Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 82 / 1 / a / i

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.