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

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.

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.