Longitudinal Sound Waves in a Fluid

Derivation

Longitudinal Sound Waves in a Fluid

D-070 Home PU-103 Threads waves · matter · energy Depends on The Wave Equation as a Continuum Limit

Statement

For small-amplitude longitudinal disturbances in an inviscid compressible fluid, the excess pressure p (and equivalently the particle displacement ξ and the density perturbation ρ′) obeys the one-dimensional wave equation ∂²p/∂t² = c² ∂²p/∂x², propagating with speed c = √(B/ρ₀), which for an ideal gas whose compressions are adiabatic becomes c = √(γ P₀/ρ₀).

Why it matters

Sound is the archetype of a mechanical wave in a continuum: unlike the loaded string, where a discrete restoring tension was imposed by hand, here the restoring force emerges from the fluid's own compressibility. The derivation shows how the same wave equation drops out of the two most basic statements one can make about a fluid — that mass is conserved and that Newton's second law holds locally.

The result c = √(γ P₀/ρ₀) is also a small historical landmark: Newton's isothermal estimate √(P₀/ρ₀) was about 16% low, and the discrepancy was not resolved until Laplace recognised that the compressions in a sound wave are adiabatic, not isothermal. The factor γ is the physical fingerprint of that insight.

Assumptions

Small amplitude.Density and pressure perturbations are much smaller than their equilibrium values (ρ′ ≪ ρ₀, p ≪ P₀), so products of perturbations may be dropped. Drop this and the equation becomes nonlinear (finite-amplitude acoustics), admitting steepening and shock formation.

Inviscid fluid.Viscosity and heat conduction are neglected. Drop this and the wave equation gains dissipative terms, so sound attenuates as it travels and high frequencies die out fastest.

One spatial dimension.Fields depend only on x and t (plane waves). Drop this and one obtains the full 3D scalar wave equation with the same c; the physics of the speed is unchanged.

Adiabatic, reversible compressions.Each fluid element is compressed fast enough that no appreciable heat flows to its neighbours, yet slowly and smoothly enough to stay in local thermodynamic equilibrium, so Pρ^−γ is constant. Drop this toward the isothermal limit and c = √(P₀/ρ₀) instead — Newton's underestimate.

Continuum, local equilibrium.The wavelength greatly exceeds the molecular mean free path, so pressure, density and velocity are well-defined smooth fields. Drop this (rarefied gas, ultrasound at very low pressure) and hydrodynamics fails; a kinetic-theory treatment is required.

Derivation

Work in one dimension. Let ρ₀ and P₀ be the uniform equilibrium density and pressure, and write the perturbed fields as ρ = ρ₀ + ρ′, P = P₀ + p, with fluid velocity u(x,t). All of ρ′, p, u are first-order small.

∂ρ/∂t + ∂(ρ u)/∂x = 0

Conservation of mass (continuity equation) for a 1D compressible flow: the rate of change of density in a slab equals minus the divergence of mass flux. A

ρ(∂u/∂t + u ∂u/∂x) = −∂P/∂x

Euler's equation — Newton's second law per unit volume for an inviscid fluid; the only force on a fluid element is the net pressure force −∂P/∂x. A

p = (dP/dρ)₀ ρ′ ≡ c² ρ′, with c² ≡ (dP/dρ)₀

Equation of state closes the system: pressure is a function of density, expanded to first order about equilibrium. The derivative is evaluated along the actual thermodynamic path (adiabat), fixing the constant c². B

∂ρ′/∂t + ρ₀ ∂u/∂x = 0

Linearise continuity (step 1): substitute ρ = ρ₀ + ρ′, and discard the second-order term ∂(ρ′ u)/∂x since it is a product of two small quantities. Small-amplitude assumption. B

ρ₀ ∂u/∂t = −∂p/∂x

Linearise Euler (step 2): the convective term u ∂u/∂x is second-order and dropped, and ρ → ρ₀ in the leading factor; ∂P/∂x = ∂p/∂x since P₀ is uniform. B

∂²ρ′/∂t² + ρ₀ ∂²u/∂x∂t = 0

Differentiate the linearised continuity equation (step 4) with respect to t; fields are smooth so mixed partials commute. B

