ASTR 5110 Atomic & Molecular Physics Fall 2013. Stat Mech Midterm. This is an open book, take home, 24 hour exam. When you have finished, put your answers in the envelope provided, mark the envelope with the time you started and the time you finished, and put in my JILA box (the JILA box is more secure than the APS box). You may quote without proof or penalty any result you like from the notes on Statistical Mechanics. Youmayuseacalculatororacomputer ifyouwish. Donotcollaborate. Attempt three out of four questions. Each question is worth 33 points, making a total of 100 for the exam (everybody gets 1 free point). Read the question before you answer it. A partial answer receives partial credit; no answer receives no credit. Here are some useful physical constants, in cgs units, as preferred by old-fashioned astronomers. You are welcome to use mks units, as recommended by wise authorities. c = 2.998 10 10 cms 1. h = 6.626 10 27 ergs. G = 6.67 10 8 c.g.s. k = 1.381 10 16 ergk 1. m e = 9.110 10 28 gm. m p = 1.673 10 24 gm. M J = 1.9 10 30 gm. 1eV = 11605K. 1
1. Metallic Hydrogen in a Planetary Core (33 points) The characteristic properties of metals their high electrical and thermal conductivity, and high reflectivity can be traced to the fact that the outer electrons of the atoms form an electron degenerate gas. (a) (4 points) Consider an ideal gas of free electrons in thermodynamic equilibrium at effectively zero absolute temperature, T = 0. Argue that the occupation number of each single-particle state i is N i = 1 if the energy of the single-particle state is below some level, ǫ i < µ, while N i = 0 if the energy is above this level, ǫ i > µ. The energy µ is called the Fermi energy. (b) (8 points) The mean number of electrons N and electron pressure P in a volume V can be expressed as usual in terms of the occupation numbers N i as N = i N i and PV = 1 v i p i Ni (1.1) 3 i where for free particles the sum over single-particle states i translates into an integral over free states, i g spin d 3 xd 3 p/h 3. Use these equations and your answer to (a) to show that the electron number density n N/V and pressure P in a fully degenerate gas of nonrelativistic electrons are given by (what is g spin for electrons?) n N V = g spin4π(2mµ) 3/2 3h 3 (1.2) P = g spin8π(2m) 3/2 µ 5/2 15h 3 (1.3) [these are equations (36.11) and (36.12) in the notes]. (c) (6 points) The equations of stellar/planetary structure lead to the following approximate relations between the central density and pressure ρ and P, and the mass M and radius R, of a planet: ρ M R, P 3 ρ GM (1.4) R [the second of these equations comes from balancing pressure and gravity]. Eliminate R from these equations to derive an approximate expression for the mass M of the planet in terms of the gravitational constant G, and the central density and pressure ρ and P. (d) (8 points) Suppose that core of the planet is in fact electron degenerate, and furthermore is made of hydrogen, so that the mass density is ρ = m p n. Use equations (1.2) and (1.3) to rewrite your expression for the mass M from part (c) in terms of the Fermi energy µ and a bunch of constants. (e) (7 points) Hydrogen will become metallic roughly when the Fermi energy of the electrons exceeds the ionization energy of hydrogen, µ > χ H = 13.6eV. Estimate how massive a planet should be, in Jupiter masses M J, in order that hydrogen at its core be metallic. 2
2. Solar Opacity and H (33 points) The negative hydrogen ion H, which contains a protonnucleus and two bound electrons, H H+e has a single bound state with binding energy χ H = 0.754eV. The negative ion H is the dominant source of opacity in the atmospheres of stars like the sun. The extra electron needed to form H comes mainly from metals with weakly bound electrons, such as sodium Na, whose ionization potential is χ Na = 5.139eV. So consider thermodynamic equilibrium between neutral hydrogen H, the negative hydrogen ion H, neutral sodium Na, singlyionized Na +, and electrons e. Assume that all species are in their ground states (i.e., ignore the contribution of excited states), and ignore all other ionic species. The degeneracy (number of states) of the ground electronic states of H and Na is 2, and that of H and Na + is 1. (a) (4 points) Write down the Saha equation for X n H n H in terms of electron density n e and temperature T. (b) (4 points) Write down the Saha equation for Y n Na n Na + in terms of electron density n e and temperature T. (c) (4 points) Using your answers to (a) and (b), reexpress Y in terms of X and T. (d) (4 points) Write down the condition for charge neutrality (an equation relating the number densities of the various species). (e) (4 points) The abundance of sodium in the sun is very much smaller than the abundance of hydrogen. Using this assumption and your answer to (d), show that the abundance of H must be very much less than that of neutral H, that is, n H n H. (f) (3 points) Let α = n Na +n Na + n H denote the abundance of sodium relative to neutral hydrogen. Rewrite the charge neutrality condition (d) as an equation for n e in terms of X, Y, α, and n H. (g) (6 points) Substituting your result for n e from (f) into your equation for X from (a), and using (c) to eliminate Y, find a quadratic equation for X in terms of α, n H, and T. Write down the numerically stable solution for X. (Yes, I know, the expression is a bit of a mess if you write it out in full, but you can shorten it by defining some subsidiary quantities.) (h) (4 points) Argue physically (don t try to get heavy into the mathematics) that X must become small at both low temperature and high temperature, and that X must therefore reach a maximum at some temperature, which depends somewhat on the density n H, but which is of the order of several thousand K. 3
