Saturday, October 30, 2010

Introducing Some Basic Astrophysics

Let's now start examining a number of basic problems typically posed in astrophysics. We'll start out with one of the most fundamental, the nature of "gray atmospheres" in stellar modelling. Stellar modelling itself is a topic that can occupy whole monographs (e.g. 'An Introduction to the Study of Stellar Structure' by S. Chandrasekhar) but we will confine current interest strictly to one simplified aspect of stellar atmospheres for now. (At a later point, we can examine how a full stellar model is developed from the interior to the outer envelope).

Modeling stellar atmospheres is similarly a very complex undertaking that often requires we make basic assumptions. The “gray atmosphere” is one such simplifying assumption.

First, some preliminaries on essential technical terms, etc.

The Planck function describes the distribution of radiation for a black body, and can be expressed:

B(L) = {(2 hc^2)/ L^5} * [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 defines 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
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This is opposed to employing curved layers (as would technically be the case, but for which the math is many times more complex!)

Now as a beam of radiation passes through 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 coefficent, and k(L) the absorption coefficient.

In the case of simple radiation transfer in a 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) = k(L)[S(L) – I(L)] - 0

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 for any stellar atmosphere – implies LOCAL THERMODYNAMIC EQUILIBRIUMor LTE

Note: 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:ke.g. k has the same value at ALL wavelengths!

This is what is meant by the “gray atmosphere” approximation.

I’ll now give a specific application of the gray atmosphere approximation. In a particular integral, the surface flux

(pi F(O)) = 2 pi (I(cos (theta)) =[a(L) + 2(b(L)/3 ] pi

and F_L(0) = 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 important ‘Eddington-Barbier’ relation that paves the way for the 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 andF_L(0) = B_L(T(t = 2/3)

Thus, the energy distribution of F_L is that of a black body corresponding to the temperature at an optical depth t = 2/3.

From this, with some simple substitutions and integrations (hint: look at the Stefan-Boltzmann law!) the interested reader can easily determine:

pi F(O) = o(T_eff)^4 and T_eff = T(t = 2/3)

where o is the Stefan-Boltzmann constant. Thus, the temperature at optical depth 2/3 must equal the effective temperature!