ρ₀ ∂²u/∂t∂x = −∂²p/∂x²

Differentiate the linearised Euler equation (step 5) with respect to x. This produces the same mixed derivative of u that appears in step 6, ready for elimination. B

∂²ρ′/∂t² = ∂²p/∂x²

Eliminate the velocity: subtract step 7 from step 6 so the mixed u-derivative cancels. Velocity has been removed, leaving a closed relation between ρ′ and p. B

∂²p/∂t² = c² ∂²p/∂x²

Use the equation of state p = c²ρ′ (step 3) to replace ρ′ by p/c² on the left of step 8 and multiply through by c². Identical in form to the loaded-string wave equation. A

c² = (dP/dρ)_adiabatic = B_ad/ρ₀

Identify the propagation speed with the adiabatic bulk modulus B_ad = ρ(dP/dρ): the general result for any fluid, valid before specialising to an ideal gas. B

P ρ^−γ = const ⇒ dP/P = γ dρ/ρ ⇒ (dP/dρ)₀ = γ P₀/ρ₀

Adiabatic ideal-gas relation: take the logarithmic differential of P = (const)ρ^γ and evaluate at equilibrium. Adiabatic (Laplace), not isothermal (Newton), assumption. C

c = √(γ P₀/ρ₀)

Substitute step 11 into step 10 and take the positive root. Using the ideal-gas law P₀ = ρ₀R T/M gives the equivalent form c = √(γ R T/M). A

Result

∂²p/∂t² = c² ∂²p/∂x², c = √(B_ad/ρ₀) = √(γ P₀/ρ₀) = √(γ R T/M)

Reading. The pressure perturbation propagates undistorted at a single speed c set by how stiff the fluid is (bulk modulus B_ad, the numerator) against how heavy it is (density ρ₀, the denominator) — stiffer means faster, denser means slower, exactly the tension-over-mass-density pattern of the string. For an ideal gas the stiffness is γ P₀, so c depends only on temperature and molecular mass, not on the pressure or density separately.

Units check. [P₀/ρ₀] = (Pa)/(kg m⁻³) = (N m⁻²)/(kg m⁻³) = (kg m⁻¹ s⁻²)/(kg m⁻³) = m² s⁻², and γ is dimensionless, so √(γ P₀/ρ₀) has units m s⁻¹. Likewise [R T/M] = (J mol⁻¹ K⁻¹·K)/(kg mol⁻¹) = J kg⁻¹ = m² s⁻². ✓

Limiting cases

Isothermal limit (Newton). If heat equilibrated instantly, γ → 1 and c → √(P₀/ρ₀), about 16% lower for air (γ = 1.4). This is the historically wrong answer, showing the adiabatic assumption is essential.
Temperature scaling. Since c = √(γ R T/M), the speed rises as √T: air at 0 °C gives ≈ 331 m/s, at 20 °C ≈ 343 m/s. It is independent of ambient pressure at fixed T.
Molecular mass scaling. Lighter gas, faster sound: helium (M ≈ 4 g/mol, γ = 5/3) carries sound at ≈ 970 m/s, roughly three times air — the "squeaky voice" effect.
Incompressible limit. As B_ad → ∞ (a rigid, incompressible medium), c → ∞: disturbances propagate instantaneously, and the wave description collapses to constraint dynamics.
Liquid/solid. The general form c = √(B/ρ) still holds; water (B ≈ 2.2 GPa, ρ ≈ 1000 kg/m³) gives c ≈ 1480 m/s. Only the ideal-gas specialisation γ P/ρ is lost.

Breaks when

Finite amplitude. When perturbations are not small (loud sound, blast waves), the dropped nonlinear terms matter: crests travel faster than troughs, the waveform steepens, and a shock forms across which the linear theory and single c are invalid.
Strong dissipation / high frequency. When viscosity and thermal conduction are not negligible — ultrasound in gases, sound at very low ambient pressure, or when the acoustic period approaches the molecular relaxation time — the wave attenuates and becomes dispersive, so a single non-attenuating speed no longer describes it.
Breakdown of continuum. When the wavelength approaches the molecular mean free path (rarefied gas, very high-frequency ultrasound), pressure and density cease to be smooth fields; hydrodynamics fails and kinetic theory is required.
Departure from adiabatic behaviour. In media where heat conducts efficiently on the timescale of a period (some porous or metallic contexts, very slow low-frequency limits), the compressions are no longer adiabatic and the factor γ is wrong.

