= Solution
For instantaneous complete <hydrogen> <reionization>, take $n_{e0}=\Omega_b\rho_{{\rm crit},0}/(m_pc^2)=3\Omega_bH_0^2/(8\pi Gm_p)$ and $n_e(z)=n_{e0}(1+z)^3$. Helium or incomplete <ionization> would multiply this by the appropriate free-electron count per baryon. In the matter-plus-curvature model,
$$
H(z)=H_0(1+z)\sqrt{1+\Omega_mz},\qquad
|dt|=\frac{dz}{(1+z)H(z)}.
$$
The <reionization optical depth in an open matter universe> is therefore
$$
\kappa=\frac{3\Omega_b\sigma_TcH_0}{8\pi Gm_p}
\int_0^{z_{\rm re}}\frac{1+z}{\sqrt{1+\Omega_mz}}dz.
$$
Set $u=\sqrt{1+\Omega_mz}$ and $\beta^2=1-\Omega_m$. The integral is $2\Omega_m^{-2}[u^3/3-\beta^2u]_1^{u_{\rm re}}$, giving
$$
\boxed{\kappa=\frac{\Omega_b C_H}{\Omega_m^2}
\left[u_{\rm re}^3-1-3(1-\Omega_m)(u_{\rm re}-1)\right],\quad
u_{\rm re}=\sqrt{1+\Omega_mz_{\rm re}},\quad C_H=\frac{\sigma_TcH_0}{4\pi Gm_p}.}
$$
At high redshift where matter dominates over curvature, this approaches $\kappa\simeq\Omega_bC_H(1+z_{\rm re})^{3/2}/\sqrt{\Omega_m}$. The exact expression should be retained when $\Omega_mz_{\rm re}$ is not large.
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