11.7.1 Compact sound sources: monopoles, dipoles, quadrupoles

In this section we give a brief but general account of sound generation, concentrating particularly on compact sources (i.e. the source region is small compared to the wavelength of sound, or equivalently the Helmholtz number is small). This description draws heavily on Chapter 7 of the textbook by Dowling and Ffowcs Williams [1].

The first step is to recall a result from section 4.3.1: the sound pressure radiated by a small sphere vibrating sinusoidally at frequency $\omega$ is

$$p=\dfrac{i \omega \rho_0 q}{4 \pi r} e^{i \omega(t-r/c)} \tag{1}$$

where $r$ is the distance to the listener from the centre of the sphere, $q$ is the total volume flux displaced by the pulsating sphere, and $\rho_0$ is the density of air. Noting that $i \omega q$ is simply the frequency-domain expression for the rate of change $\dot{q}$, we can see that all the frequency components of a general waveform $\dot{q}(t)$ have identical behaviour: an outgoing spherical wave at speed $c$, with amplitude decreasing inversely with distance. We can deduce the time-domain equivalent formula for the observed pressure at distance $r$:

$$p(t)=\dfrac{\rho_0 }{4 \pi r} \dot{q}(t-r/c) . \tag{2}$$

We will shortly use this result in a similar manner to the Rayleigh integral (see section 4.3.2), to determine the total sound pressure generated by a spatially distributed sound source region by performing a suitable integration.

The next step is to revisit the derivation of the linearised wave equation from section 4.1.1, and add in any possible source terms. The derivation involved two fundamental equations: mass conservation and momentum balance (or Newton’s law). Each of those could have an associated source term.

The equation of mass conservation then reads

$$\dfrac{\partial \rho}{\partial t} + \rho_0 \nabla \cdot \underline{v} = m \tag{3}$$

where $\underline{v}$ is the air velocity vector. This is equation (15) of section 4.1.1, with the addition of the term $m$ on the right-hand side. This allows for the possibility of mass being added to the system: it describes the rate at which mass is added, per unit volume, at a given position in the source region. The kind of application we are interested in involves air flowing out of a tone-hole, in which case if the volume flow rate is $q$ per unit volume, then $m=\rho_0 q$.

In a similar vein, the equation of momentum balance now reads

$$\nabla p + \rho_0 \dfrac{\partial \underline{v}}{\partial t} = \underline{f} , \tag{4}$$

which is the same as equation (16) from section 4.1.1 with the addition of the term $\underline{f}$ describing external force per unit volume that may be applied to the air at a given position.

Now we proceed much as before. We can substitute $\rho = p/c^2$ in equation (3), then eliminate $\underline{v}$ between equations (3) and (4) to leave an equation for $p$. To achieve this we take $\partial/\partial t$ of equation (3) and we take the divergence of equation (4), then subtract one from the other. The result is

$$\dfrac{1}{c^2} \dfrac{\partial^2 p}{\partial t^2} \mathrm{~}-\mathrm{~} \nabla^2 p = \rho_0 \dot{q} \mathrm{~}-\mathrm{~} \nabla \cdot \underline{f} .\tag{5}$$

We see the familiar wave equation, with two source terms on the right-hand side. We can conclude that within the approximation of linearised acoustics these are the only two possible ways in which sound can be generated. To see what they mean, we can consider them in turn. The first term, $\rho_0 \dot{q}$, can be related directly to the result in equation (2) derived from the pulsating sphere. The sphere problem had a concentrated excitation involving $\rho_0 \dot{q}$, while equation (5) has a distributed version of exactly the same thing, expressed per unit volume. So, just as we did with the Rayleigh integral in section 4.3.2, we can integrate equation (2) over the volume $V$ containing the source region, where $\rho_0 \dot{q}$ is non-zero. The result is

$$p(\underline{x},t)=\rho_0 \int \int \int_V{\dfrac{\dot{q}(\underline{y},t-|\underline{x} – \underline{y}|/c)}{4 \pi |\underline{x} – \underline{y}|} d^3 \underline{y}} \tag{6}$$

where the position vectors $\underline{x}$ and $\underline{y}$ point to the receiver position and an element within the source region, as indicated in Fig. 1, and $d^3 \underline{y}$ denotes the element of volume with respect to the position $\underline{y}$. The equation looks at first sight more complicated than equation (2), but this is simply because we need the more cumbersome expression $|\underline{x} – \underline{y}|$ in place of the radial distance $r$: each element of the source is at a different distance from the receiver.

Figure 1. Sketch of a source region, an element within it (green cube), and the position vectors of this element and of the receiver where the pressure is to be observed.

If the source region is small compared to the wavelength of sound and receiver is very distant from the source, the variation in $|\underline{x} – \underline{y}|$ for different positions $\underline{y}$ is very small. Under those circumstances, equation (6) reduces to a simple approximate form:

$$p(\underline{x},t) \approx \dfrac{\rho_0}{4 \pi R} \dot{Q}(t-R/c) \tag{7}$$


$$\dot{Q}=\int \int \int_V{\dot{q} \mathrm{~} d^3 \underline{y}} \tag{8}$$


$$R=|\underline{x} – \underline{Y}| \tag{9}$$


$$\underline{Y}=\dfrac{1}{V} \int \int \int_V{\underline{y} \mathrm{~} d^3\underline{y}} \tag{10}$$

is the mean position vector indicating the middle of the source region. Equation (7) is exactly the same as equation (2): it describes omnidirectional monopole sound radiation with a strength $\dot{Q}$ given by simply adding all the distributed contributions (equation (8)).

Returning to equation (5), we now look at the effect of the second source term. We can use the same integration approach as in equation (6), to obtain

$$p(\underline{x},t)=- \int \int \int_V{\dfrac{\nabla \cdot \underline{f}(\underline{y},t-|\underline{x} – \underline{y}|/c)}{4 \pi |\underline{x} – \underline{y}|} d^3 \underline{y}} . \tag{11}$$

If we then looked for a similar approximation to equations (7) and (8), we would hit a snag. The equivalent of equation (8) would involve the integral

$$\int \int \int_V{\nabla \cdot \underline{f} \mathrm{~}d^3 \underline{y}}, \tag{12}$$

and we can easily show that this integral is zero. We apply the divergence theorem (see section 4.1.4 equation (11)) to a surface $S$ which entirely encloses the source volume $V$, far enough out that $\underline{f}=0$ everywhere on the surface. Then

$$\int \int \int_V{\nabla \cdot \underline{f} \mathrm{~}d^3 \underline{y}} = \int \int_S{\underline{f} \cdot d\underline{A}} = 0 . \tag{13}$$

So the net monopole strength of this kind of sound source is zero: in fact, it describes a dipole source. To see this clearly, we use a trick — closely related to the trick we used in section 4.3.1 when we first met dipoles. Suppose the equation we are trying to solve had been

$$\dfrac{1}{c^2} \dfrac{\partial^2 p_1}{\partial t^2} – \nabla^2 p_1 = \underline{f} \tag{14}$$

rather than equation (5). Then we would know the solution by our integration procedure: it would be

$$p_1(\underline{x},t)=\int_V{\dfrac{\underline{f}(\underline{y},t-|\underline{x} – \underline{y}|/c)}{4 \pi |\underline{x} – \underline{y}|} d^3 \underline{y}} . \tag{15}$$

Now we can take the divergence of both sides of equation (14) to deduce the solution we really want:

$$\dfrac{1}{c^2} \dfrac{\partial^2 p}{\partial t^2} – \nabla^2 p = – \nabla \cdot \underline{f} \tag{16}$$


$$p(\underline{x},t) = -\nabla \cdot p_1(\underline{x},t) = -\nabla \cdot \int_V{\dfrac{\underline{f}(\underline{y},t-|\underline{x} – \underline{y}|/c)}{4 \pi |\underline{x} – \underline{y}|} d^3 \underline{y}} \tag{17} .$$

From here, we can obtain an approximate form for a compact source region, similar to equations (7) and (8):

$$p(\underline{x},t) \approx -\nabla \cdot \left[ \dfrac{\underline{F}(t-R/c)}{4 \pi R} \right] \tag{18}$$


$$\underline{F} = \int \int \int_V{\underline{f} \mathrm{~} d^3 \underline{y}} \tag{19}$$

and $R$ is given by equation (9).

To put this result in a more intuitive form, suppose we choose to align our $z$-axis with the direction of the net force $\underline{F}$. Write the magnitude of the force $F(t)$. Then equation (18) becomes

$$p(\underline{x},t) \approx -\dfrac{\partial}{\partial z} \left[ \dfrac{F(t-R/c)}{4 \pi R} \right] . \tag{20}$$

As we did in section 4.3.1 we can relate a small change in $z$ to a small change in $R$ via the polar angle $\theta$ with the $z$-axis: $dR=dz \cos \theta$. So finally

$$p(\underline{x},t) \approx – \cos \theta \dfrac{\partial}{\partial R} \left[ \dfrac{F(t-R/c)}{4 \pi R} \right]$$

$$ = \dfrac{\cos \theta}{4 \pi}\left[ \dfrac{1}{Rc} \dfrac{dF}{dt} + \dfrac{F}{R^2} \right] . \tag{21}$$

This is exactly the form we found for a dipole source in section 4.3.1. In the far field, only the first term in the square brackets matters.

We have seen that within the scope of linear acoustics there are only two types of sound source: anything producing a net variation in volume gives a monopole source, and anything applying a net force to the fluid gives a dipole source. But there is a third kind of source, if we take account of the nonlinearity of the full governing equations for fluid motion. The underlying theory was developed by Sir James Lighthill [2] back in the 1950s, when noise from jet aircraft was beginning to be a major concern.

