= Solution
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> as
$$
\rho'(\mathbf x,t)=\frac1{4\pi c_0^2}\partial_{x_i}\partial_{x_j}\int\frac{T_{ij}(\mathbf y,t-|\mathbf x-\mathbf y|/c_0)}{|\mathbf x-\mathbf y|}\,d^3y.
$$
This 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>, $k_0\ell\ll1$, 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 $k_0r\gg1$, 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 is
$$
\boxed{\rho'(\mathbf x,t)\sim\frac{n_in_j\ddot S_{ij}(t-r/c_0)}{4\pi c_0^4r}=\frac{x_ix_j\ddot S_{ij}(t-r/c_0)}{4\pi c_0^4r^3}.}
$$
The 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 $T_{ij}=O(\rho_0U^2)$, $S_{ij}=O(\rho_0U^2\ell^3)$, and two time derivatives of order $(U/\ell)^2$. Therefore
$$
\boxed{\frac{\rho'}{\rho_0}=O\left[\left(\frac U{c_0}\right)^4\frac\ell r\right]=O(m^4\ell/r).}
$$
The 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.
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