Mean Molecular Weight the Thermodynamic Relations Between P

Mean Molecular Weight

The thermodynamic relations between P , ρ, and T , as well as the calculation of stellar opacity requires knowledge of the system’s mean molecular weight (deﬁned as the mass per unit mole of material, or, alternatively, the mean mass of a particle in Atomic Mass Units). Recall that a mole of any substance contains NA = 6.02252 1023 atoms. Thus, the number density of ions is related to the mass× density, ρ, by ρ ρN n = = A (5.1.1) µma µ where ma is the mass that is equivalent to 1 A.M.U. If the mass fraction of species i is xi, then its number density is

xiρNA ni = (5.1.2) Ai where Ai is the atomic weight of the species. The number density of all ions in a volume of gas is then x n = n = ρN i I i A A i i i X X or −1 ρN x n = A where µ = i (5.1.3) I µ I A I Ã i i ! X To compute the contribution of (massless) electrons to the mean molecular weight, let Zi be the atomic number of species i, and fi be the species’ ionization fraction, i.e., the fraction of electrons of i that are free. The number density of electrons is therefore x n = ρN i f Z (5.1.4) e A A i i i i X µ ¶ or −1 ρN Z x f n = A where µ = i i i (5.1.5) e µ e A e Ã i i ! X Note that in the case of total ionization (fi = 1), this equation simpliﬁes greatly. Since Zi/Ai = 1 for hydrogen, and 1/2 for everything else, ∼

−1 −1 1 (1 X) 2 µ = X + (Y + Z) = X + − = e 2 2 1+ X µ ¶ µ ¶ (5.1.6)

From the deﬁnitions above, the total number density of particles is ρN n = n + n = A I e µ where the mean molecular weight is deﬁned as

−1 1 1 µ = + (5.1.7) µ µ · I e ¸ The Ionization Fraction

The calculation of mean molecular weight requires knowledge of the chemical composition of the material and the ionization fraction. To calculate ionization fraction, one needs the Saha equation. In general, the Saha equation can be used to compute ionization fractions over most of the star. It does, however, require that the gas be in thermodynamic equilibrium. This is true through- out the star, as at high densities, collisions will control the level populations. This approximation only breaks down in the solar corona, where the densities become very low. The Saha equation also breaks down in the centers of stars, where high densities cause the ionization energies of atoms to be re- duced. (Obviously, if the mean distance between atoms is d, then there can be no bound states with radii greater than d/2.) In the case of the hydrogen atom, the Bohr radius of level∼ n is

2 2 ¯h −9 2 an =(n + 1) 2 = 5.28 10 (n + 1) cm mee ×

Thus, if the particle density is

µma −3 ρ 3 0.3 µ g cm ∼ 4/3π(2a0) ∼ then all the hydrogen is necessarily pressure ionized. In practice, the Saha equation begins to break down at nuclear distances of −3 −3 10 a0, which corresponds to 2.7 10 µ g-cm . To correct ∼for this eﬀect, the Saha equation∼ is normally× used until it begins to show decreasing ionization fractions toward the center of the star. When this happens, complete ionization is assumed. To derive the Saha equation, begin by considering the Boltzmann equation, which states that the number of atoms in level i relative to level j is n ω − i = i e χij /k T (5.2.1) nj ωj where ωi is the statistical weight of the level i (i.e., the number of separate, individual states that are degenerate in energy), and χij is the diﬀerence in energy between the two levels. The number of atoms in level i relative to the number in all levels is thus

ni ωi = + + + n ω0e χi0/kT + ω1e χi1/kT + ω2e χi2/kT + . . .

−χi/kT ωie = − − ω0 + ω1e χ1/kT + ω2e χ2/kT + . . .

− n e χi/kT i = ω (5.2.2) n i u

th where χi is the energy diﬀerence between the i level and the ground state. The variable u is the partition function for the atom (or ion). Because u is a function of temperature, it is sometimes written u(T ). Now let’s generalize this equation to electrons in the continuum. Let ni be the number of atoms in all levels (deﬁned as n above), and let state i+1 be that where an excited electron is in the continuum with momentum between p and p + dp. The Boltzmann equation then gives

2 dn +1 dω +1 χ + p /2m i = i exp i e n u − kT i i µ ¶ where χi is the energy needed to ionize the ground state of the atom, and dωi+1 is the statistical weight of the ionized state. Now consider that dωi has two components: one from the ion (ωi+1), and other from the free electron (dωe). The former is just the statistical weight of the ground state of the ion, while the latter can be computed using the exclusion rule. Since each quantum cell in phase space can have only two electrons in it (spin up and spin down), then the number of degenerate states in a volume h3 is d3xd3p dV d3p 2 dω = 2 = 2 = dV 4πp2dp (5.2.3) e h3 h3 h3 Thus 2 2 dn +1 8πp ω +1 χ + p /2m i = i exp i e dV dp n h3 u (T ) − kT i i µ ¶

The number of electrons in volume dV = 1/ne, so the total number of electrons in all continuum states is therefore R ∞ 2 n +1 ω +1 8π − p i = i e χi/kT p2 exp dp n u (T ) n h3 −2m kT i i e Z0 µ e ¶ 2 2 or, if we let x = p /2mekT , then ∞ n +1 ω +1 8π − − 2 i = i e χi/kT (2m kT ) x2 e x (2m kT )1/2 dx n u (T ) n h3 e · e i i e Z0 ∞ ω +1 8π − − 2 = i e χi/kT (2m kT )3/2 x2e x dx u (T ) n h3 e i e Z0

ωi+1 8π −χi/kT 3/2 √π = 3 e (2mekT ) ui(T ) neh · 4

3/2 ni+1 2 ωi+1 (2πmekT ) −χi/kT = 3 e (5.2.4) ni ne ui(T ) h Finally, note that for the calculation above ni+1 represents those atoms of species ni that have one electron in the continuum state, i.e., ionized. It does not consider atoms of ni+1 that are them- selves excited. (In other words, ni+1 in (5.2.4) only includes ionized atoms in their ground state.) To include all the excited states of ni+1, we must again sum the contributions in exactly the same way as we did in (5.2.2). Thus, the statistical weight in (5.2.4) should be replaced by the partition function, and

3/2 n +1 2 u +1(T ) 2πm kT − i = i e e χi/kT (5.2.5) n n u (T ) h2 i e i µ ¶ This is the Saha equation, which relates the number of atoms in ionization state i + 1 to the number in ionization state i. Note that if need be, we can substitute the electron pressure for the electron density using Pe = nekT , and write the Saha equation as

3/2 n +1 u +1(T ) 2πm − i P = 2 i e (kT )5/2 e χi/kT (5.2.6) n e u (T ) h2 i i µ ¶ The sense of these equations is intuitive: the higher the temperature, the greater the ratio, but the higher the density (or pressure), the lower the ratio due to the greater possibility for recombinations).