Detailed Balance

Detailed Balance

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.