 Solution

Let the observer have cylindrical coordinates $(r,\theta ,z)$ and total distance $R=\sqrt{r^2+z^2}$. At retarded emission time $\tau$, the source-observer distance isThe retarded time relation $t=\tau +|\mathbf{x}-\mathbf{y}(\tau )|/c_0$ consequently becomesSet $\Omega \tau =\theta -\pi /2+\Theta$. ThenThe observer-direction Mach number at emission is $M_r=\dot{\mathbf{y}}(\tau )\cdot \mathbf{n}/c_0=m\cos \Theta$. For a constant strength $q$ and $m<1$, the single retarded root has $1-M_r>0$. Substituting into the moving-source field gives the rotating acoustic point source resultIn particular, on the rotation axis $m=0$ and the result is $q/(4\pi c_0^{2}R)$.

For a genuinely varying source strength, replace $q$ by $q(\tau )$ in the last formula. If $m>1$, the retarded relation need not be one-to-one: each retarded root contributes with $|1-M_r|$ in the denominator. The general result is a sum of $q({\tau }_j)|{\dot{\Theta }}_j|/(4\pi c_0^2\Omega R)$. Dropping the absolute value and the root sum requires the subsonic assumption. At $1-M_r=0$ the ideal point-source expression has a caustic; finite source size and bandwidth regularize that idealization.