Eigenvalue Problem for Normal Modes

The generalized eigenvalue problem $(\mathbf K - \omega^2\mathbf M)\mathbf a = 0$ is, by volume of computation, one of the most-solved equations in engineering. Every finite-element model of a bridge, aircraft wing, engine block, or skyscraper is exactly equation (2) with $N$ in the millions: mesh the structure, assemble $\mathbf M$ and $\mathbf K$, and extract the lowest few hundred modes with Lanczos-type iterations (nobody expands a million-degree characteristic polynomial — the determinant is a proof device, not an algorithm). The mode list is the structure's dynamic identity: automotive engineers shift body modes away from engine firing frequencies; launch providers run "coupled loads analysis," re-solving the eigenproblem for rocket-plus-satellite to certify that liftoff won't find a resonance; and a spacecraft cannot book a ride until its modal survey — measured frequencies and shapes, shaken out on a vibration table — matches the model. When any of these systems is then driven at a mode frequency, the response is governed by the single-oscillator physics of the Resonance article, one mode at a time. The same mathematics runs far from mechanics. Quantum chemistry's workhorse Hartree–Fock–Roothaan equations, $\mathbf F\mathbf C = \mathbf S\mathbf C\varepsilon$, are a generalized symmetric eigenproblem with the overlap matrix $\mathbf S$ playing the role of $\mathbf M$ — same theorems, same whitening trick, same warning about the wrong inner product. Phonon band structures, the normal modes of crystals computed from density-functional $\mathbf K$-matrices, decide materials' thermal conductivity and superconducting properties. And gravitational-wave astronomers fit black-hole ringdowns to normal-mode frequencies of spacetime itself. Wherever two quadratic forms meet — one inertia-like and positive, one stiffness-like — this article's machinery is already there.

Ask one question of any coupled system: which starting shapes does the motion preserve? Displace a system of masses in an arbitrary pattern and the pattern immediately falls apart — different parts oscillate at different effective rates and the configuration smears. But certain special patterns hold their shape forever, every coordinate rising and falling in unison like a photograph being rescaled, and each such pattern rings at one pure frequency. Those are the normal modes. The eigenvalue problem is nothing more than this question written in algebra: a shape $\mathbf a$ is preserved when the restoring forces $-\mathbf K\mathbf a$ it generates are distributed across the masses in exactly the proportions of $\mathbf a$ itself, i.e. $\mathbf K\mathbf a = \omega^2\mathbf M\mathbf a$ — the force pattern is the inertia pattern scaled by a constant. That constant is the squared frequency. The chain of three masses — light, heavy, light, joined by two equal springs, a caricature of the CO₂ molecule — makes the answer concrete. It has exactly three shape-preserving patterns, and you can guess all three before touching a determinant:

Slide all three masses together and no spring changes length: zero restoring force, zero frequency — a "mode" that is really the free molecule drifting through space. Stretch the two outer masses symmetrically outward and the center, pulled equally both ways, stays put while each outer mass rides its own spring. Push both outer masses the same way and the center must recoil to keep the center of mass fixed; both springs now work on every mass and the frequency is the highest of the three. Three degrees of freedom, three modes — and, as this article proves, that counting is not luck but theorem. The second picture to install is geometric, because it explains the strangest feature of the subject: mode shapes of systems with unequal masses look skewed — plotted as vectors, they visibly fail to be perpendicular — yet the theory keeps calling them orthogonal. The resolution is that "perpendicular" is being measured with the wrong ruler. The physically meaningful geometry of configuration space is the one set by kinetic energy: the length of a velocity vector is $\dot{\mathbf q}^{\mathsf T}\mathbf M\dot{\mathbf q}$, in which heavy coordinates count for more. Rescale each coordinate by the square root of its mass — stretch the heavy axes — and in the rescaled space the kinetic energy becomes the ordinary Euclidean length, and the mode shapes snap to exact right angles.

