# Collisional Excitations

## Collisional Excitation Cross-sections

The Einstein analog:

${\displaystyle \overbrace {A} ^{low\ E\ particle}+\overbrace {B_{fast}} ^{high\ E\ e^{-}}\to \overbrace {A^{*}} ^{excited}+B_{slow}\,\!}$

So the Rate of Excitations ${\displaystyle R_{ex}}$ is given by:

${\displaystyle R_{ex}=n_{A}n_{B}\sigma _{12}f(v_{rel})v_{rel}\,\!}$

Suppose we have some distribution of relative velocities given by ${\displaystyle f(v)dv}$, where ${\displaystyle f}$ is the fraction of collisions occurring with relative velocities ${\displaystyle [v_{rel},v_{rel}+dv]}$. Then:

${\displaystyle R_{ex}=n_{A}n_{B}\int _{0}^{\infty }f(v)dv\sigma _{12}(v)v\,\!}$
${\displaystyle =n_{A}n_{B}\left\langle \sigma _{12}v\right\rangle \,\!}$
${\displaystyle =n_{A}n_{B}q_{12}\,\!}$

where ${\displaystyle q_{12}}$ is the “collisional rate coefficient” ${\displaystyle [cm^{3}s^{-1}]}$. Then the Rate of de-excitation is given by:

${\displaystyle R_{deex}=n_{A}n_{B}\int _{0}^{\infty }{f(v)v\,dv\sigma _{21}(v)=n_{A}^{*}n_{B}q_{21}}\,\!}$

We recognize now that ${\displaystyle \sigma _{12}(v)n_{A}n_{B}f(v)dv\,v}$ is the rate of excitations of A using B moving at relative velocity ${\displaystyle v}$. If we have detailed balance, then this has to be the same as the rate of de-excitation ${\displaystyle n_{A}^{*}n_{B}f(v^{\prime })v^{\prime }\sigma _{21}(v^{\prime })}$.

${\displaystyle \overbrace {{\frac {1}{2}}m_{r}v^{2}} ^{center\ of\ mass\ E}=hv_{12}+{\frac {1}{2}}m_{r}v^{\prime 2}\,\!}$

Where ${\displaystyle m_{r}}$ is the reduced mass ${\displaystyle m_{A}m_{B} \over m_{A}+m_{B}}$. However many ${\displaystyle B_{slow}}$ are created by collisional excitation, the same number are used for the reverse de-excitation. This is detailed balance.

Second, under thermal equilibrium, particles have a Maxwellian velocity distribution:

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

(Maxwellian velocity distribution) In thermal equilibrium,

${\displaystyle {n_{A}^{*} \over n_{A}}={g_{2} \over g_{1}}e^{-h\nu _{21} \over kT}\,\!}$

Now, assuming detailed balance and thermal equilibrium,

${\displaystyle n_{A}n_{B}f(v)dv\,v\sigma _{12}=n_{A}^{*}n_{B}f(v^{\prime })dv^{\prime }v^{\prime }\sigma _{21}\,\!}$
${\displaystyle \sigma _{12}v^{2}e^{-m_{r}v^{2} \over 2kT}={g_{2} \over t_{1}}e^{-h\nu _{21} \over 2kT}\nu ^{\prime 2}e^{-m_{r}v^{\prime 2} \over 2kT}v^{\prime }dv^{\prime }\sigma _{21}\,\!}$
${\displaystyle \sigma _{12}dv\,v^{3}e^{-m_{r}v^{2} \over 2kT}={g_{2} \over t_{1}}e^{-h\nu _{21} \over 2kT}e^{-h\nu _{21} \over 2kT}e^{-m_{r}v^{2} \over 2kT}v^{\prime 3}dv^{\prime }\sigma _{21}\,\!}$
${\displaystyle {g_{1}v^{2}\sigma _{12}(v)=g_{2}v^{\prime 2}\sigma _{21}(v^{\prime })}\,\!}$

This is the analog of the relationship between the Einstein coefficients ${\displaystyle B_{12}}$ and ${\displaystyle B_{21}}$.

For a specific case, ${\displaystyle B=e^{-}}$, ${\displaystyle A=}$ion with bound electron.

### More on Einstein Analog

Forgot something for the Einstein Analog. Recall:

${\displaystyle g_{1}v^{2}\sigma _{12}(v)=g_{2}v^{\prime 2}\sigma _{21}(v^{\prime })\,\!}$
${\displaystyle {\frac {1}{2}}m_{r}v^{2}=h\nu _{21}+{\frac {1}{2}}m_{r}(v^{\prime })^{2}\,\!}$

For the special case of ${\displaystyle f(v)}$ being Maxwellian, then:

${\displaystyle {q_{12} \over q_{21}}\equiv {\left\langle \sigma _{12}v\right\rangle \over \left\langle \sigma _{21}v\right\rangle }={g_{2} \over g_{1}}e^{-h\nu _{21} \over kT}\,\!}$

This has a Boltzmann factor, which makes you thing we’re assuming LTE, but we’re not.

Last time, we were talking about electron-ion collisional excitation, given by:

${\displaystyle q_{12}\equiv \left\langle \sigma _{12}v\right\rangle \propto {1 \over v}\propto T^{-{\frac {1}{2}}}\,\!}$

We can extend this for neutral-ion collisional excitation:

${\displaystyle q_{12}\propto T^{\frac {1}{2}}T^{-{\frac {1}{2}}}\propto T^{0}\,\!}$
${\displaystyle \sigma _{12}\equiv {\pi \hbar ^{2} \over m_{e}v^{2}}{\Omega (1,2) \over g}\,\!}$

where ${\displaystyle \Omega (1,2)}$ scales as ${\displaystyle T^{\frac {1}{2}}}$. Notice this means that for some neutral-ion collisional excitation, it is temperature independent. For neutral-neutral collisional excitation:

${\displaystyle q_{12}=\left\langle \sigma _{12}v\right\rangle \propto T^{\frac {1}{2}}\,\!}$
${\displaystyle \sigma _{12}\propto T^{0}\sim \pi (Fa_{0})^{2}\,\!}$

This is all we’ll talk about bound-bound transitions.