4.2
What is sound?
§4 · The acoustic wave equation

4.2 Conservation of mass for a fluid slab

Mass is conserved. If the mass inside a fixed control volume is changing, then more mass is flowing in than flowing out (or vice versa). This is the first of the three fluid-mechanics laws that combine to make the wave equation.

The slab picture

Consider a thin slab of fluid between xx and x+Δxx + \Delta x, of cross-sectional area AA. The mass inside the slab is ρ(x,t)⋅A Δx\rho(x, t) \cdot A\, \Delta x. The mass flowing in through the left face per unit time is ρ(x,t) v(x,t) A\rho(x, t)\, v(x, t)\, A; the mass flowing out through the right face is ρ(x+Δx,t) v(x+Δx,t) A\rho(x + \Delta x, t)\, v(x + \Delta x, t)\, A. Conservation of mass says

ddt[ρ A Δx]  =  [ρv]xA  −  [ρv]x+ΔxA.\frac{d}{dt}\big[\rho\, A\, \Delta x\big] \;=\; \big[\rho v\big]_x A \;-\; \big[\rho v\big]_{x + \Delta x} A.

Divide by A ΔxA\, \Delta x and take Δx→0\Delta x \to 0:

∂ρ∂t  =  −∂(ρv)∂x.\frac{\partial \rho}{\partial t} \;=\; -\frac{\partial (\rho v)}{\partial x}.

In three dimensions the same argument with a divergence on the right gives

    ∂ρ∂t  +  ∇⋅(ρ v)  =  0.    \boxed{\;\;\frac{\partial \rho}{\partial t} \;+\; \nabla \cdot (\rho\, \mathbf{v}) \;=\; 0.\;\;}

This is the continuity equation: density changes by the negative divergence of the mass flux. We will use the 1-D version below and the full 3-D version in lesson 4.5.

Interactive

v_in = 0.50v_out = 0.50density ρcontrol volume — area A, thickness Δxρ = 1.00 ∂ρ/∂t = +0.00
0.50
0.50
v_in = v_out → density parked. This is the *incompressible* case ∇·v = 0.

Slide the inflow and outflow velocities. When inflow exceeds outflow the slab’s density rises (compression). When outflow exceeds inflow the density falls (rarefaction). When they match, the density is parked — the incompressible case ∇⋅v=0\nabla \cdot \mathbf{v} = 0.

Linearised form

For sound we are interested in small perturbations on top of equilibrium. Write ρ=ρ0+ρ′\rho = \rho_0 + \rho' and v=0+v′\mathbf{v} = \mathbf{0} + \mathbf{v}'. Substitute into the continuity equation:

∂(ρ0+ρ′)∂t+∇⋅[(ρ0+ρ′) v′]  =  0.\frac{\partial (\rho_0 + \rho')}{\partial t} + \nabla \cdot \big[(\rho_0 + \rho')\, \mathbf{v}'\big] \;=\; 0.

The first term is ∂tρ′\partial_t \rho' since ρ0\rho_0 is constant. The second term, expanded, contains ρ0∇⋅v′\rho_0 \nabla \cdot \mathbf{v}' (linear in perturbations) and ∇⋅(ρ′v′)\nabla \cdot (\rho' \mathbf{v}') (quadratic — both factors are first-order small). Throwing away the quadratic term:

    ∂ρ′∂t  +  ρ0 ∇⋅v′  =  0.    \boxed{\;\;\frac{\partial \rho'}{\partial t} \;+\; \rho_0\, \nabla \cdot \mathbf{v}' \;=\; 0.\;\;}

This is the linearised continuity equation. It is the first of the three equations we will combine into the wave equation in lesson 4.5.

▶ Why we can throw away $\nabla \cdot (\rho' \mathbf{v}')$

In the small-perturbation regime, ρ′/ρ0\rho' / \rho_0 and ∣v′∣/c|\mathbf{v}'| / c are both numerically tiny (around 10−710^{-7} for conversational speech). Each is “first-order small” — keeping only first-order terms is the linearisation. The product ρ′v′\rho' \mathbf{v}' is second-order small; the divergence of a second-order quantity is also second-order. We are throwing away 10−1410^{-14}-level effects in favour of 10−710^{-7}-level effects.

This is not a trick or a sleight-of-hand — it is exactly the assumption ∣p′∣≪p0|p'| \ll p_0 from lesson 1.4. Every term we drop is one we will recover in chapter 10, where the nonlinear corrections give us wave steepening, shock formation, and the entry to cavitation.

Next: Newton’s second law for the same fluid slab.