2.$$

The positive-semidefiniteness of $K$ guarantees ${\omega }_{\alpha }^2={\mathbf{e}}_{\alpha }^{\mathsf{T}}{\Omega }^2{\mathbf{e}}_{\alpha }={\mathbf{e}}_{\alpha }^{\mathsf{T}}{Q}^{\mathsf{T}}BQ{\mathbf{e}}_{\alpha }\ge 0$. $\square$

Corollary (Reality of frequencies). Contract $K\mathbf{a}={\omega }^{2}M\mathbf{a}$ with ${\mathbf{a}}^{\ast \mathsf{T}}$:

$${\omega }^2=\frac{{\mathbf{a}}^{\ast \mathsf{T}}K\mathbf{a}}{{\mathbf{a}}^{\ast \mathsf{T}}M\mathbf{a}}.$$

The numerator is non-negative (positive-semidefinite $K$) and the denominator is strictly positive (positive-definite $M$), so ${\omega }^2\ge 0$. All normal-mode frequencies are real.

Exercises Intermediate+

Lean formalization Intermediate+

The simultaneous diagonalisation of two symmetric bilinear forms $M$ and $K$ is the mathematical core of this unit. Mathlib contains the prerequisites: Matrix.PosDef, LinearAlgebra.Eigenspace, and LinearAlgebra.Cholesky for the decomposition of positive-definite matrices. What is missing is the physics-specific theorem that the Taylor expansion of the Lagrangian about a stable equilibrium produces two quadratic forms whose generalised eigenvalues are the squared normal-mode frequencies, together with the guarantee that these eigenvalues are real and non-negative. The normal-coordinate transformation as a congruence diagonalising both forms simultaneously would build on existing diagonalisation results but requires a physics-specific API. This unit ships without a Lean module.

Advanced results Master

Perturbation of degenerate frequencies

When a symmetry produces degenerate normal-mode frequencies, a small symmetry-breaking perturbation splits the degeneracy. Suppose ${\omega }_A^2={\omega }_B^2$ for two modes ${\mathbf{a}}^{(A)}$ and ${\mathbf{a}}^{(B)}$ of the unperturbed system, and the perturbation modifies $K\to K+\delta K$. To first order, the degenerate subspace produces a $2\times 2$ sub-problem:

$$\left(\begin{array}{cc}\delta K_{AA}& \delta K_{AB}\\ \delta K_{BA}& \delta K_{BB}\end{array}\right)\left(\begin{array}{c}{c}_A\\ c_B\end{array}\right)=\delta {\omega }^2\left(\begin{array}{c}{c}_A\\ c_B\end{array}\right),$$

where $\delta K_{\alpha \beta }={\mathbf{a}}^{(\alpha )\mathsf{T}}(\delta K){\mathbf{a}}^{(\beta )}$. The frequency splitting is $\delta {\omega }^2=\frac{1}{2}(\delta K_{AA}+\delta K_{BB})\pm \frac{1}{2}\sqrt{(\delta K_{AA}-\delta K_{BB}{)}^2+4\delta K_{AB}^2}$. The "correct" linear combinations of the degenerate modes are the eigenvectors of this $2\times 2$ matrix, and the splitting is first-order in the perturbation strength. This is the classical analogue of degenerate perturbation theory in quantum mechanics.

Molecular vibrations and the Wilson GF method

The normal-mode analysis of polyatomic molecules is central to infrared and Raman spectroscopy. For a molecule of $N$ atoms, there are $3N$ Cartesian displacement coordinates, of which 6 (or 5 for linear molecules) are zero-frequency translations and rotations. The remaining $3N-6$ (or $3N-5$) are the genuine vibrational modes.

The Wilson GF method (also called the FG matrix method) is the standard computational procedure. Define the $3N\times 3N$ G matrix (inverse mass matrix in internal coordinates) and the corresponding F matrix (force-constant matrix in internal coordinates). The secular equation is $\det (GF-\lambda I)=0$. The advantage of internal coordinates (bond stretches, angle bends, torsions) over Cartesian coordinates is that the force constants have direct chemical meaning and the translational-rotational modes are automatically separated.

The method proceeds by: (1) choosing a set of internal coordinates $\mathbf{s}$ related to Cartesians by the $B$ matrix: $\mathbf{s}=B\mathbf{x}$; (2) computing $G=B{M}^{-1}{B}^{\mathsf{T}}$ where $M$ is the diagonal Cartesian mass matrix; (3) solving $GF\mathbf{L}=\mathbf{L}\Lambda$ where $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_{3N-6})$. The eigenvectors $\mathbf{L}$ give the normal modes in internal coordinates. Group theory (point-group symmetry) block-diagonalises $GF$ into irreducible representations, greatly reducing the computational cost and predicting which modes are IR active (transform as the dipole operator) or Raman active (transform as the polarisability tensor).

