Difference between revisions of "Milne Relation"

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* [http://scienceworld.wolfram.com/physics/MilneRelations.html Milne Relation (Wolfram)]
 
* [http://scienceworld.wolfram.com/physics/MilneRelations.html Milne Relation (Wolfram)]
 
* [http://home.strw.leidenuniv.nl/~michiel/ismclass_files/ism10_lecture2.pdf The ISM (Hogerheijde, Leiden)]
 
* [http://home.strw.leidenuniv.nl/~michiel/ismclass_files/ism10_lecture2.pdf The ISM (Hogerheijde, Leiden)]
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===Need a Review?===
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* [[Einstein Coefficients]]
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* [[Saha Equation]]
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* [[Boltzmann distribution]]
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* [[Maxwellian velocity distribution]]
  
 
<latex>
 
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Note also that $\sigbf$ depends on $\nu$, and $n_+\over n_0$ is evaluated
 
Note also that $\sigbf$ depends on $\nu$, and $n_+\over n_0$ is evaluated
 
at the relative velocity $v$ such that $h\nu=\hf m_ev^2+\chi$, where $\chi$
 
at the relative velocity $v$ such that $h\nu=\hf m_ev^2+\chi$, where $\chi$
is the threshold ionization energy.  In thermal equilibrium, we know that:
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is the threshold ionization energy.  In thermal equilibrium, we know that (see [[Boltzmann distribution]]):
 
$${n_+\over n_0}={g_+\over g_0}e^{-E\over kT}$$
 
$${n_+\over n_0}={g_+\over g_0}e^{-E\over kT}$$
 
where $E$ is the energy difference between state 1 (proton + unbound $e^-$)
 
where $E$ is the energy difference between state 1 (proton + unbound $e^-$)
 
and state 2 (bound proton/electron pair).  Thus $E$ is given by:
 
and state 2 (bound proton/electron pair).  Thus $E$ is given by:
 
$$E=\hf m_ev^2-(-\chi)=h\nu$$
 
$$E=\hf m_ev^2-(-\chi)=h\nu$$
So we can make our n's and g's go away.  For $f(v)$, we'll use our Maxwellian:
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So we can make our n's and g's go away.  For $f(v)$, we'll use our [[Maxwellian velocity distribution]]:
 
$$f(v)=4\pi\left({m_e\over 2\pi kT}\right)^{3\over 2}v^2e^{-m_ev^2\over 2kT}$$
 
$$f(v)=4\pi\left({m_e\over 2\pi kT}\right)^{3\over 2}v^2e^{-m_ev^2\over 2kT}$$
Finally, Saha tells us in thermal equilibrium:
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Finally, the [[Saha Equation]] tells us in thermal equilibrium:
 
$${n_+n_e\over n_0}=\left[{2\pi m_ekT\over h^2}\right]^{3\over 2}{2g_+\over g_n}
 
$${n_+n_e\over n_0}=\left[{2\pi m_ekT\over h^2}\right]^{3\over 2}{2g_+\over g_n}
 
e^{-\chi\over kT}$$
 
e^{-\chi\over kT}$$

Revision as of 10:41, 5 December 2017

Short Topical Videos

Reference Material

Need a Review?

Bound-Free Transitions (Photoionization)

We’ll calculate the cross-section of a bound-free transition :

It turns out that is about . , so scaling from Lyman-alpha:

It turns out that the real answer is . In general:

That exponent (-3) is actually near the edge and goes to far from it. So you see 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 is the cross-section for an ion recapturing its electron and emitting a photon. We’ll relate to . 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:

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

where is the # density of neutrals. Note this has units of # flux. Note also that depends on , and is evaluated at the relative velocity such that , where is the threshold ionization energy. In thermal equilibrium, we know that (see Boltzmann distribution):

where is the energy difference between state 1 (proton + unbound ) and state 2 (bound proton/electron pair). Thus is given by:

So we can make our n’s and g’s go away. For , we’ll use our Maxwellian velocity distribution:

Finally, the Saha Equation tells us in thermal equilibrium:

So now we’re essentially done:

and plugging in all of our relations we get:

(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).