Relation between Einstein's Coefficients$^*$

Analysis using the Einstein's coefficients is applicable to the state in which the thermal equilibrium is not achieved. However, if this is applied to the thermal equilibrium, we can obtain relations between these coefficients.

Assume the detailed balance is achieved between the spontaneous emission, the induced emission and the absorption as

N_{n_u}\left(A_{n_u\rightarrow n}+B_{n_u\rightarrow n}J_{n_...
...arrow n}\right)=N_{n}B_{n\rightarrow n_u}J_{n_u\rightarrow n}.
\end{displaymath} (2.144)

In the thermal equilibrium, the population of atoms is given by the Boltzmann distribution as
\frac{N_{n_u}}{N_n}=\frac{g_{n_u}}{g_n}\exp\left(-\frac{E_{n_u n}}{kT}\right),
\end{displaymath} (2.145)

where $g_n$ represents the statistical weight for the state n.

Equation (2.144) gives

\frac{N_{n_u}A_{n_u\rightarrow n}}{N_n}
=J_{n_u\rightarrow ...
...ightarrow n_u}-\frac{N_{n_u}}{N_n}B_{n_u\rightarrow n}\right).
\end{displaymath} (2.146)

In the case of thermal equilibrium, the average specific intensity agrees with the Planck function $B(\nu,T)$ as
\end{displaymath} (2.147)

Using equations (2.145) and (2.147), equation (2.144) reduces to
\frac{g_{n_u}}{g_n}A_{n_u\rightarrow n}=
\frac{2 h \nu^3}{c...
...laystyle B_{n\rightarrow n_u}}}
\end{displaymath} (2.148)

Lefthand-side of the equation is not dependent on the temperature. For this equation to be valid for various temperature,
\frac{g_{n_u}}{g_n}\frac{B_{n_u\rightarrow n}}{B_{n\rightarrow n_u}}=1,
\end{displaymath} (2.149)

\frac{g_{n_u}}{g_n}A_{n_u\rightarrow n}=
\frac{2 h \nu^3}{c^2}B_{n\rightarrow n_u}.
\end{displaymath} (2.150)

Kohji Tomisaka 2009-12-10