The vectors drawn are the actual mode shapes of the first worked example below ($m_2 = 2m_1$). This picture is the entire content of "$\mathbf M$-orthogonality": the modes are perpendicular — in the geometry where kinetic energy is the ruler. Everything that follows (the decoupling of energies, the projection of initial conditions, the completeness of the mode basis) works because of that hidden right angle. The phenomenon these modes orchestrate — beating, energy exchange — lives in the Coupled Oscillators article; the mode concept itself is developed in Normal Modes; and what you do with the answer — turn each mode into an independent coordinate — is the business of Normal Coordinates. This article owns the machinery that finds the modes and the theorems that guarantee it always works.

Every small-oscillation problem, whatever its origin, arrives in the same standard form. Expand the Lagrangian about a stable equilibrium to second order in the displacements $\mathbf q = (q_1, \dots, q_N)$ and their velocities, and it reads

where: $\mathbf q$ — column vector of generalized displacements from equilibrium; $\mathbf M$ — the mass (kinetic-energy) matrix, $M_{ij} = \partial^2 T/\partial\dot q_i\partial\dot q_j$ evaluated at equilibrium; $\mathbf K$ — the stiffness matrix, $K_{ij} = \partial^2 V/\partial q_i\partial q_j$ at equilibrium. Both matrices are constant, real, and symmetric. $\mathbf M$ is positive definite — $\dot{\mathbf q}^{\mathsf T}\mathbf M\dot{\mathbf q} > 0$ for any nonzero velocity, because kinetic energy is positive whenever anything moves. $\mathbf K$ is positive semidefinite when the equilibrium is a true minimum of $V$: displacements cost energy, or at worst (symmetry directions) cost nothing. The Euler–Lagrange equations of (1) — one derivative through each quadratic form, the symmetry of the matrices absorbing the factors of two — give the coupled equations of motion in matrix form:

where: each row says: the mass-weighted acceleration of coordinate $i$ is driven by every displacement that appears in row $i$ of $\mathbf K$. Off-diagonal entries of $\mathbf K$ (or $\mathbf M$) are the coupling. This is $N$ second-order equations tangled together; the next derivation untangles them into algebra.

The characteristic polynomial is worth actually looking at once, because its graph is the frequency spectrum. For the m–M–m chain of the worked example below ($m = 1.0$ kg, $M = 2.0$ kg, $k = 100$ N/m), the determinant expands to $P(\lambda) = -2\lambda(\lambda - 100)(\lambda - 200)$ with $\lambda = \omega^2$ — a cubic whose three crossings are the three squared frequencies, zero mode included:

For a generic polynomial, real coefficients guarantee nothing — roots pair off into the complex plane whenever they please. The remarkable claim of the figure is that this polynomial's roots always land on the real axis, and never below zero when the equilibrium is stable. That is not an accident of the example; it is the first structure theorem.

Two theorems carry the whole subject on their backs. Both are one-page consequences of symmetry and positivity, and both proofs run on the same fuel: the $\mathbf M$-weighted inner product $\langle\mathbf u, \mathbf v\rangle_{\mathbf M} = \mathbf u^{\dagger}\mathbf M\,\mathbf v$, the mass-metric geometry of the physical picture.

Together the theorems deliver the complete package promised in the overview. Reality plus symmetry guarantees a full set of $N$ modes spanning configuration space (repeated roots included — see below), $\mathbf M$-orthogonality makes them independent, and mass normalization makes them an orthonormal basis in the kinetic-energy geometry. Transforming to that basis is precisely what turns the Lagrangian into $N$ independent oscillators — the construction carried out in the Normal Coordinates article.

Two special spectral features are so common in real systems that they deserve their own treatment, and both are opportunities rather than pathologies. A zero mode is a root $\omega^2 = 0$ of the secular equation, which by (3) means $\mathbf K\mathbf a^{(0)} = 0$: a displacement pattern that generates no restoring force at all, a flat direction of the potential. Free-floating systems always have them — translate the whole m–M–m chain and neither spring changes length. For a zero mode the equation of motion is not oscillation but $\ddot\xi_0 = 0$, integrating to

where: $c_1, c_2$ — constants set by the initial displacement and velocity along the mode. Uniform drift: the pattern slides at constant rate forever. This is momentum conservation wearing mode clothing — the zero mode of translation is the center-of-mass coordinate, and $c_2$ is its conserved velocity. Count them by symmetry: a molecule free in three dimensions has three translational zero modes and three rotational ones (two if linear, since spinning a line about its own axis moves nothing), which is exactly the origin of the $3N - 6$ (or $3N - 5$) count of true vibrations. A zero mode in a structure that is supposed to be anchored is a design alarm: some direction is unbraced.

