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, conservation of mass together with Newton's second law and an adiabatic equation of state yield the one-dimensional wave equation \(\partial_{tt}s = c^2\,\partial_{xx}s\) for the particle displacement \(s(x,t)\), with propagation speed \(c=\sqrt{\gamma P_0/\rho_0}\), where \(P_0\) and \(\rho_0\) are the equilibrium pressure and mass density and \(\gamma=C_P/C_V\) is the ratio of heat capacities.

Why it matters

Sound is the paradigm of a mechanical wave in a continuous medium, and this derivation is where the abstract wave equation first meets bulk thermodynamic material properties rather than the ad hoc tension-and-linear-density of a string. The result \(c=\sqrt{\gamma P_0/\rho_0}\) predicts the speed of sound in any ideal gas from first principles and, historically, exposed a subtle physics error: Newton's isothermal estimate was about \(16\%\) too low until Laplace recognised that sound compressions are adiabatic.

The same continuity-plus-momentum-plus-equation-of-state template scales up to acoustics, shock physics, stellar oscillations, and the whole of fluid dynamics. Understanding exactly which assumptions produce a linear, non-dispersive wave clarifies when and how real sound departs from the ideal.

Assumptions

Small amplitude (linearisation).If displacements are not small, \(\partial_x s\) is not negligible, the density and pressure relations become nonlinear, harmonics are generated and compressions steepen into shocks; a single wave speed \(c\) is then ill-defined.

One-dimensional plane motion.Drop it and the scalar equation must be replaced by the full vector Euler system with \(\nabla\) operators; transverse gradients allow diffraction and geometric spreading not captured here.

Inviscid, non-conducting fluid (no dissipation).With viscosity or thermal conduction the wave equation acquires damping terms, the speed becomes complex and frequency-dependent, and high-frequency sound is attenuated.

Adiabatic, reversible compressions.Compressions and rarefactions happen too fast for heat to diffuse between them, so entropy is conserved and \(P\propto\rho^{\gamma}\). Drop this for the isothermal law \(P\propto\rho\) and the speed falls to \(\sqrt{P_0/\rho_0}\) — Newton's erroneous value.

Local thermodynamic equilibrium with a well-defined \(\gamma\).If the compression rate rivals internal molecular relaxation times (rotational, vibrational), the effective \(\gamma\) becomes frequency-dependent and the medium is dispersive and absorptive.

Derivation

\[ x' = x + s(x,t) \]

Define the displacement field: a fluid element whose equilibrium coordinate is \(x\) sits at \(x'\) at time \(t\). This Lagrangian label is the exact analogue of the string displacement in the loaded-string derivation. A

\[ \rho_0\,A\,dx = \rho\,A\left(1+\frac{\partial s}{\partial x}\right)dx \]

Conservation of mass. A slab of cross-section \(A\) and equilibrium width \(dx\) has its faces at \(x\) and \(x+dx\) displaced by \(s(x)\) and \(s(x+dx)\), so its width becomes \(dx\,(1+\partial_x s)\). The enclosed mass cannot change. A

\[ \rho = \rho_0\left(1+\frac{\partial s}{\partial x}\right)^{-1}\approx \rho_0\left(1-\frac{\partial s}{\partial x}\right)\;\Rightarrow\; \delta\rho \equiv \rho-\rho_0 = -\rho_0\frac{\partial s}{\partial x} \]

Solve for density and linearise using \((1+\varepsilon)^{-1}\approx 1-\varepsilon\) for \(|\partial_x s|\ll 1\). The condensation \(\delta\rho\) is first order in the displacement gradient. A

\[ P = P_0\left(\frac{\rho}{\rho_0}\right)^{\gamma} \]

Adiabatic equation of state. Sound oscillations are fast compared with heat diffusion, so each element evolves at constant entropy and an ideal gas obeys \(P\rho^{-\gamma}=\text{const}\). This is the physically decisive assumption (Laplace, not Newton). B

\[ \delta P \equiv P-P_0 \approx \left.\frac{dP}{d\rho}\right|_0\,\delta\rho = \frac{\gamma P_0}{\rho_0}\,\delta\rho = -\gamma P_0\,\frac{\partial s}{\partial x} \]

Linearise the equation of state about equilibrium: \(dP/d\rho = \gamma P_0/\rho_0\) evaluated at \(\rho=\rho_0\), then substitute \(\delta\rho\) from Step 3. The combination \(B_{\text{ad}}=\gamma P_0\) is the adiabatic bulk modulus. B

\[ \rho_0\,A\,dx\;\frac{\partial^2 s}{\partial t^2} = \big[P(x)-P(x+dx)\big]A = -\frac{\partial P}{\partial x}\,A\,dx \]

