11.2: Statistics of the Gas and the Equation of State

For normal stellar atmospheres, the effective temperatures range from a few thousand degrees to perhaps 50,000 K. The pressures are such as to permit the existence of spectral lines and so the densities cannot be great enough to cause departures from the ideal-gas law. The assumption of LTE implies that the material components of the gas making up the stellar atmosphere behave as if they were in thermodynamic equilibrium characterized by the local value of the kinetic temperature. In Chapter 1 [equations \ref{1.1.16} and \ref{1.1.17}] we found that as long as the density of available cells in phase space is much greater than the particle phase density, such a gas should obey Maxwell-Boltzmann statistics. If this is true, the fraction of particles that have a certain kinetic energy \(w_\mathrm{i}\) is \[\frac{N_i}{N}=\frac{g_i e^{-w_i /(k T)}}{U(T)}\label{11.1.1}\]

Since the pressure is the second velocity moment of the density, the Maxwell-Boltzmann distribution formula [equation \ref{11.1.1}] leads to the ideal-gas law [equation \ref{1.3.4}], namely, \[P_g=n k T\label{11.1.2}\]

Later [equation \ref{4.1.6}] we found it convenient to represent the total number of particles in terms of the density and the corresponding mass of hydrogen atoms \(\mu m_\mathrm{h}\) that would yield the required number of particles, so that \[P_g=\frac{\rho k T}{\mu m_h}\label{11.1.3}\]

The parameter \(\mu\) is called the mean molecular weight.

a. Boltzmann Excitation Formula

Under the assumption of LTE, the energy distribution of all particles represents the most probable macrostate for the system. This state is arrived at through random collisions between the gas particles themselves. Such a gas is said to be collisionally relaxed and is in stationary equilibrium. This means that all aspects of the gas will exhibit the same energy distribution, including those energy aspects of the gas which do not allow for a continuum distribution of energy states − specifically those states described by the orbital electrons. Thus, an ensemble of atoms will exhibit a distribution of states of electronic excitation that follows the Maxwell-Boltzmann distribution law. Were the energy not shared between the excitation energy and the kinetic energy of the particles, collisions would ensure that energy differences were made up in the deficit population at the expense of the population that had the relative surplus.

Such a situation would not represent a time-independent distribution until equilibrium between the two populations was established and therefore would not be the most probable macrostate. Since the various states of atomic excitation will be distributed according to the Maxwell-Boltzmann distribution law, the number of particles in any particular state of excitation will be \[\frac{N_j}{N}=\frac{g_j e^{-\epsilon_j /(k T)}}{U(T)}\label{11.1.4}\]

Here, \(\mathrm{g_j}\) is the statistical weight, and it has the same meaning as it did in Chapter 1 [equation \ref{1.1.17}]. The parameter \(\varepsilon_\mathrm{j}\) is the excitation energy above the ground state, and U(T) is the partition function. U(T) is nothing more than a normalization parameter that reflects the total number of particles available for distribution among the various energy states. It also has the same meaning as it did for the continuum distribution of energies discussed in Chapter 1, but now will be determined from the sum over the discrete states of excitation. Its role as a normalization parameter of the distribution is most clearly demonstrated by summing equation \ref{11.1.4} over all particles and their energy states so that the left-hand side is unity. This then confirms the form of the partition function as \[U(T)=\sum_j g_j e^{-\epsilon_j /(k T)}\label{11.1.5}\]

b. Saha Ionization Equilibrium Equation

A completely rigorous derivation of the Saha equation from first principles is long and not particularly illuminating. So instead of performing such a derivation, we appeal to arguments similar to those for the Boltzmann excitation formula. The object now is to find the equilibrium distribution formula for the distribution of the various states of ionization for a collection of atoms. Again, we assume that a time-independent equilibrium exists between the electrons and the ions. However, now the electron population will depend on the equilibrium established for all elements, which appears to make this case quite different from the Boltzmann excitation formula. However, we proceed in a manner similar to that for the Boltzmann excitation formula. Let us try to find the probability of excitation, not for a bound state, but of a state in the continuum where the electron can be regarded as a free particle.

