# Saha Equation

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<latex> \documentclass[11pt]{article} \def\inv#1Template:1 \over \def\ddtTemplate:D \over dt \def\mean#1{\left\langle #1\right\rangle} \def\sigot{\sigma_{12}} \def\sigto{\sigma_{21}} \def\eval#1{\big|_{#1}} \def\tr{\nabla} \def\dce{\vec\tr\times\vec E} \def\dcb{\vec\tr\times\vec B} \def\wz{\omega_0} \def\ef{\vec E} \def\ato{{A_{21}}} \def\bto{{B_{21}}} \def\bot{{B_{12}}} \def\bfieldTemplate:\vec B \def\apTemplate:A^\prime \def\xp{{x^{\prime}}} \def\yp{{y^{\prime}}} \def\zp{{z^{\prime}}} \def\tp{{t^{\prime}}} \def\upxTemplate:U x^\prime \def\upyTemplate:U y^\prime \def\e#1{\cdot10^{#1}} \def\sigfb{\sigma_{fb}} \def\sigbf{\sigma_{bf}} \def\hf{\frac12} \usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document} \subsection*{More on Milne } Recall that we were deriving the Milne relation: $${\sigfb(v)\over\sigbf(\nu)}={g_0\over g_+}\left({h\nu\over m_ecv}\right)^2$$ $(\hf m_ev^2+\chi=h\nu)$. In the following equation, Eugene has replaced the $g_e$ of Rybicki and Lightman with the number $2$, because that is the number of spin states of the electron. This is just to be more clear. So the {\bf Saha} equation is: $${n_+n_e\over n_0}=\left[{2\pi m_ekT\over h^2}\right]^{3\over2}{2g_+\over g_0}e^{-\chi\over kT}$$ \centerline{(Saha Equation)} Note that the number of internal degrees of freedom for the proton $g_+=2$, and the number for neutral hydrogen: $$g_0=\overbrace{n^2\cdot2}^{deg\ of\ freedom\ bound\ e^-}\overbrace{\cdot2} ^{proton}$$ $$g_0=4n^2$$ Beware that Shu says $g_0=2n^2$ and $g_+=2$. We are now going to write the Recombination Coefficient $\alpha$: \begin{aligned}\alpha&=\sum{\int_0^\infty{\sigfb(n,v)v\,f(v)dv}}\\ &=\sum_n{\mean{\sigfb(n,v)v}}\\ \end{aligned} $$\vartheta(\alpha)\simeq\vartheta\left(\sigbf({h\nu\over m_ecv})^2v\right)$$ We'll estimate that for hydrogen, $\sigbf\sim10^{-18} cm^2$, $h\nu\approx 13.6eV$, and $v\sim\sqrt{2kT\over m_e}$, where $T\sim10^4K$. Thus $$\vartheta(\alpha)\simeq10^{-13}{cm^3\over s}\left({10^4K\over T}\right)^\hf$$ You can look this up in Osterbrock: at $T=10^4K$, the sum over all bound states (recombination): $$\alpha_A=4\e{-13}{cm^3\over s}$$ and the sum over all bound states except $n=1$ (we might be interested in omitting the free-to-ground transition because it is likely to ionize a nearby atom): $$\alpha_B=2\e{-13}{cm^3\over s}$$
Saha makes use of LTE. This equation tells us what the \# density ratio is between states 1 and 2. We'll say that $n_{0,1}$ is the \# density of neutral atoms with an $e^-$ in energy level 1. $n_{+,1}$ is the \# density of ionized atoms which still have an $e^-$ in energy level 1. Saha says collisional ionizations match the rate of collisional recombination: $${n_+n_e\over n_0}=\left[{2\pi m_ekT\over h^2}\right]^{3\over2}{2g_+\over g_0}e^{-\chi\over kT}$$ The Boltzmann relation says: \def\npo{n_{+,1}} \def\noo{n_{0,1}} \def\gpo{g_{+,1}} \def\goo{g_{0,1}} $${\npo\over\noo}={\gpo\over\goo}e^{-E\over kT}$$ where $E=E_{state\ 2}-E_{state\ 1}=\hf m_ev^2-(-\chi)=\hf