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For compact Lighthill stress tensor and applied-force distributions, define $S_{ij}=\int T_{ij}{d}^{3}y$ and $F_i=\int f_i{d}^{3}y$. Differentiating the retarded acoustic Green function at far distance gives the displayed leading radiation field, evaluated at $t-r/c_0$. Source compactness requires its size to be small compared with the acoustic wavelength, not merely with the observer distance.

For a fixed impermeable object, the Ffowcs Williams-Hawkings equation has a volume acoustic quadrupole and a surface acoustic dipole, with no mass-injection term. If $F$ is the total force exerted on the fluid and $Q=\int TdV$ the integrated Lighthill stress tensor, the acoustic compact-source approximation in the acoustic far field gives the displayed pressure at retarded time. A force convention directed onto the body reverses the loading sign. The retarded acoustic Green function carries $1/c_0^2$ when the wave operator is ${\partial }_t^2-c_0^2\Delta$.

The Lighthill acoustic analogy rewrites fluid mass and momentum conservation as a constant-speed wave equation driven by a nonlinear Lighthill stress tensor. It is an exact rearrangement before making source approximations. The outgoing field can be constructed with a retarded acoustic Green function.

Use the conservative continuity equation and momentum balance, with ${\sigma }_{ij}$ the viscous stress:Differentiate the first equation in time and take the divergence of the second. Eliminating momentum givesSubtract $c_0^2\Delta \rho$, and define the Lighthill stress tensor byThe constant reference subtraction has no double divergence. The exact Lighthill acoustic analogy isThis is an exact rearrangement; linearization is not used to eliminate the nonlinear source tensor.

Convolving with the retarded acoustic Green function and transferring the source derivatives by integration by parts givesThe acoustic compact-source approximation requires source size $\ell$ to satisfy $\Omega \ell /c_0\ll 1$ for its characteristic frequency $\Omega$; the acoustic far field additionally requires $r\gg \ell$ and $\Omega r/c_0\gg 1$. Internal retardation is then negligible at leading order, and the radiating parts of the two observer derivatives act on the common retarded time. With $n_i=x_i/r$ and $S_{ij}=\int T_{ij}{d}^{3}y$, the far-field acoustic force and stress moments giveThe two spatial derivatives supply two factors of $1/c_0$, in addition to the $1/c_0^2$ in the Green function.

For a subsonic fluctuating source of velocity scale $V$, size $\ell$ and turnover time $\ell /V$, suppose its stress scale is $T_{ij}=O({\rho }_0{V}^2)$. Then $S_{ij}=O({\rho }_0{V}^2{\ell }^3)$ and ${\ddot{S}}_{ij}=O({\rho }_0{V}^4\ell )$. The compact acoustic quadrupole Mach-number scaling is thereforeThis fourth-power statement concerns the acoustic density amplitude under the specified eddy-time and stress scaling. With an independently imposed frequency, the general estimate instead retains $V^2{\Omega }^2{\ell }^3/c_0^{4}r$.

With a mass source, the continuity equation becomes ${\rho }_t+{\partial }_i(\rho u_i)=M$. If the conservative momentum equation has no added source, repeating the elimination gives the mass-injection term in the acoustic analogy:More generally, injection can carry momentum. If its momentum source is $S_i$, the additional term is ${\partial }_{t}M-{\partial }_i{S}_i$; prescribing $M$ alone does not determine this dipole contribution. The displayed monopole formula assumes no separately imposed momentum source.

Let $\mathcal{M}(t)=\int M(\mathbf{y},t)d^{3}y$ be the total mass rate. Convolution with the retarded acoustic Green function, followed by the acoustic compact-source approximation, gives the extra acoustic monopoleThe stress quadrupole from the preceding part remains present. If required, a compact injected momentum rate ${\mathcal{S}}_i=\int S_i{d}^{3}y$ adds the dipole $n_i{\dot{\mathcal{S}}}_i/(4\pi c_0^{3}r)$.

A small pulsating bubble displaces liquid with volume flux $\dot{\mathcal{V}}(t)$ and acts acoustically like a mass rate $\mathcal{M}={\rho }_0\dot{\mathcal{V}}$. Spherical symmetry makes this a monopole, withFor volume oscillation $\mathcal{V}={\mathcal{V}}_0+\Delta \mathcal{V}\cos \omega t$, the density amplitude is ${\rho }_0{\omega }^2|\Delta \mathcal{V}|/(4\pi c_0^{2}r)$. At fixed volume-displacement amplitude, it scales as ${\omega }^2$. Holding volume-flux amplitude fixed would instead give an $\omega$ scaling. The bubble must remain small compared with the wavelength, and a frequency-dependent dynamical bubble amplitude is a separate effect.

