Notes for Ast 4001 by K. Davidson – Univ of Minnesota, 2013
The Diffusion Approximation in Radiative Transfer
--- a.k.a. the Eddington Approximation --Imagine radiation moving through a gas which can absorb, scatter, and emit photons. The
basic equation for specific intensity along a ray is
dI / ds
=  ( k abs + k sc ) I + k sc J +
 / 4 ,
(eqn. 1)
where s is path length along the ray, a straight line. Absorption and scattering coefficients
k abs and k sc as well as emissivity  may be functions of position s , but we’ll assume
that they’re isotropic -- i.e., they don’t depend on direction in space. Specific intensity is
usually defined per unit frequency or wavelength, I  or I  , so, if you feel in a pedantic
mood, here I signifies I  or I  (which are equivalent).
As a first, simple case, imagine a plane-parallel stellar atmosphere where k abs , k sc , and
 are functions of the vertical height z . By symmetry, radiation generated in this atmosphere
cannot depend on x or y , and the specific intensity can’t depend on the azimuthal direction .
So the specific intensity depends only on vertical position and on direction relative to vertical:
I ( z ,  ) . Since the upward component of direction matters most, our equations will be a
little simpler if we adopt  = cos  as our second independent variable instead of  .
Thus I = I ( z ,  ) in a plane-parallel geometry.
(Remember:  = + 1 means straight upward,  is straight downward.)
We should alter equation (1) to use z , the fundamental position coordinate, instead of s
which refers to a particular ray. Since d z = cos  d s =  d s (see figure), it becomes
  I /  z
=  ( k abs + k sc ) I
+ k sc J
+
 / 4 .
(2)
Formally this is a partial derivative  /  z instead of d / d z , because I is a function of
two independent variables z and  . But don’t worry about this mathematical distinction;
along any ray it’s the same as d / d z .
----- continued on next page -----
radtr01 - p2
Traditionally the next step is to define three quantities which are weighted averages of specific
intensity over all directional solid angle  or, in this case, direction cosines  . Why? -Because they’ll turn out to be easier to work with than the complete radiation field I .
J(z) =
H(z) =
K(z) =
 I d
( 1 / 4  )   I d
( 1 / 4  )    I d
(1/ 4  )
=
( 1 / 2 )  I d ,
(3a)
=
( 1 / 2 )   I d ,
(3b)
= ( 1 / 2 )   I d .

(3c)
The integrals cover all 4 steradians of solid angle  , or the entire range of  from 
to  . (Recall that d  = sin  d  d  =  d  d  . ) J ( z ) , for example, is simply
the average intensity at location z , averaged over all directions. In a plane-parallel geometry
J ( z ) , H ( z ) , and K ( z ) describe the most important characteristics of the radiation field
I ( z ,  ) . Each of them is mathematically simpler than I ( z ,  ) , because they’re functions
only one variable. The most essential characteristics of the direction dependence are given
by the ratios H / J and K / J. And usually J, H, K are smooth and well-behaved.
So far all this stuff is abstract, but later below we’ll notice simple interpretations of
J , H , K which are easy to visualize. For instance, J represents energy density.
For later reference, note that the first half of each definition above, with d integrals,
can be applied to a more general case where the gas is not plane-parallel.
Now look again at equation (2). By averaging it over  , ( 1 / 2 ) eqn. 2  d  , we can
convert it into an equation with J and H instead of I . The left side becomes
(1/ 2 )

  I /  z } d ,
i.e., conceptually we differentiate  I /  z and then we integrate over d . Note, however,
that the only z –dependence in this expression is in I ( z ,  ) . Therefore we can reverse the
order of operations, integrating  I d first and then differentiating d / d z . The term thus
becomes simply
d / dz
{(1/ 2 ) 
 I d } = d H / d z .
This conversion is legitimate for any I ( z ,  ) that is a nice continuous function of z and  .
Next we do the first term on the right side of eqn. (2): if k abs + k sc doesn’t depend on
direction, then the average over  is ( k abs + k sc ) J . The final two terms with J and
have no  –dependence, so their averages are themselves. Altogether, we’ve converted
eqn. (2) into a form that has only one independent variable z ( it’s “one-dimensional” ):
dH / dz
=
 ( k abs + k sc ) J
=
 k abs J
+
+ k sc J
+
 / 4
 / 4 

radtr01 - p3
Viewing eqn. (2) again, suppose we multiply it by  before integrating d . Then we’ll
get a different expression containing d K / d z and H instead of d H / d z and J . The
reasoning isn’t much different from what led to eqn. (4) above. Left side first:
(1/ 2 )

