Outline

4. Equation of State and Related Descriptions

From Hansen and Kawaler.

4.4C. Boltzmann-Saha statisitics

Elementary statistical mechanics leads to the following expression for the number of particles of species i in phase-space volume d 3r d 3p :

ni(p) d 3r d 3p = (d 3r d 3p / h3) ∑ j gj {e [−μi + εj + ε(p)] / kT ± 1}−1 .

Consider some collection of distinguishable particles related via a reaction of the form

Aj+ + eAio + &chi

where &chi is the energy required to unbind the electron. Notice that it doesn't matter whether the initial and final states of the species A are ground or excited states; we only care about the energy difference.

a. The description in H&K

For a gas of distinguishable particles, we may use the general equations for the total densities of each species as derived in the section on Ideal gasses:

n = (g / h3 ) (2πmkT )3/2 eε / kT e μ / kT

Now, we set the zero-point of energy such that εe is zero. Then the Saha equation is the ratio:

ne n+/no = (geg+/go) (2πkT )3/2 h−3 (mem+/mo)3/2 e(μe + μ+μo) / kT eχ / kT ,

but we have already seen that chemical equilibrium demands that μe + μ+μo = 0, and we can assume with pretty high accuracy that m+ = mo. Finally, the statistical weight of an electron ge = 2, so

ne n+/no = (2g+/go) (2πmekT / h2 )3/2 eχ / kT .

This derivation by H&K sweeps some considerations under the rug, but comes out with the right answer. It would be more rigorous to specify the specific momentum (or energy) of the electron, and then integrate over the available electron states. See the next section.

Notice that the Saha equation may be written ne n+ = no (g+/go) Φ(χ,T).

Homework: write the equations (using Φ's) for a given mixture of hydrogen and helium. Include species and charge conservation. If ne is known, the equations are a linear set and can be solved as a matrix equation for the vector of species densities. Write a program to solve for the fractional densities as functions of ne and T. Assume all ions are in the ground state (not a bad approximation for H and He, but other elements with low-lying excitations must include those states).

b. Partition functions and pressure ionization

In this subsection I describe an alternative approach.
In the chemical description, we imagine an ensemble of different ions and electrons as an ensemble of different 'chemicals' and calculate rates of 'chemical' reactions between them. In strict thermodynamic equilibrium (hereafter STE), all processes are in detailed balance; that is, the rate of reactions per unit volume is the same in both directions at all levels of detail. We have seen that a particular variety of ion, say C IV + e, will be distributed over its available states (including bound C III) according to

n(ε) = g(ε) e μ/kT eε /kT

in thermodynamic equilibrium. Here g(ε) (bold-face g) is the number of available states in the energy interval , per unit volume. Conservation of number n (total number of C IV + e, bound and unbound) requires that

n = e μ/kTg(ε) eε/kT ≡ e μ/kT n U(T).

Here U(T) is called the partition function per nucleon, and is a function only of T for a given available state function g(ε).

Since each nucleus has the same available states, we may replace g(ε), which is the number of available states per unit volume, with n g(ε) , where g (normal-face) is the number of available states per nucleon. Then e μ/kT = 1/U(T), and

n(ε) = n g(ε) eε /kT/ U(T).

All independent degrees of freedom of an ion are separable. Thus we can consider the distribution over excitation states of all C IV + e ions moving with a specific translational momentum p. And, except for a constant factor given by the Maxwell distribution ratio, this internal distribution will be the same for all other momenta. In STE, fast-moving ions are not likely to be more excited than slow-moving ones.

