4.2: The Ionization, Abundances, and Opacity of Stellar Material

We have now described the manner in which nuclear energy is produced in most stars, but before we can turn to the methods by which it flows out of the star, we must quantitatively discuss the processes which impede that flow. Each constituent of the gas will interact with the photons of the radiation field in a way that is characterized by the unique state of that particle. Thus, the type of atom, its state of ionization, and excitation will determine which photons it can absorb and emit. It is the combination of all the atoms, acting in consort that produces the opacity of the gas. The details which make up this combination can be extremely complicated. However, several of the assumptions we have made, and justified, will make the task easier and certainly the principles involved can be demonstrated by a few examples.

a. Ionization and the Mean Molecular Weight

Our first task is to ascertain how many of the different kinds of particles that make up the gas are present. To answer this question, we need to know not only the chemical makeup of the gas, but also the state of ionization of the atoms. We have already established that the temperatures encountered in the stellar interior are very high, so we might expect that most of the atoms will be fully ionized. While this is not exactly true, we will assume that it is the case. A more precise treatment of this problem will be given later when we consider the state of the gas in the stellar atmosphere where, the characteristic temperature is measured in thousands of degrees as opposed to the millions of degrees encountered in the interior.

We will find it convenient to divide the composition of the stellar material into three categories. \[\begin{array}{l}
\mathbf{X}&= \text{mass fraction of gas which is hydrogen}\\
\mathbf{Y}&= \text{mass fraction of gas which is helium}\\
\mathbf{Z}&= \text{mass fraction of gas which is everything else}\\
\end{array}\label{4.1.1}\]

It is common in astronomy to refer to everything which is not hydrogen or helium as "metals". For complete ionization, the number of particles contributed to the gas per element is just \[n_i=\frac{Z_i+1}{A_i}\label{4.1.2}\]

where \(\mathrm{Z_i}\) is the atomic number and \(\mathrm{A_i}\) is the atomic weight of the element. Thus, the number of particles contributed by hydrogen will just be twice the hydrogen abundance, and for helium, three-fourths times the helium abundance. In general, \[n_i \approx \frac{Z_i+1}{2 Z_i}\label{4.1.3}\]

The limit of equation \ref{4.1.3} for the heavy metals is ½. However, even at 10⁷ K the inner shells of the heavy metals will not be completely ionized and so ½ will be an overestimate of the contribution to the particle number. This error is somewhat compensated by the 1 in the numerator of equation \ref{4.1.3} for the light elements where it provides an underestimate of the particle contribution. Thus we take the total number of particles contributed by the metals to be ½ Z. The total number of particles in the gas from all sources is then \[N=\frac{\rho}{m_h} \sum_i n_i x_i=\frac{\rho}{m_h}\left(2 \mathbf{X}+\frac{3}{4} \mathbf{Y}+\frac{1}{2} \mathbf{Z}\right)\label{4.1.4}\]

Since everything in the star is classed as either hydrogen, helium, or metals, X + Y + Z = 1 and we may eliminate the metal abundance from our count of the total number of particles, to get \[N=\frac{1}{2}\left(\frac{\rho}{m_h}\right)\left(3 \mathbf{X}+\frac{1}{2} \mathbf{Y}+1\right)\label{4.1.5}\]

Throughout the book we have introduced the symbol m as the mean molecular weight of the gas without ever providing a clear definition for the quantity. It is clearly time to do so, and we can do it easily given the ideal-gas law and our expression for the total number of particles N. Remember \[P=N k T=\frac{\rho k T}{\mu m_h}\label{4.1.6}\]

The mean molecular weight \(\mu\) must be defined so that this is a correct expression.

Thus, \[\mu=\frac{2}{1+3 \mathbf{X}+\frac{1}{2} \mathbf{Y}}\label{4.1.7}\]

b. Opacity

In general, a photon can interact with atoms in three basic ways which result in the photon being absorbed:

1. Bound-bound absorption (atomic line absorption)
2. Bound-free absorption (photoionization)
3. Free-free absorption (bremsstrahlung)

For us to calculate the impeding effect of these processes on the flow of radiation, we must calculate the cross section for the processes to occur for each type of particle in the gas. For a particular type of atom, this parameter is known as the atomic absorption coefficient \(\alpha_\mathrm{n}\). The atomic absorption coefficient is then weighted by the abundance of the particle in order to produce the mass absorption coefficient \(\boldsymbol{\kappa}_{\boldsymbol{\nu}}\), which is the opacity per gram of stellar material at a particular frequency \(\nu\). Since we have assumed that the gas and photons are in STE, we can average the opacity coefficient over frequency to determine a mean opacity coefficient \(\overline{\kappa}\), which will represent the average effect on the diffusion of energy through the star of the material itself. However, to calculate this mean, we must calculate \(\kappa_\nu\) itself. As an example of how this is done, we shall consider the atomic absorption coefficient of hydrogen and "hydrogen-like" elements.