Failure modes

Using isothermal B = P₀. Forgetting Laplace's correction and writing c = √(P₀/ρ₀) — the classic 16% error. The bulk modulus in a sound wave is the adiabatic one, γ P₀.
Thinking c depends on pressure. Concluding that pumping up the pressure at fixed temperature raises the sound speed. At fixed T, P₀/ρ₀ is constant, so c does not change.
Mixing molar and specific gas constants. Using c = √(γ R T/M) with R = 8.314 J mol⁻¹K⁻¹ but M in g/mol instead of kg/mol, giving an answer √1000 ≈ 31.6× too large.
Retaining nonlinear terms inconsistently. Keeping u ∂u/∂x but dropping ∂(ρ′u)/∂x, or vice versa. Both are second order; a consistent linearisation drops all of them together.
Confusing particle velocity u with wave speed c. The fluid oscillates at speed u (millimetres per second for ordinary sound); the disturbance pattern travels at c ≈ 343 m/s. They are unrelated in magnitude.
Using γ for the wrong gas. Applying γ = 1.4 (diatomic) to a monatomic gas like helium or argon, where γ = 5/3, or to a polyatomic gas.

Discussion

The deep lesson is that the wave equation and its speed came from combining three independent inputs: conservation of mass, Newton's second law, and a thermodynamic equation of state. The first two are pure mechanics and would hold for any compressible fluid; the third is where the physics of the specific medium enters. The propagation speed is therefore not a mechanical quantity alone — it is fixed by an equilibrium thermodynamic derivative (dP/dρ) evaluated along the correct thermodynamic path. That path is adiabatic, and choosing it correctly is the whole content of the Newton–Laplace story.

Compare with the loaded string, whose wave equation we already have. There the speed was √(T/μ) — a restoring quantity (tension) over an inertial quantity (linear mass density). Here it is √(B/ρ) — restoring stiffness over volumetric inertia. The two are the same structural statement, "restoring ÷ inertia", realised in different physical hardware. This is why the string is a legitimate stepping-stone: once you trust that a linear restoring force plus inertia gives ∂_tt = c²∂_xx, sound is just the identification of the fluid's own restoring stiffness.

A subtler point is that adiabaticity here is not an assumption of "no heat ever" but of a separation of timescales. Heat diffuses a distance √(χ/f) in one acoustic period (χ the thermal diffusivity, f the frequency), which for audible sound in air is far smaller than a wavelength; so each compression is essentially thermally isolated. The same reasoning, run in reverse, tells us exactly where the adiabatic picture must fail — at frequencies high enough that thermal diffusion keeps pace with the oscillation, the compressions become isothermal and c drifts from √(γ P/ρ) toward √(P/ρ), with dissipation in between. The single number γ thus quietly encodes an entire assumed hierarchy of timescales.

Common misconceptions. Sound speed is often thought to increase with loudness — it does not, to leading order; amplitude affects the wave only through the nonlinear corrections that produce shocks. It is also commonly believed that denser air (say, humid or high-pressure air) transmits sound faster; in fact humid air is less dense and slightly faster, and at fixed temperature pressure has no effect at all because P/ρ is pinned by temperature.

Worked examples

Example 1 — Speed of sound in air at 20 °C.

c = √(γ R T/M)

Ideal-gas form of the result; more convenient than γ P/ρ when T is known. A

γ = 1.40, R = 8.314 J mol⁻¹K⁻¹, T = 293.15 K, M = 0.02896 kg mol⁻¹

Insert values for dry air: diatomic γ, mean molar mass 28.96 g/mol converted to kg/mol, and T = 20 °C + 273.15. A

γ R T/M = (1.40 × 8.314 × 293.15)/0.02896 = 1.1783×10⁵ m²s⁻²

Evaluate the argument; units J kg⁻¹ = m²s⁻². A

c = √(1.1783×10⁵) ≈ 343 m s⁻¹

Reading. The standard textbook value for air at 20 °C, matching measurement to three figures. Had we (wrongly) used γ = 1, we would have obtained 290 m/s — the size of Newton's error.

