Download Lecture 6: Continuum Opacity and Stellar Atmospheres

Lecture 6: Continuum Opacity and Stellar Atmospheres
To make progress in modeling and understanding stellar atmospheres beyond the
gray atmosphere, it is necessary to consider the real interactions between
photons and “particles” (atoms, electrons, and sometimes molecules and dust
grains). Clearly, the full calculation of opacity is incredibly difficult because there
are so many “species” to deal with (92 elements, each with a variety of ionization
stages! plus the electrons.) And that doesn’t even count the molecules! Here we
will consider only the dominant opacity sources in stellar atmospheres.
Basic Definitions:
We have introduced the “volumetric” opacity (k) which is the crosssection per cubic cm (or m) of the material. It has the units of area/volume
which, of course is a linear dimension. Hence optical depth (which is a unitless
variable -- i.e. the number of mfp through a medium) is given by
!
τν =
kν ds
path
An alternate definition is often used, including in the textbook, which is the
“mass” opacity (kappa), namely the cross-section per gm (or kg).
In this case,
τν =
!
κν ρds
path
where rho is the density of the matter along the path.
For any particular type of particle, having a cross-section sigma and number
density n (i.e. number of particles per cubic cm (or m)), we have kν = nσν
Principal Opacity Mechanisms:
As discussed in the book, the principal mechanisms for continuum opacity
are bound-free and free-free atomic transitions and electron
scattering (usually from free electrons). Bound-bound transitions produce line
absorption which we talk about in coming lectures, but are not usually
considered a continuum opacity source, although there is the phenomenon
known as “line blanketing” to consider.
Electron Scattering:
We start with this because it is the easiest. The free electron simply
scatters an incoming photon with a well known phase function that depends on
the polarization of the incoming light wave. It does not change the frequency of
the radiation. The cross-section is independent of wavelength. A classical
description of the process is that the incoming wave accelerates the electron
This kind of scattering is called “Thomson scattering and the cross-section of
the electron is σes = 6.65x10−25 cm2 = 0.665 barns (A barn is a unit of
cross-section equal to 10^-24 cm^2 that derives from the phrase “you couldn’t
hit the broad side of a barn” -- probably an old hunting taunt.
Electron scattering opacity is very important in the hottest stars where
there are plenty of free electrons. In cooler stars, the electrons are bound to
atoms so they are not able to scatter. Electron scattering was also key in creating
the opaque conditions of the early Universe that provides the “surface of last
scattering” from which the Cosmic Background Radiation comes. Once the
Universe cooled sufficiently for the number of free electrons to drop
precipitously (as they became bound to Hydrogen atoms), the opacity dropped
and the Universe “cleared”. Since this happens at a temperature of around a few
thousand degrees, that is the temperature of the CBR (redshifted, of course by
about a 1000!). Also note that electron scattering opacity is independent of
frequency, but it does produce polarization of light.
Note on Polarization by Scattering:
One of the interesting features of scattering is that it also polarizes light.
(This is why Polaroid sunglasses are good for driving, since they reduce glare
from light scattered of your windshield.) It is often electron scattering that
produces polarized light in astrophysical contexts. Perhaps the most important is
polarization of the surface of last scattering that occurs in the Cosmic
Background Radiation. Other examples include reflection nebulae and scattering
off disks and envelopes around stars or in AGN’s.
In the classical description of
E&M, the scattered light is simply the
radiation induced by the acceleration
of the electron. The electron oscillates
parallel to the the E-field of the
“driving wave” and acts like a dipole
radiator. The phase function for dipole
radiation looks like this:
Note that it is symmetric in
azimuth but there is no radiation along
the poles, since the electron moves
back and forth along your line of sight
in that direction.
The result is to produce a polarization pattern that has the dominant E-field
oriented perpendicular to the direction between the source and the scatterer.
For example the polarization of a reflection nebula will look like this:
This general form of scattering applies to
Rayleigh scattering and to scattering off of dust
grains (Mie scattering) as well. But in those
cases the phase function may not be
symmetric in azimuth. More likely it will be
strong asymmetric in the forward scattering
direction, and could be quite complex. An
example is the Zodiacal light and the
“Gegenschein”. Note that the daylight sky
is highly polarized. This is why a polarizing filter
can give cool photographs!
Bound-Free Absorption: This is the dominant form of opacity in most
stellar atmosphere and the dominant species is, of course, the HI atom. HeI is so
tightly bound it is not important except for the very hottest stars. Other
elements have such low abundance that they are generally negligible, especially at
optical wavelengths. One other species (the H-minus ion) is actually quite
important in some stars, including the Sun, as we shall see. First, however, we
look at the HI atom.
HI: The cross-section of HI to incoming radiation depends on which excited
state the atom is in. Consider, first the atom in its ground state. To go from the
neutral to the ionized state, starting from the ground state, requires an energy
input of 13.6 electron volts (the “ionization potential” of the HI atom)
corresponding to a photon with a wavelength shorter than 912 angstroms. This is
in the extreme UV region of the spectrum, so there is no influence of HI atoms
in the ground state on the opacity at visible wavelengths. Ground state HI atoms
are completely unaffected by visible light photons... these do not have enough
energy to get absorbed. However, a photon with just enough energy to ionize
the atom from the ground state will see a large cross-section for the HI atom.
Higher energy photons can also ionize the atom, of course, but they see
progressively smaller cross-sections. The cross-section goes as frequency^-3. The
sharp rise in opacity at 912 angstroms is an example of an “ionization edge”.
The HI atom can also be ionized from any excited state, as well, in which
case less energy is required. For example, an HI atom in the n=2 state is already
at a level of over 10.x ev above ground so it only requires an additional 3.x ev to
ionize it. This requires a photon of 3646 angstroms or shorter, in the visible UV.