Classical field normal modes

The normal-mode analysis extends to continuous systems. A vibrating string of length $\ell$ with fixed endpoints has displacement field $u(x,t)$ satisfying the wave equation $\mu \ddot{u}=\kappa u^{\prime \prime }$. The normal modes are $u_n(x,t)=\sin (n\pi x/\ell )\cos ({\omega }_{n}t)$ with frequencies ${\omega }_n=n\pi c/\ell$ where $c=\sqrt{\kappa /\mu }$. The general solution is

$$u(x,t)=\sum _{n=1}^{\infty }{A}_n\sin \left(\frac{n\pi x}{\ell }\right)\cos ({\omega }_{n}t+{\delta }_n).$$

This is the continuum limit of the discrete lattice problem 09.02.06. The sum over normal modes becomes a Fourier series. For a system in two or three dimensions, the normal modes become functions of spatial position satisfying the Helmholtz equation $({\nabla }^2+k^2)\varphi =0$ with appropriate boundary conditions. The Sturm-Liouville theory of differential equations is the infinite-dimensional generalisation of the finite-dimensional eigenvalue problem studied here.

Stability analysis via the Hessian

The condition for stable equilibrium is that the Hessian $K_{ij}={\partial }^{2}V/\partial q^i\partial q^j$ be positive-semidefinite. If $K$ has a negative eigenvalue, the equilibrium is unstable: small displacements along that direction grow exponentially. If $K$ has a zero eigenvalue, the equilibrium is marginally stable along that direction -- the system can drift without restoring force. The number of zero eigenvalues equals the number of continuous symmetries of the potential. A molecule free in space has six zero-frequency modes: three translations and three rotations.

The Rayleigh quotient

The normal-mode frequencies can be characterised variationally via the Rayleigh quotient. For any trial displacement $\mathbf{q}$:

$$R[\mathbf{q}]=\frac{{\mathbf{q}}^{\mathsf{T}}K\mathbf{q}}{{\mathbf{q}}^{\mathsf{T}}M\mathbf{q}}.$$

The stationary values of $R$ are the eigenvalues ${\omega }_{\alpha }^2$, and the stationarity condition $\delta R=0$ recovers $K\mathbf{q}={\omega }^{2}M\mathbf{q}$. The minimum of $R$ gives ${\omega }_1^2$ (the fundamental frequency), and the Courant-Fischer min-max principle gives the higher frequencies. This variational characterisation is the basis of the Rayleigh-Ritz method for computing approximate normal modes in engineering and of the functional analysis of Sturm-Liouville problems in the continuum limit.

Forced oscillations and resonance

When a harmonic driving force $\mathbf{F}(t)={\mathbf{F}}_0\cos (\Omega t)$ is applied, the response in normal coordinates is

$${\eta }_{\alpha }(t)=\frac{f_{\alpha }}{{\omega }_{\alpha }^2-{\Omega }^2}\cos (\Omega t),$$

where $f_{\alpha }={\mathbf{a}}^{(\alpha )\mathsf{T}}{\mathbf{F}}_0$ is the projection of the driving force onto mode $\alpha$. Resonance occurs when $\Omega$ approaches any ${\omega }_{\alpha }$. A force orthogonal to a mode ($f_{\alpha }=0$) does not excite it -- the principle behind vibration isolation.

Connection to quantum mechanics

The normal-mode Hamiltonian is $H={\sum }_{\alpha }\frac{1}{2}(p_{\alpha }^2+{\omega }_{\alpha }^2{\eta }_{\alpha }^2)$. Upon quantisation each mode becomes an independent quantum harmonic oscillator with energy levels $E_{n_{\alpha }}=\hslash {\omega }_{\alpha }(n_{\alpha }+\frac{1}{2})$. The quantised excitations of the vibrational normal modes of a crystal lattice are phonons, the quasiparticles of solid-state physics. The normal-mode decomposition is the prerequisite for the second-quantised phonon Hamiltonian.

The diatomic chain and optical modes

When the lattice has two different masses $M$ and $m$ alternating, the dispersion relation splits into two branches. The acoustic branch has $\omega \to 0$ as $q\to 0$ (long-wavelength translations). The optical branch has $\omega \to {\omega }_0>0$ as $q\to 0$ (the two sublattices oscillate against each other even at long wavelength). In ionic crystals the optical mode carries an oscillating electric dipole that couples to electromagnetic radiation -- hence the name "optical." The gap between the branches is a band gap in the phonon spectrum.

Synthesis. The normal-mode decomposition puts together the Lagrangian quadratic expansion, the generalised eigenvalue problem, and simultaneous diagonalisation of two quadratic forms; the central insight is that stability of an equilibrium point guarantees the positive-semidefiniteness of both $M$ and $K$, and the spectral theorem for symmetric matrices then ensures the existence of real normal frequencies and orthogonal mode shapes. This generalises directly to the continuum limit where the discrete eigenvalue problem becomes a Sturm-Liouville problem for wave modes, and the quantum-mechanical promotion of each normal mode to an independent oscillator is the foundational reason that phonons, photons, and all quantum-field-theoretic quasiparticles share the harmonic-oscillator structure. The normal-mode decomposition is the bridge from classical mechanics to quantum field theory.