   I /  z } d = d / d z { ( 1 / 2 )

On the right side, first we see
 ( k abs + k sc )


 I d } = d K / d z .
 I d =  ( k abs + k sc ) H .
Next the k sc J term gives zero, because J doesn’t depend on  ; consequently the integral
in that term is merely

 d = 0.
Likewise the  / 4  term disappears too. Result:
d K / d z =  ( k abs + k sc ) H =  k tot H .
(5)

Equations (4) and (5) are impressively simple, but there are only two of them for three
unknown functions J ( z ) , H ( z ) , K ( z ) . In many problems we can “close the loop”
by invoking an approximate relation between J and K . Imagine a place deep in the
atmosphere where radiation cannot easily escape. I (  ) should be almost isotropic there,
because there’s almost as much downward-moving radiation as upward. But there must
be a small upward preference, and we employ a power series:
I (  ) = A + B  + C  2 + ...
A and B should be positive, and, more important, we expect A >> B >> | C | >> ...
if the location is deep enough. Coefficient B disappears when we evaluate J and K
via integrals (3a) and (3c): J = A + C / 3 + ... and K = A / 3 + C / 5 + ...
Their ratio is K / J = 1 / 3 + ( 4 / 45 ) ( C / A ) + ... If | C | << A , then we can
safely assume that K  J / 3 , so eqn. (5) becomes
dJ / dz
  3 k tot H .
(6) 
Now (4) and (6) are two equations for two unknowns, enabling us to solve many radiative
transfer problems. Remember, though, that the factor 3 is only an approximation.
K  J / 3 is sometimes called the diffusion approximation or Eddington approximation.
It’s not very accurate in regions where radiation can escape easily; for example, in one
class of problem, K  0.41 J at the top of the atmosphere. But the ratio falls below
0.35 at optical depth  1 (see below) in the same type of problem, and 1 / 3 is quite a
good approximation at deeper levels.
Equations (4) and (6) have clear interpretations if we see that J is related to radiation energy
density and H is related to the vertical energy flux. Consider a textbook-style beam of radiation,
carrying energy flux F ( W / m 2 ) and having energy density U ( J / m 3 ) . The well-known
relation between them is U = F / c . If we identify a narrow ray sample I  with F ,
and then add up the radiation densities for all directions, it isn’t hard to show that
U
=

I d / c = 4 J / c .
(7)
radtr01 - p4
What about H ? Consider a flux that is somewhat differently defined. In order to measure the
vertical energy flux F z (i.e., flux in the z –direction, at a given location), imagine a sampling
area A that is horizontal in the plane-parallel atmosphere. Conceptually, we measure the net
rate at which energy passes upward through that area – it’s a “net rate” because we must subtract
the downward rays. If you draw a sketch and think about the effect of I  in each direction,
you can show that
Fz =

 I d
= 4 H .
(8)
Knowing interpretations (7) and (8), we can rewrite eqns. (4) and (6) as
dFz / dz
=
 k abs c U
+

F z   ( c / 3 k tot ) d U / d z .
(10) 
Look at (10): an energy flux is proportional to an energy gradient. Where have we seen
that before? Everyday heat transfer! Flux   gradient is a diffusion equation. This thought
inspires a way to generalize the equations, so they’re not restricted to plane-parallel cases.
Density U is obviously a scalar function, and F z looks like the z –component of a vector
function; F x and Fy are zero here only because the geometry has plane-parallel symmetry.
If this surmise is true, then eqn. (10) must be the z -component of a vector equation.
What’s the easiest way to generalize the right side so d U / d z is a component of a vector?
 U , of course!
F   ( c / 3 k tot )  U .
(11) 
Indeed this is the classic steady-state diffusion equation in three dimensions. And, unlike most
examples that you can read about, we even have an approximate expression for the diffusion
coefficient, c / 3 k tot . Warning, however: This is sometimes a poor approximation for
locations where radiation can escape easily. But it’s fairly good at optical depths larger than
1 or 2, and quite accurate where radiation is strongly trapped at optical depths larger than 4.
Now let’s try to guess a three-dimensional equation that might have (9) as a special case.
The right side with U and  is obviously scalar rather than vector. Thus we have to guess
a formula that makes a scalar quantity out of expressions that resemble d F z / d z . The
simplest possibility is the divergence,  F =  Fx /  x +  Fy /  y +  F z /  z . This
idea is surely right, because the resulting equation
 F =  k abs c U +

