# Saha Equation

### More on Milne

Recall that we were deriving the Milne Relation:

${\displaystyle {\sigma _{fb}(v) \over \sigma _{bf}(\nu )}={g_{0} \over g_{+}}\left({h\nu \over m_{e}cv}\right)^{2}\,\!}$

${\displaystyle ({\frac {1}{2}}m_{e}v^{2}+\chi =h\nu )}$. In the following equation, Eugene has replaced the ${\displaystyle g_{e}}$ of Rybicki and Lightman with the number ${\displaystyle 2}$, because that is the number of spin states of the electron. This is just to be more clear. So the Saha equation is:

${\displaystyle {n_{+}n_{e} \over n_{0}}=\left[{2\pi m_{e}kT \over h^{2}}\right]^{3 \over 2}{2g_{+} \over g_{0}}e^{-\chi \over kT}\,\!}$

(Saha Equation) Note that the number of internal degrees of freedom for the proton ${\displaystyle g_{+}=2}$, and the number for neutral hydrogen:

${\displaystyle g_{0}=\overbrace {n^{2}\cdot 2} ^{deg\ of\ freedom\ bound\ e^{-}}\overbrace {\cdot 2} ^{proton}\,\!}$
${\displaystyle g_{0}=4n^{2}\,\!}$

Beware that Shu says ${\displaystyle g_{0}=2n^{2}}$ and ${\displaystyle g_{+}=2}$. We are now going to write the “Recombination Coefficient” ${\displaystyle \alpha }$:

{\displaystyle {\begin{aligned}\alpha &=\sum {\int _{0}^{\infty }{\sigma _{fb}(n,v)v\,f(v)dv}}\\&=\sum _{n}{\left\langle \sigma _{fb}(n,v)v\right\rangle }\\\end{aligned}}\,\!}
${\displaystyle \vartheta (\alpha )\simeq \vartheta \left(\sigma _{bf}({h\nu \over m_{e}cv})^{2}v\right)\,\!}$

We’ll estimate that for hydrogen, ${\displaystyle \sigma _{bf}\sim 10^{-18}cm^{2}}$, ${\displaystyle h\nu \approx 13.6eV}$, and ${\displaystyle v\sim {\sqrt {2kT \over m_{e}}}}$, where ${\displaystyle T\sim 10^{4}K}$. Thus

${\displaystyle \vartheta (\alpha )\simeq 10^{-13}{cm^{3} \over s}\left({10^{4}K \over T}\right)^{\frac {1}{2}}\,\!}$

You can look this up in Osterbrock: at ${\displaystyle T=10^{4}K}$, the sum over all bound states (“recombination”):

${\displaystyle \alpha _{A}=4\cdot 10^{-13}{cm^{3} \over s}\,\!}$

and the sum over all bound states except ${\displaystyle n=1}$ (we might be interested in omitting the free-to-ground transition because it is likely to ionize a nearby atom):

${\displaystyle \alpha _{B}=2\cdot 10^{-13}{cm^{3} \over s}\,\!}$

### Derivation of Saha

Saha makes use of LTE. This equation tells us what the # density ratio is between states 1 and 2. We’ll say that ${\displaystyle n_{0,1}}$ is the # density of neutral atoms with an ${\displaystyle e^{-}}$ in energy level 1. ${\displaystyle n_{+,1}}$ is the # density of ionized atoms which still have an ${\displaystyle e^{-}}$ in energy level 1. Saha says collisional ionizations match the rate of collisional recombination:

${\displaystyle {n_{+}n_{e} \over n_{0}}=\left[{2\pi m_{e}kT \over h^{2}}\right]^{3 \over 2}{2g_{+} \over g_{0}}e^{-\chi \over kT}\,\!}$

The Boltzmann distribution says:

${\displaystyle {n_{+,1} \over n_{0,1}}={g_{+,1} \over g_{0,1}}e^{-E \over kT}\,\!}$

where ${\displaystyle E=E_{state\ 2}-E_{state\ 1}={\frac {1}{2}}m_{e}v^{2}-(-\chi )={\frac {1}{2}}m_{e}v^{2}+\chi }$. Now we need to figure out our numbers of internal degrees of freedom:

${\displaystyle g_{+,1}=g_{+,1}{\big |}_{internal}\cdot g_{e}{\big |}_{internal}\cdot g_{e}{\big |}_{translational\ motion}\cdot g_{+,1}{\big |}_{translational\ motion}\,\!}$

