Modeling stellar atmospheres is typically a very complex undertaking that often requires we have detailed knowledge of stellar spectra and its behavior in different conditions. However, it is possible to approach the topic at a rudimentary (but still useful level) by appealing to simplified assumptions.
The
“gray atmosphere” is one such simplifying assumption. First, we need to present some preliminaries.
The
Planck function describes the distribution of radiation for a black
body, and can be expressed:
B(l) = {(2 hc2)/
l5} [1/ exp (hc/lkT) - 1)]
where h is Planck’s constant, c is the speed of light, T is the absolute
temperature, k is the Boltzmann constant, and l, the wavelength. In the plane-parallel
treatment, we take layers of the gases in a stellar atmosphere to be like
layers of a “sandwich”, where ds is an element of length or path perpendicular
to the layers.
-------------------
-------------------
-------------------] --------------------------------------- ds
This is opposed to employing curved layers (as would technically be the case), for which the math is many times more complex! A more detailed plane-parallel model atmosphere is depicted below:
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. However, what follows should not be beyond the ability of anyone who’s done first year 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.
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
Problems:
1) Find the effective temperature of the Sun and the boundary temperature (To) and account for any difference.
(Take the Stefan -Boltzmann constant s = 5.67 x 10-8 W m-2 K-4 )
2) Prove J = ½(I1 + I2)
By integrating I in the forward and backward directions.
No comments:
Post a Comment