Degeneracy is a repeated root: two (or more) modes sharing one frequency. Then the null space of $\mathbf K - \omega^2\mathbf M$ is a plane rather than a line, and any vector in that plane is a legitimate mode shape — the eigenvalue problem refuses to pick favorites. The freedom is real and harmless: choose any $\mathbf M$-orthogonal basis within the degenerate subspace (Gram–Schmidt, using the $\mathbf M$-inner product) and all the formulas of this article proceed unchanged. The overwhelmingly common cause is symmetry: CO₂ can bend in the vertical plane or the horizontal one, and the molecule's axial symmetry makes the two bends physically indistinguishable, hence exactly degenerate. Symmetry-adapted combinations are the natural choice — and the deep version of this remark, that degenerate subspaces carry representations of the symmetry group, is the road from this article into molecular spectroscopy and solid-state physics.

The modes are building blocks; the general motion is the building. Because equation (2) is linear, any superposition of mode solutions solves it, and because the $N$ mode shapes span configuration space, every solution is such a superposition:

where: $C_r, D_r$ — two constants per mode, $2N$ in all, exactly matching the initial data of $N$ positions and $N$ velocities (for a zero mode the bracket is replaced by the drift $c_1 + c_2 t$ of equation (6)). A second-order system with $N$ coordinates needs $2N$ constants, and the mode expansion supplies them in the most organized possible way: two per pure tone, amplitude and phase. Extracting the constants from initial conditions is where theorem two pays its way. Contract $\mathbf q(0)$ and $\dot{\mathbf q}(0)$ with $\mathbf a^{(r)\mathsf T}\mathbf M$ and orthonormality (5) collapses the sums:

where: the mass matrix sits inside every projection — the single most forgotten factor in the subject. Without $\mathbf M$-orthogonality each initial condition would spray across all $N$ constants and you would face a $2N \times 2N$ linear solve; with it, each constant is one inner product. That is the practical content of "the modes form an orthonormal basis": initial-value problems in coupled oscillations are exactly as easy as Fourier analysis — indeed Fourier analysis is this procedure applied to the translation-invariant chain, as the continuum limit makes explicit.

1 · Two unequal masses: the mass metric in action (confidence builder)

A chain runs wall–spring–mass–spring–mass–spring–wall along a line: $m_1 = 1.0$ kg and $m_2 = 2.0$ kg, with spring constants $k_a = 1.0$, $k_b = 2.0$, $k_c = 2.0$ N/m (wall–$m_1$, $m_1$–$m_2$, $m_2$–wall). Find the normal frequencies and mode shapes, and test the shapes for orthogonality — both kinds.

Step 1. Build the matrices. $\mathbf M = \mathrm{diag}(1, 2)$ kg. Each diagonal stiffness entry collects every spring touching that mass; the off-diagonal entry is minus the coupling spring: $K_{11} = k_a + k_b = 3$, $K_{22} = k_b + k_c = 4$, $K_{12} = -k_b = -2$, all in N/m. So $\mathbf K = \begin{pmatrix} 3 & -2 \\ -2 & 4 \end{pmatrix}$.

Step 2. Secular equation (4): $\det\begin{pmatrix} 3-\omega^2 & -2 \\ -2 & 4-2\omega^2 \end{pmatrix} = (3-\omega^2)(4-2\omega^2) - 4 = 2\omega^4 - 10\omega^2 + 8 = 0$, i.e. $\omega^4 - 5\omega^2 + 4 = (\omega^2 - 1)(\omega^2 - 4) = 0$. The frequencies: $\boxed{\omega_1 = 1.0\ \mathrm{rad/s}}$ and $\boxed{\omega_2 = 2.0\ \mathrm{rad/s}}$ — periods $6.28$ s and $3.14$ s. Both real, both positive: theorem one, on duty.

