4.6 Route 2 — from a lattice of oscillators

We saw in chapter 3 that a 1-D chain of mass-springs, in the limit of large $N$ and small spacing, becomes a continuous string obeying the wave equation. The same construction works for an acoustic wave — except now the springs are compressional (along the chain), not transverse, and the displacement $u(x, t)$ is longitudinal. The route is short and clean.

Setup

A row of point masses $m$ at equilibrium positions $x_n = n a$ along a line, connected to neighbours by identical springs of stiffness $\kappa$ . Let $u_n(t)$ be the longitudinal displacement of mass $n$ from $x_n$ . The spring on the left of mass $n$ is stretched by $u_n - u_{n-1}$ ; the one on the right by $u_{n+1} - u_n$ . Newton’s second law gives

m\, \ddot u_n \;=\; \kappa\,(u_{n+1} - 2 u_n + u_{n-1}).

Identical in form to the transverse-string case from chapter 3.

The continuum limit

Take $N \to \infty$ , $a \to 0$ , keeping $\mu = m/a$ (mass per unit length) and $K = \kappa a$ (a bulk modulus with units of force) fixed. As before,

u_{n+1} - 2 u_n + u_{n-1} \;=\; a^2\, \partial_x^2 u \;+\; O(a^4),

so the equation of motion becomes

\mu\, \partial_t^2 u \;=\; K\, \partial_x^2 u,

\boxed{\;\;\partial_t^2 u \;=\; c^2\, \partial_x^2 u, \qquad c^2 \;=\; K / \mu \;=\; \kappa a / m \cdot a \;=\; \kappa a^2 / m.\;\;}

The wave equation, with propagation speed $c = a\sqrt{\kappa/m}$ — exactly the dimensional-analysis guess from chapter 3.1.

Comparing the constants

Route 1 (fluid mechanics) gave $c^2 = (\partial p / \partial \rho)_s$ — an adiabatic compressibility divided by a density. Route 2 (lattice) gives $c^2 = K / \mu$ — a bulk modulus divided by a linear density. For the lattice to represent a real fluid, these have to match: $K / \mu = (\partial p / \partial \rho)_s$ , which fixes $\kappa$ and $a$ in terms of the fluid’s macroscopic properties.

This is the dictionary: a lattice with the right stiffness and spacing is an acoustic medium. The same wave equation governs both. The lattice picture is microscopic; the fluid picture is macroscopic; the equation is the same.

What this route adds

Three things:

Independence from “fluid”. The wave equation falls out of a chain of oscillators, period. There is nothing in the derivation that requires the medium to be a gas, a liquid, or even continuous. The wave equation describes any linearly-restoring continuum.
Dispersion in the discrete case. Before the continuum limit is taken, the discrete chain has dispersion — different wavelengths propagate at different speeds. Specifically, with normal-mode wavenumber $q$ ,

\omega(q) \;=\; 2 \sqrt{\kappa/m}\, |\sin(qa/2)|.

For long wavelengths ( $q a \ll 1$ ), $\omega \approx c\, q$ with $c = a\sqrt{\kappa/m}$ — non-dispersive, exactly the wave equation’s prediction. For short wavelengths comparable to $a$ , dispersion appears. This is one of the deep insights of solid-state physics: continuum mechanics is the long-wavelength limit of a discrete underlying structure. The wave equation is exact only at long wavelengths. At wavelengths comparable to the molecular mean free path in air, additional physics enters.

A direct preview of Brillouin zones. The wavenumber $q$ is periodic with period $2\pi/a$ in the discrete chain; modes with $q$ and $q + 2\pi/a$ are physically identical. This is the first Brillouin zone — relevant for periodic acoustic metamaterials, photonic crystals, and phonon dispersion in solids. Out of scope for this book, but worth noting that this route opens that door.

▶ Dispersion of the discrete chain — quick version

Look for normal-mode solutions $u_n(t) = U e^{i(qna - \omega t)}$ . Substituting into $m\ddot u_n = \kappa(u_{n+1} - 2u_n + u_{n-1})$ :

-m\omega^2 \;=\; \kappa(e^{iqa} - 2 + e^{-iqa}) \;=\; -2\kappa(1 - \cos qa) \;=\; -4\kappa \sin^2(qa/2).

Therefore $\omega = 2\sqrt{\kappa/m}\,|\sin(qa/2)|$ . For $qa \ll 1$ , $\sin(qa/2) \approx qa/2$ , giving $\omega \approx (a\sqrt{\kappa/m})\, q = cq$ — linear dispersion, the wave equation’s signature.

What this route does not add

It doesn’t connect $c$ to thermodynamics (you need route 1 for that), and it doesn’t say anything about the molecular kinematics that produce the spring-like restoring force (you need route 3 for that). Route 2 is the continuum-mechanics perspective: take a discrete oscillating system, take the appropriate limit, and the wave equation appears.

Next: the same equation from the molecular side.