(12)
has a clear physical meaning! According to Gauss’ theorem,  F V is the net flow of
radiation coming out of a small localized volume V . (See chapters about  E and  B
in a good E & M textbook.) On the right side of (12),  is the supply of radiation per unit
volume while k abs c U is the rate of absorption per unit volume. So the equation says
“net amount emerging from a small volume = amount emitted minus amount absorbed.”
This is so natural and sensible that the foregoing scalar-vector guesses must be correct.
radtr01 - p5
____________________________________________________________________________
A cultural digression related to other branches of physics
Look again at eqns. (7) and (8). If J and H are related to density and flux, then what is K ?
We can get a hint from eqn. (3b) which defines H . It contains a direction cosine  relative
to the z – axis; we could signify this by labeling it  z . Then an obvious generalization is
to use two other direction cosines relative to the other axes,  x and  y . Then  x ,  y ,
and  z are the components of a unit vector signifying direction. Thus, by inserting  x ,
 y , and  z in place of  , we can convert (3b) into three equations for components of a
vector field H ( x , y , z ) . In math language, “H transforms like a vector” if we rotate the
coordinate axes. Eqn. (8) indicates that F = 4  H .
Following this line of thought, next look at eqn. (3c) which defines K in the plane-parallel
case. With three direction cosines  x ,  y ,  z , the integral in (3c) can obviously have
 x2 ,  y2 , or z2 instead of  2 . Viewed this way, each component of K is proportional
to the rate of momentum transfer in that direction, which is pressure P . (In a physics
textbook chapter on kinetic theory of particle motions, pressure is related to the squares of
velocity components, v x2 etc. Those components involve the directional quantities  x2 ,
 y2 , z2 .) Thus we find that P = 4  K / c , and the Eddington approximation amounts
to P  U / 3 for photons -- a fact that’s worth remembering.
But the concept of K is more general that that! The broader form of eqn. (3c) can also have
two-component terms with  x  y ,  x  z , etc. ... It can be converted into a matrix K i j
with i , j signifying the x , y , z directions. Apart from a factor of 4  / c , this turns out
to be the stress tensor which plays a major role in fluid physics, plasma physics, and even
general relativity. The diagonal xx , yy , zz terms can be identified with pressure while
the off-diagonal terms with xy , xz , etc. represent viscosity.
____________________________________________________________________________
Now, back to the matter at hand. The coefficients in equations (9) and (10) involve quantities
that may be functions of position: k abs ( z ) , k tot ( z ) ,  ( z ) . As an example of what to do
about this, suppose that k abs ( z ) =  k tot ( z ) where  is a constant. In this case, we define
optical depth  ( z ) by d  =  k tot d z , choosing whichever sign ( or) seems to fit the
problem best. In principle we know the relation between  and z by integrating d  . Here
let’s choose d  =  k tot d z . Then eqns. (9) and (10) become
dFz / d
=
  cU
  c S (  ) ,
Fz   ( c / 3 ) dU / d ,
(13)
(14)
where S (  ) =  ( z ) / c k abs ( z ) is sometimes called the “source function.” These equations
are fairly easy to integrate if we know S (  ) , since each coefficient is constant. The situation
becomes more complicated if  = k abs / k tot is not a constant, but even then we usually employ
optical depth  .
radtr01 - p6
Boundary conditions
In astrophysics, the most common plane-parallel type of problem concerns a stellar atmosphere
where k abs ( z ) , k tot ( z ) ,  ( z ) become small for large positive z . In that case we can set
 = 0 at the top of the atmosphere and  increases downward, attaining large values deep in
the atmosphere. By substituting eqn. (14) into (13) we get a second-order differential equation:
d 2 U / d  2 =  3  U  3 S (  ) ,
(15)
presumably spanning the range 0 <  <  . A second-order equation needs two boundary
conditions. One of them is merely that U (  ) must not become absurd for large  ; this leads
us to exclude a term with a rapidly growing  -dependence like exp { + ( 3   } .
But here we’re more concerned with the boundary condition at  = 0 , the top of the atmosphere
where radiation escapes. There, the absence of downward radiation implies a close relation
between F ( 0 ) and U ( 0 ) . If the radiation could emerge as a vertical beam then we’d have
F = c U . A beam in any other direction would have its upward flux diminished by its direction
cosine  – for instance, the middle example in the figure shows a beam 60 degrees from vertical.
In any realistic case, I ( 0,  ) extends through the whole range of outward directions 0 <  < +1,
Thus we’ll have F =  av c U at the boundary, with  av = average direction cosine of the
emergent radiation field I ( 0,  ) .
What value shall we use for  av ? It’s the same thing as H / J , the ratio {eqn. 3b} / {eqn. 3a}
evaluated at the boundary.  av probably can’t be smaller than 1/ 2, which is the value for
isotropic outward radiation ( I = constant for 0 <  < +1 ). [Why? Think about it.] Other
examples suggest the range 0.5 <  av < 0.7 , and 0.6 is generally a fair guess. This
statement isn’t as sloppy as it seems! For instance, the approximate solution for a simple
pure-scattering problem turns out to be U (  ) = { 3  + 1 /  av } F / c . A 15 % error
in our choice of  av causes an error in U of only about 5 % at  = 1 and 1.5 % at
 = 5 . These are good enough for many purposes.
Not having an exact value for  av , sometimes we choose a value that makes our solution
formula as simple as possible -- typically 0.5, or 1 / 3 , or 2 / 3 , depending on the details.
Using fancy methods not described here, the exact solution to the simple pure-scattering
problem is known to be U (  ) = { 3  + q (  ) } F / c with a slowly varying function q (  ) ;
and q ( 0 ) = exactly  3 . Comparing this to the approximate solution noted above, we see
that  av = 1 / 3 in that particular problem. This doesn’t apply in general, however.
radtr01 - p7
More accurate methods
At small optical depths  , the diffusion or Eddington approximation isn’t accurate enough for
professional-grade atmosphere models. The classic book on more sophisticated methods is
Radiative Transfer by S. Chandrasekhar, written more than 60 years ago. That was before the
computer age, but some of the concepts are well suited to modern numerical computation.
For numerical work we have to represent I ( z ,  ) with a discrete set of rays having particular
directions. Recall that radiation density and flux are integrals of I d  through the range
 = 1 to +1, eqns. (3a, 3b). We replace each integral with a sum:

