Friday, June 21, 2013

Introducing Some Advanced Astrophysics (4)

    Modeling stellar atmospheres, just as computing self-consistent stellar models, is similarly a very complex undertaking that often requires we make basic assumptions. However, it is possible to approach the topic at a rudimentary (but still useful level) by appealing to simplified assumptions and not being so rigorous as to expectations.

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 the sketch diagram above.   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 specific 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) :  i.e.  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.

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 Teff   is the effective temperature and  s is the Stefan-Boltzmann constant (5.67 x 10-8 W m-2 K-4 ).  Thus, the temperature at optical depth 2/3 must equal the effective temperature!

Problems for the budding astrophysicst:

1.  Estimate the specific intensity I (q=p/4) if the surface flux from the Sun is  6.3 x 10 7 Jm-2 s-1.

2.  Find the effective temperature of the Sun and the boundary temperature (To) and account for any difference. (Hint: The effective temperature is related to the boundary temperature by: Teff  = (2)1/4 To   )

No comments: