5.3
What is sound?
§5 · Energy, momentum, impedance

5.3 Intensity and time-averaged flux

The energy in an acoustic field doesn’t just sit there — it flows. The rate at which it flows, per unit area perpendicular to the flow direction, is the acoustic intensity:

I(r,t)  =  p′(r,t) v′(r,t).\mathbf{I}(\mathbf{r}, t) \;=\; p'(\mathbf{r}, t)\, \mathbf{v}'(\mathbf{r}, t).

This is a vector (because v′\mathbf{v}' is) with units of W/m².

Why pressure-times-velocity is the energy flux

Consider a surface element dAdA with normal n^\hat{\mathbf{n}}. The fluid on one side of dAdA exerts a force p′ dA n^p'\, dA\, \hat{\mathbf{n}} on the fluid on the other side. The fluid on the other side moves at velocity v′\mathbf{v}'. The rate at which work is done by the first fluid on the second is force times velocity in the direction of force:

dPwork  =  p′ dA (n^⋅v′)  =  (p′v′)⋅n^ dA.dP_\text{work} \;=\; p'\, dA\, (\hat{\mathbf{n}} \cdot \mathbf{v}') \;=\; (p' \mathbf{v}') \cdot \hat{\mathbf{n}}\, dA.

So the energy flux vector is exactly I=p′v′\mathbf{I} = p' \mathbf{v}' — work done per unit area per unit time. This satisfies the energy continuity equation ∂tE+∇⋅I=0\partial_t \mathcal{E} + \nabla \cdot \mathbf{I} = 0 identically for the linearised acoustic system (a direct calculation from Euler + continuity).

Time-averaged intensity

Most measurements care about the average intensity, not the instantaneous value. For a plane wave with p′=P0cos⁡(ωt−kx)p' = P_0 \cos(\omega t - k x) and v′=(P0/ρ0c)cos⁡(ωt−kx)v' = (P_0/\rho_0 c) \cos(\omega t - k x):

⟨I⟩  =  ⟨p′v′⟩  =  P02ρ0c ⟨cos⁡2⟩  =  P022ρ0c.\langle I \rangle \;=\; \langle p' v' \rangle \;=\; \frac{P_0^2}{\rho_0 c}\, \langle \cos^2 \rangle \;=\; \frac{P_0^2}{2 \rho_0 c}.

In complex notation, using the rule ⟨Re⁡[A~eiωt]Re⁡[B~eiωt]⟩=12Re⁡[A~B~∗]\langle \operatorname{Re}[\tilde A e^{i\omega t}] \operatorname{Re}[\tilde B e^{i\omega t}]\rangle = \tfrac12 \operatorname{Re}[\tilde A \tilde B^*] from lesson 2.2:

    ⟨I⟩  =  12Re⁡[p~′ v~′∗]  =  P022ρ0c k^.    \boxed{\;\;\langle \mathbf{I} \rangle \;=\; \tfrac12 \operatorname{Re}[\tilde p'\, \tilde{\mathbf{v}}'^*] \;=\; \frac{P_0^2}{2 \rho_0 c}\, \hat{\mathbf{k}}.\;\;}

For a plane wave, intensity points in the propagation direction, and its magnitude is P02/(2ρ0c)P_0^2 / (2 \rho_0 c).

▶ The factor $\tfrac12$ is the time-average of $\cos^2$

For any pair of real sinusoidal signals A(t)=A0cos⁡(ωt+α)A(t) = A_0 \cos(\omega t + \alpha) and B(t)=B0cos⁡(ωt+β)B(t) = B_0 \cos(\omega t + \beta), the time average over one period is

⟨AB⟩  =  1T∫0TA0B0cos⁡(ωt+α)cos⁡(ωt+β) dt.\langle A B \rangle \;=\; \frac{1}{T} \int_0^T A_0 B_0 \cos(\omega t + \alpha) \cos(\omega t + \beta)\, dt.

Using cos⁡xcos⁡y=12[cos⁡(x−y)+cos⁡(x+y)]\cos x \cos y = \tfrac12[\cos(x - y) + \cos(x + y)], the integrand has a constant piece 12A0B0cos⁡(α−β)\tfrac12 A_0 B_0 \cos(\alpha - \beta) and an oscillating piece 12A0B0cos⁡(2ωt+α+β)\tfrac12 A_0 B_0 \cos(2\omega t + \alpha + \beta). The oscillating piece integrates to zero over a full period. The constant piece gives

⟨AB⟩  =  12A0B0cos⁡(α−β)  =  12Re⁡[A~B~∗].\langle AB \rangle \;=\; \tfrac12 A_0 B_0 \cos(\alpha - \beta) \;=\; \tfrac12 \operatorname{Re}[\tilde A \tilde B^*].

For A=BA = B (same signal), this is 12A02\tfrac12 A_0^2 — the time-average of cos⁡2\cos^2 is 1/21/2.

Intensity, energy density, and the speed of sound

Look at the ratio:

⟨I⟩⟨E⟩  =  P02/2ρ0cP02/2ρ0c2  =  c.\frac{\langle I \rangle}{\langle \mathcal{E} \rangle} \;=\; \frac{P_0^2 / 2 \rho_0 c}{P_0^2 / 2 \rho_0 c^2} \;=\; c.

The intensity is the energy density times the propagation speed. This is the only thing it could be — energy density is energy per volume, intensity is energy per area per time; their ratio must have units of velocity, and for a non-dispersive wave the only velocity in the problem is cc. The energy density is transported at the wave’s group velocity, which for acoustic plane waves equals the phase velocity, which equals cc.

Numerical scale

For conversational pressure (P0=10−2P_0 = 10^{-2} Pa):

⟨I⟩  =  (10−2)22⋅1.2⋅343  ≈  1.2×10−7 W/m2.\langle I \rangle \;=\; \frac{(10^{-2})^2}{2 \cdot 1.2 \cdot 343} \;\approx\; 1.2 \times 10^{-7}\, \text{W/m}^2.

Tenths of a microwatt per square metre. For comparison:

The dynamic range from threshold to pain is 13 orders of magnitude in intensity — or 26 orders in pressure squared. This is why we use decibels.

Next lesson: the impedance ρ0c\rho_0 c that keeps appearing, and what it means.