Convolution with the retarded acoustic Green function, followed by two integrations by parts in the source coordinate, yieldsFor the acoustic compact-source approximation, replace the distance and the source retarded time by $r=|x|$ and $\tau =t-r/c_0$ in the leading moment. Thus the integral is $S_{ij}(\tau )/(4\pi c_0^{2}r)$, where $S_{ij}=\int T_{ij}dV$. In the acoustic far field, each spatial derivative acts predominantly on retarded time: ${\partial }_{x_i}\tau =-n_i/c_0$, where $n_i=x_i/r$. Derivatives of $1/r$ and of $n_i$ give lower powers of $r$. The two retarded derivatives therefore giveThis is the compact acoustic quadrupole field, retaining its angular stress projection.

For the compact acoustic quadrupole Mach-number scaling, let the source length be $\ell$, typical fluctuation speed be $U$, and typical time be $\ell /U$. The low-Mach number stress is $T=O({\rho }_0{U}^2)$, so $S=O({\rho }_0{U}^2{\ell }^3)$ and $\ddot{S}=O({\rho }_0{U}^4\ell )$. ConsequentlyWith the geometric ratio separated, this is the stated fourth-power dependence. It assumes an advective source time scale and compactness $\omega \ell /c_0=O(m)\ll 1$. An externally driven source with an independent frequency does not acquire the same Mach power automatically.

Put $r=|\mathbf{x}|$, $\mathbf{n}=\mathbf{x}/r$, and take a compactly supported Lighthill stress tensor. The isotropic term in that tensor is $(p-c_0^2\rho ){\delta }_{ij}$: the Kronecker delta is implicit in the printed shorthand. Convolving the Lighthill acoustic analogy with the retarded acoustic Green function and integrating twice by parts in the source coordinates givesThe boundary terms vanish because of compact support. For source size $\ell$, the acoustic compact-source approximation requires $\omega \ell /c_0\ll 1$. At $r\gg \ell$, we may therefore replace the denominator by $r$ and the retarded argument by $t-r/c_0$ in the leading source integral. Define $S_{ij}(t)=\int T_{ij}(\mathbf{y},t)d^{3}y$. In the radiation region $\omega r/c_0\gg 1$, the leading two spatial derivatives act on the retarded time rather than on $1/r$ or $\mathbf{n}$:Thus the leading acoustic quadrupole density isFor the compact acoustic quadrupole Mach-number scaling, let $U$ be the source fluctuation speed, $m=U/c_0\ll 1$, and use the advective source time $\ell /U$. With $T_{ij}=O({\rho }_0{U}^2)$, one has $S_{ij}=O({\rho }_0{U}^2{\ell }^3)$ and ${\ddot{S}}_{ij}=O({\rho }_0{U}^4\ell )$. ConsequentlyThe fourth power refers to the radiation coefficient with geometric spreading separated. It assumes that the source frequency scales as $U/\ell$; an independently imposed frequency or an independently strong thermal source would change that Mach number estimate.

Let $R=|\mathbf{x}|$, $\mathbf{n}=\mathbf{x}/R$ and $t_R=t-R/c_0$. Convolving the source terms in the Lighthill acoustic analogy with the retarded acoustic Green function, and moving their spatial derivatives outside the integral, givesThe integrals have compact spatial support. For an acoustically compact source, its size $l$ satisfies $\omega l/c_0\ll 1$, so the source-dependent retardation is negligible at leading order. Also take $R\gg l$. Define $S_{ij}=\int T_{ij}{d}^{3}y$ and $F_i=\int f_i{d}^{3}y$. In the radiation terms the spatial derivatives act on $t_R$, with ${\partial }_{x_i}{t}_R=-n_i/c_0$. Derivatives of $1/R$ produce smaller near-field terms. Thus the far-field acoustic force and stress moments giveUsing $n_i=x_i/R$ gives the equivalent powers of $R$ requested. The force term is an acoustic dipole; the Lighthill stress tensor produces an acoustic quadrupole.

