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Acoustic directivity describes the angular dependence of radiation in the acoustic far field. For a time-harmonic source in $d$ spatial dimensions, separating the geometric spreading and oscillatory phase leaves a complex amplitude pattern $\mathcal{D}$. The angular intensity pattern is proportional to $|\mathcal{D}{|}^2$. Directivity can be normalized by a reference direction or total radiated power; a stated convention is needed when comparing absolute directivity factors.

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 ${\partial }_t^2-c_0^2\Delta$.

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 $V^2{\Omega }^2{\ell }^3/c_0^{4}r$.

Fluid-loaded plate. Use complex amplitudes with time dependence $e^{-i\omega t}$ and take the velocity potential in the fluid above the elastic plate to be $\Phi$. Let ${\rho }_0$ and $c_0$ be the undisturbed mass density and speed of sound. The linearized Euler equations give $p=i{\rho }_0\omega \Phi$; the kinematic boundary condition is ${\Phi }_y(x,0)=-i\omega \eta (x)$. We model loading from one fluid half-space, with no fluid loading on the other face. Loading by identical fluids on both faces instead doubles the fluid term below.

Define the Fourier transform and its inverse byThe Sommerfeld radiation condition selects $\hat{\Phi }(k,y)=\hat{\Phi }(k,0)e^{-\gamma y}$, with ${\gamma }^2=k^2-k_0^2$ and $k_0=\omega /c_0$. Here and in the diffraction calculation below, the outgoing square root is specified by first taking $\mathrm{Im}\omega >0$. Draw a vertical cut upwards from $+k_0$ and another downwards from $-k_0$, and putThus ${\gamma }_u$ is analytic in the upper half-plane and ${\gamma }_l$ in the lower half-plane. In the real-frequency limit,The Fourier transform contour $C$ follows the real axis with the limiting-absorption pole prescriptions. In particular, an outgoing propagating component has positive vertical wavenumber.

The kinematic boundary condition gives $\hat{\Phi }(0)=i\omega \hat{\eta }/\gamma$, so $\hat{p}(0)=-{\rho }_0{\omega }^2\hat{\eta }/\gamma$. Eliminating the fluid pressure from the bending balance gives the line-forced fluid-loaded bending plate kernelFor harmonic line force amplitude $F_0$ at the origin, the plate displacement and fluid velocity potential are thereforeAn arbitrary localized loading replaces $F_0$ by its Fourier transform. This representation both calculates the response and identifies its different wave components through poles and branch points.

For an observer $x=r\cos \theta$, $y=r\sin \theta$, with $0<\theta <\pi$, the method of steepest descent uses the saddle point $k_s=k_0\cos \theta$. At fixed angle away from grazing, the acoustic far field isThe $r^{-1/2}$ spreading is that of cylindrical acoustic waves. The phase velocity parallel to the plate is $\omega /|k|$: components with $|k|<k_0$ are supersonic in that sense and radiate, whereas those with $|k|>k_0$ are subsonic and exponentially decay normally to the plate. The latter have reactive fluid loading and zero mean normal acoustic energy flux.

There is also a real guided root $\kappa >k_0$. Indeed, $K(k_0+)=-\infty$, $K(+\infty )=+\infty$, andThe residue theorem then gives an outgoing displacement contribution $i{F}_0{e}^{i\kappa |x|}/K^{\prime }(\kappa )$. It transports energy along the plate while its fluid field decays as $e^{-\gamma (\kappa )y}$. Other zeros on continued sheets can describe leaky or resonant waves; branch point contributions describe near and grazing fields. A fixed-angle acoustic far field expansion is not uniform near the plate, where these components can matter.

For a propagating spectral component, write $\ell =(k_0^2-k^2{)}^{1/2}$. Its upward normal velocity is $i\ell \hat{\Phi }$ and its pressure is $i{\rho }_0\omega \hat{\Phi }$. Applying Parseval's identity to the time-averaged acoustic energy flux gives the upward power per unit length of the forcing line:Equivalently, the integrand is $|\hat{\eta }{|}^2/\ell$ with prefactor ${\rho }_0{\omega }^3/(4\pi )$. Integrating the radial acoustic intensity of the far field over a semicircle gives the same expression. The input power $\omega \mathrm{Im}(F_0^{\ast }\eta (0))/2$ also feeds guided waves along the plate, so it must not be identified with $P_y$ alone.

Rigid half-plane diffraction. Suppress the time factor and assume $0<{\theta }_0<\pi$. Write $a=k_0\cos {\theta }_0$, $b=k_0\sin {\theta }_0$, and denote the incident spatial field by $I=e^{-iax-iby}$. The Helmholtz equation, Sommerfeld radiation condition, and square-root branches are as above. Across the open positive axis the scattered field and its normal derivative are continuous. Across the rigid negative axis its normal derivative is also continuous, while its potential can jump. Let $J(x)$ be this jump, supported on $x<0$. The transformed scattered field isThe jump transform is analytic in the upper half-plane. The absence of an even scattered component follows from derivative continuity and the outgoing condition. Its normal derivative on either face is $-\gamma {\hat{J}}_u/2$.

On the negative axis this derivative must cancel $-ib{e}^{-iax}$. Since the Half-range Fourier transform of $H(-x)e^{-iax}$ is $i/(k+a)$, the Wiener-Hopf equation has the formwhere $G_l$ is the transform of an unknown function supported on $x>0$ and analytic in the lower half-plane. The incident pole $k=-a$ lies below the Fourier contour. Divide by ${\gamma }_l$ and perform pole subtraction in a Wiener-Hopf equation:The first term is upper analytic and the second lower analytic, since the latter's apparent pole cancels. Equality of the two continued sides gives an entire function. The finite-energy edge condition, with $J(x)=O(\sqrt{-x})$, and the decay of the transforms at infinity force that function to vanish by Liouville's theorem. ConsequentlyThis is the Wiener-Hopf solution of rigid half-plane diffraction with the present transform and time conventions.

