Decoupling into Normal Coordinates

Derivation

Decoupling into Normal Coordinates

D-132 Home PU-201 Threads waves · energy Depends on Normal Modes as a Generalized Eigenproblem

Statement

For a conservative system of n coupled oscillators about a stable equilibrium, with symmetric positive-definite mass matrix M and symmetric positive-semidefinite stiffness matrix K, there exists a single linear change of coordinates q = A ξ that simultaneously diagonalizes both matrices — A^TM A = I and A^TK A = Ω = diag(ω₁², …, ω_n²) — reducing the coupled equations M q̈ + K q = 0 to n independent scalar harmonic oscillators ξ̈_r + ω_r² ξ_r = 0. The ξ_r are the normal coordinates.

Why it matters

Coupling is a statement about coordinates, not about physics. The masses appear to push and pull on one another only because we described them in laboratory coordinates that are not aligned with the system's natural motions. Decoupling into normal coordinates exhibits the true dynamical content: any small motion whatsoever is a superposition of n independent modes, each oscillating at its own fixed frequency and never exchanging energy with the others.

This is the organizing principle behind lattice phonons, molecular vibrational spectra, coupled electrical circuits, and the mode expansion that ultimately underlies field quantization. Once a problem is cast in normal coordinates, every hard question — energy content, response to forcing, thermal averages — separates into n textbook one-dimensional problems.

Assumptions

Small oscillations about a stable equilibrium.If displacements are not small, the potential cannot be truncated at quadratic order, K ceases to be constant, and the equations become nonlinear — no constant matrix diagonalizes them and mode frequencies become amplitude-dependent. Mass matrix M is symmetric and positive-definite.Positive-definiteness guarantees the kinetic energy ½ q̇^TM q̇ is strictly positive for any nonzero velocity; drop it and M^1/2 fails to exist, so no M-orthonormal basis and no simultaneous diagonalization is guaranteed. Stiffness matrix K is symmetric.Symmetry follows from K_ij = ∂²V/∂q_i∂q_j for a potential V; without a potential (e.g. velocity-dependent or gyroscopic forces) K is not symmetric, eigenvectors need not be M-orthogonal, and the modes generally cannot be decoupled by a real congruence. The eigenvectors span the full configuration space (completeness).Guaranteed here because M^−1/2K M^−1/2 is real symmetric; if it failed (as for genuinely non-normal or defective systems) the modal matrix A would be singular and q = A ξ would not be an invertible change of variables. No dissipation.A damping term C q̇ generally cannot be diagonalized by the same A that diagonalizes M and K; the modes stay coupled through A^TC A unless C is a linear combination of M and K (Rayleigh/proportional damping).

Derivation

We take as given (from small-oscillations-eigenproblem) the equations of motion M q̈ + K q = 0 and the generalized eigenproblem K a^(r) = ω_r² M a^(r), whose eigenvectors are M-orthonormal: a^(r)TM a^(s) = δ_rs.

A = [ a⁽¹⁾ a⁽²⁾ ⋯ a⁽ⁿ⁾ ] (the modal matrix; columns are the eigenvectors)

Purely a definition: collect the n generalized eigenvectors as the columns of an n×n matrix. Legitimate because the eigenproblem supplies exactly n of them. A

(A^TM A)_rs = a^(r)TM a^(s) = δ_rs ⟹ A^TM A = I

The (r,s) entry of A^TM A is exactly the M-inner product of columns r and s. Substituting the given M-orthonormality of the eigenvectors returns the identity matrix. B

(A^TK A)_rs = a^(r)TK a^(s) = ω_s² a^(r)TM a^(s) = ω_s² δ_rs ⟹ A^TK A = Ω = diag(ω₁²,…,ω_n²)

Use the eigen-relation K a^(s) = ω_s² M a^(s) to replace K a^(s), then apply Step 2. Both matrices are now diagonal in the same basis — this is the simultaneous diagonalization. B

