= Solution
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 $Q=\rho_0v_n$. For a <rigid body>, the leading <acoustic compact-source approximation> to that source has
$$
\int_SQ\,dS=\rho_0\int_Sv_n\,dS=\rho_0\frac{d\mathcal V}{dt}=0.
$$
Thus 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 $\omega R/c_0\gg1$. For a source of size $\ell$ with $\omega\ell/c_0\ll1$ 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> $\mathcal F(\tau)=\int_S F\,dS$. Differentiating its <retarded time>, rather than its $R^{-1}$ spreading factor, gives the radiating term
$$
\boxed{\rho'(x,t)\sim\frac{n\cdot\dot{\mathcal F}(\tau_0)}{4\pi c_0^3R},\qquad
p'(x,t)=c_0^2\rho'\sim\frac{n\cdot\dot{\mathcal F}(\tau_0)}{4\pi c_0R}.}
$$
The 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.
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