Sound as displacement and pressure

On a string the displacement is transverse: y(x, t) points across the wave. A gas has no restoring force against a sideways shear, so the only mechanical wave it supports is longitudinal, and the wave variable is the displacement s(x, t) of a thin layer of air along the very axis the wave travels. Positive s means that layer has slid a little way toward +x; a moment later it slides back. Write it as s(x, t) = sₘₐₓ cos(kx − ωt), exactly the form used for a string, with k = 2π/λ, ω = 2πf and ω/k = v still holding. The scale is startling. A 1.00 kHz tone at ordinary conversational loudness has λ = 0.343 m in air but sₘₐₓ of order 10⁻⁸ m: the air shifts by about 3 × 10⁻⁸ of a wavelength. Sound is a very small disturbance riding on a very large background pressure, and that ratio is exactly what makes a linear treatment work.

Take a slab of air between x and x + Δx with cross-section A, so its volume is A Δx. The face at x moves to x + s(x); the face at x + Δx moves to x + Δx + s(x + Δx). The new thickness is Δx + [s(x + Δx) − s(x)], so the fractional volume change is [s(x + Δx) − s(x)]/Δx, which becomes ∂s/∂x in the limit. Notice what that says: if every layer shifts by the same amount, ∂s/∂x = 0 and nothing is compressed — a wholesale translation of the air makes no sound at all. The bulk modulus is defined by B = −ΔP/(ΔV/V), the minus sign making B positive because squeezing raises the pressure. Substituting gives p(x, t) = −B ∂s/∂x, where p is the departure from ambient, not the absolute pressure. This is the partial derivative used operationally: hold t fixed, differentiate along x, and the result is a quantity a microphone can measure.

Differentiate: ∂s/∂x = −k sₘₐₓ sin(kx − ωt), so p(x, t) = B k sₘₐₓ sin(kx − ωt). A cosine has become a sine, which is a shift of a quarter cycle — a quarter period in time, a quarter wavelength in space. Where the displacement is at a crest the pressure is exactly ambient; where the displacement passes through zero the pressure is at a compression or a rarefaction. Now differentiate with respect to time instead: u = ∂s/∂t = ω sₘₐₓ sin(kx − ωt) is the air's own velocity, and it is in step with the pressure. Eliminating sₘₐₓ gives p = ρ v u, so pressure keeps company with velocity, not with displacement. It is the pressure the eardrum and the microphone diaphragm answer to. Comparing amplitudes, pₘₐₓ = B k sₘₐₓ = ρ v ω sₘₐₓ, and in air at 1.00 kHz that factor is 2.59 × 10⁶ Pa per metre. At the threshold of hearing, pₘₐₓ = 2.83 × 10⁻⁵ Pa, the air moves 1.09 × 10⁻¹¹ m — a tenth of an atomic diameter.

Apply Newton's second law to the same slab. The net force on it is A[p(x) − p(x + Δx)] = −A Δx ∂p/∂x, its mass is ρ A Δx, and its acceleration is ∂²s/∂t². Substituting p = −B ∂s/∂x gives B ∂²s/∂x² = ρ ∂²s/∂t², the linear wave equation with v² = B/ρ. Check the dimensions: [B]/[ρ] = (M L⁻¹ T⁻²)/(M L⁻³) = L² T⁻², a speed squared. Newton did this in 1687 and, assuming each compression stayed at ambient temperature, used B = P. That gives √(101325/1.204) = 290 m s⁻¹, some 15% below the measured value — a famous and honest failure. Laplace supplied the fix: an audio-frequency compression is far too quick for heat to leak out of it. In one period at 1.00 kHz heat diffuses only √(α/f) ≈ 0.15 mm in air, taking the thermal diffusivity α = 2.2 × 10⁻⁵ m² s⁻¹, against a wavelength of 343 mm, so each compression is thermally sealed. The right modulus is the adiabatic one, B = γP = 1.42 × 10⁵ Pa, and √(B/ρ) = 343 m s⁻¹.

Substituting B = γP into v = √(B/ρ) gives v = √(γP/ρ), and the ideal-gas law written as ρ = PM/RT removes the pressure entirely: v = √(γRT/M). Raise the atmospheric pressure and the air stiffens and thickens in the same proportion, so the speed does not move. What is left is temperature and molar mass. Air at 0 °C gives √(1.40 × 8.314 × 273.15 / 0.02896) = 331 m s⁻¹, and at 20 °C, 343 m s⁻¹. Differentiating, dv/dT = v/2T = 0.59 m s⁻¹ per kelvin near room temperature, which is the familiar rule of about 0.6 m s⁻¹ per degree. Molar mass bites just as hard: helium, with γ = 5/3 and M = 4.00 g mol⁻¹, carries sound at 1.01 × 10³ m s⁻¹ at that same 20 °C. Note what is absent — the frequency. Within this model air is non-dispersive, which is why a chord played 30 m away still arrives as a chord.

Every step above assumed the strain |∂s/∂x| = k sₘₐₓ is small enough for B to count as constant. Put a number on it: a pressure amplitude of 200 Pa, loud enough to be painful, gives a strain of 200/141855 = 1.41 × 10⁻³, and a fractional pressure change of only 0.20%. Linearity looks safe, and locally it is. It still fails over distance. A compression is slightly hotter than ambient and is also carried forward on air that is itself moving, so a point on the waveform travels at v + βu with β = (γ + 1)/2 = 1.20 for air. Crests overtake the zero crossings, the waveform steepens, and after xₛ = ρv³/(βω pₘₐₓ) ≈ 32 m at 1.00 kHz the front goes vertical: a shock, whose own existence has destroyed the assumption that produced it. Two further limits sit outside the model — the continuum picture needs λ ≫ the 68 nm mean free path, and the lossless assumption ignores the viscous and thermal absorption that strips high frequencies over long ranges.