M = M^1/2M^1/2, C ≡ M^−1/2K M^−1/2 = C^T ⟹ C = O Ω O^T, A = M^−1/2O

Existence proof, not merely bookkeeping. Because M is symmetric positive-definite it possesses a symmetric positive-definite square root M^1/2. Then C = M^−1/2K M^−1/2 is real symmetric, so the spectral theorem gives a real orthogonal O with O^TC O = Ω. Setting A = M^−1/2O reproduces Steps 2–3 and guarantees A is invertible. C

q(t) = A ξ(t) ⟺ ξ(t) = A⁻¹q(t)

Define the normal coordinates ξ by this linear map. It is invertible because the columns of A are linearly independent (they are M-orthonormal, hence a basis) — so no information is lost and the two descriptions are equivalent. B

M q̈ + K q = 0 ⟹ M A ξ̈ + K A ξ = 0

Substitute q = A ξ into the equations of motion. Since A is a constant matrix it commutes with the time derivative: q̈ = A ξ̈. A

A^TM A ξ̈ + A^TK A ξ = 0 ⟹ I ξ̈ + Ω ξ = 0

Left-multiply the whole equation by A^T (legal: multiplying both sides of a linear identity by a fixed matrix). Insert the diagonalizations from Steps 2 and 3. B

ξ̈_r + ω_r² ξ_r = 0 (r = 1, …, n)

Read off row r of the diagonal system. Because I and Ω are diagonal, row r involves ξ_r alone — the equations no longer share variables. A

Result

A^TM A = I, A^TK A = Ω ⟹ ξ̈_r + ω_r² ξ_r = 0, ξ_r(t) = C_r cos(ω_r t + φ_r)

Reading. One linear change of variables turns n coupled oscillators into n uncoupled ones. Each normal coordinate ξ_r obeys the equation of a single free harmonic oscillator at its own frequency ω_r, with amplitude C_r and phase φ_r fixed by initial conditions. The physical motion is the superposition q(t) = Σ_r a^(r) ξ_r(t): the columns of A are the mode shapes, the ξ_r are how strongly each mode is excited. Modes with ω_r = 0 are free translations/rotations obeying ξ̈_r = 0.

Units check. A^TM A = I is dimensionless, and [M] = kg, so [A] = kg^−1/2 and each normal coordinate carries [ξ_r] = [q]/[A] = m·kg^1/2. Then [Ω] = [A]²[K] = kg⁻¹·(N/m) = s⁻², so ω_r is in rad·s⁻¹ and both terms of ξ̈_r + ω_r² ξ_r carry m·kg^1/2·s⁻². Consistent.

Limiting cases

Uncoupled to begin with. If K (and M) are already diagonal, the eigenvectors are the Cartesian unit vectors, A is diagonal, and normal coordinates coincide (up to scale) with the original coordinates — the transformation does nothing, as it must.
Degenerate frequencies. When two modes share ω_r = ω_s, any M-orthonormal combination within that eigenspace is equally valid; the decoupling still holds but the mode shapes are non-unique.
Zero mode. A vanishing eigenvalue ω_r² = 0 (rigid translation/rotation of the whole system) gives ξ_r(t) = a_r + b_rt — free drift, still decoupled from the vibrational modes.
Identical masses, symmetric couplings. Modes reduce to symmetric/antisymmetric combinations and the frequencies take the familiar closed forms (e.g. ω² = k/m and 3k/m for two masses with three equal springs).

Breaks when

Amplitudes leave the linear regime. Anharmonic terms in V couple the normal coordinates back together (ξ_r-ξ_s mixing), producing energy transfer between modes, frequency pulling, and — in lattices — thermal conductivity that the harmonic decoupling cannot describe.
Non-conservative or gyroscopic forces are present. Damping C q̇ or Coriolis/magnetic terms make the governing matrices non-symmetric or introduce a term the same real congruence cannot diagonalize; modes acquire complex frequencies and, in general, cannot be simultaneously decoupled by real normal coordinates.
The mass matrix loses positive-definiteness. At a constraint or singular limit M^1/2 fails to exist, the M-orthonormal basis collapses, and the construction of A is invalid.
Explicit time dependence. If M(t) or K(t) vary in time, no constant A diagonalizes them for all t; one must resort to adiabatic or Floquet methods instead.