Classical View of Absorption Imagine a classical electromagnetic wave encountering an atom. The time-varying electromagnetic field will cause the electron to be accelerated so that it oscillates at the same frequency as the wave. However, just as an accelerated charge radiates energy, to accelerate a charge requires energy, and in this instance it is the energy of the electromagnetic wave. If the electron is bound in an atom, the energy may just be sufficient to raise the electron to a higher orbit, and we say that a bound-bound transition has taken place. If the energy is sufficient to remove the electron and ionize the atom, we say that a bound-free transition, or photoionization, has taken place. Finally, if a free electron is passing an ion in an unbound orbit, it is possible for the electron to absorb the energy from an electromagnetic wave that happens to encounter this system. Energy and momentum are conserved among the two particles and the photon with the result that the electron is moved to a different unbound orbit of higher energy relative to the ion. This is known as a free-free absorption.

Quantum Mechanical View of Absorption In quantum mechanics the classical view of a finite cross section of an atom for electromagnetic radiation is replaced by the notion of a transition probability. That is, one calculates the probability that an electron will make a transition from some initial state to another state while in the presence of a photon. One calculates this probability in terms of the wave functions of the two states, and it usually involves a numerical integration of the wave functions over all space. Instead of becoming involved in the detail, we shall obtain a qualitative feeling for the behavior of this transition probability.

Within the framework of quantum mechanics, the probability that an electron in an atom will have a specific radial coordinate is \[\langle\vec{r}\rangle=\int_V \Psi_i \vec{r} \Psi_i^* d V\label{4.1.8}\]

where i denotes the particular quantum state of the electron (that is, n,j,l) and \(\Psi_{\mathrm{i}}\) is the wave function for that state. In classical physics, the dipole moment \(\vec{\mathrm{P}}\) of a charge configuration is \[\overrightarrow{\mathrm{P}}=\int_v \vec{r} \rho_c(\vec{r}) d V\label{4.1.9}\]

where \(\rho_\mathrm{c}\) is the charge density. The quantum mechanical analog is \[\left\langle\vec{P}_{i j}\right\rangle=e \int_v \Psi_i \vec{r} \Psi_j d V\label{4.1.10}\]

where i denotes the initial state and j the final state.

Now, within the context of classical physics, the energy absorbed or radiated per unit time by a classical oscillating dipole is proportional to \(\overrightarrow{\mathrm{P}} \bullet \overrightarrow{\mathrm{P}}\), and this result carries over to quantum mechanics. This classical power is \[\mathbf{P} \propto \nu^4 P^2\label{4.1.11}\]

Since the absorbed power P is just the energy absorbed per second, the number of photons of energy \(\mathrm{h}\nu\) that are absorbed each second is \[N_p \propto \nu^3 p^2\label{4.1.12}\]

However, the number of photons absorbed per second will be proportional to the probability that one photon will be absorbed, which is proportional to the collision cross section. Thus, we can expect the atomic cross section to have a dependence on frequency given by \[\alpha_\nu=\frac{\text { const }}{\nu^3 P^2}\label{4.1.13}\]

In general, we can expect an atomic absorption coefficient to display the \(\nu^{-3}\) dependence while the constant of proportionality can be obtained by finding the dipole moment from equation \ref{4.1.10}. The result for the bound-free absorption of hydrogen and hydrogenlike atoms is \[\begin{aligned}
& \left.\alpha_\nu^{\mathrm{b}-\mathrm{f}}(i, n)=\frac{64 \pi^4 Z_i^4 m_e e^{10}}{3 \sqrt{3} h^6 c} \frac{1}{n^5} S_{n i}^4 \delta_\nu^{\mathrm{b}-\mathrm{f}}(i, n)\right]\left(\frac{1}{\nu^3}\right) \\
& \alpha_\nu^{\mathrm{b}-\mathrm{f}}(i, n)=\left(2.815 \times 10^{29}\right) \frac{Z_i^4 S_{n i}^4 g_\nu(i, n)}{n^5 \nu^3} \mathrm{~cm}^2
\end{aligned}\label{4.1.14}\]

where \[\begin{aligned}
i & =\text { state of ionization } \\
n & =\text { principal quantum number of electron } \\
Z_i & =\text { atomic number } \\
S_{n i} & =\text { screening factor resulting from interior electrons } \\
g_\nu(i, n) & =\text { gaunt factor } \approx 1
\end{aligned}\label{4.1.15}\]

