# Difference between revisions of "Milne Relation"

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### Bound-Free Transitions (Photoionization)

We’ll calculate the cross-section of a bound-free transition ${\displaystyle \sigma _{bf}}$:

${\displaystyle \sigma _{bf}\sim {\lambda ^{2} \over 8\pi }{A_{21} \over \Delta \nu }\,\!}$

It turns out that ${\displaystyle \Delta \nu }$ is about ${\displaystyle \nu }$. ${\displaystyle \lambda \approx 912\mathrm {\AA} }$, so scaling from Lyman-alpha:

${\displaystyle \sigma _{bf}\sim {(912\mathrm {\AA} )^{3} \over c\cdot 8\pi }A_{21,Ly\alpha }\left({1216\mathrm {\AA} \over 912\mathrm {\AA} }\right)^{3}\sim 10^{-18}cm^{2}\,\!}$

It turns out that the real answer is ${\displaystyle \sigma _{bf}\sim 6\cdot 10^{-18}cm^{2}}$. In general:

${\displaystyle \sigma _{bf}=\sigma {\big |}_{edge}\left({E_{photon\,in} \over E_{edge}}\right)^{-3}\,\!}$

That exponent (-3) is actually ${\displaystyle -{\frac {8}{3}}}$ near the edge and goes to ${\displaystyle -{\frac {7}{2}}}$ far from it. So you see ${\displaystyle \sigma _{bf}}$ spike up as the photon reaches the ionization energy, and then decrease exponentially as energy increases. However, you can see new spikes from ionizing electrons in inner shells.

### Radiative Recombination

This is the inverse process of photoionization, so ${\displaystyle \sigma _{fb}}$ is the cross-section for an ion recapturing its electron and emitting a photon. We’ll relate ${\displaystyle \sigma _{fb}}$ to ${\displaystyle \sigma _{bf}}$. This is called the Milne Relation. In this derivation, we’ll start by assuming complete thermal equilibrium and derive a result which will end up being independent of thermal equilibrium. Let’s start calculating the rate of radiative recombinations. Thermal equilibrium dictates that this must equal the rate of photoionization. For radiative recombination:

${\displaystyle rate\ of\ recombination=n_{+}n_{e}\sigma _{fb}(v)v[f(v)dv]={\#\ of\ recombinations \over volume\ time}\,\!}$

We’ll set this equal to the rate of photoionization. This rate is:

${\displaystyle rate\ of\ photoionization={B_{\nu }4\pi d\nu \over h\nu }n_{0}\sigma _{bf}\overbrace {\left(1-{g_{0} \over g_{+}}{n_{+} \over n_{0}}\right)} ^{{correction\ for \atop stimulated} \atop recombination}\,\!}$

where ${\displaystyle n_{0}}$ is the # density of neutrals. Note this has units of # flux. Note also that ${\displaystyle \sigma _{bf}}$ depends on ${\displaystyle \nu }$, and ${\displaystyle n_{+} \over n_{0}}$ is evaluated at the relative velocity ${\displaystyle v}$ such that ${\displaystyle h\nu ={\frac {1}{2}}m_{e}v^{2}+\chi }$, where ${\displaystyle \chi }$ is the threshold ionization energy. In thermal equilibrium, we know that (see Boltzmann distribution):

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

where ${\displaystyle E}$ is the energy difference between state 1 (proton + unbound ${\displaystyle e^{-}}$) and state 2 (bound proton/electron pair). Thus ${\displaystyle E}$ is given by:

${\displaystyle E={\frac {1}{2}}m_{e}v^{2}-(-\chi )=h\nu \,\!}$

So we can make our n’s and g’s go away. For ${\displaystyle f(v)}$, we’ll use our Maxwellian velocity distribution:

${\displaystyle f(v)=4\pi \left({m_{e} \over 2\pi kT}\right)^{3 \over 2}v^{2}e^{-m_{e}v^{2} \over 2kT}\,\!}$

Finally, the Saha Equation tells us in thermal equilibrium:

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

So now we’re essentially done:

${\displaystyle 1={n_{+}n_{e}f(v)v\sigma _{fb}(v)dv \over {4\pi B_{\nu } \over h\nu }d\nu (1-e^{-h\nu _{21} \over 2kT})n_{0}\sigma _{bf}(v)}\,\!}$

and plugging in all of our relations we get:

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

(Milne Relation) Notice how all of the T’s vanished. This result is independent of thermal equilibrium. However, you still have to pay attention to your statistical weights (g’s).