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Chapter 9
Stellar Atmospheres
Specific Intensity, I
I ( or I ) is a vector (units: W m-2 Hz-1 sterad-1 )
Specific Intensity, I
I is a vector (units: W m-2 Hz-1 sterad-1 )
Illustration of an Anisotopic Radiation Field
Specific Intensity, I
Define: outward Flux, F, luminosity, L, mean intensity, J and
radiation pressure, P. The main objective of all this math is to
explain limb darkening!
The Radiative Transfer Equation
dI=- I ds + jds
Where is the opacity (note the minus sign) and
j is the emissivity. is the density and ds is the
path length along the beam.
Sources of Opacity
• Bound-Bound absorption, ,bb produces spectral absorption
lines, consequently the opacity is small except at discrete
frequencies corresponding to transitions in the atom. There is
no simple expression for ,bb
• Bound-Free absorption, ,bf also known as photoionization,
occurs when a photon has enough energy to ionize an atom.
An approximate estimate for the photoionization crosssection of an H atom is given by
σbf = 1.31 x 10-15 n-5 ( λ/5000 Å)3 cm2
Sources of Opacity (continued)
• Free-free absorption, ,ff is a scattering process, a free
electron absorbs a photon in the vicinity of an ion, increasing
the speed of the electron. This process can occur over a range
of wavelengths and so contributes to the continuum opacity.
Free-free absorption is also known as Bremstrahlung which is
German for “breaking” as in “acceleration radiation.
• Electron (Thomson) scattering, es a photon is scattered, but
not absorbed. The cross-section for Thomson scattering is
independent of frequency;
σT = 8 π (e2 /mec2)2 = 6.65 x 10-25 cm2
3
Sources of Opacity (continued)
The total opacity, = ,bb + ,bf + ,ff + es
For the purposes of stellar atmosphere calculations, a frequency
average opacity, , is employed, called the Rosseland mean
where;
bf ~ ρT-3.5 m2 kg-1 see eqn. 9.22
ff ~ ρT-3.5 m2 kg-1 see eqn. 9.23
es ~ 0.02 (1 + X) m2 kg-1 (for a gas of H mass fraction X)
see eqn. 9.27
These opacity functions are plotted in Fig. 9.10
Mean molecular mass and mass fractions.
Consider a star made of Hydrogen and Helium, then the mass, m,
of an average atom,
m = (NH mH + NHe mHe) / (NH + NHe )
…..(1)
Where NH and NHe are the number of H and He atoms, and mH
and mHe are the masses of a single H and He atom, respectively.
However, it is more useful to express the composition of a star in
terms of mass fractions, where X = MH/MTot and Y = MHe/Mtot
So, invert equation 1, above, to get
1/m = (NH + NHe )/ (NH mH + NHe mHe)
1/m = (NH + NHe )/ (NH mH + NHe mHe)
the denominator is the total mass of gas
1
m
=
NH
MTot
+
NHe
MTot
Multiply each term on the R.H.S by 1, by forming a ratio mH/mH for the first term and
mHe/mHe for the second term;
1=
m
NHmH +
MTot mH
1=
m
X +
mH
NHemHe
MTot mHe
Y
mHe
By definition, m = mH, so
1
mH
=
X +
mH
Y
mHe
1=
mH
X +
mH
Y
mHe
Now cross multiply by mH
1=
X
+
mH Y
mHe
where mH /mHe = ¼, is just the ratio of atomic weights, A.
So, that
1=
X
+
Y
4
Additional terms may be added for additional elements, but usually one additional term, Z,
which refers to the mass fraction of everything, other than H and He, collectively referred
to as “metals”.
An equivalent expression for the mean electron molecular weight is given by
1=
e
1 X
+
2Y
4
Where the numerator is the mass fraction times the number of electrons for each element.
The Source Function, S
1/ is the ‘mean free path’, so that is the number of mean free paths
in distance s. Then the radiative transfer equation;
dI = - I ds + jds
becomes
dI= -I +
ds
j
Usually written as
- dI= I ds
Where
S = j
S
is the Source Function
Optical Depth,
d = ds
1/ is the ‘mean free path’, so that is the number of mean free paths
in s.
If z is the vertical displacement, then z = -s Cos (see Fig 9.14)
ds = - dz
Cos
Then the radiative transfer equation
dI=- I ds + jds
becomes
Cos dI= I - j
d’
where ’ = dz, is the vertical optical depth.
Integrate R.T.E. for the Eddington Atmosphere
the radiative transfer equation, integrated over d, is
Cos dI = I - S
d’
where ’ is the vertical optical depth. The frequency dependence in
is taken care of by the Rossland Mean opacity.
Take two moments of the R.T.E. which means integrate over solid angle,
d, to get the first moment. Then multiply by Cos and integrate again to
get the second moment. After some algebra and using some previous
results you can show that
T4 = ¾ Te4 (’ + 2/3) where T is the temperature at a vertical optical depth,
’ and Te is the effective temperature, characterizing the light we see from
the star.
The result shows that the light we see from the star emerges not from ’ =
0 as you would expect, but rather from ’ = 2/3. This explains limb
darkening as at large , you see higher in the atmosphere where the
atmosphere is cooler and hence emitting less light.
Spectral Line Formation
Many of the same concepts are encountered in the formation of spectral lines.
The line opacity is greatest at the center of the line. Consequently, the radiation
at the line center emerges from higher in the photosphere.
This would make a good research project……