Lighthill’s key insight was that you could take the full, nonlinear equations governing fluid flow, and manipulate them into a form

$$<\mathrm{Wave~equation}>\mathrm{~}= \mathrm{~}<\mathrm{Other~stuff}>$$

and then interpret the $<\mathrm{Other~stuff}>$ as a source term, creating sound waves. This formulation is usually called “Lighthill’s acoustical analogy”. We will give a brief account here, but we need not go deeply into the rather complicated mathematics associated with this subject. We have laid the groundwork for the analysis in section 11.2.1. The steps are very similar to the derivation of the wave equation given above: we will take the equations of mass conservation and momentum balance and combine them to give a single equation, in this case one governing the behaviour of fluid density rather than pressure.

We will use the notation of section 11.2.1, denoting the three Cartesian coordinates by $(r_1,r_2,r_3)$ rather than $(x,y,z)$ so that we can express the equations in a compact form. Equation (5) from that section expressed conservation of mass in the form

$$\dfrac{\partial \rho}{\partial t} = – \nabla \cdot (\rho \underline{u}) = -\sum_{j=1}^3\dfrac{\partial}{\partial r_j} \left( \rho u_j \right) . \tag{22}$$

Equation (20) from that section expressed momentum balance in the form

$$\rho \dfrac{Du_i}{Dt}=\sum_{j=1}^3{\dfrac{\partial \sigma_{ij}}{\partial r_j}}\mathrm{~~~for~~}i=1,2,3 \tag{23}$$

where $\sigma_{ij}$ is the stress tensor, and for our present purpose we have not included the term expressing a body force — we have already dealt with external forces in the earlier discussion. Expanding out the convective derivative $D/Dt$, this equation gives

$$\rho \dfrac{\partial u_i}{\partial t}=- \rho \sum_{j=1}^3{u_j \dfrac{\partial u_i}{\partial r_j}} + \sum_{j=1}^3{\dfrac{\partial \sigma_{ij}}{\partial r_j}}\mathrm{~~~for~~}i=1,2,3 . \tag{24}$$

Now we can combine equations (22) and (24) to show

$$\dfrac{\partial}{\partial t}(\rho u_i) =\dot{\rho} u_i + \rho \dot{u_i}$$

$$= -\sum_{j=1}^3\dfrac{\partial}{\partial r_j} \left( \rho u_j \right) u_i \mathrm{~}-\mathrm{~} \rho \sum_{j=1}^3{u_j \dfrac{\partial u_i}{\partial r_j}} + \sum_{j=1}^3{\dfrac{\partial \sigma_{ij}}{\partial r_j}}$$

$$= \sum_{j=1}^3 \dfrac{\partial}{\partial r_j} \left[\sigma_{ij} \mathrm{~}-\mathrm{~} \rho u_i u_j \right] \mathrm{~~~for~~}i=1,2,3 \tag{25}$$

Now we can take $\partial/\partial t$ of equation (22) and subtract the divergence of equation (25) to obtain

$$\dfrac{\partial^2 \rho}{\partial t^2} = -\sum_{i=1}^3 \sum_{j=1}^3 \dfrac{\partial^2}{\partial r_i \partial r_j} \left[\sigma_{ij} \mathrm{~}-\mathrm{~} \rho u_i u_j \right] . \tag{26}$$

Finally, if we subtract from both sides of this equation a term

$$c^2 \nabla^2 \rho = c^2 \sum_{i=1}^3 \dfrac{\partial^2 \rho}{\partial r_i^2} \tag{27}$$

we can express the result in the form we are looking for:

$$\dfrac{\partial^2 \rho}{\partial t^2} \mathrm{~}-\mathrm{~} c^2 \nabla^2 \rho = \sum_{i=1}^3 \sum_{j=1}^3 \dfrac{\partial^2 T_{ij}}{\partial r_i \partial r_j} \tag{28}$$

where Lighthill’s stress tensor $T_{ij}$ is defined by

$$T_{ij} = \rho u_i u_j \mathrm{~}-\mathrm{~} \sigma_{ij} \mathrm{~}-\mathrm{~} c^2 \rho \delta_{ij} \tag{29}$$

where $\delta_{ij}$ denotes the unit matrix as usual.

We need a slightly more general expression for $\sigma_{ij}$ than the one given in equation (21) of section 11.2.1, because we do not want to assume incompressible flow. We need not go into details, but it is always possible to write

$$\sigma_{ij} = -p \delta_{ij} + \tau_{ij} \tag{30}$$

where $\tau_{ij}$ is a symmetric matrix (or, more correctly, tensor) called the deviatoric viscous stress tensor. It vanishes for inviscid flow. If we assume the usual relation $p=c^2 \rho$, the final form of the Lighthill tensor is

$$T_{ij} = \rho u_i u_j \mathrm{~}-\mathrm{~} \tau_{ij} . \tag{31}$$

Looking back at equation (18), we saw that a dipole field was given by the divergence of a monopole field. Equation (28) now shows that the sound field generated by the nonlinearity of flow involves a double divergence, which immediately tells us that it will be quadrupole in character.

[1] A. P. Dowling and J. E. Ffowcs Williams; “Sound and sources of sound”, Ellis Horwood (1983).

[2] M. J. Lighthill; “On sound generated aerodynamically. I and II”. Proceedings of the Royal Society of London, Series A 211, 564-587 (1952) and 222, 1-32 (1954).