Deforming to a steepest descent contour gives a residue from the incident pole precisely when $x+|y|\cot {\theta }_0<0$. At that pole $\gamma (-a)=-ib$; its residue producesThus the total geometrical optics field is $I$ plus a reflected wave in the upper reflection region; in the lower shadow region the scattered residue cancels $I$, and outside that shadow the direct incident wave remains. This gives both the amplitude and the spatial support of the geometrical optics fields rather than merely naming a reflected wave.

For $x=r\cos \theta$, $y=r\sin \theta$, $-\pi <\theta <\pi$, the saddle contribution uses ${\gamma }_l(-a)=-i\sqrt{2k_0}\cos ({\theta }_0/2)$ and ${\gamma }_u(k_s)=\sqrt{2k_0}\cos (\theta /2)$. The diffracted wave isThe total field is $I+{\varphi }_{\mathrm{GO}}+{\varphi }_d$ to this order. This far-field expression applies away from the shadow/reflection boundaries $|\theta |=\pi -{\theta }_0$. There the pole and saddle coalesce, and a uniform Fresnel integral approximation replaces the divergent separate terms by a smooth transition. The exact contour integral remains well defined.

Interpret the surface delta as ${\delta }_s=|\nabla F|\delta (F)$, so integration against it gives a surface integral. If $F$ is signed distance, this equals the notation $\delta (F)$ directly. This convention is needed because an arbitrary rescaling of the level-set function must not change the physical source.

Define the volume flux $\mathcal{V}(t)={\int }_S{V}_{n}dS$ and total pressure force $f(t)={\int }_{S}pndS$. Here $\mathcal{V}$ is the rate of change of body volume, not the volume itself. The surface acoustic monopole contributes, in the acoustic compact-source approximation,The surface acoustic dipole contributesThe second term is a nonradiating near-field term and is smaller in the acoustic far field. Writing $\mathcal{F}(t)=x\cdot f(t)$ with the observation point fixed, the leading augmentation is thereforeThe printed dipole coefficient is missing a factor $1/c_0$. The extra inverse speed follows from differentiating $\tau =t-r/c_0$, and is required dimensionally: the printed expression with $c_0^2$ has the units of density times speed. This corrected acoustic loading noise term also agrees with the far-field acoustic force and stress moments. The monopole coefficient is unchanged.

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.

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.

In the Ffowcs Williams-Hawkings equation, the other source types are a surface acoustic monopole associated with acoustic thickness noise, and a volume acoustic quadrupole involving the Lighthill stress tensor. On an impermeable material surface, fluid and surface normal velocities agree. There is no through-surface mass-flux source; the remaining thickness source is $Q={\rho }_0{v}_n$. For a rigid body, the leading acoustic compact-source approximation to that source hasThus there is no leading net-volume acoustic monopole. To neglect thickness radiation beyond that leading cancellation, assume negligible volume displacement, as for ideal thin blades, or that its higher multipoles are small compared with the retained acoustic loading noise. Rigidity alone does not make a moving finite-volume body's local thickness source identically zero.

The volume acoustic quadrupole may be neglected for low Mach number motion when exterior turbulent or nonlinear stresses do not provide a competing strong source. We also assume small linear acoustics perturbations, a uniform reference sound speed, and negligible relevant viscous and entropy sources. These are source-strength approximations, particularly important if a loading contribution itself cancels by symmetry. Under them, the retained acoustic dipole is the force exerted by the object on the fluid, with the sign used in the previous solution.

Let $R=|x|$, $n=x/R$, and ${\tau }_0=t-R/c_0$. In the acoustic far field, $R$ is large compared with the object and $\omega R/c_0\gg 1$. For a source of size $\ell$ with $\omega \ell /c_0\ll 1$ and small surface Mach number, source-dependent delays and the Doppler factor can be neglected to leading order. The surface integral then contains just the total force $\mathcal{F}(\tau )={\int }_{S}FdS$. Differentiating its retarded time, rather than its $R^{-1}$ spreading factor, gives the radiating termThe sign follows from ${\partial }_{x_i}{\tau }_0=-n_i/c_0$. Differentiating $R^{-1}$ or the direction $n$ instead produces the lower-order near field. A constant total force does not radiate at this leading compact order.

The rotation introduces angular frequency $\Omega$ and, after combining the two blades, harmonics such as $2\Omega$. The acoustic compact-source approximation requires the propagation time $a/c_0$ to be small compared with ${\Omega }^{-1}$. ThusIt also bounds every blade element's Mach number by $\mu$. Consequently $|1-M_r|=1+O(\mu )$, and its leading value is one. The separate acoustic far field condition is $\Omega R/c_0\gg 1$.

Choose the positive rotation sense so that a first blade at phase $\alpha$ has radial and tangential unit vectorsIntegrating the given line force from $r=0$ to $a$ yields ${\mathcal{F}}_1=F{e}_{\varphi }(\Omega \tau )+D{e}_x$. Its axial component is constant, while ${\dot{\mathcal{F}}}_1=-F\Omega e_r$. Since $n=\cos \theta e_x+\sin \theta e_y$, the compact acoustic dipole sound from this blade isThe other blade has phase $\alpha +\pi$ and contributes the opposite rotating force. At a common compact retarded time, their total force is $2D{e}_x$, soThis cancellation calls for the next source-delay correction; it does not mean that the complete moving-source field vanishes.