# Detailed Balance

## Detailed Balance

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.