5.2
What is sound?
§5 · Energy, momentum, impedance

5.2 Acoustic energy density

The Lagrangian derivation in lesson 4.8 quietly produced an energy density for the acoustic field:

E  =  12ρ0∣v′∣2⏟kinetic  +  p′22ρ0c2⏟potential.\mathcal{E} \;=\; \underbrace{\tfrac12 \rho_0 |\mathbf{v}'|^2}_{\text{kinetic}} \;+\; \underbrace{\frac{p'^2}{2 \rho_0 c^2}}_{\text{potential}}.

Kinetic plus potential. The first term is the obvious one — the moving mass per unit volume carrying kinetic energy 12ρ0v′2\tfrac12 \rho_0 v'^2. The second is the potential energy of the local compression, analogous to the 12kx2\tfrac12 k x^2 of a spring.

▶ Where the potential-energy term comes from

In adiabatic compression, work done compressing the fluid is stored as internal energy. For a small volume change δV\delta V at pressure pp, the work done on the fluid by surroundings is −p δV-p\, \delta V. For a perturbation around equilibrium, expand to second order:

δW  =  −p0δV  −  12(∂p/∂V) (δV)2+⋯\delta W \;=\; -p_0 \delta V \;-\; \tfrac12 (\partial p / \partial V)\, (\delta V)^2 + \cdots

The first-order term integrates to zero over a cycle (it’s a linear oscillation about equilibrium). The second-order term is the stored potential energy. Using δV/V=−ρ′/ρ0\delta V / V = -\rho'/\rho_0 and (∂p/∂V)s=−c2ρ02/V(\partial p / \partial V)_s = -c^2 \rho_0^2 / V at constant entropy,

δU  =  12⋅c2ρ02V V⋅(ρ′ρ0)2V  =  12(ρ′)2c2ρ0V.\delta U \;=\; \tfrac12 \cdot \frac{c^2 \rho_0^2}{V}\, V \cdot \left(\frac{\rho'}{\rho_0}\right)^2 V \;=\; \tfrac12 \frac{(\rho')^2 c^2}{\rho_0} V.

So per unit volume, the potential energy is 12(ρ′)2c2/ρ0\tfrac12 (\rho')^2 c^2 / \rho_0. Using p′=c2ρ′p' = c^2 \rho', this is (p′)2/(2ρ0c2)(p')^2 / (2 \rho_0 c^2) — matching what we wrote down.

Energy in a plane wave

For a plane harmonic wave with pressure amplitude P0P_0:

Substituting,

E(r,t)  =  12ρ0⋅P02ρ02c2cos⁡2(⋯ )  +  P022ρ0c2cos⁡2(⋯ )  =  P02ρ0c2cos⁡2(ωt−kx).\mathcal{E}(\mathbf{r}, t) \;=\; \tfrac12 \rho_0 \cdot \frac{P_0^2}{\rho_0^2 c^2} \cos^2(\cdots) \;+\; \frac{P_0^2}{2 \rho_0 c^2} \cos^2(\cdots) \;=\; \frac{P_0^2}{\rho_0 c^2} \cos^2(\omega t - k x).

Note that the kinetic and potential energy densities are equal at every point at every instant — this is the equipartition property of plane waves. (For a standing wave, they’re 90° out of phase: kinetic peaks at the antinodes, potential peaks at the moments of full compression.)

Time-averaging cos⁡2\cos^2:

    ⟨E⟩  =  P022ρ0c2.    \boxed{\;\;\langle \mathcal{E} \rangle \;=\; \frac{P_0^2}{2 \rho_0 c^2}.\;\;}

The mean energy density is proportional to the square of the pressure amplitude. Doubling the amplitude quadruples the energy density. This is why energy and intensity are quadratic in p′p', and why decibels are logarithmic in p′2p'^2 rather than in p′p'.

Conservation

Energy density is locally conserved: ∂tE+∇⋅S=0\partial_t \mathcal{E} + \nabla \cdot \mathbf{S} = 0, where S\mathbf{S} is the energy flux. We work out S\mathbf{S} in the next lesson and call it the acoustic intensity.

A plane wave’s energy density propagates rigidly with the wave at speed cc. For a finite wave packet, the total energy is conserved exactly — no losses, no dispersion — until the packet hits a boundary or until nonlinear effects matter (chapter 10).

Numerical scale

For a conversational pressure amplitude P0≈10−2P_0 \approx 10^{-2} Pa, ρ0=1.2\rho_0 = 1.2 kg/m³, c=343c = 343 m/s:

⟨E⟩  ≈  (0.01)22⋅1.2⋅3432  ≈  3.5×10−10 J/m3.\langle \mathcal{E} \rangle \;\approx\; \frac{(0.01)^2}{2 \cdot 1.2 \cdot 343^2} \;\approx\; 3.5 \times 10^{-10}\, \text{J/m}^3.

A third of a nanojoule per cubic metre. Sound is energetically tiny. The interesting thing about sound is not how much energy it carries but the structure it carries.