Statistical Equilibria: Saha EquationInitial questions: What sets the temperature of the warm interstellar medium? Whatdetermines the epoch from which we see the CMB?We are now going to consider the statistics of equilibrium. Specifically, suppose we leta system stand still for an extremely long time, so that all processes can come into balance.We then measure the fraction of particles that are in each of several states (e.g., if we havea pure hydrogen gas, we might want to know the ionization fraction). How can we computethis fraction? We’ll take a look at this in several different ways. We will then step backand ask ourselves when these equilibrium formulae do not apply. This is important! Toparaphrase Clint Eastwood, “An equation’s got to know its limitations”. In virtually allastronomical problems we deal with approximations, so we always need a sense of whenthose approximations fail. This serves a double purpose: it keeps us from making errors, butit also keeps us from applying too complex an approach to a simple problem.Let’s start out with a simple case. We consider a column of isothermal gas (temperatureT ) in a constant gravitational acceleration g. Each molecule has mass m. If we can treat thislike an ideal gas, Ask class: how do we figure out how the pressure vary with height h? Anideal gas has pressure P = nkT , where n is the number density. In equilibrium the pressuregradient balances the weight of the gas, so dP/dh = −ρg. Using ρ = nm = P m/kT andthe constancy of T , we have dP/dh = −(mg/kT )P , so P = P0exp(−mgh/kT ), where P0isdefined as the pressure at h = 0. This is the familiar Boltzmann formula. As usual, we cancheck various limits: when h increases the pressure decreases (correct); when T decreasesthe pressure decreases at a fixed h (correct); and so on. Note that the decrease is faster forheavier molecules. This is one reason why at high altitudes we have an oxygen deficit: ournatural breathing rate is controlled by the level of CO2in the bloodstream, and this decreasesrapidly at high altitude so we don’t breathe as often as we should until we acclimate.This is one special case of a much more general formula of statistical distribution inthermal equilibrium. I highly recommend you look at the Feynman Lectures on Physics if youwant physical insight on just about anything; Feynman had a remarkably original perspectiveon physics, which is different enough that it often provides the key to understanding. In it,he shows that in any potential, the distribution of states is proportional to exp(−∆E/kT ),where ∆E is the difference in energy from some arbitrary level (it’s the ratio that matters,so we don’t need to specify the reference level). There may be other factors (e.g., themultiplicity of the state), but this basic one is always obtained in a classical equilibrium.Another example is the Maxwell-Boltzmann distribution. At temperature T , the fraction in avolume d3p around momentum p is ∝ exp(−E/kT )d3p, which (for nonrelativistic motion andisotropic distribution in momentum space) is ∝ exp(mv2/2kT )p2dp. Yet another exampleis the population of states in an atom. Compared to some reference state (e.g., the groundstate), the fraction of atoms in a state of energy E is ∝ exp(−E/kT ).Now let’s return to the subject of ionization. This is important because the ionizationfraction has consequences for the equation of state (pressure related to density and temper-ature) and opacity (e.g., which sources matter most), among other things. Let us assumethat we have a cavity in which atoms and radiation are in thermal equilibrium with the sametemperature T . For simplicity, consider a hydrogen atom with a single energy level (groundstate), and consider the ionization reactionH++ e−←→ H0+ χH, (1)where χH=13.6 eV is the ground state binding energy. In equilibrium, what can be saidabout the abundances? In this case, let the reference energy (E = 0) be for the free electronand proton. Statistical weights: 2 for the hydrogen atom, two combined for the proton andelectron (spins in same direction, spins in opposite directions). From the previous lecturenotes, integration of the distribution function over the momenta yields the number densitiesne=2[2πmekT ]3/2h3eµ−/kT, (2)n+=[2πmpkT ]3/2h3eµ+/kT, (3)andn0=2[2π(mp+ me)kT ]3/2h3eµ0/kTeχH/kT. (4)Ask class: if we were being really accurate, would the actual mass in the third equationbe mp+ me? (No, because we’d have to subtract the binding energy. But this is only13.6/511,000 of even the electron mass, which would be a 1/40,000 correction, and can bedumped. In the same spirit, the electron mass is only a 1/1,800 correction and can be ignoredhere.)Forming the product n+ne/n0, we findn+nen0=µ2πmekTh2¶3/2e(µ−+µ+−µ0)/kTe−χH/kT. (5)Ask class: what is (µ−+ µ+− µ0) in equilibrium, which we’ve been assuming? It’s zero.The numerical factor is about 2.4 × 1015T3/2cm−3. Ask class: is this equation sufficient todetermine the abundances of electrons, protons, and neutral hydrogen? No, because we haveonly one equation for three unknowns. Ask class: what are reasonable auxilliary conditions?Charge balance implies n+= n−, and conservation of nucleons means that n0+ n+= n, thetotal number density of protons (which equals the total number density of electrons). Withthese extra conditions, the degree of ionization y ≡ n+/n = ne/n is, numerically,y21 − y=4 × 10−9ρT3/2e−1.6×105/T, (6)where as usual quantities such as ρ and T are referenced to the appropriate cgs unit.An important thing to remember about this equation and many like it is that theexponential usually dominates. What this means specifically is that for “interesting” levelsof ionization (say, around 50%), T is so much lower than 1.6 × 105K that a tiny change inT makes a huge difference in the ionization fraction. Operationally, this means that you canget a remarkably good first guess at the temperature for a specified value of y by ignoringthe T3/2factor, which doesn’t change as rapidly. If you like, you can then substitute yourvalue of T in the T3/2factor and solve again by iteration. This is an easy way to convergerapidly to a solution without having to write a computer program.In fact, let’s do a practice example. Suppose we take a number density of 1 cm3, whichis about average for our interstellar medium. Then ρ = 1.7 × 10−24g cm−3, so the Sahaequation becomesy21 − y≈