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Lecture 8
Radiative transfer
Solar abundances from absorption lines
From an analysis of spectral lines, the following are the most abundant
elements in the solar photosphere
Element
Hydrogen
Atomic
Number
1
Log Relative
Abundance
1
Column Density
kg m-2
11
Helium
2
-1.01
43
Oxygen
8
-3.07
0.15
Carbon
6
-3.4
0.053
Neon
10
-3.91
0.027
Nitrogen
7
-4
0.015
Iron
26
-4.33
0.029
Magnesium
12
-4.42
0.01
Silicon
14
-4.45
0.011
Sulfur
16
-4.79
0.0057
Emission coefficient
The emission coefficient is the opposite of the opacity: it quantifies
processes which increase the intensity of radiation,
dI   j ds
The emission coefficient has units of W/m/str/kg
Thus, accounting for both processes:
dI     I  ds  j ds
The source function
The intensity of radiation is therefore determined by the relative
importance of the emission coefficient and the opacity

1 dI  dI 

 I   S
   ds d 
where we have defined the source function:
•
•
S 
j

The source function has units of intensity, Wm-3sr-1
As the ratio of two inverse processes (emission and absorption), the
source function is relatively insensitive to the detailed properties of
the stellar material.
Radiative transfer
This is the time independent radiative transfer equation
1 dI  dI 


 I   S
   ds d 
For a system in thermodynamic equilibrium (e.g. a blackbody), every process
of absorption is perfectly balanced by an inverse process of emission.
Since the intensity is equal to the blackbody function and therefore
constant throughout the box:
dI 
0
ds
S   I   B
Radiative transfer: general solution
dI 
 I   S
d 
I  (0)  I  ,0 e
  , 0
  ,0

 
S
e
  d 
0
i.e. the final intensity is the initial intensity, reduced by absorption, plus
the emission at every point along the path, also reduced by absorption
Example: homogeneous medium
Imagine a beam of light with I,0 at s=0 entering a
volume of gas of constant density, opacity and source
function.
I  (0)  I  ,0e
  , 0
In the limit of high optical depth
In the limit of
  ,0  1
 S 1  e 

  , 0
I  (0)  S
I  (0)  I  ,0  S  ,0
Break
Approximate solutions
Approximation #1: Plane-parallel atmospheres
We can define a vertical optical depth such that
0
  ,v ( z )     dz
where
z
i.e.
  ,v ( z )    cos 
and the transfer equation
becomes
dI 
cos 
 I   S
d  ,v
dz  ds cos
Approximate solutions
Approximation #2: Gray atmospheres
Integrating the intensity and source function over all


wavelengths,
S   S d
I   I  d
0
0
We get the following simplified transfer equation
Integrating over all solid angles,
d
I cos d   Id  S  d

d v
dFrad
 4  I  S 
d v
where Frad is the radiative flux through unit
area
cos 
dI
 I S
d v
The photon wind
cos 
dI
 I S
d v
In a spherically symmetric star with the origin at the centre
dPrad


Frad
dr
c
So the net radiative flux (i.e. movement of photons through the
star) is driven by differences in the radiation pressure
Approximate solutions
Approximation #3: An atmosphere in radiative equilibrium
Frad  Te4
Prad 

c
Te4 v  C
The Eddington approximation
To determine the temperature structure of the
atmosphere, we need to establish the temperature
dependence of the radiation pressure to solve:
 4
Prad  Te  v  C
c
Since
Prad
1
  I cos 2 d
c
We need to assume something about the angular distribution of the
intensity
The Eddington approximation
1
I 
4
2
Frad 
2

1
 0 0I sin dd  2 I out  I in 

  I cos sin dd   I
out
 I in 
0 0
2
Prad

1
4
   I cos 2  sin dd 
I
c  0  0
3c
Prad 

c
Te4 v  C
4

I  Te4 v  C
3c
c
3 4 
2
I 
Te  v  
4 
3
4 
2
Fsurface  I
 v  
3 
3
 S ( v  2 / 3)
1
This is the Eddington-Barbier relation: the
surface flux is determined by the value of the
source function at a vertical optical depth of 2/3