Newton's second law (the Euler momentum balance) on the slab. The net longitudinal force is the pressure difference across its two faces; mass times acceleration equals that force. To leading order the mass uses the equilibrium density \(\rho_0\). A

\[ \rho_0\,\frac{\partial^2 s}{\partial t^2} = -\frac{\partial (\delta P)}{\partial x} = \gamma P_0\,\frac{\partial^2 s}{\partial x^2} \]

Cancel \(A\,dx\), write \(\partial_x P = \partial_x(\delta P)\) since \(P_0\) is uniform, and insert \(\delta P=-\gamma P_0\,\partial_x s\) from Step 5. This one substitution of the constitutive law into the momentum equation closes the system. B

\[ \frac{\partial^2 s}{\partial t^2} = c^2\,\frac{\partial^2 s}{\partial x^2}, \qquad c^2 = \frac{\gamma P_0}{\rho_0} \]

Divide by \(\rho_0\). The equation is identical in form to the loaded-string wave equation, so its d'Alembert solutions \(s=f(x-ct)+g(x+ct)\) carry over directly and the coefficient of \(\partial_{xx}s\) is the squared speed. A

Result

\[ \boxed{\,c=\sqrt{\dfrac{\gamma P_0}{\rho_0}}=\sqrt{\dfrac{\gamma R T}{M}}\,} \]

Reading. The speed of sound is set by the fluid's stiffness against compression (the adiabatic bulk modulus \(\gamma P_0\)) divided by its inertia (density \(\rho_0\)). A stiffer or lighter gas carries sound faster. Using the ideal-gas law \(P_0=\rho_0 RT/M\), the pressure and density dependence cancels: at fixed temperature \(c\) depends only on \(\gamma\) and the molar mass \(M\), not on pressure — which is why the sound speed in air is essentially altitude-independent at fixed temperature.

Units check. \([\gamma P_0/\rho_0]=\dfrac{\text{Pa}}{\text{kg}\,\text{m}^{-3}}=\dfrac{\text{N}\,\text{m}^{-2}}{\text{kg}\,\text{m}^{-3}}=\dfrac{\text{kg}\,\text{m}^{-1}\text{s}^{-2}}{\text{kg}\,\text{m}^{-3}}=\text{m}^2\,\text{s}^{-2}\). The square root gives \(\text{m}\,\text{s}^{-1}\), a speed; \(\gamma\) is dimensionless.

Limiting cases

Isothermal limit (\(\gamma\to 1\)): \(c\to\sqrt{P_0/\rho_0}\), Newton's original underestimate — recovered if heat had time to equilibrate during each cycle.
Monatomic vs diatomic: \(\gamma=5/3\) (noble gases) gives a faster wave than \(\gamma=7/5\) (air) at equal \(T\) and \(M\), a purely thermodynamic effect.
Light gas (\(M\to 0\)): \(c\to\infty\) as \(M^{-1/2}\); helium's low mass, not any change in stiffness per mole, gives the "squeaky voice".
Incompressible limit (\(\gamma P_0\to\infty\), i.e. bulk modulus large): \(c\to\infty\); a truly incompressible fluid transmits pressure instantaneously and supports no acoustic wave.
Liquids and solids: replace \(\gamma P_0\) by the material's adiabatic bulk modulus \(B\); the same \(c=\sqrt{B/\rho}\) form governs, giving \(\sim 1500\ \text{m/s}\) in water.

Breaks when

Finite amplitude / shocks. When \(\partial_x s\) is not small the nonlinear terms dropped in Steps 3 and 5 dominate: crests travel faster than troughs, waveforms steepen, and a discontinuous shock forms across which the linear speed is meaningless (sonic booms, blast waves).
Dissipative or dispersive media. Viscosity and thermal conduction (ignored here) add a term \(\propto \partial_{txx}s\); at ultrasonic frequencies, or when the compression rate approaches molecular relaxation times, \(\gamma\) becomes effective and frequency-dependent, so the wave attenuates and different frequencies travel at different speeds.
Rarefied gas. When the acoustic wavelength approaches the molecular mean free path the continuum equation of state fails; sound cannot be treated as a fluid wave and kinetic theory is required.

