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Normal Modes of Oscillation

Source lecture(s): PHY621 Ch. 2

Intuition

This is the payoff of the eigenvalue chapter. A system of coupled oscillators is a tangle of coupled ODEs — each mass pushed by its neighbours, nothing solvable in isolation. Diagonalise the problem and the tangle becomes \(N\) independent oscillators. The eigenvalues are the frequencies; the eigenvectors are the shapes.

Setting up

For small oscillations about equilibrium, kinetic and potential energies are quadratic forms:

\[T = \tfrac12\dot{\mathbf{q}}^{\mathsf T}A\,\dot{\mathbf{q}}, \qquad V = \tfrac12\mathbf{q}^{\mathsf T}B\,\mathbf{q}\]

with \(A\) the (positive-definite) mass matrix and \(B\) the stiffness matrix, both symmetric. Lagrange's equations give

\[A\ddot{\mathbf{q}} = -B\mathbf{q}\]

Seek \(\mathbf{q} = \mathbf{a}e^{i\omega t}\) and you get the generalised eigenvalue problem

\[B\mathbf{a} = \omega^2 A\,\mathbf{a}\]

Why "generalised" matters

This is not the standard problem \(B\mathbf{a} = \lambda\mathbf{a}\) unless \(A \propto I\) (equal masses). Two consequences:

  1. The right inner product is mass-weighted. Modes satisfy \(\mathbf{a}_i^{\mathsf T}A\,\mathbf{a}_j = 0\) for \(i \neq j\), not the naive \(\mathbf{a}_i\cdot\mathbf{a}_j = 0\). This is the first place in the course where the inner product is something the physics chooses for you.
  2. Reduce it, don't hack it. Substituting \(\mathbf{a} = A^{-1/2}\mathbf{y}\) turns it into the symmetric standard problem \(\left(A^{-1/2}BA^{-1/2}\right)\mathbf{y} = \omega^2\mathbf{y}\), which keeps the symmetry and hence guarantees real \(\omega^2\) and orthogonal modes. Computing \(A^{-1}B\) instead destroys the symmetry and is numerically worse.

Since \(A\) and \(B\) are symmetric with \(A\) positive definite, all \(\omega^2\) are real — the Hermitian guarantee. Stable equilibrium additionally needs \(B\) positive semi-definite; a negative \(\omega^2\) means imaginary \(\omega\), i.e. exponential growth, i.e. the equilibrium was unstable.

The worked example: a linear triatomic

Masses \(m\), \(\mu m\), \(m\) in a line joined by two springs \(k\) — a CO₂ molecule, with \(\mu = 12/16 = 0.75\). The three modes:

Mode \(\omega\) Shape Depends on μ?
1 \(0\) all three translate together no
2 \(\sqrt{k/m}\) outer masses opposite, centre stationary no
3 \(\sqrt{(k/m)(1 + 2/\mu)}\) outer masses together, centre opposite yes

Three things worth extracting:

  • The zero mode is physics, not a bug. Uniform translation costs no energy, so \(\omega = 0\). It is a null vector of \(B\), and it exists because momentum is conserved. Every unconstrained system has one zero mode per continuous symmetry.
  • Mode 2 has no \(\mu\) in it — because the centre mass does not move, so its mass cannot matter. The algebra discovers this on its own; you can watch it happen in the normal modes lab, where the centre amplitude comes out as \(10^{-16}\).
  • Only mode 3 involves \(\mu\), and it is the one that moves the centre. In real CO₂ these are the modes that absorb infrared, which is why the molecule is a greenhouse gas — and the antisymmetric stretch at \(\omega_2\) is the strong absorber near 15 µm.

Normal coordinates

Expanding \(\mathbf{q}(t) = \sum_n Q_n(t)\mathbf{a}_n\) turns the coupled system into

\[\ddot{Q}_n = -\omega_n^2 Q_n\]

\(N\) independent simple harmonic oscillators. The general motion is a superposition with amplitudes and phases fixed by initial conditions — and it is generally not periodic, since the \(\omega_n\) are usually incommensurate.

Common mistakes

  • Using the ordinary dot product for orthogonality. With unequal masses the modes are \(A\)-orthogonal. Checking with the wrong product makes correct modes look wrong.
  • Forgetting the zero modes. If your \(N\)-mass free chain returns \(N\) non-zero frequencies, you have over-constrained the stiffness matrix.
  • Computing \(A^{-1}B\). It has the right eigenvalues but is not symmetric, so standard symmetric solvers do not apply and the eigenvectors are not orthogonal in any obvious product.
  • Assuming the motion is periodic. Only single-mode motion is.

Knowledge graph position

Prerequisites: eigenvalues, diagonalization, Hermitian matrices. Leads to: molecular spectroscopy, phonons, the continuum limit (Fourier series), stability analysis.

Quiz

Q1 (conceptual). Why does the antisymmetric stretch frequency not depend on the central mass?

Answer

In that mode the centre mass has exactly zero amplitude — the two outer masses move oppositely and the middle stays put. A stationary mass contributes no kinetic energy, so its value cannot enter the frequency. The eigenvector encodes this automatically.

Q2 (computational). For the triatomic with \(\mu = 0.75\), find \(\omega_3\) in units of \(\sqrt{k/m}\).

Answer

\(\omega_3 = \sqrt{1 + 2/0.75} = \sqrt{1 + 2.667} = \sqrt{3.667} = 1.9149\).

Q3 (MCQ). In a system with unequal masses, normal modes are orthogonal with respect to:

  • (a) the ordinary dot product
  • (b) the mass-weighted product \(\mathbf{a}_i^{\mathsf T}A\mathbf{a}_j\)
  • (c) the stiffness-weighted product only
  • (d) no inner product
Answer

(b). The generalised eigenproblem \(B\mathbf{a} = \omega^2A\mathbf{a}\) makes \(A\) the natural metric. (They are also \(B\)-orthogonal, which follows, but \(A\) is the one the kinetic energy defines.)