${\displaystyle g_{e}}$ is easy: ${\displaystyle g_{e}=2}$. ${\displaystyle g_{e}{\big |}_{translational}}$ is harder:

${\displaystyle g_{e}{\big |}_{translational}={(vol\ in\ conf\ space)(vol\ in\ momentum\ space) \over h^{3}}\,\!}$

The volume in configuration space is ${\displaystyle {1 \over n_{e}}}$. The volume in momentum space is ${\displaystyle 4\pi p^{2}dp=4\pi m_{e}^{3}v^{2}dv}$. Now on to ${\displaystyle g_{0,1}}$:

${\displaystyle g_{0,1}=g_{0,1}{\big |}_{internal}\cdot g_{0,1}{\big |}_{translational}\,\!}$
${\displaystyle {n_{+,1} \over n_{0,1}}=\int _{0}^{\infty }{g_{+,1}{\big |}_{internal}\cdot g_{+,1}{\big |}_{trans} \over g_{0,1}{\big |}_{internal}\cdot g_{0,1}{\big |}_{trans}}{2 \over n_{e}}{4\pi m_{e}v^{2}dv \over h^{3}}e^{-({\frac {1}{2}}m_{e}v^{2}+\chi _{1}) \over kT}\,\!}$
${\displaystyle {n_{+,1}n_{e} \over n_{0,1}}=\left[{2\pi m_{e}kT \over h^{2}}\right]^{3 \over 2}{2g_{+,1}{\big |}_{int} \over g_{0,1}{\big |}_{int}}e^{-\chi _{1} \over kT}\,\!}$

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 ${\displaystyle g_{0,1}}$ with ${\displaystyle g_{0,2}}$ and ${\displaystyle \chi _{1}}$ with ${\displaystyle \chi _{2}=\chi _{1}-E_{0,12}}$. Thus:

${\displaystyle {n_{+,1}n_{e} \over n_{0,2}}=\left[{2\pi m_{e}kT \over h^{2}}\right]^{3 \over 2}{2g_{+,1}e^{-\chi _{2} \over kT} \over g_{0,2}e^{E_{0,12} \over kT}}\,\!}$

Where the rightmost, bottom factor used to be ${\displaystyle g_{0,1}}$. In general,

${\displaystyle {n_{0,j} \over t_{j}e^{-E_{0,1j} \over kT}}={n_{+,1}n_{e} \over \left[{2\pi m_{e}kT \over h^{2}}\right]^{3 \over 2}2g_{+,1}e^{-\chi _{1} \over kT}}\,\!}$

We’ll define the right-hand side above to be ${\displaystyle R}$. Then:

${\displaystyle {R \over n_{0}}={1 \over U_{0}(T)}={n_{+,1}n_{e} \over \left[{2\pi m_{e}kT \over h^{2}}\right]^{3 \over 2}2g_{+,1}e^{-\chi _{1} \over kT}n_{0}}\,\!}$
${\displaystyle {n_{+,1}n_{e} \over n_{0}}=\left[{2\pi m_{e}kT \over h^{2}}\right]^{3 \over 2}{2g_{+,1} \over U_{0}(T)}e^{-\chi _{1} \over kT}\,\!}$
${\displaystyle {{n_{p}n_{e} \over n_{o}}=\left[{2\pi m_{e}kT \over h^{2}}\right]^{3 \over 2}{2U_{+}(T) \over U_{0}(T)}e^{-\chi _{1} \over kT}}\,\!}$

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 ${\displaystyle 10^{-3}n_{B}}$ (where ${\displaystyle n_{B}}$ is the number of baryons). We can estimate why this is. The time for a proton to find an electron is given by:

${\displaystyle t={1 \over n_{e}\left\langle \sigma _{fb}v\right\rangle }\ll {a \over {\dot {a}}}\,\!}$

That is, collisional ionizing equilibrium just starts to fail when:

${\displaystyle {1 \over n_{e}\left\langle \sigma _{fb}v\right\rangle }\sim {a \over {\dot {a}}}\,\!}$

Since we know the Hubble time (${\displaystyle 2\cdot 10^{5}yrs}$), we can actually estimate the number of free electrons in the universe at the time of recombination:

${\displaystyle n_{e}\sim {1 \over (2\cdot 10^{-13})(2\cdot 10^{5})(\pi \cdot 10^{7})}\sim 1\,\!}$