m_ev^2+\chi$. Now we need to figure out our numbers of internal degrees of freedom: $$\gpo =\gpo\eval{internal}\cdot g_e\eval{internal}\cdot g_e\eval{translational\ motion}\cdot \gpo\eval{translational\ motion}$$ $g_e$ is easy: $g_e=2$. $g_e\eval{translational}$ is harder: $$g_e\eval{translational} = {(vol\ in\ conf\ space)(vol\ in\ momentum\ space) \over h^3}$$ The volume in configuration space is $\inv{n_e}$. The volume in momentum space is $4\pi p^2dp=4\pi m_e^3v^2dv$. Now on to $\goo$: $$\goo=\goo\eval{internal}\cdot\goo\eval{translational}$$ $${\npo\over\noo}=\int_0^\infty{\gpo\eval{internal}\cdot\gpo\eval{trans} \over\goo\eval{internal}\cdot\goo\eval{trans}}{2\over n_e} {4\pi m_ev^2dv\over h^3}e^{-(\hf m_ev^2+\chi_1)\over kT}$$ $${\npo n_e\over \noo}=\left[{2\pi m_ekT\over h^2}\right]^{3\over2} {2\gpo\eval{int}\over\goo\eval{int}}e^{-\chi_1\over kT}$$ Now what if we were talking about going from neutral bound state in energy level 2 (instead of 1) to an ionized atom with electron in energy state 1. It turns out, we just need to replace $\goo$ with $g_{0,2}$ and \def\got{g_{0,2}} \def\nzt{n_{0,2}} $\chi_1$ with $\chi_2=\chi_1-E_{0,12}$. Thus: $${\npo n_e\over\nzt}=\left[{2\pi m_ekT\over h^2}\right]^{3\over2}{2\gpo e^{-\chi_2\over kT}\over\got e^{E_{0,12}\over kT}}$$ \def\tpmekth{\left[{2\pi m_ekT\over h^2}\right]^{3\over2}} Where the rightmost, bottom factor used to be $\goo$. In general, $${n_{0,j}\over t_je^{-E_{0,1j}\over kT}}={\npo n_e\over\tpmekth 2\gpo e^{-\chi_1\over kT}}$$ We'll define the right-hand side above to be $R$. Then: $${R\over n_0}=\inv{U_0(T)}={\npo n_e\over\tpmekth 2\gpo e^{-\chi_1\over kT}n_0}$$ $${\npo n_e\over n_0}=\tpmekth{2\gpo\over U_0(T)} e^{-\chi_1\over kT}$$ $$\boxed{{n_pn_e\over n_o}=\tpmekth{2U_+(T)\over U_0(T)} e^{-\chi_1\over kT}}$$ This is the full Saha equation. Remember that Saha assumes that we are in LTE: the rate of collisional ionizations must equal the rate of collisional recombination. This predicts why we had to wait until the universe got to 3000K until recombination occurred. We would have naively expected that as soon as the temperature of the universe dropped below the ionizing energy of ground-state hydrogen, we would have recombination. However, Remember that we can't follow Saha out too far. Soon, collisions stop happening often enough to maintain LTE, and Saha becomes invalid. For example, Saha would say that after recombination we would continue to lose free electrons at the same (logarithmic) rate. In truth, the number of free electrons asymptotically approaches $10^{-3}n_B$ (where $n_B$ is the number of baryons). We can estimate why this is. The time for a proton to find an electron is given by: $$t=\inv{n_e\mean{\sigfb v}}\ll{a\over \dot a}$$ That is, collisional ionizing equilibrium just starts to fail when: $$\inv{n_e\mean{\sigfb v}}\sim{a\over \dot a}$$ Since we know the Hubble time ($2\e5 yrs$), we can actually estimate the number of free electrons in the universe at the time of recombination: $$n_e\sim\inv{(2\e{-13})(2\e5)(\pi\e7)}\sim1$$