Difference between revisions of "The Saha equation"

The Saha equation for hydrogen can be presented in the following form: $$ \frac{1 - X}{X^2} = n\lambda _{Te}^3\exp \left( \frac{I_0}{kT}\right), $$ where $X = {n_e}/n$ is equilibrium ionization degree, $n_e$ and $n$ are concentrations of electrons and atoms (both neutral and ionized) respectively, \[\lambda _{Te}^2 = \frac{2\pi \hbar ^2}{m_ekT}\] is the electron's thermal de Broglie wave length, $I_0 = 13.6\,\mbox{eV}$ ionization energy for hydrogen. Concentration of hydrogen on the last scattering surface is approximately equal to to concentration of baryons, which in turn is connected to concentration of photons in the Universe: $n_{_B} \simeq 6 \cdot 10^{ -10}n_\gamma.$ Explicit form of the latter relation can be found from the Planck distribution: $$ n_\gamma = \frac{\rho _\gamma }{E_\gamma} = \frac{\alpha T^4}{3kT}= \frac{\pi ^2}{45}{\left( \frac{kT}{\hbar c}\right)^3}. $$ Introducing the dimensionless variable \[y = \frac{I_0}{kT},\] one finally obtains: $$ \frac{1 - X}{X^2} = Ay^{ - 3/2}e^y. $$ where $$ A = \frac{2^{3/2}\pi ^{7/2}}{45} \frac{n_{_B}}{n_\gamma }\left(\frac{I_0}{m_ec^2} \right)^{3/2} \approx 3 \cdot 10^{ - 16}. $$ Numerical solution of the obtained equation for $X = 0.1$ leads to $y \approx 46,$ which corresponds to temperature value $T \approx 3431\,K.$

The Saha equation can be conveniently rewritten in the form: $$ y = \frac{3}{2}\ln y + \ln \frac{1 - X}{AX^2}. $$ Consecutive iterations of the latter equation for $X = 0.1$ and $A = 3 \cdot 10^{ - 16}$ give the following results: $$ y_0 = \ln \frac{1 - X}{AX^2} \approx 40.24; \Rightarrow T_0 \approx 3920\,K; $$ $$ y_1 = \frac{3}{2}\ln y_0 + \ln \frac{1 - X}{AX^2} \approx 45.8; \Rightarrow T_1 \approx 3446\,K; $$ $$ y_2 = \frac{3}{2}\ln y_1 + \ln \frac{1 - X}{AX^2} \approx 46.0;\Rightarrow T_2 \approx 3431\,K. $$ Result of the third iteration already coincides with the exact one with accuracy of one degree. It can be analytically presented in the following form: $$ kT \approx \frac{I_0} {\ln \left( \frac{1 - X} {AX^2}{\left\{ \ln \left[ \frac{1 - X} {AX^2}{\left(\ln \frac{1 - X} {AX^2} \right)}^{3/2}\right]\right\}}^{3/2}\right)}.$$

Numerical solution of the Saha equation for $X=0.9,$ gives the corresponding temperature $T_1 = 4\ 029\,K,$ and full recombination (corresponding to $X = 0.1$) takes place at $T_2 = 3\ 431\,K,$ which corresponds to the scale factor values $a_1 = 6.8 \cdot 10^{ - 4}$ and $a_2 = 7.9 \cdot 10^{ - 4}$ respectively (setting $a_0=1$ in the present time). The recombination took place in the matter dominated epoch ( $a = \left( t/{t_0} \right)^{2/3},\;t_0 = 14 \cdot 10^9 $ years), therefore $t_1 \approx 250\ 000$ and $t_2 \approx 310\ 000$ years, i.e. the recombination endured approximately $60\cdot 10^3$ years.

The Saha equation with given concentration of the gas can be conveniently presented in the following form $$ \frac{1 - X} {X^2} = n\lambda_0^3y^{3/2}e^y,$$ where $$ \lambda _0 \equiv \lambda _{Te} \left.\right|_{kT = I_0} = \frac{\sqrt {2\pi } \hbar c}{\sqrt m_ec^2I_0}\simeq 1.877 \cdot 10^{ - 8}\,cm. $$ Thus for hydrogen density in the center of the Sun one obtains $ n \simeq \rho /m_p \approx 6 \cdot 10^{25}\,cm^{-3}.$ It then follows $$ X = \frac{1}{\sqrt{\frac{1}{4} + n\lambda _0^3\left( \frac{I_0}{kT} \right)^{3/2}\exp \left( \frac{I_0}{kT} \right)} + \frac{1}{2}}\approx 0.754 $$ The unreasonable result (almost 25\% of neutral hydrogen in the center of the Sun) appeared because of inapplicability of the Saha equation in such dense and highly ionized plasma: in the considered case the Debye radius and interatomic spacing are of the same order $r_D \approx a \approx 10^{- 9}\,cm.$