Small Oscillations and Normal Modes

At a stable equilibrium, $V$ has a local minimum. Expanding around equilibrium ($q_j = q_j^0 + \eta_j$):

$L \approx \frac{1}{2}\sum_{j,k} T_{jk}\dot{\eta}_j\dot{\eta}_k - \frac{1}{2}\sum_{j,k} V_{jk}\eta_j\eta_k$

Where $T_{jk} = \frac{\partial^2 T}{\partial \dot{q}_j \partial \dot{q}_k}\bigg|_0$ is the (constant) Mass matrix and $V_{jk} = \frac{\partial^2 V}{\partial q_j \partial q_k}\bigg|_0$ is the (constant) Stiffness matrix.

Both $T_{jk}$ and $V_{jk}$ are symmetric matrices. $T$ is positive definite (kinetic energy is always positive), and $V$ is positive definite at a stable equilibrium.

Writing the Lagrangian in matrix form:

$L = \frac{1}{2}\dot{\boldsymbol{\eta}}^T \mathbf{T}\, \dot{\boldsymbol{\eta}} - \frac{1}{2}\boldsymbol{\eta}^T \mathbf{V}\, \boldsymbol{\eta}$

The Euler-Lagrange equations become:

$\mathbf{T}\, \ddot{\boldsymbol{\eta}} + \mathbf{V}\, \boldsymbol{\eta} = \mathbf{0}$

Assuming solutions of the form $\eta_j = a_j e^{i\omega t}$The eigenvalue problem is:

$(\mathbf{V} - \omega^2 \mathbf{T})\mathbf{a} = \mathbf{0}$

The secular equation (characteristic equation) is:

$\det(\mathbf{V} - \omega^2 \mathbf{T}) = 0$

This is a polynomial of degree $n$ in $\omega^2$ whose roots are the squared normal mode frequencies $\omega_\alpha^2$.

Theorem 7.1. For a stable system, all normal mode frequencies are real and positive. The normal Modes are orthogonal with respect to both $\mathbf{T}$ and $\mathbf{V}$.

Proof. Since $\mathbf{T}$ is positive definite, we can write $\mathbf{T} = \mathbf{L}\mathbf{L}^T$ (Cholesky decomposition). Defining $\boldsymbol{\xi} = \mathbf{L}^T\mathbf{a}$ and $\mathbf{W} = \mathbf{L}^{-1}\mathbf{V}\mathbf{L}^{-T}$The eigenvalue problem becomes $\mathbf{W}\boldsymbol{\xi} = \omega^2\boldsymbol{\xi}$. Since $\mathbf{W}$ is symmetric and $\mathbf{V}$ is positive definite, all eigenvalues $\omega^2$ are real and positive. Orthogonality follows from the symmetry of $\mathbf{W}$. $\blacksquare$

Theorem 7.2. The normal mode vectors $\mathbf{a}^{(\alpha)}$ satisfy:

$\sum_{j,k} a_j^{(\alpha)} T_{jk} a_k^{(\beta)} = T_\alpha\, \delta_{\alpha\beta}, \quad \sum_{j,k} a_j^{(\alpha)} V_{jk} a_k^{(\beta)} = \omega_\alpha^2 T_\alpha\, \delta_{\alpha\beta}$

Where $T_\alpha$ is a normalisation constant.

This allows us to expand any motion as a superposition of normal modes.

Problem. Two identical simple pendulums of length $l$ and mass $m$ are coupled by a spring of constant $k$ connecting the bobs. Find the normal modes and frequencies.

For two equal masses $m$ on massless rods of length $l$:

The kinetic and potential energy matrices (to second order) give the eigenvalue problem with solutions $\omega_1 = \sqrt{g/l}(2 - \sqrt{2})^{1/2}$ and $\omega_2 = \sqrt{g/l}(2 + \sqrt{2})^{1/2}$.

The corresponding normal modes are:

• Mode 1: both pendulums swing in the same direction (in phase).
• Mode 2: the pendulums swing in opposite directions (out of phase).

• Assuming the mass matrix is always diagonal; in generalised coordinates it can have off-diagonal terms.
• Forgetting that both $T$ and $V$ must be positive definite for all eigenvalues to be real and positive.
• Confusing the secular equation $\det(\mathbf{V} - \omega^2\mathbf{T}) = 0$ with $\det(\mathbf{V} - \lambda\mathbf{I}) = 0$; the latter only applies when $T = I$.
• Neglecting to normalise eigenvectors with respect to the mass matrix, which leads to incorrect mode superpositions.
• Misidentifying degenerate modes: when $\omega_\alpha = \omega_\beta$, any linear combination of the corresponding eigenvectors is also a normal mode.
• Assuming that small oscillation analysis captures the full dynamics; it only describes motion near equilibrium and breaks down for large amplitudes.