For a small body in low-speed flow, put $M=u/c_0\ll 1$. A typical unsteady force has size $F=O({\rho }_0{u}^2{l}^2)$ and time scale $l/u$, whereas $S=O({\rho }_0{u}^2{l}^3)$. ThereforeThese are the acoustic dipole Mach-number scaling and compact acoustic quadrupole Mach-number scaling. Since $p^{\prime }=c_0^2{\rho }^{\prime }$ and outgoing acoustic intensity is of order $p^{\prime 2}/({\rho }_0{c}_0)$, their sound powers scale asThus an unsteady force on the body can produce the dominant low-Mach number sound. If the integrated force or its time derivative vanishes, that dipole contribution is absent. Force on the body and force on the fluid have opposite signs.

Let $\chi =H(F)$ use the Heaviside step function, and ${\delta }_s=|\nabla F|\delta (F)$, so that ${\mathbf{n}}_s=\nabla F/|\nabla F|$ points from the body into the fluid. The kinematic boundary condition gives $F_t+\mathbf{u}\cdot \nabla F=0$ on the surface. In the exterior, mass conservation and the Euler equations for an inviscid fluid areProducts with $\chi$ are understood distributionally, using ${\partial }_t\chi =F_t\delta (F)$ and ${\partial }_i\chi =F_i\delta (F)$. With ${\rho }^{\prime }=\rho -{\rho }_0$, the extended continuity equation isThe corresponding extended momentum equation isbecause its advective surface term is $\rho u_i(F_t+\mathbf{u}\cdot \nabla F)\delta (F)=0$. Differentiate continuity in time and substitute the divergence of momentum. Subtracting $c_0^2{\nabla }^2({\rho }^{\prime }\chi )$ gives the Ffowcs Williams-Hawkings equationwhere the inviscid Lighthill stress tensor isIf $p$ denotes total pressure, this tensor includes the constant ambient pressure $p_0$. Its contribution $p_0{\nabla }^2\chi$ cancels the corresponding surface term exactly. For radiation integrals it is convenient to subtract $p_0$ from both terms, making $T_{ij}=\rho u_i{u}_j+(p^{\prime }-c_0^2{\rho }^{\prime }){\delta }_{ij}$ with $p^{\prime }=p-p_0$. The displayed source identity is unchanged by this consistent pressure-gauge subtraction.

Assume no incoming sound and no free initial acoustic field. Put $R_y=|\mathbf{x}-\mathbf{y}|$, ${\tau }_y=t-R_y/c_0$, $Q={\rho }_0{u}_n{\delta }_s$, and $P_i=p^{\prime }{n}_{s,i}{\delta }_s$. Convolution with the retarded acoustic Green function gives the three source integrals, for an observer outside the body:These are the surface thickness, surface loading and volume quadrupole contributions. The retarded surface Dirac delta distributions must remain inside the integrals: integrating them over a moving retarded surface produces the moving-surface retarded Jacobian, and is not generally the same as simply substituting one retarded time into an instantaneous surface integral.

For an acoustic compact-source approximation, suppose the dominant sources have size $\ell$, source frequency scale $\omega$, and $\omega \ell /c_0\ll 1$. Also require $r=|\mathbf{x}|\gg \ell$ and the acoustic far field $\omega r/c_0\gg 1$. Then $R_y\simeq r-\mathbf{n}\cdot \mathbf{y}$, $1/R_y\simeq 1/r$, and the leading source integrals use the common time $\tau =t-r/c_0$. Slow source motion relative to $c_0$ is implicit in this leading retardation approximation. DefineThus $m$ is net displaced mass per unit time, $\mathbf{f}$ is the fluid pressure force on the body, and $S_{ij}$ is the integrated excess Lighthill stress tensor. The minus sign in $\mathbf{f}$ fixes the requested loading convention; the force on the fluid is opposite. On the radiating $1/r$ terms, ${\partial }_{x_i}\simeq -(n_i/c_0){\partial }_t$, givingSpatial derivatives of $1/r$ produce near-field terms of orders $r^{-2}$ and $r^{-3}$, omitted here. Neglecting source delays is a separate approximation from discarding those observer near-field terms. If a leading integrated source vanishes by symmetry, its first nonzero delay correction can be as important as another retained multipole.

For the translating sphere, write its center as $\mathbf{X}(t)=a_1\cos (\omega t)\mathbf{e}$, with $\mathbf{e}$ a unit vector, and its velocity as $\mathbf{V}=\dot{\mathbf{X}}$. Let $\mathcal{V}=4\pi a^3/3$, $\eta =a\omega /c_0\ll 1$ and $d=a_1/a\ll 1$. Impermeability gives $u_n=\mathbf{V}\cdot {\mathbf{n}}_s$. Since its volume is constant,in the leading compact formula. The individual local thickness sources have scale ${\rho }_0(a_1/r){\eta }^2$, but their net monopole cancels.

