PHYS 633: Introduction to Stellar Astrophysics

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1 PHYS 633: Introduction to Stellar Astrophysics Spring Semester 2006 Rich Townsend Ionization During our derivation of the equation of state for stellar material, we saw that the partial pressure P e due to free electrons depends on the electron number density n e. In the limit where the material is completely ionized, we wrote n e in the form n e = Z i X i ρ, (1) m i i where X i, Z i and m i are the mass fraction, atomic number and nuclear mass of element i. But what about cases where the ionization isn t complete? In these cases, we need to devise relationships describing the ionization state of all elements making up the stellar material, so that we can calculate how many free electrons are contributed by each atom. Even when in thermal equilibrium, the atoms of each element constituting stellar material are in a range of different ionization stages, and for each ionization state in a range of excitation states above the ground state. However, although the configuration of a individual atom changes quite rapidly with time, as it emits and absorbs photons, the average number of atoms in each state remains well defined can be calculated using statistical mechanics. For a unit volume, let n r,s denote the number of atoms of a given element in ionization stage r and excitation state s. We adopt the convention that r = 0 corresponds to no ionization (i.e., a neutral atom), and s = 0 corresponds to the ground excitation state. Boltzmann demonstrated that, for such a system in equilibrium at temperature T, n r,s can be related to the number in the ground state of the same ionization stage,, via the formula n r,s = g r,s e ψr,s/kt. (2) Here, ψ r,s is the energy of the excited state relative to the ground state, and g r,s and are the statistical weights of the excited and ground states, respectively. These weights represent the number of differing substates that together make up an excitation state of a given energy. A similar formula can be derived to relate the number of particles in the ground states of two adjacent ionization stages, n r+1,0 and. But in this case, in addition to accounting for the statistical weight g r+1,0 of the upper ionization state, we must account for the statistical weight of the free electron created during the process of ionization. Likewise, we must include the kinetic energy 1

2 of this electron, when calculating the energy separation of the two ionization states. So, let s suppose the free electron ends up with a momentum in the interval [p e, p e + dp e ], and suppose the number of substates with energy in this interval is dg(p e ). Then, the number of atoms per unit volume with electron momenta in the interval will be given by dn r+1,0 = g [ r+1,0dg(p e ) exp χ r + p 2 ]. (3) kt Here, χ r is the r th ionization potential of the element the amount of energy required to remove an additional electron from the ground state of an atom that has already been ionized r times. The second term in the exponential, which depends on the electron mass m e, represents the electron s kinetic energy. By integrating equation (3) over all possible values of the free electron momentum, we can obtain a value for n r+1,0. First, however, we need to calculate the differential statistical weight dg(p e ) of the free electron, and for this we to apply some quantum-mechanical principles. The Heisenberg uncertainty principle tells us that 6-dimensional phase space is quantized into cells with volume h 3. With two possible spin orientations for each cell in phase space, the statistical weight of the free electron is therefore dg(p e ) = 2d3 p e d 3 r e h 3, (4) where d 3 p e is the momentum volume occupied by the electron, and d 3 r e is the corresponding space volume. For electrons with momenta in the interval [p e, p e + dp e ], d 3 p e is given by the volume of a spherical shell with radius p e and thickness dp e. That is, d 3 p e = 4πp 2 e dp e, (5) so that the statistical weight is dg(p e ) = 4πp2 edp e d 3 r e h 3. (6) To calculate the space volume d 3 r e occupied by the electron, we make use of the Pauli exclusion principle, which requires that no two electrons be in exactly the same momentum/space/spin state. This means that, with n e electrons per unit volume, each individual electron can access only a volume 1/n e without running foul of the exclusion principle. Therefore, with d 3 r e = 1/n e, we have dg(p e ) = 4πp2 edp e h 3 n e. (7) We can substitute this result back into eqn. (3), to obtain dn r+1,0 = g r+1,0 4πp 2 [ e h 3 exp χ r + p 2 ] dp e (8) n e kt 2

