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## Collisional Excitation Cross-sections

The Einstein analog:

$\overbrace {A} ^{low\ E\ particle}+\overbrace {B_{fast}} ^{high\ E\ e^{-}}\to \overbrace {A^{*}} ^{excited}+B_{slow}\,\!$
So the Rate of Excitations $R_{ex}$ is given by:

$R_{ex}=n_{A}n_{B}\sigma _{12}f(v_{rel})v_{rel}\,\!$
Suppose we have some distribution of relative velocities given by $f(v)dv$, where $f$ is the fraction of collisions occurring with relative velocities $[v_{rel},v_{rel}+dv]$. Then:

$R_{ex}=n_{A}n_{B}\int _{0}^{\infty }f(v)dv\sigma _{12}(v)v\,\!$
$=n_{A}n_{B}\left\langle \sigma _{12}v\right\rangle \,\!$
$=n_{A}n_{B}q_{12}\,\!$
where $q_{12}$ is the “collisional rate coefficient” $[cm^{3}s^{-1}]$. Then the Rate of de-excitation is given by:

$R_{deex}=n_{A}n_{B}\int _{0}^{\infty }{f(v)v\,dv\sigma _{21}(v)=n_{A}^{*}n_{B}q_{21}}\,\!$
We recognize now that $\sigma _{12}(v)n_{A}n_{B}f(v)dv\,v$ is the rate of excitations of A using B moving at relative velocity $v$. If we have detailed balance, then this has to be the same as the rate of de-excitation $n_{A}^{*}n_{B}f(v^{\prime })v^{\prime }\sigma _{21}(v^{\prime })$.

$\overbrace {{\frac {1}{2}}m_{r}v^{2}} ^{center\ of\ mass\ E}=hv_{12}+{\frac {1}{2}}m_{r}v^{\prime 2}\,\!$
Where $m_{r}$ is the reduced mass $m_{A}m_{B} \over m_{A}+m_{B}$. However many $B_{slow}$ 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:

${f(v)=4\pi \left({m_{r} \over 2\pi kT}\right)^{3 \over 2}v^{2}e^{-m_{r}v^{2} \over 2kT}}\,\!$
(Maxwellian velocity distribution) In thermal equilibrium,

${n_{A}^{*} \over n_{A}}={g_{2} \over g_{1}}e^{-h\nu _{21} \over kT}\,\!$
Now, assuming detailed balance and thermal equilibrium,

$n_{A}n_{B}f(v)dv\,v\sigma _{12}=n_{A}^{*}n_{B}f(v^{\prime })dv^{\prime }v^{\prime }\sigma _{21}\,\!$
$\sigma _{12}v^{2}e^{-m_{r}v^{2} \over 2kT}={g_{2} \over t_{1}}e^{-h\nu _{21} \over 2kT}\nu ^{\prime 2}e^{-m_{r}v^{\prime 2} \over 2kT}v^{\prime }dv^{\prime }\sigma _{21}\,\!$
$\sigma _{12}dv\,v^{3}e^{-m_{r}v^{2} \over 2kT}={g_{2} \over t_{1}}e^{-h\nu _{21} \over 2kT}e^{-h\nu _{21} \over 2kT}e^{-m_{r}v^{2} \over 2kT}v^{\prime 3}dv^{\prime }\sigma _{21}\,\!$
${g_{1}v^{2}\sigma _{12}(v)=g_{2}v^{\prime 2}\sigma _{21}(v^{\prime })}\,\!$
This is the analog of the relationship between the Einstein coefficients $B_{12}$ and $B_{21}$.

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

### More on Einstein Analog

Forgot something for the Einstein Analog. Recall:

$g_{1}v^{2}\sigma _{12}(v)=g_{2}v^{\prime 2}\sigma _{21}(v^{\prime })\,\!$
${\frac {1}{2}}m_{r}v^{2}=h\nu _{21}+{\frac {1}{2}}m_{r}(v^{\prime })^{2}\,\!$
For the special case of $f(v)$ being Maxwellian, then:

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

$q_{12}\equiv \left\langle \sigma _{12}v\right\rangle \propto {1 \over v}\propto T^{-{\frac {1}{2}}}\,\!$
We can extend this for neutral-ion collisional excitation:

$q_{12}\propto T^{\frac {1}{2}}T^{-{\frac {1}{2}}}\propto T^{0}\,\!$
$\sigma _{12}\equiv {\pi \hbar ^{2} \over m_{e}v^{2}}{\Omega (1,2) \over g}\,\!$
where $\Omega (1,2)$ scales as $T^{\frac {1}{2}}$. Notice this means that for some neutral-ion collisional excitation, it is *temperature independent*. For neutral-neutral collisional excitation:

$q_{12}=\left\langle \sigma _{12}v\right\rangle \propto T^{\frac {1}{2}}\,\!$
$\sigma _{12}\propto T^{0}\sim \pi (Fa_{0})^{2}\,\!$
This is all we’ll talk about bound-bound transitions.