Example 2 — Sound speed in helium, and the voice effect.

c_He = √(γ_He R T/M_He)

Same result; helium is monatomic so γ = 5/3, not 1.4. A

γ_He = 1.667, T = 293.15 K, M_He = 0.004003 kg mol⁻¹

Helium molar mass 4.003 g/mol; room temperature as before. A

γ R T/M = (1.667 × 8.314 × 293.15)/0.004003 = 1.0148×10⁶ m²s⁻²

Evaluate; the small M in the denominator drives the large value. A

c_He = √(1.0148×10⁶) ≈ 1007 m s⁻¹ ≈ 2.9 c_air

Reading. Sound travels almost three times faster in helium. The resonant frequencies of the vocal tract (fixed by its geometry, f ∝ c/L) rise by the same factor, raising the formant pitch — the familiar cartoon voice. Note the pitch of the vocal-cord vibration itself is unchanged; only the resonances shift.

Problems

Temperature dependence. By what percentage does the speed of sound in air change between a cold morning (−10 °C) and a hot afternoon (35 °C)?
Solution
c ∝ √T with T in kelvin. T₁ = 263.15 K, T₂ = 308.15 K. Ratio c₂/c₁ = √(308.15/263.15) = √1.1710 = 1.0821. So the speed rises by about 8.2%. In absolute terms c₁ ≈ 325 m/s, c₂ ≈ 352 m/s.
Newton vs Laplace. Compute the sound speed in air at 0 °C using (a) the isothermal formula and (b) the adiabatic formula, and give the percentage by which Newton's value falls short. Take P₀ = 1.013×10⁵ Pa, ρ₀ = 1.293 kg/m³, γ = 1.40.
Solution
(a) Isothermal: c = √(P₀/ρ₀) = √(1.013×10⁵/1.293) = √(78 344) = 280 m/s. (b) Adiabatic: c = √(1.40 × 78 344) = √(109 682) = 331 m/s. Shortfall = 1 − 280/331 = 0.154, about 15.4% (≈ 1 − 1/√1.4). The measured value at 0 °C is 331 m/s, vindicating Laplace.
Bulk modulus of water. Sound travels at 1480 m/s in water of density 1000 kg/m³. Using the general result, find water's bulk modulus and compare with air's γ P₀ ≈ 1.42×10⁵ Pa.
Solution
c = √(B/ρ) ⇒ B = ρ c² = 1000 × (1480)² = 1000 × 2.190×10⁶ = 2.19×10⁹ Pa = 2.19 GPa. This is about 1.5×10⁴ times air's effective modulus. Water is far denser than air (×~1000) but vastly stiffer (×~15 000), and stiffness wins, so sound is ~4.3× faster in water than in air.
Particle velocity vs wave speed. A plane sound wave in air (ρ₀ = 1.21 kg/m³, c = 343 m/s) has pressure amplitude p₀ = 28 Pa (about 120 dB, painfully loud). Find the peak particle velocity u₀ and compare it with c. Use u₀ = p₀/(ρ₀c).
Solution
The acoustic impedance is Z = ρ₀c = 1.21 × 343 = 415 Pa·s/m. Then u₀ = p₀/Z = 28/415 = 0.0675 m/s ≈ 6.7 cm/s. Ratio u₀/c = 0.0675/343 = 2.0×10⁻⁴. Even at the threshold of pain the fluid oscillates at ~7 cm/s while the wave pattern travels at 343 m/s — a ratio of ~1/5000, confirming the small-amplitude assumption and the distinction between u and c.
Identifying an unknown gas. At 300 K the speed of sound in an unknown diatomic gas (γ = 1.40) is measured to be 323 m/s. Estimate its molar mass and identify the gas.
Solution
From c = √(γ R T/M), solve for the mass: M = γ R T/c² = (1.40 × 8.314 × 300)/(323)² = 3492/1.0433×10⁵ = 0.03347 kg/mol = 33.5 g/mol. This is very close to 32 g/mol, and with γ = 1.40 (diatomic) the gas is oxygen, O₂. Cross-check: for O₂ exactly, c = √(1.40×8.314×300/0.032) = √(1.0912×10⁵) = 330 m/s, within 2% of the measurement — the small gap is the sort of error real thermometry and molar-mass rounding introduce. The method is: invert the formula for M, then match to a known molecule of the right γ.