Small oscillations as a generalised eigenvalue problem

Statement

Any conservative system near a stable equilibrium behaves as a set of independent harmonic oscillators, and finding them is a matrix eigenvalue problem.

Why it matters

This is the reason the harmonic oscillator is not one system among many but the universal small-amplitude behaviour of everything. It is also the first place in a physics degree where linear algebra stops being a tool and becomes the content: normal modes are eigenvectors. The same structure returns as phonons in PU-303 and as field modes in PU-402.

Assumptions

Derivation

Result

Limiting cases

• Equal masses (M = m I) — the generalised problem collapses to the ordinary eigenproblem det(K − ω²m I) = 0.
• Zero coupling — K is already diagonal, the modes are the coordinates themselves and no diagonalisation is needed.
• n → ∞ on a chain — the discrete spectrum becomes the phonon dispersion relation of PU-303.

Breaks when

Failure modes

• Solving det(K − ω²I) = 0 instead of using M. Correct only when all masses are equal; silently wrong otherwise.
• Expanding about a point that is not an equilibrium, leaving a linear term that then gets dropped anyway.
• Reporting n frequencies but forgetting that a zero-frequency root is a real and important solution — it is a free translation or rotation of the whole system.

Worked number — two identical pendulums coupled by a spring

With M = m I and K giving diagonal mg/ℓ + k and off-diagonal −k, the modes are the symmetric one at ω₁² = g/ℓ (the spring never stretches, so it does not appear) and the antisymmetric one at ω₂² = g/ℓ + 2k/m. For ℓ = 0.50 m, m = 0.20 kg, k = 2.0 N m⁻¹: ω₁ = 4.43 rad s⁻¹ and ω₂ = 6.26 rad s⁻¹. The familiar beating occurs at the difference frequency, 1.83 rad s⁻¹ — a beat period of 3.4 s, exactly what the classroom demonstration shows.

Run the check → scipy.linalg.eigh(K, M) on a 5-mass chain; compare to the analytic dispersion

Discussion

The single most important feature of the result ( K − ω²M )a = 0 is that two matrices appear, not one. Both the potential energy and the kinetic energy are quadratic forms near equilibrium — V = ½η^TKη and T = ½η̇^TMη̇ — and a normal mode is the special direction in configuration space along which these two ellipsoids share a common axis. That is exactly what a generalised eigenvalue problem finds: the vectors a^(r) that are simultaneously principal axes of the stiffness form and the inertia form. In that basis the two forms are diagonal at once, the coordinates decouple, and the system is n genuinely independent oscillators. The apparent coupling in the original coordinates was never physical; it was a statement that our chosen coordinates were not aligned with the natural axes of the problem.

Read this way, the structure of the answer explains several of its own features. There are exactly n frequencies because det(K − ω²M) is a degree-n polynomial in ω². Every ω_r² is real and non-negative precisely because M is positive definite (kinetic energy is never negative) and K is positive definite at a stable equilibrium. Let one eigenvalue of K go to zero or negative — a saddle rather than a minimum — and the corresponding ω² follows it, so the “oscillation” at imaginary frequency becomes exponential runaway. Linear-stability analysis of any equilibrium, from a spinning top to a plasma, is just the sign audit of this same spectrum.

The cleanest way to see why the roots are real is to remove M by a congruence transformation. Because M is symmetric positive definite it has a unique symmetric positive-definite square root, so writing a = M^−1/2b turns the generalised problem into the ordinary symmetric one ( M^−1/2KM^−1/2 )b = ω²b. The mass-weighted matrix M^−1/2KM^−1/2 is symmetric, so D-018 (real spectrum, orthogonal eigenvectors of a symmetric matrix) applies verbatim. Transforming back, the modes are orthogonal not in the naive dot product but with respect to M: a^(r)TMa^(s) = 0 for ω_r ≠ ω_s. That weighted orthogonality is what makes the total energy split into a clean sum, one uncoupled term per mode, with no cross terms — the algebraic content of “independent oscillators.”