Unless otherwise specified, all further reference to the function g(ε) will imply the availability of internal states per nucleon. We may write the ratio of populations at different energies, whether internal or translational, as

n(ε) / n(ε′) = [g(ε) / g(ε′)] e(ε′−ε) / kT,

as long as we properly account for all available states. Notice that the partition function has cancelled out. So far we have made no statement about the form of g(ε) or the zero of the energy scale. It is customary to set the latter at the ionization limit of the bound configuration, so stationary electrons infinitely far from any nucleus have zero energy. Then all bound states (well, all bound states of a singly excited valence electron) have ε < 0, and all free states (well, all states of electrons capable of reaching infinity) have ε > 0. Going back to the carbon ions, we can consider C IV ions as C III's with positive internal energy, as long as we include correctly the number of free electrons. Tightly bound C III can exist only in discrete eigenstates at particular energies. Thus, over the negative ε range, the form of the internal availability function per C III ion is

g(ε <0) = ∑i gi δ(εεi) ,

where gi is the degeneracy (statistical weight) of the ith eigenstate, εi is its energy, and δ(x) is the delta function. The states really have finite lifetimes, so the delta function is an idealization.

For unbound C III (specifically ground state C IV with a free electron in its vicinity) we must be careful in how we enumerate the available states per carbon nucleus. The number of states per carbon nucleus per unit volume, with the free electron in the energy range about ε is

gV(ε) = gk (2/h3) 4π m(2)1/2 .

Here gk is the statistical weight of the parent ion in its ground state (that's where the 'per carbon nucleus' part of the units comes from) and the 2 enters due to the two possible spin states of the electron. The mass m is technically the reduced mass between ion and electron, but in effect is that of the electron alone. To cancel the 'per unit volume' units, we must multiply by the volume associated with each electron, namely 1/ne. So, unbound C III has the internal availability function

g(ε >0) = (2gk/neh3) 4π m(2)1/2 ,

per carbon nucleus. Now all carbon particles are on an equal footing. Thus, the ratio of any two bound state populations is

nj / ni = (gj / gi) e(εiεj) / kT,

The ratio of any unbound state population to a bound state population is

n(ε >0) / ni = (2gk/gineh3) 4π m(2)1/2 e(εiε)/kT

where the n(ε >0) indicates populations per unit energy interval, and it should be remembered that εi < 0.
Rarely do we care about the energy state of the free electron; if needed, we can always recover it from the Maxwell distribution. If we integrate the above equation over all positive ε, we get:

nk / ni = (2gk/gine) (2π mkT/h2)3/2 eεi /kT .

This equation is the specific form of the Saha equation, relating the populations in two states separated by an ionization. The same expression could be derived for any ionized eigenstate k, not just the ground state, as long as one remembers to include an excitation energy in the exponent. We can also write this equation in the inverted form

ni = nenk 2.07×10−16 (gi/gk) T −3/2 eεi /kTnenk Φik(T).

Further, by summing over all bound levels in both ions, we can write:

nI = nenI+1 2.07×10−16 (UI /UI+1) T −3/2 e &chi /kT

where the U's are the appropriate bound state partition functions and χ is the (positive) ionization potential of the I th ion.

All of the above equations are ratios. To close the set and calculate the number of electrons, we employ the conservation equations for nuclei number and charge.

Before closing this section, we should address one 'problem' conveniently hidden in the symbolism above. The number of available states per nucleus is not an easily determined quantity for states having nearly zero energy. Tightly bound states are not much perturbed by neighbors, and high-energy free states are also not much perturbed by the raggedness of the underlying potential. But electrons finding themselves only loosely bound to nuclei are no longer in time-invariant potentials, and can skip from one nucleus to another, or be bound to small, temporary, clusters of nuclei. The wave functions of such electrons rapidly become delocalized. Thus we must be careful in using the formulae above for g(ε) when ε ≈ 0. Books on atomic structure have no qualms about telling us that highly excited atoms become hydrogenic (one electron far outside a core producing a Coulomb potential), and that the statistical weights of such excited states within some energy interval become proportional to − ε−3 as ε → 0. However, no atom, even one lonely enough to find itself in one of the giant cosmic voids which fill our universe, is truly isolated. Stationary eigenstates cease to exist if they extend beyond neighboring nuclei. For hydrogen atoms at typical photospheric densities (n ≈ 1014 cm−3), there are no stationary states with principle quantum number n > 30. If we increase the number density to 1024 (ρ ≈ 1 g cm−3) atoms become about 1 Bohr radius apart, and even quantum number n = 1 becomes untenable. This phenomenon is called pressure ionization and it must be accounted for in the cores of all stars.