3. Nuclear Statistical Equilibrium (33 points) During the explosive nuclear burning of a carbon-oxygen white dwarf, all nuclei consisting of a whole number of α-particles ( 4 He nuclei) are liable to reach mutual thermodynamic equilibrium. (a) (8 points) Write down the Saha equation which relates the number density n N of nuclei consisting of N α-particles to the number density n α of α-particles. Ignore excited nuclear states. (b) (12 points) 56 Ni is the most tightly bound nucleus consisting of a whole number (N = 14) of α-particles. At approximately what temperature, in MeV, are the number densities of 56 Ni and 4 He equal? [Hints: From the condition N N = N α you should derive an equation of the form ρ = f(t), where for simplicity you can take ρ = ρ N +ρ α (N +1)ρ α. To evaluate the Boltzmann factor in the equation, you will need to use the table of binding energies below. Make a short table of ρ as a function of T, and thereby show that the transition temperature is rather insensitive to the density. As a practical matter, you may assume that the relevant range of mass (not number!) density is ρ 10 6 10 8 gmcm 3.] (c) (6 points) What happens if the temperature exceeds the temperature in part (b)? Argue that the transition acts as a sort of thermostat that ensures that the temperature of the burning material is likely to be close to the transition temperature in part (b). (d) (7 points) What is the ratio n N /(n N 1 n N+1 ) 1/2 of the number density of species N relative to the geometric mean of its two neighbors? In the light of your answer, and the table below, which species are likely candidates as the most abundant product of explosive nucleosynthesis? [No need to be terribly sophisticated here.] Here is a table of the binding energies E N relative to carbon, and of the differences E N E N (E N 1 + E N+1 )/2 between the binding energy of species N and the mean of its two neighbors: N Nuc E N E N N Nuc E N E N (MeV) (MeV) (MeV) (MeV) 1 He 2.42 9 Ar 30.23 0.20 2 Be 4.94 3.73 10 Ca 34.85 0.90 3 C 0 0.10 11 Ti 37.66 1.17 4 O 4.74 1.22 12 Cr 42.81 0.18 5 Ne 7.04 2.29 13 Fe 48.33 0.03 6 Mg 13.93 0.33 14 Ni 53.90 2.64 7 Si 21.49 1.52 15 Zn 54.19 8 S 26.01 0.15 4
4. Cosmic Neutrino Temperature (33 points) There is believed to be a cosmic neutrino background today, but its temperature should be slightly different from the temperature of the cosmic photon (microwave) background. This question discovers why. Back when the Universe was about 1 second old, its contents could be well described as an adiabatically expanding gas of non-number-conserving (particle-antiparticle pairs were constantly being created and destroyed) relativistic particles in thermodynamic equilibrium: photons γ, electrons eand positrons e +, and neutrinos ν andantineutrinos ν. Nonrelativistic particles contributed negligibly to the energy density and other thermodynamic quantities. (a) (8 points) Using the elementary fact that 1 z 2 1 = 1 (z 1)(z +1) = 1 2(z 1) 1 2(z +1) (4.1) prove that, for n > 1, [this is equation (A36.9) in the notes]. 0 x n 1 dx e x +1 = ( 1 2 1 n) x n 1 dx 0 e x 1 (4.2) (b) (7 points) Arguefrom part (a) that the ratio of energies Ū in fermionic (f) versus bosonic (b) species in an ideal relativistic gas of non-number-conserving (so chemical potential µ = 0 for all species) free particles in thermodynamic equilibrium is Ū f = 7 g f (4.3) Ū b 8 g b where g f and g b are the number of fermionic and bosonic types. [Hint: The energy of a relativistic gas of free particles with µ = 0 is given by the usual formula Ū = V in which ǫ = pc for relativistic species.] ǫ g4πp 2 dp (4.4) e ǫ/kt ±1 h 3 (c) (8 points) Given that Ū = avt4 for photons where a is the radiation constant [equation (36.29) in the notes], obtain an expression for the entropy S in a volume V of the Universe, in terms of its temperature T and the number of relativistic fermionic and bosonic types g f and g b. [Hint: Photons are of course bosons, and have two spin types, so g γ = 2. The entropy is given by S = Ū +PV = 4Ū (4.5) T 3T in which the first equality is the second fundamental equation of thermodynamics (for µ = 0), and the second equality is valid for relativistic free particles, Ū = 3PV.] (d) (8 points) After a minute or so, the temperature of the expanding Universe fell to the point where electrons and positrons annihilated (ultimately leaving a negligible residual of 5
electrons). Entropy was conserved during this process. However by this time neutrinos had become decoupled from the photon-electron-positron plasma, so the entropy of the electrons and positrons was converted into entropy of photons, the entropy of the neutrinos and antineutrinos being conserved separately. In other words, the governing equations relating the entropy before e e + annihilation (initial) to the entropy after annihilation was complete (final) are (S γ +S e +S e +) init = S γ fin (S ν +S ν ) init = (S ν +S ν ) fin. (4.6) If all temperatures were equal before annihilation, in particular (T ν /T γ ) init = 1, then what was (and remains today) the ratio (T ν /T γ ) fin of neutrino to photon temperatures at the end of annihilation? [Hint: The Universe expanded during annihilation, so V init V fin. Electrons and positrons are fermions, with 2 spin states each. Neutrinos come in 3 types with 1 spin state each, but you should find that this fact is irrelevant to the question.] (e) (2 points) The temperature of the cosmic microwave background today is 2.725 K. What is the temperature of the cosmic neutrino background? 6