The Saha equation

Degree of ionization of atomic hydrogen in thermal equilibrium can be described by the Saha equation \[\frac{1-X}{X^2}=n\lambda_{Te}^3e^{\frac{I}{kT}},\] where $X=n_e/n$ is the degree of ionization, $n_e$ and $n$ are concentrations of electrons and atoms (both neutral and ionized) respectively, \[\lambda_{Te}^2=\frac{2\pi\hbar^2}{m_e kT}\] is the electron's thermal de Broglie wave length and $I=13.6eV$ is the ionization energy for hydrogen. It is often used in astrophysics for description of stellar dynamics.

6.6.1

Problem 6.6.1

Using the Saha equation, determine the hydrogen ionization degree

a) 100 seconds after the Big Bang;

b) at the epoch of recombination;

c) at present time.

Assume for simplicity $\Omega=1$.

solution

6.6.2

Problem 6.6.2

Assuming that the ionization degree at the last scattering was equal to $10\%$, determine the decoupling temperature using the Saha equation.

solution

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 $T \approx 3431\,K.$

6.6.3

Problem 6.6.3

How many iterations in the Saha equation are needed in order to obtain the decoupling temperature with accuracy $1K$? Write down analytically the approximate result.

solution

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: $$ kT \approx \frac{I_0} {\ln \left( \frac{1 - X} {AX^2}{\left\{ \ln \left[ \frac{1 - X} {AX^2}{\left(\ln \frac{1 - X} {AX^2} \right)}^{3/2}\right]\right\}}^{3/2}\right)}.$$

6.6.4

Problem 6.6.4

Estimate the duration of the epoch of recombination: how long did it take for hydrogen ionization degree to change from $90\%$ to $10\%$ according to the Saha equation?

solution

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.

6.6.5

Problem 6.6.5

Using the Saha equation, determine the hydrogen ionization degree in the center of the Sun ($\rho=100g/cm^3$, $T=1.5\cdot10^7K$).

solution

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 $r_D \approx a \approx 10^{- 9}\,cm.$