Failure modes

Newton's isothermal blunder: using the isothermal bulk modulus \(P_0\) instead of \(\gamma P_0\), giving \(343/\sqrt{1.4}\approx 290\ \text{m/s}\) for air — wrong by about \(16\%\).
Confusing particle speed with wave speed: the fluid particle velocity \(\partial_t s\) is tiny (mm/s for ordinary sound) and is not \(c\).
Thinking \(c\) rises with pressure: forgetting that \(P_0\propto\rho_0\) at fixed \(T\), so the ratio and hence \(c\) are pressure-independent for an ideal gas.
Molar-mass unit slip: using \(M\) in grams or grams-per-mole instead of \(\text{kg/mol}\), a factor-\(\sqrt{1000}\) error.
Using \(T\) in Celsius in \(c=\sqrt{\gamma RT/M}\); the ideal-gas law requires absolute (kelvin) temperature.
Wrong \(\gamma\): applying \(\gamma=7/5\) to a monatomic gas, or to a diatomic gas hot enough to excite vibration, giving the wrong stiffness.

Discussion

The structural lesson is that any restoring mechanism linear in a displacement gradient, divided by an inertia, produces the same non-dispersive wave equation. For the loaded string the restoring agent was tension; here it is the fluid's resistance to compression. The replacement \(T/\mu \to \gamma P_0/\rho_0\) is the whole physical content — everything downstream (d'Alembert solutions, superposition, standing waves, reflection) transfers verbatim. This is why one derivation of the wave equation pays dividends across acoustics, optics analogues, and elasticity.

The appearance of \(\gamma\) rather than a purely mechanical quantity is what makes sound a genuinely thermodynamic phenomenon. Measuring the speed of sound is one of the cleanest ways to measure \(\gamma\), and hence to count a molecule's active degrees of freedom \(f\) via \(\gamma=1+2/f\): air gives \(\gamma\approx 1.40\Rightarrow f=5\) (three translational plus two rotational, vibration frozen out at room temperature). Sound is thus a probe of molecular structure.

Equivalently, the derivation can be phrased entirely in the pressure perturbation \(\delta P\) or the condensation \(\delta\rho\); all three fields obey the same wave equation and travel together, but with characteristic phase relationships. In a travelling wave the pressure and particle velocity are in phase, and their ratio defines the acoustic impedance \(Z=\rho_0 c\), the quantity that governs how much sound reflects at an interface — the reason so little airborne sound enters water.

At the next level of rigour the linear acoustic system is the low-amplitude limit of the compressible Euler equations linearised about a uniform background; \(c^2=\left(\partial P/\partial\rho\right)_S\) is exactly the isentropic derivative, which for an ideal gas evaluates to \(\gamma P_0/\rho_0\). Retaining the next order in amplitude yields Burgers-type equations whose characteristics cross to form shocks, and including the neglected transport coefficients yields the Navier–Stokes acoustic dispersion relation with a complex wavenumber; the ideal result here is the leading term of both expansions.

Common misconceptions. Sound speed does not depend on the loudness (amplitude) in the linear regime, nor on frequency (air is very nearly non-dispersive), nor on the ambient pressure at fixed temperature. It does depend on temperature — through \(c\propto\sqrt{T}\) — and on the gas's identity through \(\gamma\) and \(M\).

Worked examples

Speed of sound in dry air at \(20^\circ\text{C}\)

Symbolic first: \(c=\sqrt{\gamma P_0/\rho_0}\). Diatomic air: \(\gamma=1.40\). At \(20^\circ\text{C}\), \(P_0=1.013\times10^{5}\ \text{Pa}\), \(\rho_0=1.204\ \text{kg m}^{-3}\). A

\[ c=\sqrt{\frac{(1.40)(1.013\times10^{5}\ \text{Pa})}{1.204\ \text{kg m}^{-3}}}=\sqrt{1.178\times10^{5}\ \text{m}^2\text{s}^{-2}} \]

Insert numbers with units after the symbolic form. Numerator \(=1.418\times10^{5}\ \text{Pa}\); divide by the density. A

\[ c\approx 343\ \text{m s}^{-1} \]

Reading. The textbook room-temperature value. Cross-check via \(c=\sqrt{\gamma RT/M}=\sqrt{(1.40)(8.314)(293)/(0.02896)}\approx 343\ \text{m s}^{-1}\), confirming the two forms agree.