Consider an atom in a particular state of ionization, and denote the number of such atoms in a particular state of excitation, say the ground state, by the quantity \(n_{00}\). The number of these atoms excited to the ionized state where the electron can be regarded as a free particle is then given by the Boltzmann excitation formula as \[\frac{n_{01}}{n_{00}}=\frac{g_f}{g_{00}} e^{-\left(\chi_0+\frac{1}{2} m v^2\right) /(k T)}\label{11.1.6}\]

Here \(\mathscr{g}_\mathrm{f}\) is the statistical weight of the final state of the ionized atom, and \(\chi_0\) is the ionization potential of the atom in question. The second term in the exponential is simply the energy of the electron that has been elevated to the continuum. Now we can relate the total number of ionized atoms in the ground state to the total number of ionized atoms through the repeated use of the Boltzmann excitation formula and the partition function as \[n_{0 i}=\frac{N_i g_{0 i} e^{-\epsilon_0 /(k T)}}{U(T)}\label{11.1.7}\]

where \(\mathrm{N_i}\) is the total number of atoms in the ith state of ionization. However, since the resultant ionized atoms we are considering are in their ground state, \(\varepsilon_0\) is zero by definition. In addition, the statistical weight of the final state \(g_\mathrm{f}\), can be written as the product of the statistical weight of the ground state of the ionized atom and that of a free electron. We may now use this result and equation \ref{11.1.6} to write \[\frac{n_{01}}{n_{00}}=\frac{N_1 g_{01} / U_1(T)}{N_0 g_{00} / U_0(T)}=\frac{g_{01} g_e}{g_{00}} e^{-\left(\chi_0+\frac{1}{2} m v^2\right) /(k T)}\label{11.1.8}\]

which simplifies to \[\frac{N_1}{N_0}=\frac{U_1(T)}{U_0(T)} e^{-\chi_0 /(k T)} g_e e^{-\frac{1}{2} m v^2 /(k T)}\label{11.1.9}\]

Although we picked a specific state of excitation − the ground state − to arrive at equation \ref{11.1.9}, that choice was in no way required. It only provided us with a way to use the Boltzmann excitation formula for atoms in two differing states of ionization.

The statistical weight of a free electron is really nothing more than the probability of finding a given electron in a specific cell of phase space, so that \[g_e=\frac{2 d x d y d z d p_x d p_y d p_z}{h^3}\label{11.1.10}\]

Again, for electrons, the familiar factor of 2 arises because the spin of an electron can be either "up" or "down". Assuming that the microscopic velocity field is isotropic, we can replace the "momentum volume" by its spherical counterpart and express it in terms of the velocity: \[d p_x d p_y d p_z=4 \pi p^2 d p=4 \pi m^3 v^2 d v\label{11.1.11}\]

The "space volume" of phase space occupied by the electron can be expressed in terms of the inverse of the electron number density, so that the statistical weight of an electron becomes \[g_e=\frac{8 \pi m^3 v^2 d v}{N_e h^3}\label{11.1.12}\]

and equation \ref{11.1.9} can be written as \[\frac{N_e N_1}{N_0}=\frac{U_1(T)}{U_0(T)} e^{-\chi_0 /(k T)} \frac{8 \pi m_e^3 v^2}{h^3} e^{-\frac{1}{2} m_e v^2 /(k T)} d v\label{11.1.13}\]

So far we have assumed that the ionization produced an electron in a specific free state, that is, with a specific velocity in the energy continuum. However, we are interested in only the total number of ionizations, so we must integrate equation \ref{11.1.13} over all allowed velocities for the electrons that result from the ionization process. Thus, \[\frac{N_e N_1}{N_0}=\frac{U_1(T)}{U_0(T)} e^{-\chi_0 /(k T)} \frac{8 \pi m_e^3 v^2}{h^3} e^{-\frac{1}{2} m_e v^2 /(k T)} d v\label{11.1.14}\]

For convenience, we also assumed that the states of ionization of interest were the neutral and first states of ionization. However, the argument is correct for any two adjacent states of ionization so we can write with some generality \[\frac{N_{i+1} N_e}{N_i}=\frac{U_{i+1}(T)}{U_i(T)} \frac{2\left(2 \pi k T m_e\right)^{3 / 2}}{h^3} e^{-\chi_i /(k T)}\label{11.1.15}\]

This expression is often written in terms of the electron pressure as \[\frac{N_{i+1}}{N_i} P_e=\frac{2\left(2 \pi m_e\right)^{3 / 2}(k T)^{5 / 2}}{h^3} \frac{U_{i+1}(T)}{U_i(T)} e^{-\chi_i /(k T)}\label{11.1.16}\]

Both expressions [equations \ref{11.1.15} and \ref{11.1.16}] are known as the Saha ionization equation. Its validity rests on the Boltzmann excitation formula and the velocity distribution of the electrons produced by the ionization being described by the Boltzmann distribution formula. Both these conditions are met under the conditions of LTE. Indeed, many authors take the validity of the Saha and Boltzmann formulas as a definition for LTE.

The Boltzmann excitation equation and the Saha ionization equation can be combined to yield the fraction of atoms in a particular state of ionization and excitation. Knowing that fraction, we are in a position to describe the extent to which those atoms will impede the flow of photons through the gas.