We first look at the simplest case, of a one-dimensional sound field in which the variations of pressure, density and particle displacement all depend only on the $x$-coordinate. We will assume that the air properties are the same throughout space, and that the fluctuations in pressure and density due to the sound wave are small compared to the steady atmospheric values. For the moment, we simply look at the behaviour of sound waves propagating in empty space: we will think about sound sources and the influence of walls and obstacles a bit later.

We can write density as $\rho_0 + \rho'(x)$ and pressure as $p_0 + p'(x)$ where $\rho_0$ and $p_0$ are the steady density and pressure of air, and the quantities with primes are the small deviations from those steady values due to the sound wave. Now consider a small element of the air, lying between positions $x$ and $x+\delta x$, sketched in Fig. 1 with the wave inside a pipe with cross-sectional area $A$. As a result of the sound wave, the particles that were initially at the ends of this element are displaced by a small distance, to $x+\xi(x)$ and $x + \delta x + \xi(x + \delta x)$ respectively.

We can apply conservation of mass to this element of air: the product of density and volume must remain constant, so

$$\rho_0 A \delta x = (\rho_0 + \rho’) A [\delta x – \xi(x) + \xi(x+ \delta x)] $$

$$\approx (\rho_0 + \rho’) A \left[\delta x + \frac{\partial \xi}{\partial x} \delta x \right] . \tag{1}$$

Cancelling $A \delta x$ and ignoring products of small quantities because we want a linearised relation, this reduces to

$$0 \approx \rho’ + \rho_0 \frac{\partial \xi}{\partial x} .\tag{2}$$

Now we can express $\rho’$ in terms of the pressure change $p’$. At typical acoustic frequencies, sound waves fluctuate sufficiently quickly that the air behaves *adiabatically* (or *isentropically*). This means that the thermodynamic relation

$$p V^\gamma = \mathrm{constant} \tag{3}$$

can be assumed for any small fluid volume $V$, where $\gamma$ is the ratio of specific heats for air, a constant with the value 1.4. In terms of density, this means

$$p \rho^{-\gamma} = \mathrm{constant} \tag{4}$$

so that for our element,

$$p_0 \rho_0^{-\gamma} = (p_0 +p’) (\rho_0 + \rho’)^{-\gamma} = (p_0 +p’) \rho_0 ^{-\gamma} (1+ \rho’/\rho_0)^{-\gamma} \tag{5}$$

so that

$$0 \approx p’ – p_0 \gamma \frac{\rho’}{\rho_0} \tag{6}$$

using the binomial theorem and ignoring products of small terms. Substituting in eq. (2) then gives

$$\frac{\partial \xi}{\partial x} = – \frac{p’}{p_0 \gamma} . \tag{7}$$

We can obtain a second relation between $p’$ and $\xi$ by applying Newton’s law to our small volume. The pressure acts on the two end faces, giving a net force which must be balanced by mass times acceleration. So

$$A p(x) -A p(x+ \delta x) = \rho_0 A \delta x \frac{\partial^2 \xi}{\partial t^2} \tag{8}$$

so that

$$- \frac{\partial p’}{\partial x} \approx \rho_0 \frac{\partial^2 \xi}{\partial t^2} . \tag{9}$$

Differentiating this with respect to $x$ and then using eq. (7), we obtain

$$\frac{\partial^2 p’}{\partial x^2} = -\rho_0 \frac{\partial^2}{\partial t^2} \left[ \frac{\partial \xi}{\partial x} \right]= \rho_0 \frac{\partial^2}{\partial t^2} \left[ \frac{p’}{p_0 \gamma} \right] \tag{10}$$

So finally, if we define a constant $c$ satisfying

$$c^2 =\frac{\gamma p_0}{\rho_0} \tag{11}$$

we obtain the governing equation

$$\frac{\partial^2 p’}{\partial t^2}= c^2 \frac{\partial^2 p’}{\partial x^2} .\tag{12}$$

We can recognise this as exactly the same equation we found for the ideal stretched string in section 3.1.1, and so we already know that $c$ is the wave speed, in other words in this case it is the speed of sound. Standard values for the properties of air at sea level and 15$^\circ$C are $\rho_0=$1.225 kg/m$^3$ and $p_0=$1.01 MPa, leading to $c=340$ m/s, the expected value for the speed of sound.

It is not difficult to generalise this derivation to the general case of sound waves in three dimensions. We now consider a small rectangular element of air, with sides $\delta x$, $\delta y$ and $\delta z$. The pressure is still a scalar quantity, but the particle displacement becomes a vector $\underline{\xi}(x,y,z)$ with components $(\xi_1,\xi_2,\xi_3) $ in the three axis directions. We follow the same sequence of steps. For the equivalent of eq. (1), we need to allow for the volume change associated with each of the three pairs of faces of the fluid element: the result is

$$\rho_0 \delta x \delta y \delta z \approx (\rho_0 + \rho’) \left[1 + \frac{\partial \xi_1}{\partial x} \right] \delta x \left[1 + \frac{\partial \xi_2}{\partial y} \right] \delta y \left[1 + \frac{\partial \xi_3}{\partial z} \right] \delta z \tag{13}$$

and so

$$0 \approx \rho_0 \left[\frac{\partial \xi_1}{\partial x} + \frac{\partial \xi_2}{\partial y} + \frac{\partial \xi_3}{\partial z} \right] + \rho’ \tag{14}$$

or, in the notation of vector calculus involving the divergence operator,

$$-\rho_0 \nabla \cdot \underline{\xi} = \rho’ = \frac{p’}{c^2} \tag{15}$$

using eqs. (6) and (11). (If you are hazy about vector calculus, the next link gives an introduction to the main concepts.)

To apply Newton’s law to the fluid element, we simply have the equivalent of eqs. (8) and (9) for each component separately, leading to

$$\rho_0 \ddot{\underline{\xi}} = – \nabla p’ \tag{16}$$

in terms of the gradient of pressure. Taking the divergence of this equation and then using eq. (15), we obtain

$$\nabla \cdot \nabla p’ = -\rho_0 \nabla \cdot \ddot{\underline{\xi}} = \frac{\partial^2}{\partial t^2} \left[ \frac{p’}{c^2} \right] . \tag{17}$$

The result is the wave equation in three dimensions:

$$\frac{\partial^2 p’}{\partial t^2} = c^2 \nabla^2 p’ .\tag{18}$$