= Solution
Write $\alpha=\Omega\tau_0^*$ and $s=\sin\theta$, where $\tau_0^*=t-R/c_0$. Label the two arms by $\sigma=\pm1$. Their positions are $y_\sigma=\sigma r e_r(\Omega\tau_\sigma^*)$. To radiating far-field accuracy,
$$
|x-y_\sigma|=R-\sigma rs\cos(\Omega\tau_\sigma^*)+O(a^2/R).
$$
The <retarded time> equation therefore gives, with $\mu_r=\Omega r/c_0$,
$$
\Omega\tau_\sigma^*=\alpha+\sigma\mu_rs\cos(\Omega\tau_\sigma^*)+O(\Omega a^2/(c_0R))
=\alpha+\sigma\mu_rs\cos\alpha+O(\mu^2,\Omega a^2/(c_0R)).
$$
For the first arm, $\sigma=1$, this is the requested phase expansion. The more explicit geometric error also makes the dimensional meaning of the printed $O(1/R)$ term clear.
Apply the <Taylor theorem> to the rotating line <force>. Since $de_\phi/d\alpha=-e_r$, its first two orders, expressed in a common reference frame, are
$$
F_\sigma(r,\tau_\sigma^*)=\frac{2r}{a^2}
\left[F\sigma e_\phi(\alpha)+D e_x-F\mu_rs\cos\alpha\,e_r(\alpha)\right]+O\left(\frac{r}{a^2}|F|\mu^2\right).
$$
Also $v_\sigma=\sigma\Omega r e_\phi(\Omega\tau_\sigma^*)$, so the <radial Mach number> is
$$
M_{r,\sigma}=-\sigma\mu_rs\sin\alpha+O(\mu^2),\qquad
\boxed{|1-M_{r,\sigma}|^{-1}=1-\sigma\mu_rs\sin\alpha+O(\mu^2).}
$$
The absolute value causes no change of sign in this subsonic limit. Multiplying these two expansions before summing is essential: both the shifted <force> and the <moving-surface retarded Jacobian> contribute at the same order.
Denote the integrated numerator, including that Jacobian, by $\mathcal B$. Pairing the two blades cancels all terms odd in $\sigma$, including the first Doppler correction to the axial load. Hence
$$
\begin{aligned}
\mathcal B(\alpha,n)
&=\sum_{\sigma=\pm1}\int_0^a\frac{F_\sigma(r,\tau_\sigma^*)}{1-M_{r,\sigma}}\,dr\\
&=2D e_x-\frac{4F\Omega a}{3c_0}s\left(\cos\alpha\,e_r(\alpha)+\sin\alpha\,e_\phi(\alpha)\right)+O\big((|F|+|D|)\mu^2\big)\\
&=2D e_x-\frac{4F\Omega a}{3c_0}s\left(\cos2\alpha\,e_y+\sin2\alpha\,e_z\right)+O\big((|F|+|D|)\mu^2\big).
\end{aligned}
$$
The factor $a/3$ comes from $a^{-2}\int_0^a r^2dr$. Only the time-dependent term radiates at order $R^{-1}$. Applying $-\partial_{x_i}$ to the integral now gives $\rho'_{\rm rad}=(4\pi c_0^3R)^{-1}n\cdot\partial_t\mathcal B$, and thus
$$
\boxed{\rho'_{\rm rad}=\frac{2F\Omega^2a}{3\pi c_0^4R}\sin^2\theta\,
\sin\big(2\Omega(t-R/c_0)\big)+O\left(\frac{(|F|+|D|)\Omega\mu^2}{c_0^3R}\right).}
$$
There are also nonradiating terms of order $R^{-2}$. Multiply by $c_0^2$ for the acoustic <pressure>. The stated coefficient uses the rotation convention fixed above; reversing the rotation reverses the corresponding signed load and phase convention.
This is a <compact rotating two-blade loading source> acting as an <acoustic quadrupole>. The compact total rotating <force> cancels, leaving the first spatial moment of the loading. Its two factors of the observer's projection into the rotor plane produce $\sin^2\theta$: there is no leading sound on the rotation axis and the density amplitude is maximal in the rotor plane. The configuration repeats after half a rotation, explaining frequency $2\Omega$. These statements concern amplitude; the corresponding <acoustic intensity> has a $\sin^4\theta$ factor.
If $F=0$, the displayed first-order contribution vanishes as well. For completeness, expanding the axial Jacobian to its next even order gives
$$
(1-M_{r,\sigma})^{-1}=1-\sigma\mu_rs\sin\alpha-\mu_r^2s^2\cos2\alpha+O(\mu_r^3).
$$
The first remaining axial-load radiation is then
$$
\rho'_{D,\rm rad}=\frac{D\Omega^3a^2}{2\pi c_0^5R}\cos\theta\sin^2\theta\sin2\alpha+\text{higher orders}.
$$
Thus the constant axial total <force> does not produce the lower-order term, even though its moving spatial distribution can radiate at a higher order.
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