Step 3. Mode shapes from the null vectors of (3). For $\omega^2 = 1$: the first row reads $(3-1)a_1 - 2a_2 = 0$, so $a_2 = a_1$: $\mathbf a^{(1)} \propto (1, 1)$ — the masses swing together. For $\omega^2 = 4$: $(3-4)a_1 - 2a_2 = 0$, so $a_1 = -2a_2$: $\mathbf a^{(2)} \propto (2, -1)$ — opposition, with the light mass moving twice as far. (Check the second rows: $-2(1) + (4-2)(1) = 0$ ✓ and $-2(2) + (4-8)(-1) = 0$ ✓.)

Step 4. Test orthogonality. Euclidean: $\mathbf a^{(1)}\!\cdot\mathbf a^{(2)} = (1)(2) + (1)(-1) = 1 \neq 0$ — not perpendicular. Mass metric: $\mathbf a^{(1)\mathsf T}\mathbf M\,\mathbf a^{(2)} = (1)(1)(2) + (1)(2)(-1) = 2 - 2 = \boxed{0}$ — exactly perpendicular, as theorem two demands. This pair of vectors is the pair drawn in the geometry figure above.

Step 5. Mass-normalize for later use: $\mathbf a^{(1)\mathsf T}\mathbf M\mathbf a^{(1)} = 1 + 2 = 3$ and $\mathbf a^{(2)\mathsf T}\mathbf M\mathbf a^{(2)} = 4 + 2 = 6$, so the normalized shapes are $\tilde{\mathbf a}^{(1)} = (1,1)/\sqrt3$ and $\tilde{\mathbf a}^{(2)} = (2,-1)/\sqrt6$ in $\mathrm{kg^{-1/2}}$.

2 · The m–M–m triatomic chain, end to end (exam level)

Three masses on a line — $m = 1.0$ kg, $M = 2.0$ kg, $m = 1.0$ kg — joined by two springs of $k = 100$ N/m, free at both ends (a stand-in for a linear molecule like CO₂). Solve the eigenvalue problem completely: frequencies, shapes, orthogonality check, and the physical meaning of every mode.

Step 1. Build the matrices. $\mathbf M = \mathrm{diag}(1, 2, 1)$ kg. Only the interior springs contribute (no walls): $\mathbf K = \begin{pmatrix} 100 & -100 & 0 \\ -100 & 200 & -100 \\ 0 & -100 & 100 \end{pmatrix}$ N/m. Note each row of $\mathbf K$ sums to zero — the fingerprint of a free-free system, promising a zero mode.

Step 2. Expand the secular determinant along the first row, with $\lambda = \omega^2$: $\det(\mathbf K - \lambda\mathbf M) = (100-\lambda)\big[(200-2\lambda)(100-\lambda) - 10^4\big] + 100\big[-100(100-\lambda)\big]$. Factoring $(100-\lambda)$ out of both terms: $(100-\lambda)\big[(200-2\lambda)(100-\lambda) - 2\times10^4\big] = (100-\lambda)(2\lambda^2 - 400\lambda) = 2\lambda(100-\lambda)(\lambda-200)$. The roots: $\boxed{\lambda = 0,\ 100,\ 200\ \mathrm{(rad/s)^2}}$, matching the general formulas $0$, $k/m$, and $(k/m)(1 + 2m/M) = 100(1 + 1) = 200$. Frequencies: $\omega_2 = 10.0$ rad/s ($1.59$ Hz) and $\omega_3 = \sqrt{200} = 14.1$ rad/s ($2.25$ Hz).

Step 3. Null vector at $\lambda = 0$: $\mathbf K\mathbf a = 0$ gives $a_1 = a_2 = a_3$, so $\mathbf a^{(1)} \propto (1, 1, 1)$ — rigid translation, no spring stretched, the zero mode of equation (6): struck along this pattern the chain simply drifts, conserving momentum.

Step 4. Null vector at $\lambda = 100$: $\mathbf K - 100\,\mathbf M = \begin{pmatrix} 0 & -100 & 0 \\ -100 & 0 & -100 \\ 0 & -100 & 0 \end{pmatrix}$, forcing $a_2 = 0$ and $a_3 = -a_1$: $\mathbf a^{(2)} \propto (1, 0, -1)$ — the symmetric stretch. The heavy center sits still (pulled equally both ways), and each outer mass oscillates on its own spring against an effectively fixed anchor, hence exactly $\omega^2 = k/m$ with no trace of $M$.

