4.3
What is sound?
§4 · The acoustic wave equation

4.3 Euler’s equation — Newton’s second law in a fluid

Newton said mass times acceleration equals force. For a fluid, we apply this law to a small slab and write force per unit volume. The pressure difference across the slab is what supplies the force.

The slab picture, again

Consider the same slab as in the last lesson — between xx and x+Δxx + \Delta x, area AA. The pressure on the left face pushes the slab rightward with force p(x,t) Ap(x, t)\, A. The pressure on the right face pushes it leftward with force p(x+Δx,t) Ap(x + \Delta x, t)\, A. The net force is

Fnet  =  [ p(x,t)−p(x+Δx,t) ] A  =  −∂p∂x A Δx  +  O(Δx2).F_\text{net} \;=\; \big[\, p(x, t) - p(x + \Delta x, t)\,\big]\, A \;=\; -\frac{\partial p}{\partial x}\, A\, \Delta x \;+\; O(\Delta x^2).

Mass times acceleration is ρ A Δx⋅v˙\rho \, A \, \Delta x \cdot \dot v (where v˙\dot v is the rate of change of velocity following the slab — the material derivative; more on this below). Equating and dividing by AΔxA \Delta x:

ρ v˙  =  −∂p∂x.\rho\, \dot v \;=\; -\frac{\partial p}{\partial x}.

In 3-D with vector velocity and gradient of pressure,

    ρ DvDt  =  −∇p.    \boxed{\;\; \rho\, \frac{D \mathbf{v}}{D t} \;=\; -\nabla p.\;\;}

This is Euler’s equation — Newton’s second law for an ideal (inviscid, incompressible-friction-free) fluid. The material derivative D/Dt=∂t+v⋅∇D/Dt = \partial_t + \mathbf{v} \cdot \nabla accounts for both the change in v\mathbf{v} at a fixed point and the slab’s motion through the field.

Interactive

m = ρ₀ A Δxp_left = 0.50p_right = 0.50∂p/∂x = +0.00 v_slab = +0.00
0.50
0.50
∂p/∂x = 0 → no net force → slab coasts at constant velocity.

Slide the left and right face pressures. When the gradient ∂p/∂x\partial p / \partial x is non-zero the slab accelerates in the opposite direction of the gradient — high pressure pushes low. Equilibrium is the special case of zero gradient.

Linearised form

For sound, ρ=ρ0+ρ′\rho = \rho_0 + \rho', v=v′\mathbf{v} = \mathbf{v}', p=p0+p′p = p_0 + p'. The material derivative has a v⋅∇\mathbf{v} \cdot \nabla term that is quadratic in perturbations (one v′\mathbf{v}' times one gradient of v′\mathbf{v}'), so to first order D/Dt≈∂tD/Dt \approx \partial_t. The density on the left side becomes ρ0+ρ′\rho_0 + \rho', but since the right side has ∇p0=0\nabla p_0 = 0 (equilibrium pressure is uniform), the leading ρ′\rho' multiplies an already-first-order quantity and contributes only at second order. Keeping first-order terms:

    ρ0 ∂v′∂t  =  −∇p′.    \boxed{\;\; \rho_0\, \frac{\partial \mathbf{v}'}{\partial t} \;=\; -\nabla p'.\;\;}

This is the linearised Euler equation. It is the second of the three equations we are about to combine.

▶ Why the linearised Euler equation looks so simple

The full Euler equation is ρ(∂tv+v⋅∇v)=−∇p\rho(\partial_t \mathbf{v} + \mathbf{v} \cdot \nabla \mathbf{v}) = -\nabla p. Three places to linearise:

  1. ∂tv\partial_t \mathbf{v} in the equilibrium-plus-perturbation expansion: ∂t(0+v′)=∂tv′\partial_t(\mathbf{0} + \mathbf{v}') = \partial_t \mathbf{v}' — already first order.
  2. v⋅∇v\mathbf{v} \cdot \nabla \mathbf{v}: substitute v′\mathbf{v}', get v′⋅∇v′\mathbf{v}' \cdot \nabla \mathbf{v}' — second order, dropped.
  3. ρ\rho multiplying everything: ρ0+ρ′\rho_0 + \rho'; the ρ′\rho' piece multiplies a first-order term, so ρ′⋅∂tv′\rho' \cdot \partial_t \mathbf{v}' is second order, dropped. Leaves ρ0⋅∂tv′\rho_0 \cdot \partial_t \mathbf{v}'.
  4. −∇p=−∇(p0+p′)=−∇p′-\nabla p = -\nabla(p_0 + p') = -\nabla p' since ∇p0=0\nabla p_0 = 0.

Result: ρ0∂tv′=−∇p′\rho_0 \partial_t \mathbf{v}' = -\nabla p'. The linearity of acoustics buys us a clean equation; the nonlinear terms return only when ∣p′∣/p0|p'|/p_0 is no longer tiny.

Two equations down, one to go: we need a relation between p′p' and ρ′\rho' to close the system. That is the equation of state, next.