3 Integrating over all possible free-electron momenta, we find that n r+1,0 is given by n r+1,0 dnr+1,0 g r+1,0 4πp 2 [ e = = 0 h 3 exp χ r + p 2 ] dp e. (9) n e kt Using the substitution this can be written as z = p e, (10) (2m e kt ) 1/2 n r+1,0 = 8π g r+1,0 (2m e kt ) 3/2 n e h 3 e χr/kt 0 z 2 e z2 dz. (11) It is straightforward but tedious to demonstrate that the integral in this expression evaluates to π 1/2 /4; hence, we obtain n r+1,0 = 1 n e g r+1,0 f r (T ), (12) where the function f r (T ) is defined by f r (T ) 2 (2πm ekt ) 3/2 h 3 e χr/kt. (13) Equation (12) relates the number densities of atoms in the the ground states of two neighboring ionization stages. But what about the number densities of atoms in any excitation state of the two stages? To calculate these values, we just sum over the different excitation states; for instance, the total number of atoms in ionization stage r is n r = s=0 n r,s. (14) Using the Boltzmann formula (2), we can write this sum in the form n r = s=0 With a little rearrangement, this becomes g r,s e ψr,s (15) n r = u r, (16) where we have introduced the partition function per unit volume of the ionization stage r (a function only of T and the atomic structure) as u r = s=0 g r,s e ψr,s. (17) 3

4 We can write down similar expressions for n r+1 in terms of n r+1,0, so that eqn. (12) may be rewritten in the more-useful form n r+1 n r = 1 n e u r+1 u r f r (T ). (18) This famous equation is known as the Saha Equation, named after the physicist Meghnad Saha; it is the most general expression for the ionization state of a system in equilibrium. So, how do we go about using the Saha equation? Our ultimate goal is to calculate the electron number density n e a quantity which already appears in the Saha equation itself! If we assume for a moment that the stellar material comprises a single element i, then the number of electrons is just given by the number of ions in the gas: Z i n e = n r. (19) r=1 (recall that Z i is the atomic number of the element, and hence represents the maximum ionization stage). To calculate the individual number densities n r of the ionization stages, we can use the Saha equation to write down Z i independent equations. These equations can be solved simultaneously to give the n r in terms of the number density of neutral atoms, n 0. To calculate n 0, we can make use of the closure relation Z i n = n r = n 0 + n e, (20) r=0 where n is, of course, the total number density (i.e., atoms and ions). This last stage seems rather opaque, so let s consider the simple case of pure hydrogen, where there are only two ionization states, with number densities n 0 and n 1. The relations introduced above let us write the number density of electrons as n e = n 1, (21) and the number density of hydrogen atoms/ions as The Saha equation (18) likewise tells us that n = n 0 + n 1. (22) n 1 n 0 = 2 n e u 1 u 0 (2πm e kt ) 3/2 h 3 e χh/kt, (23) where χ H is the ionization potential of hydrogen ( 13.6 ev). Introducing x as the ionization fraction of the hydrogen, so that x = n 1 n = n e n, (24) 4

5 we can find x by solving the quadratic equation x 2 1 x = 2 1 u 1 (2πm e kt ) 3/2 n u 0 h 3 e χh/kt. (25) If the stellar material contains more than one element (which it usually does), the situation becomes rather more complicated, and we can t use eqns. (19) (25). Nevertheless, the Saha equation (18) remains valid, and we can always obtain sufficient equations to calculate the complete ionization state at each point in the star, and thereby calculate the electron number density n e. Typically, this involves solving sets of simultaneous equations, but this can be quite easily done using a computer. There is one situation, however, where the Saha equation breaks down. Suppose that the temperature is high enough such that everything is ionized. Then, as we steadily increase the electron number density n e, the Saha equation tells us paradoxically that the ionization fraction will begin to decrease! What s going on here is that at very high densities, each ion is being bombarded with electrons, making it very difficult for the ion not to recombine. In real systems, we don t see this recombination, because there is another process (not accounted for by the Saha equation) that comes into play. At sufficiently-high densities, the ions Coulomb potentials overlap, with the consequence that their ionization potentials χ r are reduced. This compensates for the high recombination rates, and ensures that the material remains ionized even as the density is increased. 5