A similar expression can be developed for the free-free transitions of hydrogen-like atoms: \[\mathrm{\alpha_n^{f f}(i, p) d n_e(p)=\frac{4 \pi Z_i^2 e^6 S_{f i}^2}{3 \sqrt{3} \mathrm{hcm}_e^2 v(p)} g_n^{f f}\left(1 / n^3\right) \operatorname{dn}_e(p)}\label{4.1.16}\]

Here, the atomic absorption coefficient depends on the momentum of the "colliding" electron. If one assumes that the momentum distribution can be obtained from Maxwell-Boltzmann statistics, then the atomic absorption coefficient for free-free transitions can be summed over all the colliding electrons and combined with that of the bound-free transitions to give a mass absorption coefficient for hydrogen that looks like \[\begin{array}{l}
\kappa_\nu(\text {hydrogen})&=\frac{32 \pi^2 e^6 R e^{-\chi_0 /(k T)}}{3 \sqrt{3} h^3 c m_h v^3}\left(\sum_{v_n<v}^{\infty} \frac{e^{-\epsilon_n /(k T)}}{n^3} \bar{g}_n+\frac{\bar{g}_{f f} k T}{2 \chi_0}\right)\\
\text{where}&\\
\chi_0&=\text{ ionization potential of hydrogen}\\
R&=\text{ gas constant}\\
\epsilon_n&=\text{ excitation energy of }{\nu}_{\boldsymbol{n}}\text{th energy level}\\
\bar{g}_n, \bar{g}_{f f}&=\text{ gaunt factors averaged over frequency}
\end{array}\label{4.1.17}\]

The summation in equation \ref{4.1.17} is to be carried out over all \(n\) such that \(\nu_\mathrm{n}<\nu\). That is, all series that are less energetic than the frequency ν can contribute to the absorption coefficient. For us to use these results, they must be carried out for each element and combined, weighted by their relative abundances. This yields a frequency-dependent opacity per gram \(\kappa_\nu\) which can be further averaged over frequency to obtain the appropriate average effect of the material in impeding the flow of photons through matter. However, to describe the mean flow of radiation through the star, we want an estimate of the transparency of transmissivity of the material. This is clearly proportional to the inverse of the opacity. Hence we desire a reciprocal mean opacity. This frequency-averaged reciprocal mean is known as the Rosseland mean and is defined as follows: \[\frac{1}{\bar{\kappa}_\nu} \equiv \frac{\int_0^{\infty}\left(1 / \kappa_\nu\right) \partial B_\nu(T) / \partial T d \nu}{\int_0^{\infty}\left[\partial B_\nu(T) / \partial T\right] d \nu}\label{4.1.18}\]

Here, \(\mathrm{B_\nu(T)}\) is the Planck function, which is the statistical equilibrium distribution function for a photon gas in STE which we developed in Chapter 1 [equation \ref{1.1.24}]. That such a mean should exist is plausible, since we are concerned with the flow of energy through the star, and as long as we assume that the gas and photons are in STE, we know how that energy must be distributed with wavelength. Thus, it would not be necessary to follow the detailed flow of photons in frequency space since we already have that information. That there should exist an average value of the opacity for that frequency distribution is guaranteed by the mean value theorem of calculus. That the mean absorption coefficient should have the form given by equation \ref{4.1.18} will be shown after we have developed a more complete theory of radiative transfer (see Section 10.4).

Approximate Opacity Formulas Although the generation of the mean opacity coefficientk is essentially a numerical undertaking, the result is always a function of the state variables P, T, \(\rho\), and \(\mu\). Before the advent of the monumental studies of Arthur Cox and others which produced numerical tables of opacities, much useful work in stellar interiors was done by means of expressions which give the approximate behavior of the opacity in terms of the state variables. The interest in these formulas is more than historical because they provide a method for predicting the behavior of the opacity in stars and a basis for understanding its relationship to the other state variables. If one is constructing a model of the interior of a star, such approximation formulas enable one to answer the question so central to any numerical calculation: Are these results reasonable? In general, these formulas all have the form \[\begin{array}{l}
&\bar{\kappa}=\bar{\kappa}_0 \rho^n T^{-s}\\
&(a) n=1 \quad &s=3.5 \quad&\text{Kramers' law}\\
&(b) n=0.75 \quad &s=3.5 &\text{Schwarzschild's opacity}\\
&(c) n=0 \quad &s=0 &\text{electron scattering}\\
\end{array}\label{4.1.19}\]

where \(\overline{\kappa_0}\) depends on the chemical composition \(\mu\). Kramer’s opacity is a particularly good representation of the opacity when it is dominated by free-free absorption, while the Schwarzschild opacity yields somewhat better results if bound-free opacity makes an important contribution. The last example of electron scattering requires some further explanation since it is not strictly a source of absorption.