Speed of sound in helium at \(20^\circ\text{C}\)

Use the temperature form \(c=\sqrt{\gamma RT/M}\) since it isolates \(\gamma\) and \(M\). Helium is monatomic: \(\gamma=5/3\), \(M=4.00\times10^{-3}\ \text{kg mol}^{-1}\), \(T=293\ \text{K}\), \(R=8.314\ \text{J mol}^{-1}\text{K}^{-1}\). A

\[ c=\sqrt{\frac{(1.667)(8.314)(293)}{4.00\times10^{-3}}}=\sqrt{1.015\times10^{6}\ \text{m}^2\text{s}^{-2}} \]

Numerator \((1.667)(8.314)(293)=4061\ \text{J mol}^{-1}\); divide by \(M\) in \(\text{kg mol}^{-1}\) to get \(\text{m}^2\text{s}^{-2}\). B

\[ c\approx 1.01\times10^{3}\ \text{m s}^{-1} \]

Reading. Almost three times the value in air, driven mainly by the small molar mass (\(c\propto M^{-1/2}\)) with a modest boost from the larger \(\gamma\). Vocal-tract resonances scale with \(c\), so a helium-filled tract raises the formant frequencies — the "chipmunk" effect.

Problems

Show that for an ideal gas \(c=\sqrt{\gamma P_0/\rho_0}\) is equivalent to \(c=\sqrt{\gamma RT/M}\), and hence that \(c\) is independent of pressure at fixed temperature.
Solution
The ideal-gas law for \(n\) moles in volume \(V\) is \(P_0 V=nRT\). With mass \(m=nM\), the density is \(\rho_0=m/V=nM/V\), so \(P_0/\rho_0=(nRT/V)/(nM/V)=RT/M\). Substituting into \(c=\sqrt{\gamma P_0/\rho_0}\) gives \(c=\sqrt{\gamma RT/M}\). Since \(P_0\) has cancelled, \(c\) depends only on \(\gamma\), \(T\) and \(M\), not on the pressure at fixed \(T\).
By what percentage did Newton's isothermal formula \(c=\sqrt{P_0/\rho_0}\) underestimate the speed of sound in air (\(\gamma=1.40\))?
Solution
The ratio of correct (adiabatic) to isothermal speed is \(c_{\text{ad}}/c_{\text{iso}}=\sqrt{\gamma}=\sqrt{1.40}=1.183\). Newton's value is therefore \(1/1.183=0.845\) of the true value, an underestimate of \(1-0.845=0.155\), i.e. about \(16\%\). Numerically \(c_{\text{iso}}=343/1.183\approx 290\ \text{m s}^{-1}\).
The speed of sound in air is \(343\ \text{m s}^{-1}\) at \(20^\circ\text{C}\) (\(293\ \text{K}\)). Estimate it at \(-40^\circ\text{C}\) (\(233\ \text{K}\)), a typical cruising-altitude temperature.
Solution
At fixed composition \(c\propto\sqrt{T}\). Thus \(c(233)=343\sqrt{233/293}=343\sqrt{0.7952}=343\times0.8917\approx 306\ \text{m s}^{-1}\). The drop of about \(37\ \text{m s}^{-1}\) is why the Mach number for a given true airspeed is higher at altitude.
A travelling harmonic sound wave in air has pressure amplitude \(p_0=2.0\ \text{Pa}\) and frequency \(f=1.0\ \text{kHz}\). Taking \(c=343\ \text{m s}^{-1}\) and \(\rho_0=1.20\ \text{kg m}^{-3}\), find the displacement amplitude \(s_0\), using \(p_0=\rho_0 c\,\omega\, s_0\).
Solution
For a travelling wave the momentum relation gives particle-velocity amplitude \(u_0=p_0/(\rho_0 c)\) and \(u_0=\omega s_0\), so \(s_0=p_0/(\rho_0 c\,\omega)\) with \(\omega=2\pi f=6.28\times10^{3}\ \text{s}^{-1}\). Then \(s_0=2.0/[(1.20)(343)(6.28\times10^{3})]=2.0/(2.585\times10^{6})\approx 7.7\times10^{-7}\ \text{m}\), under a micrometre — illustrating how small acoustic displacements are.
Carbon dioxide has \(M=44.0\times10^{-3}\ \text{kg mol}^{-1}\). Treating it as an ideal gas with \(\gamma=1.30\), find the speed of sound at \(20^\circ\text{C}\) and comment on why it is lower than in air.
Solution
\(c=\sqrt{\gamma RT/M}=\sqrt{(1.30)(8.314)(293)/(44.0\times10^{-3})}\). Numerator \((1.30)(8.314)(293)=3167\ \text{J mol}^{-1}\); divide by \(0.0440\ \text{kg mol}^{-1}\) to get \(7.198\times10^{4}\ \text{m}^2\text{s}^{-2}\); the square root is \(\approx 268\ \text{m s}^{-1}\). It is slower than air (\(343\ \text{m s}^{-1}\)) chiefly because \(\text{CO}_2\) is heavier (\(M=44\) vs \(29\ \text{g mol}^{-1}\)), with a smaller \(\gamma\) contributing further; the greater inertia per mole outweighs any stiffness change.