 Solution

Use the Einstein coefficients convention in which $B_{ij}$ multiplies spectral radiation energy density $u_{\nu }$, with $u_{\nu }=c^{-1}\int I_{\nu }d\Omega$. A normalized spectral line profile obeys $\int \phi (\nu )d\nu =1$. The spontaneous emission probability per upper-state atom per unit time is $A_{21}$; the stimulated emission and upward absorption rates per atom are respectively $B_{21}\int u_{\nu }\phi (\nu )d\nu$ and $B_{12}\int u_{\nu }\phi (\nu )d\nu$.

For isotropic spontaneous emission, $n_2{A}_{21}$ transitions occur per unit volume and time, each supplying photon energy $h{\nu }_{12}$. Dividing among $4\pi$ steradians givesFor a small incident beam, its contribution to $u_{\nu }$ is $I_{\nu }d\Omega /c$. Absorption removes and stimulated emission adds photons in the incident direction. Matching the net loss to ${\alpha }_{\nu }{I}_{\nu }d\Omega$ therefore givesReplacing ${\nu }_{12}$ by $\nu$ across the narrow spectral line profile gives the printed expressions. The negative sign is essential: an inverted population can amplify the incident specific intensity. Thermal equilibrium and the Planck function imply $g_1{B}_{12}=g_2{B}_{21}$ and $A_{21}/B_{21}=8\pi h{\nu }_{12}^3/c^3$, where $g_i$ is the state degeneracy. A convention using mean intensity instead changes both $B$ coefficients by $4\pi /c$.