At first order in amplitude and leading order in $\eta$, the exterior incompressible velocity potential is $\varphi =-a^3\mathbf{V}\cdot {\mathbf{r}}_s/(2r_s^3)$, where ${\mathbf{r}}_s$ is measured from the mean center. It has ${\partial }_{r_s}\varphi =\mathbf{V}\cdot {\mathbf{n}}_s$ at $r_s=a$. Linear Bernoulli equation gives $p^{\prime }=-{\rho }_0{\varphi }_t=({\rho }_{0}a/2)\dot{\mathbf{V}}\cdot {\mathbf{n}}_s$ there. Since $\int n_{s,i}{n}_{s,j}dS=(4\pi a^2/3){\delta }_{ij}$,This is the added mass of a sphere. The explicit leading loading contribution isIt scales as ${\rho }_0(a_1/r){\eta }^3$. In homentropic small-amplitude flow the volume stress is quadratic to leading order, $T_{ij}=O({\rho }_0{V}^2)$ over volume $O(a^3)$, so the volume acoustic quadrupole has scaleIt is smaller than the linear dipole by $O(d\eta )=O(V/c_0)$ and can contain a second harmonic.

For completeness, the exact thickness-source integral has a nonzero first delay correction even though $m=0$. Its first spatial moment isExpand its integrand at ${\tau }_y=\tau +\mathbf{n}\cdot \mathbf{y}/c_0$. The leading nonzero translating-sphere acoustic thickness dipole is thereforeIt too scales as ${\rho }_0(a_1/r){\eta }^3$. Thus zero net-volume monopole does not mean zero thickness radiation. The full leading linear sound, summing thickness and loading, isThe leading compact equation gives the requested zero $\dot{m}$ and added-mass $\dot{f}$ terms; this extra delay calculation is necessary to obtain the complete linear field when its nominal monopole has cancelled.

Let the fluid approach a uniform, quiescent state $({\rho }_0,p_0,c_0)$ at infinity. Use the outward unit normal vector $N=\nabla f/|\nabla f|$, pointing from the object into the fluid. Write $H=H(f)$ for the Heaviside step function, $q=\rho -{\rho }_0$, and ${\delta }_S=\delta (f)|\nabla f|$. With the viscous stress tensor ${\tau }_{ij}$, defineThus $T_{ij}$ is the Lighthill stress tensor, and $L_i$ is the surface loading exerted on the fluid. The total force on the object has the opposite sign. Extend $q$ by zero inside the object. Since the boundary is fixed and impermeable, $u\cdot N=0$, and the distributional mass conservation and momentum conservation equations areFor example, the surface contribution to the momentum equation is $(\rho u_i{u}_j+P_{ij})N_j{\delta }_S=L_i{\delta }_S$. Differentiating the first equation in time and subtracting the divergence of the second gives the fixed-body Ffowcs Williams-Hawkings equation:There is no acoustic thickness noise: a fixed impermeable boundary injects no mass. Apply the retarded acoustic Green function $\delta (t-R/c_0)/(4\pi c_0^{2}R)$, where $R=|x-y|$. The supplied surface-delta identity converts the loading term to a surface integral. For an exterior observer, the required integral equation isThis acoustic analogy is an exact rearrangement of the fluid equations; source approximations enter later. In the distant linear acoustic region, $p^{\prime }=c_0^{2}q$.

The acoustic far field requires observer distance $r$ much larger than the source extent $\ell$ and $k_{0}r\gg 1$, so derivatives of the retarded phase dominate derivatives of $1/R$. The acoustic compact-source approximation requires $k_0\ell \ll 1$, so the source's differential propagation delays can be neglected. These conditions concern different lengths and neither implies the other. Put $n=x/r$, $t_r=t-r/c_0$, and define the loading and stress momentsThe leading compact acoustic far field isThe surface term is an acoustic dipole; the volume term is an acoustic quadrupole. The plus sign of the loading term follows from the minus sign of its spatial divergence and the derivative of retarded time.