Connections Master

• 09.02.01 The action principle for small oscillations is $\delta S=0$ applied to the quadratic Lagrangian $L=\frac{1}{2}{M}_{ij}{\dot{q}}^i{\dot{q}}^j-\frac{1}{2}{K}_{ij}{q}^i{q}^j$. The Euler-Lagrange equations linearise about the equilibrium.

• 09.02.02 The Euler-Lagrange equations for the quadratic Lagrangian give the matrix equation $M\ddot{\mathbf{q}}+K\mathbf{q}=0$. The normal-mode decomposition solves this by diagonalising $M$ and $K$ simultaneously.

• 09.02.05 Lagrangian perturbation theory adds cubic and quartic terms to the quadratic potential, coupling the normal modes and generating nonlinear phenomena. The normal modes are the zeroth-order basis for the perturbation expansion.

• 09.02.06 The field-Lagrangian unit takes the continuum limit of the discrete chain; the normal-mode frequencies of the chain converge to the continuum wave-mode frequencies as $N\to \infty$.

• 09.07.02 The elastic-wave unit treats the continuum limit of the lattice normal-mode problem; the discrete frequencies become the phonon dispersion relation.

• 01.01.01 The vector-space foundation (basis, linear independence, dimension) underlies the representation of the state as a vector in the space of generalised coordinates, and the eigenvalue problem is a statement about linear operators on that space.

Historical & philosophical context Master

The analysis of small oscillations and normal modes originates with Daniel Bernoulli's 1753 work on the vibrating string, which identified the harmonic overtones as the natural modes of the system. Lagrange's Mechanique analytique (1788) gave the first systematic treatment of coupled oscillators using what we now call the Lagrangian approach, establishing the general principle that any system of coupled oscillators near equilibrium can be reduced to a set of independent oscillators by a suitable change of coordinates.

The matrix formulation and the connection to the eigenvalue problem were made explicit by Rayleigh in The Theory of Sound (1877). Rayleigh introduced the quotient that bears his name as a variational tool for computing vibration frequencies, and his work remains the standard reference for acoustical applications. Routh extended the treatment to gyroscopic and dissipative systems in his Advanced Rigid Dynamics (1892).

The generalised eigenvalue problem $K\mathbf{a}={\omega }^{2}M\mathbf{a}$ is a reformulation of the simultaneous-diagonalisation problem for two quadratic forms, solved by Weierstrass and Kronecker in the 19th century. Its appearance in mechanics is an instance of a broader pattern: whenever a physical system is linearised near a stable equilibrium, the resulting equations of motion involve two symmetric forms that must be simultaneously diagonalised. This pattern repeats in quantum mechanics (the Hartree-Fock equations), structural engineering (finite-element modal analysis), and control theory (linear system diagonalisation).

A philosophical point: the normal-mode decomposition reveals that the apparent complexity of coupled motion is a consequence of looking at the system in the "wrong" coordinates. In the original generalised coordinates, the equations are coupled and the motion looks complicated (beats, energy transfer). In normal coordinates, the same motion is simply a set of independent oscillations. The physics does not change -- only the description does. This is a recurring theme in theoretical physics: the right choice of coordinates (or basis, or representation) can reduce an apparently intractable problem to a set of decoupled, solvable ones. The normal-mode decomposition is perhaps the simplest and most physically transparent instance of this principle.

Bibliography Master

• Lagrange, J. L., Mechanique analytique (1788), Seconde Partie, §VI.
• Rayleigh, J. W. Strutt, The Theory of Sound, 2nd ed. (Macmillan, 1894), Ch. 4-5.
• Routh, E. J., Advanced Rigid Dynamics, 6th ed. (Macmillan, 1905), Ch. 6-7.
• Weierstrass, K., "Zur Theorie der bilinearen und quadratischen Formen," Monatsber. Akad. Wiss. Berlin (1858), 310-338.
• Goldstein, H., Poole, C. P. & Safko, J., Classical Mechanics, 3rd ed. (Pearson, 2002), §6.1-6.3.
• Taylor, J. R., Classical Mechanics (University Science Books, 2005), Ch. 11.
• Landau, L. D. & Lifshitz, E. M., Mechanics, 3rd ed. (Course of Theoretical Physics Vol. 1, Pergamon, 1976), §23-25.
• Arnold, V. I., Mathematical Methods of Classical Mechanics, 2nd ed. (Springer, 1989), §25.
• Wilson, E. B., Decius, J. C. & Cross, P. C., Molecular Vibrations (McGraw-Hill, 1955), Ch. 4 (the GF method).