Lecture 6: Continuum Opacity and Stellar Atmospheres

Transcription

1 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 τ ν = k ν 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).

2 In this case, τ ν = path κ ν ρds 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 k ν = nσ ν 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.

3 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 σ es =6.65x10 25 cm 2 =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.

4 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.

5 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!

6 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.

7 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!

8 σ λ (bound free HI) λ3 (forλ < ionization wavelength) n5 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.

9 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: κ λ (bound free) 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.

10 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 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).

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13 Kurucz models available at: O star A star G star M star

14 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.

15 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.