This one calculation is a hub. Quantise each normal coordinate independently and every ω_r becomes a ladder of energy quanta: in a crystal these are phonons (the N→∞ limit forwarded to PU-303), in the electromagnetic field they are photons, and the whole edifice of free quantum field theory is “small oscillations” done on infinitely many coordinates. The same eigenproblem with K the Hessian of a molecular potential gives infrared and Raman spectra; with M the inertia tensor and K absent it degenerates to the principal-axes-of-rotation problem; with K a graph Laplacian it becomes spectral graph theory. Molecular vibration, structural engineering resonance, and lattice dynamics are one theorem wearing different clothes. Common misconceptions. (i) Solving det(K − ω²I) = 0 — correct only when every mass is equal, silently wrong otherwise, because it forgets M. (ii) Believing the normal modes are the original coordinates; they are eigenvectors, generally mixtures of them. (iii) Expecting the modes to be perpendicular in the ordinary sense — they are M-orthogonal. (iv) Discarding a zero-frequency root as an error: it is a real solution, a free translation or rotation carrying no restoring force. (v) Treating a degeneracy as a breakdown — it merely signals a symmetry, which then chooses the natural basis inside the degenerate subspace.

Worked examples

Example 1 — Longitudinal modes of a linear triatomic molecule (CO₂). Three atoms on a line, outer mass m (oxygen, 16u) and central mass M (carbon, 12u), joined by two identical bonds of stiffness k = 1.6×10³ N m⁻¹. Take u = 1.66×10⁻²⁷ kg. Find the longitudinal normal frequencies.

Example 2 — The double pendulum in the small-angle limit. Two equal masses m on massless rods of equal length ℓ = 0.30 m, hanging in gravity g = 9.81 m s⁻². Generalised coordinates are the angles θ₁, θ₂. Find the normal frequencies and mode shapes. This is the canonical case where M is genuinely non-diagonal.

Problems

1. Two masses, three springs. Equal masses m = 0.40 kg sit between two fixed walls, linked by three identical springs of stiffness k = 25 N m⁻¹. Find the normal-mode frequencies and describe the mode shapes.
   Solution

   Here M = mI and K = [[2k,−k],[−k,2k]]. Because M is a multiple of the identity the generalised problem reduces to an ordinary one: det(K−mω²I) = (2k−mω²)² − k² = 0, so mω² = 2k ∓ k. Thus ω₁² = k/m with a = (1,1) (in-phase; the central spring never changes length so only the wall springs act) and ω₂² = 3k/m with a = (1,−1) (out-of-phase; the central spring is worked hardest). Numerically ω₁ = √(25/0.40) = 7.91 rad s⁻¹ and ω₂ = √(75/0.40) = 13.7 rad s⁻¹.

2. A numerical generalised eigenproblem. A two-degree-of-freedom system has M = diag(2,1) kg and K = [[6,−2],[−2,3]] N m⁻¹. Find both frequencies and mode vectors, then verify the two modes are M-orthogonal.
   Solution

   det(K−ω²M) = (6−2ω²)(3−ω²) − 4 = 2ω⁴ − 12ω² + 14 = 0, i.e. ω⁴ − 6ω² + 7 = 0, giving ω² = 3 ± √2 = 1.586 or 4.414 s⁻², so ω = 1.26 and 2.10 rad s⁻¹. For each, the first row (6−2ω²)a₁ = 2a₂ gives a₂/a₁ = +√2 for the lower mode and −√2 for the upper. Take a⁽¹⁾ = (1,√2), a⁽²⁾ = (1,−√2). Then a^(1)TMa⁽²⁾ = 2(1)(1) + 1(√2)(−√2) = 2 − 2 = 0: the modes are orthogonal through M, though their ordinary dot product 1 − 2 = −1 is not zero.

