= Solution
Retain the consistent $\Theta_{\rm phys}$ of <1/d>. Across a sufficiently thin layer, the neutral hydrogen <column density> is $N_{\rm HI}=n_H(1-\zeta_{\rm in})\Delta r+O(\Delta r^2)$. The direct stellar continuum acquires <optical depth> $\tau_\nu=N_{\rm HI}\sigma_\nu$. Define the photon-weighted second moment
$$
\langle\sigma^2\rangle=\frac1Q\int_0^\infty\sigma_\nu^2\frac{L_\nu^*}{h\nu}\,d\nu.
$$
Expanding the attenuating exponential in the <photoionization rate> gives
$$
\begin{aligned}
\frac{q(r_{\rm in}+\Delta r)}{q_{\rm in}}
&=\left(\frac{r_{\rm in}}{r_{\rm in}+\Delta r}\right)^2
\frac{\int\sigma_\nu L_\nu^*e^{-N_{\rm HI}\sigma_\nu}/(h\nu)\,d\nu}{Q\langle\sigma\rangle}\\
&=1-\frac{2\Delta r}{r_{\rm in}}-n_H(1-\zeta_{\rm in})\Delta r
\frac{\langle\sigma^2\rangle}{\langle\sigma\rangle}+O(\Delta r^2).
\end{aligned}
$$
The second <cross section> moment appears because the radiation both undergoes absorption and is weighted by the <photoionization> <cross section> in computing the rate.
In a local plane-parallel approximation that suppresses geometrical dilution, set
$$
\Gamma=1-n_H(1-\zeta_{\rm in})\Delta r\frac{\langle\sigma^2\rangle}{\langle\sigma\rangle}.
$$
Then local <photoionization equilibrium> becomes $\Theta_{\rm phys}\zeta^2=\Gamma(1-\zeta)$, so
$$
\boxed{\zeta=\frac{-\Gamma+\sqrt{\Gamma^2+4\Theta_{\rm phys}\Gamma}}{2\Theta_{\rm phys}}.}
$$
This gives the requested algebraic dependence, with the corrected definition of $\Theta$. Besides $\Delta r\ll r_{\rm in}$, linear attenuation requires $N_{\rm HI}\sigma_\nu\ll1$ over the <frequencies> contributing appreciably to the rate. In particular $\Gamma$ must remain positive.
The source's $\Gamma$ neglects $-2\Delta r/r_{\rm in}$, which is also first order in layer thickness. A consistent spherical first-order answer uses $\Gamma_{\rm sph}=\Gamma-2\Delta r/r_{\rm in}$ in the same root. Neglecting dilution relative to absorption additionally requires $n_H(1-\zeta_{\rm in})\langle\sigma^2\rangle/\langle\sigma\rangle\gg2/r_{\rm in}$. As a concrete limiting test, with negligible absorption the source's $\Gamma=1$ predicts no change, while inverse-square dilution still decreases $q$ and thus $\zeta$.
Back to article page