Electron Scattering The scattering of photons at the energies encountered in the stellar interior is a fully conservative process in that the energy of the photon can be considered to be unchanged. However, its direction is changed, resulting in the photon describing a random walk through the star. This immensely lengthens the path taken by the photon and therefore increases its "stay" in the star. The longer the photon resides in the star, the greater its path, and the greater are its chances of being absorbed by an encounter with an atom. Thus, electron scattering, while not involved directly in the absorption of photons, does significantly contribute to the opacity of the gas. The photon flow is impeded by electron scattering, first, by redirecting the photon flow and, second, by lengthening the path and increasing the photon's chances of absorption.

As long as \(\mathrm{h\nu<<m_{e}c^2}\), the electron will exhibit little or no recoil as a result of its collision with a photon and the photon energy will be unchanged. This case is called Thomson scattering and we can use the classical theory of electromagnetism to estimate its cross section. The energy radiated or absorbed per unit time by an oscillating free electron is \[\frac{d E}{d t}=\frac{2 e^4 E_0^2}{3 m_e^2 c^3} \operatorname{Sin}^2 \omega t\label{4.1.20}\]

However, the power in an electromagnetic wave is given by the Poynting vector \(\overrightarrow{\mathrm{S}}=[\mathrm{c} /(4 \pi)](\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{H}})\). In vacuum, \(\overrightarrow{\mathrm{E}} \perp \overrightarrow{\mathrm{H}}\), and in Gaussian units \(\mathrm{|E|=|H|}\). Therefore, the magnitude of the Poynting vector is \[S=\frac{c}{4 \pi} E_0^2 \operatorname{Sin}^2 \omega t\label{4.1.21}\]

The ratio of the power absorbed from the wave to the power in the wave is (dE/dt)/S and is the definition of a cross section. Thus, we can write the classical cross section for electron scattering as \[\sigma_0=\frac{8 \pi}{3}\left(\frac{e^2}{m_e c^2}\right)^2=6.652 \times 10^{-25} \mathrm{~cm}^2\label{4.1.22}\]

The quantity \(\mathrm{e^2/(m_{e}c^2)}\) is known as the classical radius of the electron and is roughly that radius for which the field energy of the electron is equal to its rest energy. The square of that radius yields a geometric cross section which is 1.5 times the classical cross section. This cross section is also known as the Thomson cross section.

Note that the Thomson cross section is not a function of frequency, which makes it particularly easy to incorporate in an expression for opacity. The symbol \(\mathrm{s_e}\) usually denotes the electron scattering coefficient per gram of stellar material, so that \[\sigma_e=\frac{\sigma_0 n_e}{\rho}\label{4.1.23}\]

However, in the limit where the total number of particles contributed to the particle density by metals is ½Z and most of those are electrons, we get the electron density to be \[n_e=\frac{\rho}{m_h}\left\langle\frac{Z}{A}\right\rangle \approx \frac{1}{2}\left(\frac{\rho}{m_h}\right)(1+X)\label{4.1.24}\]

which yields an electron scattering cross section per gram of \[\sigma_e \approx 0.2004(1+\mathrm{X}) \mathrm{cm}^2 / \mathrm{g} \leq 0.4 \mathrm{~cm}^2 / \mathrm{g}\label{4.1.25}\]

Thus, there is a limit to the scattering coefficient per gram of stellar material of something less than 0.5 cm²/g.

We have now shown how the energy generation rate and opacity per gram of stellar material can be related to the state variables \(\mathrm{P}\), \(\mathrm{T}\), \(\rho\), and \(\mu\). You must not assume that this discussion is complete in every detail. The detailed calculation of these functions is extremely complex and has entertained some of the best minds of the twentieth century. What we have seen is some of the major physical principles which affect the outcome of such efforts. For the details of the modern values of these functions, you should consult the current literature as refinements continue. Nevertheless, from now on, we may assume that we have functions of the form \[\epsilon=\epsilon(P, T, \rho, \mu) \quad \kappa=\kappa(P, T, \rho, \mu)\label{4.1.26}\]

at our disposal. Now we turn to the problem of describing the flow of energy through the star.