For the power comparison, assume low Mach number $M=U/c_0$, high Reynolds number, eddy size $\ell$, characteristic time $\tau =\ell /U$, and stress magnitude ${\rho }_0{U}^2$. Keep the prescribed turbulent region and its leading volume stress comparable when inserting the object; neglect changes of source statistics and exceptional cancellations of the leading moments. Compactness follows from $\ell /(c_0\tau )=M\ll 1$. Since $Q\sim {\rho }_0{U}^2{\ell }^3$ and the outgoing acoustic intensity is $⟨p^{\prime 2}⟩/({\rho }_0{c}_0)$, the volume power isIf the object's size is comparable to $\ell$ and its fluctuating resultant loading is $F\sim {\rho }_0{U}^2{\ell }^2$, thenAngular factors and dimensionless loading coefficients have been omitted. The acoustic dipole and acoustic quadrupole interference vanishes in the leading power integrated over a full sphere, by odd angular parity; individual directions can show interference.

A size threshold also requires a loading model. For a small nonseparating object of size $a\ll \ell$ in an approximately inviscid, smoothly varying incident eddy, a spatially uniform pressure has zero resultant on its closed surface. The first surviving loading is the pressure-gradient or added mass force, $F\sim {\rho }_0{a}^{3}U/\tau ={\rho }_0{U}^2{a}^3/\ell$. Under these explicitly stated assumptions,The two mechanisms become comparable at $a\sim \ell M^{1/3}$; for smaller objects the cases with and without the object have the same leading order of power. This is the small-body loading-noise threshold. It is not a geometry-independent size law: a separated drag force $F\sim {\rho }_0{U}^2{a}^2$ fluctuating on the original eddy time instead gives ${\mathcal{P}}_D/{\mathcal{P}}_Q\sim (a/\ell {)}^4/M^2$ and $a\sim \ell \sqrt{M}$. If both force and time are set by body-scale eddies, $\tau =a/U$, their loading power is ${\rho }_0{U}^6{a}^2/c_0^3$ and comparison with the original volume source gives $a\sim M\ell$. Each estimate states which source time and force it uses.

The physical distinction is momentum conservation. Unforced turbulence in a fluid supplies internal stresses, whose net-force contribution cancels and leaves acoustic quadrupole radiation. The supported object exchanges momentum with the fluid and the external support, permitting a time-dependent resultant force and acoustic dipole radiation. One fewer retarded derivative makes the loading radiation more efficient at low Mach number.

Put ${\rho }^{\prime }=\rho -{\rho }_0$, and take the retarded acoustic Green function for ${\partial }_t^2-c_0^2{\nabla }^2$:This selects the causal, outgoing acoustic density perturbation, with no additional incoming homogeneous wave equation solution. Convolving the acoustic dipole forcing with this Green function and moving its spatial derivative outside the integral givesThe surface delta distribution converts the spatial integral to the moving surface. At fixed surface labels $(p,q)$, setThe radial Mach number enters the moving-surface retarded Jacobian, because $dr/d\tau =-\hat{r}\cdot v$ and henceLet ${\tau }^{\ast }$ be a root of the retarded time equationThe delta change-of-variable rule now givesHere $h_p{h}_q$ is the surface-area factor from the orthogonal surface coordinates. For a subsonic surface, $|v|<c_0$, the retarded equation is monotone in $\tau$ and has one root when the motion is defined for the required past times. For more general motion, sum the displayed contribution over all simple retarded roots. A root with $1-M_r=0$ requires a separate limiting treatment; the simple-root formula does not apply there.

Let $r=|\mathbf{x}|$, $n_i=x_i/r$, and let the source have size $\ell$. The retarded acoustic Green function gives the outgoing solution of Lighthill acoustic analogy asThis follows by integrating the source derivatives by parts in the retarded convolution. Assume the source is localized and the boundary terms vanish. In the acoustic compact-source approximation, $k_0\ell \ll 1$, the retardation across the source can be neglected, while $r\gg \ell$ permits replacement of the denominator by $r$. The integral becomes $S_{ij}(t-r/c_0)/r$.

In the radiation region $k_{0}r\gg 1$, derivatives of the retarded argument dominate derivatives of the spreading factor. Since ${\partial }_{x_i}(t-r/c_0)=-n_i/c_0$, the leading acoustic quadrupole field isThe two negative retardation derivatives give a positive sign. This is a far-field approximation, with smaller near-field terms omitted.

For low-Mach number aerodynamic fluctuations of speed $U$ and advective time $\ell /U$, take $T_{ij}=O({\rho }_0{U}^2)$, $S_{ij}=O({\rho }_0{U}^2{\ell }^3)$, and two time derivatives of order $(U/\ell {)}^2$. ThereforeThe quoted fourth power is the compact acoustic quadrupole Mach-number scaling, with geometric spreading shown explicitly. It assumes the source strength and time scale just stated; the source must be acoustically compact, and the observation point must remain in the radiation region.