 Solution

The specific intensity $I_{\nu }$ is energy transported per unit time, projected area, solid angle and frequency. The path coordinate $s$ increases along the ray. The emission coefficient $j_{\nu }$ is emitted energy per unit time, volume, solid angle and frequency, whereas the absorption coefficient ${\alpha }_{\nu }$ has units of inverse length. Thus $j_{\nu }ds$ adds specific intensity and ${\alpha }_{\nu }{I}_{\nu }ds$ removes it. Stimulated emission can be included as negative absorption.

Define the optical depth between two points by ${\tau }_{\nu }(s_1,s_2)={\int }_{s_1}^{s_2}{\alpha }_{\nu }(u)du$. Multiplying the radiative transfer equation by the integrating factor $\exp [{\tau }_{\nu }(s_0,s)]$ and integrating givesEach local contribution is attenuated only by the material between its emission point and the observer. Where ${\alpha }_{\nu }\ne 0$, the source function is $S_{\nu }=j_{\nu }/{\alpha }_{\nu }$ and the same formal solution of the radiative transfer equation is $I_{\nu }(\tau )=I_{\nu }(0)e^{-\tau }+{\int }_0^{\tau }{S}_{\nu }(t)e^{-(\tau -t)}dt$. A constant source function gives $I_{\nu }=I_{\nu }(0)e^{-\tau }+S_{\nu }(1-e^{-\tau })$.