I ( ) d  

ai I (  i ) ,
where the  i’ s are a set of sample values and the a i’s are associated “weights” that add up
to 2 because we’re integrating from  = 1 to +1. (Each a i is like an integration interval  .)
For instance we could choose 4 directions with  i = 0.75, 0.25, +0.25, and +0.75, each with
weight 0.5. Then eqn. (2) becomes four simultaneous differential equations in four unknowns:
quite solvable with a computer.
But an almost magical trick called Gaussian integration gives a much better choice of direction
samples. (Look it up as “gaussian quadrature.”) This involves 2 n positive and negative sample
values  i , i = 1 ... n . They coincide with the zeros of a Legendre polynomial of order 2 n ,
and there’s a formula for the weights a i . The two simplest examples are –
n = 1:
 = 1 / 3 = 0.577... , with a = 1.0 for each.
n = 2:
 = 0.33998... with a = 0.65214... and
 = 0.86113... with a = 0.34785...
Gaussian integration is wonderful because 2 n sample ’s produce a correct integral of any
polynomial up to order 4 n – 1. Thus n = 1 works for a cubic polynomial, which is good,
and n = 2 works for a seventh-order polynomial even though we use only 4 sample points -which is amazing. Basically eqn. (2) gives 2 n simultaneous equations, we solve them with
a computer, and eqns. (3a,b,c) then give accurate values for the radiation density, flux, and
pressure. The boundary conditions are simple and definite, e.g. I = 0 at  = 0 for any negative
value of  . In a computer code we can use larger values of n , although n = 2 is sufficient
for quite accurate results.
Sophisticated math gives exact solutions for a few problems. For instance, the most common
pure-scattering problem is covered in chapters III and V of Chandrasekhar’s book – take a look
at those pages for a glimpse of virtuoso classical analysis.
The above techniques work well only for plane-parallel cases. Even for a spherical problem,
accurate solutions are difficult. Sometimes the best recourse is a Monte Carlo simulation,
using millions of random numbers to follow the progress of thousands of photons.