 Solution

Take the optical depth to increase inward and let $\mu >0$ denote an outward ray. For thermal absorption and emission in local thermodynamic equilibrium, the radiative transfer source function is the Planck function, so the radiative transfer equation isThis assumes negligible scattering, or a source function genuinely thermalized to $B_{\lambda }$; LTE by itself does not turn an arbitrary scattering source into a Planck function. Multiply by $e^{-\tau /\mu }$, integrate between the two boundaries, and solve for the outward specific intensity:The bottom boundary value is the quantity denoted $I_{\lambda }(0)$ in the supplied notation; it is not an optical-depth-zero boundary. The first term is attenuated incident radiation, and the second is emission from the intervening layers. This is the formal solution of the radiative transfer equation.

For an isothermal atmosphere at temperature $T$, the Planck function is constant and the integral can be evaluated:For a bounded bottom intensity and ${\tau }_1\to \infty$, $I_{\lambda }=B_{\lambda }(T)$ in every outgoing direction, and $F_{\lambda }=\pi B_{\lambda }(T)$. Thus the emergent spectrum is a blackbody spectrum. The blackbody limit of an isothermal atmosphere does not require a temperature gradient: it follows from complete thermalization in an optically thick medium.