Failure modes

Ordinary orthonormalization. Normalizing eigenvectors by a^Ta = 1 instead of a^TM a = 1. This diagonalizes K only when M ∝ I; otherwise A^TM A ≠ I and the mass term does not clear.
Diagonalizing M⁻¹K with an orthogonal matrix. M⁻¹K is generally not symmetric, so its eigenvector matrix is not orthogonal; forcing O^T(M⁻¹K)O gives nonsense. The correct symmetric object is M^−1/2K M^−1/2.
Treating ξ_r as the physical displacement of a mass. Normal coordinates are collective and carry units m·kg^1/2; the physical motion of particle i is q_i = Σ_r A_ir ξ_r.
Reusing lab-frame initial conditions directly in the mode equations. Initial ξ(0), ξ̇(0) must be obtained by ξ = A⁻¹q = A^TM q, not copied from q(0), q̇(0).
Assuming the eigenvectors are orthogonal in the usual sense. They are orthogonal under the M-weighted inner product only; forgetting the weight corrupts the projection of any state onto the modes.

Discussion

The deep point is that "coupling" has no coordinate-independent meaning for a linear system near equilibrium. Two quadratic forms live on configuration space — the kinetic energy set by M and the potential energy set by K — and the theorem of simultaneous diagonalization (a real, symmetric analogue of the principal-axis theorem) says a single basis reduces both to sums of squares. In that basis the Lagrangian is literally a sum of independent oscillator Lagrangians, L = ½ Σ_r(ξ̇_r² − ω_r² ξ_r²), so the total energy partitions cleanly, E = Σ_r E_r, with each mode conserving its share. Nothing exchanges energy because nothing connects the modes.

Geometrically, M defines a non-Euclidean metric on configuration space; the eigenvectors are orthogonal in that metric, which is why the M-weighting is unavoidable. The transformation y = M^1/2q first straightens the metric to the ordinary one, after which the remaining problem is the familiar orthogonal diagonalization of a single symmetric matrix. Normal coordinates are just the principal axes of the potential energy ellipsoid, measured in the mass-scaled space.

This is the classical seed of much of physics. Quantizing each decoupled coordinate turns ξ_r into an independent quantum oscillator with ladder operators a_r, a_r^†; the quanta are phonons in a solid and, in the continuum limit where the oscillator index becomes a wavevector, the particles of a quantum field. The whole apparatus of second quantization is the infinite-dimensional descendant of the finite decoupling performed here. The same eigen-decomposition also governs the linear response: driving frequency near any ω_r resonantly excites that single normal coordinate, which is how spectroscopy reads off Ω directly.

Common misconceptions. Normal modes are not "which mass moves"; they are patterns in which all coordinates oscillate together at one frequency with fixed relative amplitudes. And the number of modes equals the number of degrees of freedom, including any zero-frequency rigid-body modes — a free molecule of N atoms in a line has N longitudinal modes, of which one is translation.

Worked examples

Example 1 — two equal masses, three equal springs (wall–mass–mass–wall). Each mass m; springs of stiffness k connect wall–1, 1–2, 2–wall. Show the system decouples into an in-phase and an out-of-phase mode.

