PX282 - C10b - understanding the radiative transfer equation

the radiative transfer equation (RTE):

\frac{d I_{λ}}{d τ_{λ}} = I_{λ} - S_{λ}

local thermal equilibrium (LTE)

LTE implies that in a region, the temperature is effectively constant, and the velocities of particles follow the maxwell-boltzmann distribution
at LTE, the photon mean free path is small, and they interact with particles, so, any optically thick object with a non-zero temperature emits blackbody radiation
hence, LTE equates to blackbody radiation, therefore the source function equals the planck function:

S_{λ} = B_{λ}

in equilibrium, the absorption and the emission are equal, so the flux is constant, ie: $F = σ T^{4}$
using LTE, RTE becomes analytically solvable
considering an isothermal slab with its top surface at optical depth, $τ = 0$ , and its bottom surface at $τ^{'}$ , such that $I (τ^{'}) = 0$
using LTE:

\frac{d I_{λ}}{d τ_{λ}} = I_{λ} - S_{λ} = I_{λ} - B_{λ}

multiplying by $e^{- τ} :$

\begin{matrix} \frac{d I_{λ}}{d τ_{λ}} e^{- τ} = (I_{λ} - B_{λ}) e^{- τ} \\ \frac{d I_{λ}}{d τ} e^{- τ} - I_{λ} e^{- τ} = \frac{d}{d τ} (I_{λ} e^{- τ}) = - B e^{- τ} \\ [I_{λ} e^{- τ}]_{τ^{'}}^{0} = \int_{o}^{τ^{'}} B e^{- τ} d τ \\ ∴ I (0) = - B [e^{- τ}]_{0}^{τ^{'}} = B (1 - e^{- τ^{'}}) \end{matrix}

for a dense gas, $τ^{'}$ is large, so $I (0) \to B$
this explains the emission lines as hot gases are not dense enough for pure blackbodies
absorption lines: dense blackbody emitter, with cold gas in front of it