Step 5. Null vector at $\lambda = 200$: the first row of $\mathbf K - 200\mathbf M$ reads $-100\,a_1 - 100\,a_2 = 0$ so $a_2 = -a_1$; the third row gives $a_2 = -a_3$. Then $\mathbf a^{(3)} \propto (1, -1, 1)$, the general pattern $(1, -2m/M, 1)$ at $2m/M = 1$ — the antisymmetric stretch: outer masses lunge one way, the center recoils to hold the center of mass fixed, both springs work on every mass, highest frequency. Check the middle row: $-100(1) + (200 - 400)(-1) - 100(1) = -100 + 200 - 100 = 0$ ✓.

Step 6. Verify $\mathbf M$-orthogonality numerically, all three pairs. $\mathbf a^{(1)\mathsf T}\mathbf M\mathbf a^{(2)} = (1)(1)(1) + (1)(2)(0) + (1)(1)(-1) = 0$ ✓. $\mathbf a^{(1)\mathsf T}\mathbf M\mathbf a^{(3)} = 1 - 2 + 1 = 0$ ✓. $\mathbf a^{(2)\mathsf T}\mathbf M\mathbf a^{(3)} = 1 + 0 - 1 = 0$ ✓. (Euclidean check on the last pair: $(1,0,-1)\cdot(1,-1,1) = 0$ too — an accident of the symmetry, not the rule; pair 1–3 fails Euclidean: $1 - 1 + 1 = 1 \neq 0$.) Mass-normalizing: norms$^2$ are $4$, $2$, $4$, giving $\tilde{\mathbf a}^{(1)} = \tfrac12(1,1,1)$, $\tilde{\mathbf a}^{(2)} = \tfrac{1}{\sqrt2}(1,0,-1)$, $\tilde{\mathbf a}^{(3)} = \tfrac12(1,-1,1)$.

Step 7. The molecular payoff. For real CO₂ ($m = 16$ u oxygens, $M = 12$ u carbon) the same shapes predict $\omega_3/\omega_2 = \sqrt{1 + 2m/M} = \sqrt{1 + 32/12} = 1.91$; the measured infrared ratio is about $1.76$ ($2349$ vs $1333\ \mathrm{cm^{-1}}$). A two-spring caricature lands within $9\%$ of a real molecule's spectrum — and the discrepancy itself is informative, measuring the bond–bond coupling and anharmonicity the model omits.

References. Goldstein, Poole & Safko, Classical Mechanics 3rd ed., Ch. 6 (Oscillations) · Morin, Introduction to Classical Mechanics, Ch. 4 (Oscillations) and the coupled-oscillator sections · Landau & Lifshitz, Mechanics 3rd ed., §23–24 (Free oscillations of systems with several degrees of freedom) · Taylor, Classical Mechanics, Ch. 11 (Coupled Oscillators and Normal Modes) Key checkpoints for the results on this page: the Euler–Lagrange reduction yields $\mathbf{M}\ddot{\boldsymbol{\eta}}+\mathbf{K}\boldsymbol{\eta}=0$; the Rayleigh-quotient argument establishes real, non-negative $\omega_r^2$; the two-mass secular equation $(k+k'-\omega^2 m)^2=k'^2$ gives $\omega_1=\sqrt{k/m}$ and $\omega_2=\sqrt{(k+2k')/m}$; and the $\mathbf{M}^{-1/2}$ symmetrization converts the generalized problem into a standard symmetric eigenvalue problem. Worked-example numbers ($\omega = 1.0, 2.0$ rad/s for the unequal-mass pair; $\omega^2 = 0, 100, 200\ \mathrm{(rad/s)^2}$ for the m–M–m chain with $m=1$ kg, $M=2$ kg, $k=100$ N/m) follow directly from the boxed formulas; the CO₂ comparison uses the standard band centers $\approx 1333$ and $2349\ \mathrm{cm^{-1}}$.

Builds on: Small Oscillations About Equilibrium · Simple Harmonic Motion · Complex Exponential Representation of Oscillations · Amplitude, Phase, and Angular Frequency Leads to: Resonance · Driven Oscillations · Energy in Simple Harmonic Motion