M = [[m, 0],[0, m]], K = [[2k, −k],[−k, 2k]]

Kinetic energy ½m(ẋ₁²+ẋ_2²) gives M; the two springs on mass 1 give diagonal 2k, the shared middle spring gives off-diagonal −k. A

det(K − ω²M) = (2k − mω²)² − k² = 0 ⟹ 2k − mω² = ±k

Solve the generalized eigenproblem. The characteristic determinant factors as a difference of squares. B

ω₁² = k/m (a⁽¹⁾ ∝ (1, 1)), ω₂² = 3k/m (a⁽²⁾ ∝ (1, −1))

The +k root gives the symmetric (in-phase) mode; the −k root the antisymmetric (out-of-phase) mode. A

a^(r)TM a^(r) = 2m ⟹ a⁽¹⁾ = (1,1)/√(2m), a⁽²⁾ = (1,−1)/√(2m)

M-normalize so that A^TM A = I. Then A^TK A = diag(k/m, 3k/m) and the normal coordinates are ξ₁=√(m/2)(x₁+x₂), ξ₂=√(m/2)(x₁−x₂). B

m = 0.20 kg, k = 8.0 N/m: ω₁ = √(8.0/0.20) = √40 = 6.32 rad/s, ω₂ = √(3·40) = √120 = 10.95 rad/s

Insert numbers only now that the symbolic frequencies are known. A

ξ̈₁ + 40 ξ₁ = 0, ξ̈₂ + 120 ξ₂ = 0 (s⁻²)

Reading. Two independent oscillators: the in-phase breathing mode at 6.32 rad/s (both masses move together, middle spring unstretched) and the out-of-phase mode at 10.95 rad/s (masses squeeze the middle spring, hence stiffer and faster by √3). Energy placed in one never leaks to the other. Units: ω₂/ω₁ = √3 ≈ 1.73, dimensionless as required.

Example 2 — two unequal free masses joined by one spring (a zero mode plus a stretch mode). Masses m₁, m₂ slide freely on a line, connected by a spring k. This models a diatomic molecule and demonstrates decoupling in the presence of a rigid-translation (zero-frequency) mode.

M = [[m₁, 0],[0, m₂]], K = [[k, −k],[−k, k]]

Potential ½k(x₁−x₂)² gives the stiffness; each mass free, so M diagonal. Note det K = 0 — a zero mode is coming. A

det(K − ω²M) = (k − m₁ω²)(k − m₂ω²) − k² = m₁m₂ω⁴ − k(m₁+m₂)ω² = 0

Expand the determinant; the constant term cancels, leaving a factor of ω² outside — the signature of a zero mode. B

ω₀² = 0 (a⁽⁰⁾ ∝ (1,1)), ω₁² = k(m₁+m₂)/(m₁m₂) = k/μ, μ = m₁m₂/(m₁+m₂)

Factor the biquadratic. The zero root is uniform translation; the finite root is the stretch mode governed by the reduced mass μ. Its eigenvector satisfies m₁x₁+m₂x₂=0, i.e. a⁽¹⁾ ∝ (m₂, −m₁). B

m₁ = 1.0 kg, m₂ = 3.0 kg, k = 12 N/m: μ = (1·3)/4 = 0.75 kg, ω₁ = √(12/0.75) = √16 = 4.0 rad/s

Numbers enter last. Check M-orthogonality of the two mode shapes: (1,1)·M·(3,−1) = 1·1·3 + 1·3·(−1) = 0. A

ξ̈₀ = 0, ξ̈₁ + 16 ξ₁ = 0 (s⁻²)

Reading. The centre-of-mass coordinate ξ₀ drifts freely (ξ₀(t)=a+bt) while the internal stretch coordinate ξ₁ oscillates at 4.0 rad/s — the vibrational frequency set entirely by the reduced mass. The two are completely decoupled: uniform sliding never excites vibration and vice versa. Units: √(N·m⁻¹·kg⁻¹) = √(s⁻²) = s⁻¹, correct for angular frequency.

