As a beam of radiation (Il) passes through parallel-layered stellar gases, there will be emission and absorption along the way. The “source function” specifies the ratio of one to the other and can be expressed:
S(l) = e(l) / k(l)
where l again denotes wavelength, e(l) is the emission coefficient, and k(l) the absorption coefficient.
In the case of simple radiation transfer in a static model stellar atmosphere (e.g. nothing changes with time), we have the relation of radiation intensity I(l) to source function S(l):
dI(l)/ds = -k(l) I(l) + k(l) S(l)
or I(l) = S(l)
Now, for a black body, I(l) equals the Planck function B(l) :
So, in effect, we have:
S(l) = I(l) = B(l)
And this is a condition which implies LOCAL THERMODYNAMIC EQUILIBRIUM or LTE
LTE does NOT mean complete thermodynamic equilibrium!(E.g. since in the outer layers of a star there is always large energy loss from the stellar surface)
Thus, one only assumes the emission of the radiation is the same as for a gas in thermodynamic equilibrium at a temperature (T) corresponding to the temperature of the layer under consideration. Another way to say this is that if LTE holds, the photons always emerge at all wavelengths.
Now, in the above treatment, note that the absorption coefficient was always written as: k(l) to emphasize its wavelength (l) dependence.
However, there are certain specific treatments for which we may eliminate the wavelength dependence on absorption, and simply write e.g. k – to denote having the same absorption value at ALL wavelengths!
This is what is meant by the “gray atmosphere” approximation.
Here is a specific application of the gray atmosphere approximation. In a particular integral, let the surface flux
p( Fo ) = 2 p (I(cos (q)) = p [a(l) + 2(b(l)/3 ]
and Flo = S(l) t(l) = 2/3
which states that the flux coming out of the stellar surface is equal to the source function at the optical depth t = 2/3. This is the very important ‘Eddington-Barbier’ relation that facilitates an understanding of how stellar spectra are formed.
Once one then assumes LTE, one can further assume k(l) is independent of l (gray atmosphere) so that:
k(l) = k; t (l) = t and Flo = Bl (T(t = 2/3) )
Thus, the energy distribution of Fl is that of a black body corresponding to the temperature at an optical depth t = 2/3.
From this, along with some simple substitutions and integrations (hint: look at the Stefan-Boltzmann law!) the interested reader can easily determine:
p( Fo ) = s (Teff)4 and Teff = T(t = 2/3)
where s is the Stefan-Boltzmann constant. Thus, the temperature at optical depth 2/3 must equal the effective temperature!
Intensity and Moments of Intensity:
For more detailed analyses, more complex mathematics is required. What follows should not be beyond the ability of anyone who’s done first year college calculus. What we want is to first be able to write or express the emergent intensity as it’s depicted in Fig. 1. This is done by first writing:
Il (0,q) = òo z Bl(t) exp [(-tl / cos q)] dt/ cos q
where Bl(t) is the Planck function, and the integral is taken from z = 0 to the point z in the interior.
Il (0,q) = òo¥ B(x) e (-x / cos q) dx/ cos q
Then the appropriate equation of transfer would be:
dI = j dx - s I (q) dx = (j - s I (q)) dx
or: dI/dx = j - s I (q))
After some further manipulation, and replacing x with t:
(cos q) dI/ dt = I (q) – j/ s
This is the important equation, in terms of emergent intensity, that embodies the conservation of radiant energy (i.e. no more radiation can flow out of a star’s surface than can be generated within it and which approaches that surface).
Next, we want to be able to obtain an even more improved basis for our calculations and this entails getting the moments of the intensity. These are defined as follows:
J = 1/4p ò 4p I (q) dw (Mean Intensity)
Where dw is an element of solid angle - defined as (A/r2) for a sphere, for example. Thus a sphere with surface area A = 4p r2 has solid angle (4p r2 / r2 ) = 4p steradians. If we are only dealing with a sliver of emergent beam of area 0.01p r2 then the element of solid angle is:
dw = (0.01p r2 / r2) = 0.01p sr
Or 1/ 400 the volume of a sphere. The next moment is:
H = 1/4p ò 4p I (q) cos q dw
This is defined as the “net flux” or the net energy breaching the stellar surface in units of net energy per second per unit area of that surface.
Finally, we come to the last moment of intensity:
K = 1/4p ò 4 p I (q) cos 2 q dw
This is the energy density.
At this stage, we are in a position to use each of the above moments to further manipulate the equation 0f transfer. We start by multiplying the original equation of transfer:
(cos q) dI/ dt = I (q) – j/ s
through by 1/4p ò 4p I (q) dw to get:
dH/ dt = 1/4p ò4p I (q) dw - 1/4p ò 4p j/ s dw
This makes use of the definition: dH/ dt = J - j/ s
Example Problem: Verify the above definition using the definitions of J, H and the transfer equation. Thence arrive at a new form for the equation.
To solve, we multiply the equation of transfer through by H, or:
1/4p ò 4p I (q) cos q dw i.e. to get: dK/ dt = H
Or more simply: 4p H = const.
So, dH/ dt = J - j/ s = 0, and the equation of transfer now becomes:
(cos q) dI/ dt = -I (q) + 1/4p ò 4p I (q) dw = - I + J
From here a number of specific assumptions are made in order to not have to evaluate the integral. The main one is the Eddington approximation which will apply to the quantities J, H and K, written as follows:
1) J = ½(I1 + I2)
2) H = ¼( I1 - I2)
3) K = J/3
Next we focus on the boundary in the plane parallel atmosphere, and note that here the optical depth t= 0, and we must have I2 = 0 also. Since I2 = 0 then: H = ¼ I1 and J = ½I1 so clearly: J = 2H. Further, K = J/3 or (½I1)/3 so K = Ht + const.
This follows, since we had: dK/ dC = H or dK = H dt and we know t= 0, hence Ht + const. on integration. From this it follows that:
J = H(2 + 3 t) and K = J/3 = 2H/3
At the boundary everywhere.
And since H = ¼( I1 - I2) = const.
I1 (t) = H(4 + 3t) and I1 (t) = 3Ht
A special case occurs if the mean intensity J = B, the Planck function, then (since B » sT4/ p):
J = H(2 + 3 t) = sT4/ p
Therefore, the boundary temperature (To) approaches the value of the effective (or surface) temperature when t = 0. So we have the basic relationship:
sTo 4/ p = 2H
And: sT4 = sTo 4/ 2 [a + 3t]
In the limit of this approximation, Teff 4 = 2 To 4
And hence: Teff = (2)1/4 To = 1.189 To
Problem:
Consider the solar half-sphere and the energy going into it each second. We know on average photons are absorbed after traveling a distance with optical depth t =1 in the propagation direction. Averaged over all directions this corresponds to a vertical optical depth of t = 2/3. Based on this find:
a) The energy going into the half sphere each second
b) The change in (a) over each absorption and re-emission over vertical optical depth.
c) The total absorption and total emission and the relationship between then over all space.
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