Radiative Transfer Equation
Specific intensity of radiation fields
is defined by
the radiation energy
transferred by photons with wavelength
through a unit surface placed at
whose normal is directed to
per unit time per unit wavelength, and per unit steradian.
Average intensity of radiation
is defined as
![\begin{displaymath}
J(\mbox{\boldmath${x}$},\nu)=\int I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu) \frac{d\Omega}{4\pi},
\end{displaymath}](img691.png) |
(2.125) |
where
is obtained by averaging
for the solid angle.
This is related to the energy density of radiation
as
![\begin{displaymath}
u(\mbox{\boldmath${x}$},\nu)=\frac{4\pi}{c}J(\mbox{\boldmath${x}$},\nu).
\end{displaymath}](img694.png) |
(2.126) |
If the radiation is absorbed in the displacement
as
,
must be proportional to
and
as
![\begin{displaymath}
\delta I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu) = ...
...a I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu) \delta s,
\end{displaymath}](img698.png) |
(2.127) |
where
is a coefficient and called volume absorption coefficient.
The dimension of
is
.
We can rewrite the above to the differential equation as
![\begin{displaymath}
\frac{d I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu)}{...
..._\rho \rho I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu),
\end{displaymath}](img700.png) |
(2.128) |
where we used mass absorption coefficient
which represents
the absorption per mass.
Equation (2.127) is reduced to
![\begin{displaymath}
\frac{dI(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu)}{\...
...}{d \tau}=-I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu),
\end{displaymath}](img702.png) |
(2.129) |
where
![\begin{displaymath}
\tau= \int_0^s \kappa \delta s,
\end{displaymath}](img703.png) |
(2.130) |
is called the optical depth.
This means that
is a measure for absorption as
the intensity decreases at a factor
from
to
.
If the ray runs
crossing a volume whose
volume emissivity equal to
the intensity increases
![\begin{displaymath}
\delta I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu)= \epsilon(\mbox{\boldmath${x}$},\nu)\delta s.
\end{displaymath}](img709.png) |
(2.131) |
The volume emissivity
is the energy
emitted by a unit volume at a position
per unit time
per unit solid angle per unit wavelength.
From equations (2.127) and (2.131),
the radiation transfer is written as
![\begin{displaymath}
\frac{d I(\mbox{\boldmath${x}$},\mbox{\boldmath${n}$},\nu)}{...
...mbox{\boldmath${n}$},\nu)+\epsilon(\mbox{\boldmath${x}$},\nu).
\end{displaymath}](img711.png) |
(2.132) |
Using the optical depth
,
this gives
![\begin{displaymath}
\frac{dI(\nu)}{d\tau}=-I(\nu)+{\cal S}(\nu),
\end{displaymath}](img713.png) |
(2.133) |
where
is called the source function and is defined as
.
Assuming the specific intensity
at the point of
,
that at the point
on the same ray is given
![\begin{displaymath}
I=I_0\exp\left(-\tau\right)
+\int_0^{\tau(x)} {\cal S}(s(\tau'))\exp\left(\tau'-\tau(s)\right)d\tau'.
\end{displaymath}](img717.png) |
(2.134) |
In the case of constant source term
const it reduces to
![\begin{displaymath}
I=I_0\exp\left(-\tau\right)+{\cal S}\left(1-\exp\left(-\tau\right)\right).
\end{displaymath}](img719.png) |
(2.135) |
Equation (2.134) gives
![\begin{displaymath}
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.
\end{displaymath}](img720.png) |
(2.136) |
This indicates that if we see an optically thick cloud
the specific intensity reaches us represents
,
while if we see an transparent cloud
,
represents
that of background.
Problem
Show that
equations (2.134) and (2.134)
are solutions of equation (2.133).
Kohji Tomisaka
2012-10-03