Radiative Transfer Equation

Specific intensity of radiation fields $I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu)$ is defined by the radiation energy $dE$ transferred by photons with wavelength $\nu$ through a unit surface placed at $\mbox{\boldmath${x}$}$ whose normal is directed to $\mbox{\boldmath${n}$}$ per unit time per unit wavelength, and per unit steradian. Average intensity of radiation $J(\mbox{\boldmath${x}$},\nu)$ is defined as

$$J(\mbox{\boldmath${x}$},\nu)=\int I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu) \frac{d\Omega}{4\pi},$$ (2.121)

where $J$ is obtained by averaging $I$ for the solid angle. This is related to the energy density of radiation $u(\mbox{\boldmath${x}$},\nu)$ as

$$u(\mbox{\boldmath${x}$},\nu)=\frac{4\pi}{c}J(\mbox{\boldmath${x}$},\nu).$$ (2.122)

If the radiation is absorbed in the displacement $\delta s$ as $\delta I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu)$, $\delta I$ must be proportional to $\delta s$ and $I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu)$ as

$$\delta I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu) = ... ...a I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu) \delta s,$$ (2.123)

where $\kappa$ is a coefficient and called volume absorption coefficient. The dimension of $\kappa$ is ${\rm cm}^{-1}$. We can rewrite the above to the differential equation as

$$\frac{d I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu)}{... ..._\rho \rho I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu),$$ (2.124)

where we used mass absorption coefficient $\kappa_\rho$ which represents the absorption per mass. Equation (2.123) is reduced to

$$\frac{dI(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu)}{\... ...}{d \tau}=-I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu),$$ (2.125)

where

$$\tau= \int_0^s \kappa \delta s,$$ (2.126)

is called the optical depth. This means that $\tau$ is a measure for absorption as the intensity decreases at a factor $\exp(-1)$ from $\tau=0$ to $\tau=1$.

If the ray runs $\delta s$ crossing a volume whose volume emissivity equal to $\epsilon$ the intensity increases

$$\delta I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu)= \epsilon(\mbox{\boldmath${x}$},\nu)\delta s.$$ (2.127)

The volume emissivity $\epsilon(\mbox{\boldmath${x}$},\nu)$ is the energy emitted by a unit volume at a position $\mbox{\boldmath${x}$}$ per unit time per unit solid angle per unit wavelength. From equations (2.123) and (2.127), the radiation transfer is written as

$$\frac{d I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu)}{... ...mbox{\boldmath${n}$},\nu)+\epsilon(\mbox{\boldmath${x}$},\nu).$$ (2.128)

Using the optical depth $\tau=\int_0^s \kappa ds$, this gives

$$\frac{dI(\nu)}{d\tau}=-I(\nu)+{\cal S}(\nu),$$ (2.129)

where ${\cal S}(\nu)$ is called the source function and is defined as ${\cal S}(\nu)=\epsilon(\nu)/\kappa(\nu)$. Assuming the specific intensity $I=I_0$ at the point of $\tau=0$, that at the point $\tau$ on the same ray is given

$$I=I_0\exp\left(-\tau\right) +\int_0^{\tau(x)} {\cal S}(s(\tau'))\exp\left(\tau'-\tau(s)\right)d\tau'.$$ (2.130)

In the case of constant source term ${\cal S}=$const it reduces to

$$I=I_0\exp\left(-\tau\right)+{\cal S}\left(1-\exp\left(-\tau\right)\right).$$ (2.131)

Equation (2.130) gives

$$I\simeq\left\{ \begin{array}{lc} {\cal S} \ldots& \tau \gg... ...u {\cal S}+I_0(1-\tau) \ldots& \tau \ll 1. \end{array}\right.$$ (2.132)

This indicates that if we see an optically thick cloud $\tau \gg 1$ the specific intensity reaches us represents ${\cal S}$, while if we see an transparent cloud $\tau \ll 1$, $I$ represents that of background.

Problem
Show that equations (2.130) and (2.130) are solutions of equation (2.129).