This is an important region for stellar atmospheres since much of the flux is
moving at wavelengths like these. The big jump in opacity that occurs at 3646 is
referred to as the “Balmer jump” or “Balmer discontinuity” and is very important
in stellar astrophysics. The U photometric band of the UBVRI photometric
system straddles the Balmer discontinuity. Since the “size” of the HI atom in the
n=2 excited state is larger than the size in the ground state, the cross-section of
the atom is larger at 3646 angstroms than at 912 angstroms. Again, the crosssection declines like frequency^{-3} for higher energy photons.
As you might suspect, the other excited levels of HI produce similar
effects. The next important one is from the n=3 level (the Paaschen series) and
photons of wavelength 8204 angstroms and less can ionize H here. The region
between 8204 (near IR) and 3646 (beginning of UV) is in the “visible” region so it
is particularly of interest and it is referred to as the Paaschen continuum, since
the dominant HI bf opacity here comes from n=3 HI atoms. Of course, the
Boltzmann and Saha equations tell us that there will not be all that many n=3 HI
atoms in the gas and it turns out that HI bf is NOT the dominant opacity source
at visible wavelengths in the Sun!
λ3
σλ (bound f ree HI) ∝ 5 (f orλ < ionization wavelength)
n
For wavelengths longer than the ionization wavelength the cross-section is 0.
The full bound-free opacity for HI is obtained by multiplying the cross-section
for each excited state (n) by the number density of atoms in that state, obtained
from the Boltzmann and Saha equations.
The full bound-free opacity is a sum over all elements, all ionization levels and all
excitation states! Fortunately, HI usually dominates and the other elements can
be neglected. This is not true in the UV spectral range for hot stars. A “Kramer’s
Law” approximate formula for bf opacity in a star is given in your book and has
the form:
κλ (bound − f ree) ∝ ρT −3.5
Free-Free: The main free-free emission source is also from HI. This means
transitions within the continuum levels of the HI atom. (In effect, these are
ionized atoms that have electrons temporarily in their vicinity but not bound to
them. A photon passing by can interact with this proton plus electron system and
cause the photon to be absorbed while the electron gains energy. The inverse
can, of course, also happen which is free-free emission. Since the energy
differences between the states before and after photon absorption are generally
small, free-free processes are generally most important at long wavelengths (i.e.
in the infrared or radio). The cross-sections have the same wavelength
dependence (lambda^3) as the bound-free transitions, and the Kramer’s Law
opacity has the same dependence on density and temperature.
H-minus: The H-minus ion is the main opacity source in the Sun at visible
wavelengths, producing both bound-free and free-free absorption. This was
discovered in the 1930’s by examining how the limb-darkening law varied with
wavelength, which allowed one to work out how the opacity in the solar
atmosphere varied with wavelength. It did not match what one would expect for
HI (n=3) -- no rise in opacity towards 8204 angstroms or absorption edge.
The H-minus ion is an HI atom with an extra electron loosely bound to it.
The ion can be destroyed by a photon of energy only 0.754 ev. corresponding to
a wavelength of 16,400 angstroms (in the IR). Bound-free and free-free
transitions of this species are important opacity sources in stars with solar-like
temperatures.
Putting it all together: The full continuum opacity is the sum of all the
contributions, b-f, ff and scattering for all of the various atoms, ionization states,
excitation levels and electrons. Clearly this is messy.... The resulting wavelength
dependence of the opacity has a lot of influence on the way a stellar atmosphere
appears. It also has a lot of influence on the temperature structure of an
atmosphere because the opacity controls how easily energy flows through an
atmosphere at a given wavelength. Where opacity is high, energy flow (flux) is
low. Some example model atmospheres will make the point. Line blanketing is
important in some cases (especially late-type stars).
Kurucz models available at: http://stsdas.stsci.edu/ETC/STIS/stis_models.html
O star
G star
A star
M star
Note on Mie Scattering:
Dust in the interstellar medium has a typical size of about 0.1 microns, or
1,000 angstroms. This is the size of a wavelength of ultraviolet light. More than a
century ago, the physicist Mie (pronounced “Me”) developed a classical theory of
scattering that applies to small spheres of varying refractive index. The results
may be (very briefly!) summarized as: for wavelengths short compared to the
size of the particles, the scattering cross-section is about twice as large as the
geometric cross-section. For wavelengths comparable to or larger than the
particle size, the scattering cross-section declines as 1/lambda where lambda is
the wavelength of the light being scattered.
What this means is that extinction in the ISM at optical wavelengths is usually
higher for short wavelengths (blue and UV) and lower for long wavelengths (red
and IR). The ISM characteristically reddens starlight, according to a 1/lambda law
in the optical and infrared. There is essentially no interstellar extinction at radio
wavelengths. So, IR is excellent for peering THROUGH dust clouds to see what
is inside. It is excellent for star formation studies. It is also good for peering
through the galaxy to the very center.
Always keep in mind, of course, that along any line of sight, the grains causing
the scattering could be larger (or smaller) than normal. If they are larger, the
extinction could become “grayer”, so there is less reddening for a given amount
of extinction.
Note on Rayleigh Scattering:
In the Earth’s atmosphere, and presumably in many other planetary
atmospheres, the dominant source of scattering is often (e.g. on a clear day!)
from bound electrons that are part of molecules, atoms or ions. A classical
treatment of EM scattering for such objects shows that the energy scattered
from a wave when the wavelength of incident light is far from the resonant
frequency of the electron depends on 1/lambda^4. That is, the extinction is VERY
highly biased towards short wavelengths. This, of course, is why we have a blue
sky. Note also that Rayleigh scattering leads to a polarization of the skylight.
Maximum polarization occurs when you are looking at about 90 degrees off the
direction to the Sun.