Why the source function is equal to Planck function when we have a local thermodynamic equilibrium? I understand that the source function $ S_λ $ for the special case of blackbody radiation is equal to the Planck function $B_λ $. 
However, in the broader case of a local thermodynamic equilibrium (and not the special case of a blackbody) I would expect that
$$ S_λ=εB_λ $$ where $ε$ the emissivity
and the equation of radiative transfer to be:
$$ \frac{dI_λ}{k_λρds}=-I_λ+εB_λ $$
and not
$$ \frac{dI_λ}{k_λρds}=-I_λ+B_λ $$ 
Where do I make a mistake?
 A: Remembering my lessons...
In LTE, the collisions dominate over the radiative transitions, then the probability of an emitted photon to be "destroyed" by a collision is much higher than to be scattered (abosrbed and re-emitted) by the atom. The $\epsilon$ parameter defines this concept ($\epsilon=\frac{C_{ul}}{C_{ul}+A_{ul}}$). For this reason, in the LTE regime $\epsilon=1$. 
In the case of NLTE (non-LTE), the source function is defined as $S_{\lambda}=\epsilon B_{\lambda} + (1-\epsilon)J_{\lambda}$, where the $J_{\lambda}$ term considers the scattering role of the radiation field.
A: The source function is a property of the body itself and not the radiation field it is immersed in. If the object is in local thermodynamic equilibrium then the source function only depends on temperature.
In principle you could put the body in a cavity filled with black body radiation at a given temperature and it would come into equilibrium at that same temperature. Since a blackbody radiation field is isotropic then there can be no gradient in the specific intensity and therefore the source function at temperature $T$ must equal the Planck function at the same temperature.
In your prescription the radiation field could never be isotropic since there would be a gradient, $dI_\lambda/ds \neq 0$, in the specific intensity, but this is inconsistent with a blackbody radiation field.