3. Coupled pendulums and beats. Two identical pendulums (ℓ = 0.50 m, m = 0.20 kg) are joined by a weak spring k = 2.0 N m⁻¹, as in the derivation. (a) Confirm the two mode frequencies. (b) Explain why the symmetric mode is independent of k. (c) If one pendulum is displaced and released, find the period over which the energy sloshes fully to the other and back.
   Solution

   (a) The modes are ω₁² = g/ℓ = 19.6 s⁻² and ω₂² = g/ℓ + 2k/m = 19.6 + 20.0 = 39.6 s⁻², so ω₁ = 4.43 rad s⁻¹ and ω₂ = 6.29 rad s⁻¹. (b) In the symmetric mode a = (1,1) the two bobs move identically, the connecting spring keeps its natural length and exerts no force, so k drops out entirely. (c) A single-pendulum start is an equal superposition of both modes; the two cosines beat at the difference frequency ω₂−ω₁ = 1.87 rad s⁻¹. Energy returns to the original pendulum after the beat period T = 2π/(ω₂−ω₁) = 3.4 s. (A full energy exchange to the neighbour and back is one such period; the one-way transfer takes half of it, 1.7 s.)

4. Why the frequencies are real. Given only that K and M are real symmetric and that M is positive definite, prove that every root ω² of ( K−ω²M )a = 0 is real, and that eigenvectors belonging to distinct ω² are M-orthogonal.
   Solution

   Let Ka = λMa with λ = ω² and a possibly complex. Left-multiply by the conjugate transpose a^†: λ = (a^†Ka)/(a^†Ma). Both quadratic forms are real (a real symmetric matrix satisfies (a^†Aa)^* = a^†Aa), and the denominator is strictly positive because M is positive definite, so λ is real — and non-negative whenever K is positive definite. This ratio is the Rayleigh quotient, whose stationary values are exactly the ω_r². For orthogonality, take two modes Ka^(r) = λ_rMa^(r) and Ka^(s) = λ_sMa^(s). Form a^(s)TKa^(r) two ways using symmetry: it equals λ_ra^(s)TMa^(r) and also λ_sa^(s)TMa^(r). Subtracting, (λ_r−λ_s)a^(s)TMa^(r) = 0, so for λ_r ≠ λ_s the modes satisfy a^(s)TMa^(r) = 0. Equivalently, the substitution a = M^−1/2b turns the problem into the ordinary symmetric eigenproblem for M^−1/2KM^−1/2, and D-018 finishes it.

5. From modes to phonons: the monatomic chain. An infinite line of equal masses m, spacing a, each joined to its neighbours by springs k. Using the travelling-wave ansatz η_n = A e^i(qna−ωt), derive the dispersion relation ω(q). Identify the long-wavelength sound speed, the maximum frequency, and the mode at q = 0, and say how this connects to D-003.
   Solution

   Newton’s law for site n is mη̈_n = k(η_n+1 − 2η_n + η_n−1). This is the N→∞ version of Mη̈ + Kη = 0 with a circulant K; translational symmetry lets us diagonalise it with the plane-wave ansatz instead of a determinant. Substituting η_n = A e^i(qna−ωt): −mω² = k(e^iqa − 2 + e^−iqa) = 2k(cos qa − 1) = −4k sin²(qa/2). Hence

   ω(q) = 2√(k/m) |sin(qa/2)|.

   For q → 0, sin(qa/2) ≈ qa/2, so ω ≈ a√(k/m) q = v_sq: the chain carries sound at speed v_s = a√(k/m). The spectrum saturates at the Brillouin-zone boundary q = π/a, where ω_max = 2√(k/m). At q = 0 the frequency is zero and every mass moves in unison — the free-translation zero mode of D-003, now appearing as the q=0 limit of an acoustic branch. Quantising each of these travelling modes independently gives phonons, which is precisely the forward link to PU-303.