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An acoustic density perturbation is the deviation of fluid mass density from its quiescent reference value. In a homogeneous isentropic linear acoustic field, its corresponding pressure perturbation is $p^{\prime }=c_0^2{\rho }^{\prime }$. Nonlinear and entropy-dependent departures from this relation enter the Lighthill stress tensor.

In the incompressible near field of a spherical bubble, $u_r=a^2\dot{a}/s^2$. Its volume-flux acoustic monopole is the displayed term at retarded time. The integrated surface pressure force vanishes by spherical symmetry, while the Lighthill stress tensor approximation ${\rho }_0{u}_i{u}_j$ has moment $S_{ij}=4\pi {\rho }_0{a}^3{\dot{a}}^2{\delta }_{ij}/3$. Its leading compact acoustic quadrupole contribution is ${\rho }_0(a^3{\dot{a}}^2{)}^{\prime \prime }/(3c_0^{4}r)$. For small fractional radius oscillation $\delta$ and compactness parameter $\alpha =\omega a_0/c_0$, its size relative to the leading monopole is $O(\delta {\alpha }^2)$.

Mass and momentum flux defects at a moving interface add surface acoustic monopole and acoustic dipole terms to Lighthill acoustic analogy. For a physical conservative fluid shock wave, the Rankine-Hugoniot conditions set both defects to zero; the shock still affects distributional derivatives of the Lighthill stress tensor. A discontinuity alone is not independent mass or force injection.

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

Use mass conservation and conservative momentum conservation, with viscous stress ${\sigma }_{ij}$ defined so that its divergence is the viscous force:Differentiate the first equation in time and eliminate the momentum time derivative using the second. This givesSet ${\rho }^{\prime }=\rho -{\rho }_0$ and $p^{\prime }=p-p_0$, where the reference values and $c_0$ are constant. Subtract $c_0^2{\nabla }^2{\rho }^{\prime }$ to obtain the exact Lighthill acoustic analogyThe Lighthill stress tensor includes nonlinear momentum transport, departure from the reference linear pressure-density relation, and viscous stress. For an inviscid fluid the last term is absent. The derivation is an exact rearrangement of the conservation equations; treating the tensor as a known localized source is a subsequent approximation.

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.

Use the jump convention $[f]=f_1-f_2$, consistent with region 1 occupying $x>Vt$. The one-dimensional Lighthill stress tensor is $T=\rho u^2+p-c_0^2\rho$, so $T=T_2+[T]H(x-Vt)$. Both bulk stresses are constant. The distributional jump formula for a Heaviside product therefore givesThe Rankine-Hugoniot condition for conservation of mass is $[\rho u]=V[\rho ]$, while conservation of momentum gives $[\rho u^2+p]=V[\rho u]$. Hence a physical moving shock wave also satisfiesThis sign can be checked directly. Subtracting a constant reference density does not affect derivatives, and ${\rho }^{\prime }=\text{constant}+[\rho ]H(x-Vt)$. Thus ${\rho }_{tt}^{\prime }=V^2[\rho ]{\delta }^{\prime }(x-Vt)$ and ${\rho }_{xx}^{\prime }=[\rho ]{\delta }^{\prime }(x-Vt)$, exactly reproducing the coefficient above.

Let $D_{ij}=T_{ij}^{+}-T_{ij}^{-}$, so the Lighthill stress tensor is $T_{ij}^{-}+D_{ij}H(S)$. Write $S_i={\partial }_{i}S$ and $S_{ij}={\partial }_i{\partial }_{j}S$. The distributional derivative of the Heaviside step function gives ${\partial }_{i}H(S)=S_i\delta (S)$. Applying the product rule a second time yields the surface sources of a discontinuous Lighthill stress tensor:Repeated indices are summed. The first line gives the two bulk acoustic quadrupole distributions. The second line consists of additional surface delta distributions and their derivatives, all supported on the shock wave $S=0$. For a symmetric Lighthill stress tensor, the first two coefficients of $\delta (S)$ combine to $2D_{ij,i}{S}_j$ after relabelling indices.

A discontinuity therefore adds both a surface-delta source and a surface-delta-derivative source. The level function need not be a signed distance: retaining its derivatives makes the expression invariant under a smooth reparametrization of the same surface. If the two bulk fields are only continuously differentiable, their second derivatives in the first line are read weakly; no classical second derivative is being assumed. The coefficients multiplying ${\delta }^{\prime }(S)$ are kept as extensions, not replaced prematurely by traces.

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.

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.

Apply the moving-interface conservation jump identity first to mass, then to each momentum component. Write $j_i=\rho u_i$ and ${\Pi }_{ij}=\rho u_i{u}_j+p{\delta }_{ij}-{\sigma }_{ij}$, with all quantities understood piecewise on the two sides. Define the jumps of flux relative to the moving shock wave byThe global distributional conservation equations areDifferentiate the first in time and subtract the divergence of the second. With ${\rho }^{\prime }=\rho -{\rho }_0$ and the piecewise Lighthill stress tensor $T_{ij}=\rho u_i{u}_j+(p^{\prime }-c_0^2{\rho }^{\prime }){\delta }_{ij}-{\sigma }_{ij}$, this givesDerivatives act on the complete distributions, including their moving support. The acoustic quadrupole term represents momentum-stress fluctuations throughout the volume. The time derivative of the surface mass-flux defect is an acoustic monopole, representing injection or removal of mass/volume. The divergence of the surface momentum-flux defect is an acoustic dipole, representing a force sheet. This is the distributional acoustic analogy across a moving interface.

For an actual freely propagating fluid shock wave with no singular mass or momentum supply, the Rankine-Hugoniot conditions give $Q=0$ and $L_i=0$. Such a shock does not acquire independent monopole and force-sheet sources merely because it is discontinuous. Its effects remain in the distributional derivatives of $T_{ij}$, including singular derivatives of its jump. Nonzero surface sources are appropriate for an interface with exchange/forcing or for a formulation that omits one side of the fluid.

A shock does not, by itself, justify retaining the $m^4$ scaling. That estimate required a low-Mach number stress $O({\rho }_0{U}^2)$ varying on the slow time $\ell /U$. Fast shock motion, short time scales, or thermodynamic deviations can invalidate that estimate. If these same compact, slow-source assumptions remain valid for the integrated Lighthill stress tensor, its quadrupole estimate still follows, even distributionally. There is no universal replacement power deducible from the mere presence of a shock; nor should vanished physical flux defects be treated as additional independent radiation sources.

Write a piecewise smooth Lighthill stress tensor as $T_{ij}=T_{ij}^{-}+D_{ij}H(S)$, where $D_{ij}=T_{ij}^{+}-T_{ij}^{-}$ and the regular level surface $S=0$ is a shock wave. The distributional derivative of the Heaviside step function givesHere repeated indices are summed, $S_i={\partial }_{i}S$ and $S_{ij}={\partial }_i{\partial }_{j}S$. Keep the coefficients as smooth extensions before multiplying distributions: prematurely replacing the coefficient of ${\delta }^{\prime }(S)$ by its surface trace can lose a $\delta (S)$ contribution. The extra surface delta distributions and their derivatives are supported entirely on the discontinuity.