Problems

(A) Bookkeeping. For Example 1, verify explicitly that A^TK A is diagonal by computing the off-diagonal entry a^(1)TK a⁽²⁾.
Solution
K a⁽²⁾ = [[2k,−k],[−k,2k]](1,−1)/√(2m) = (3k, −3k)/√(2m). Then a^(1)T(K a⁽²⁾) = (1,1)/√(2m) · (3k,−3k)/√(2m) = (3k − 3k)/(2m) = 0. The off-diagonal vanishes, confirming decoupling. (Equivalently it must vanish since a^(1)TK a⁽²⁾ = ω₂² a^(1)TM a⁽²⁾ = ω₂²·0.)
(B) Initial conditions. In Example 1 (m=0.20 kg, k=8.0 N/m), mass 1 is displaced by x₁(0)=2.0 cm and mass 2 held at x₂(0)=0, both released from rest. Find ξ₁(t), ξ₂(t) and the resulting x₁(t).
Solution
Using ξ_r = a^(r)TM q with a^(r)=(1,±1)/√(2m): ξ₁(0)=√(m/2)(x₁+x₂)=√(0.10)(0.020)=6.32×10⁻³ m·kg^1/2, and ξ₂(0)=√(m/2)(x₁−x₂)=6.32×10⁻³ (same value). At rest, ξ̇_r(0)=0, so ξ₁(t)=6.32×10⁻³cos(6.32t), ξ₂(t)=6.32×10⁻³cos(10.95t). Reconstruct: x₁=(ξ₁+ξ₂)/√(2m)=½[0.020cos(6.32t)+0.020cos(10.95t)] m = 0.010[cos(6.32t)+cos(10.95t)] m. Energy is shared 50/50 between the modes, producing a beat-like exchange between the two masses.
(C) Reduced mass. For a CO molecule modelled as Example 2 with m_C=12 u, m_O=16 u (1 u =1.66×10⁻²⁷ kg) and bond stiffness k=1860 N/m, compute the vibrational angular frequency and wavenumber (ν̃=ω/2πc).
Solution
μ = (12·16)/(28) u = 6.857 u = 1.138×10⁻²⁶ kg. ω = √(k/μ) = √(1860/1.138×10⁻²⁶) = √(1.634×10²⁹) = 4.04×10¹⁴ rad/s. Wavenumber ν̃ = ω/(2πc) = 4.04×10¹⁴/(2π·3.0×10¹⁰ cm/s) = 2.14×10³ cm⁻¹ ≈ 2143 cm⁻¹, matching the observed CO stretch near 2143 cm⁻¹.
(B) Energy partition. Show that in normal coordinates the total energy is E = Σ_r ½(ξ̇_r² + ω_r² ξ_r²) and that each E_r is separately conserved.
Solution
Kinetic energy T=½q̇^TM q̇=½ξ̇^TA^TM A ξ̇=½ξ̇^Tξ̇=½Σ_rξ̇_r² using A^TM A=I. Potential V=½q^TK q=½ξ^TA^TK A ξ=½ξ^TΩ ξ=½Σ_rω_r²ξ_r². Hence E=T+V=Σ_rE_r with E_r=½(ξ̇_r²+ω_r²ξ_r²). Differentiating, dE_r/dt=ξ̇_r(ξ̈_r+ω_r²ξ_r)=0 by the equation of motion. Each mode conserves its own energy independently.
(C) Why not M⁻¹K? Explain why an orthogonal matrix generally cannot diagonalize M⁻¹K, and construct the symmetric matrix that can be orthogonally diagonalized, relating its eigenvectors to the modal matrix A.
Solution
M⁻¹K is a product of two symmetric matrices, which is symmetric only if they commute — generally false — so its eigenvectors are not orthogonal and no orthogonal O diagonalizes it. Instead set y=M^1/2q; the eigenproblem K a=ω²M a becomes (M^−1/2K M^−1/2)(M^1/2a)=ω²(M^1/2a). The matrix C=M^−1/2K M^−1/2 is real symmetric, hence C=OΩO^T with orthogonal O. The eigenvectors of C are o^(r)=M^1/2a^(r), so the modal matrix is A=M^−1/2O, and its M-orthonormality A^TM A=O^TO=I is exactly the ordinary orthonormality of O pulled back through M^1/2.