Physics 576 Stellar Astrophysics Prof. James Buckley. Lecture 14 Relativistic Quantum Mechanics and Quantum Statistics

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1 Physics 576 Stellar Astrophysics Prof. James Buckley Lecture 14 Relativistic Quantum Mechanics and Quantum Statistics

2 Reading/Homework Assignment Read chapter 3 in Rose. Midterm Exam, April 5 (take home) Final Project, May 4 (end of reading period)

3 Summarizing Fermi-Dirac Distribution: n k = g k e (E k µ)/kt +1 Bose-Einstein Distribution: n k = g k e (E k µ)/kt 1 Maxwell-Bolzmann Distribution: n k = g k e (E k µ)/kt Thus we see that the probability for a given quantum state to be occupied (for fermions) or the occupation number (for bosons or classical particles) can be written in the unified form: f = 1 e (E µ)/kt ± 1(0)

4 Equation of State for Fermions We already talked quite a bit about photon distribution functions in our discussion of black-body radiation, and in this course will not be too concerned with other bosons (e.g., bose-einstein condensates inside neutron stars). We focus now on the Fermi-Dirac distribution function, in the continuum limit: In the continuum limit, we talk about density of states in some momentum interval (p, p + dp) or energy interval (E,E + de) rather than summing over discrete levels. In this case, the number density of occupied states can be obtained by taking the product of the occupation probability f and the density of states ρ(p)dp to obtain the continuum limit: n(p)dp = f 8πp2 dp n = 8π h 3 h 3 = 8π h 3 p 2 dp e (E µ)/kt +1 p 2 dp e (E µ)/kt +1 To get the equation of state for a gas of fermions, we solve for the pressure given the Fermi-Dirac distribution function P = (2p cos θ)(v cos θ)n(p)dp 1 sin θdθ 2 P = 8π 3h 3 0 similarly U = 8π h 3 p 3 vdp e (E µ)/kt +1 0 Ep 2 dp e (E µ)/kt +1

5 Chemical Potential If to any homogeneous mass in a state of hydrostatic stress we suppose an infinitesimal quantity of any substance to be added, the mass remaining homogeneous and its entropy and volume remaining unchanged, the increase of the energy of the mass divided by the quantity of the substance added is the potential for that substance in the mass considered. While we may now take the fundamental definition of chemical potential to be the Lagrange multiplier that appears from adding a constraint on the total number of particles to the entropy maximization (or adding an effective term to the entropy), a more physical interpretation, is as follows: The chemical potential of a thermodynamic system is the amount by which the energy of the system would change if an additional particle were introduced, with entropy and volume held constant. If a system contains more than one species of particles, each species has its own chemical potential. Josiah Willard Gibbs ( )

6 Chemical Potential For the Maxwell Boltzman distribution we could have written f = e (µ E)/kT rather than f = Ce E/kT when solving for the function maximizing the entropy by simply redefining the Lagrange multiplier α then normalizing the distribution gives f(p)4πp 2 dp = n 4πC C = or, using the QM density of states, for low densities this is almost the same form n = 8π h 3 f(p)p 2 dp µ/kt 8π e h 3 =4πC = 0 n (2πmkT ) 3/2 e p2 /2mkT p 2 dp = n 4πn (2πmkT ) 3/2 µ = kt ln(t 3/2 /n) 3 2 ln 2πmk h 2 kt ln 2

7 Chemical Potential If one also realizes that the energy changes as one adds or removes particles of each of n species with number densities n 1,n 2,..., n n then we can write the total internal energy as: U = U(S, V, n 1,..., n n ) and then the chemical potential is U µ i N i S,V,N j=i Including the extra contribution to the total internal energy from adding or subtracting particles of different species we have: du = TdS P dv + i µ i dn i Note that all of the natural variables of U(y i ), y i extensive quantities, that is: = {S, V, and {N i }} are U({αy i })=αu({y i }) are homogeneous functions and therefore simply integrable giving an integral form for the equation of state: U = TS PV + µ i N i

8 Thermodynamic Potentials In general, a thermodynamic potential is a scalar potential function used to represent the thermodynamic state of a system. For example, the internal energy, U, is the energy of configuration of a given system of conservative forces. Expressions for all other thermodynamic energy potentials are derivable via Legendre transforms from an expression for U. Common thermodynamic energy potentials are: Potential Formula Natural Variables Internal Energy U TS PV + i µ in i S,V,{N i } Helmholtz free energy F U TS T,V,{N i } Enthalpy H U + PV S, P, {N i } Gibbs Free Energy G U + PV TS T,P,{N i

9 Gibbs Free Energy Physically, the Gibbs free energy is defined as the maximum amount of work obtainable from an isothermal, isobaric thermodynamic system. Now define a new state variable, the Gibbs free energy G: G = U + PV TS dg = du + P dv + V dp TdS SdT substituting for du du = TdS P dv + i µ i dn i dg = V dp SdT + i µ i dn i G µ i = N i T,P,n j=i Substituting for U in the expression for the Gibbs free energy, i.e.: U = TS PV + i µ i N i one obtains : G = TS PV + i µ i N i + PV TS or G = i µ i N i

10 Classical Limit For a single species with number density n = N/V, at low density (so we can treat the particles as classical, distinguishable particles) we have G = U TS + PV Substituting the expression for the internal energy per unit volume u = 3 2 nkt, pressure P = nkt and for the entropy per unit volume s = nk ln(t 3/2 /n)+ns 0 : 3 G = 2 nkt nkt ln(t 3/2 /n) nt s 0 V + nkt V and G = µn = µnv µ = 5 2 kt kt ln(t 3/2 /n) Ts 0 The classical expression for the chemical potential should agree with the lowdensity limit of the QM expression derived previously: µ = kt ln(t 3/2 /n) 3 2πmk 2 ln h 2 kt ln 2 Which is possible, if we redefine the arbitrary constant in the entropy s 0 to match the QM calculation (which takes into account the minimum volume of phase space p 3 x 3 = h 3 from the uncertainty principle) s o = 5 2 k k ln 2πmk h 2 + k ln 2

11 NR Gas, Weak Degeneracy Weak degeneracy µ/kt 1 and µ<0 e (E µ)/kt is small, and 1 f = e (E µ)/kt (1 + e (E µ)/kt ) E = p2 2m f e (E µ)/kt (1 e (E µ)/kt ) n e = 8π h 3 fp 2 dp for nonrelativistic energies E<mc2 substitute x p/ 2mkT E/kT = x 2 n e = 8π (2mkT )3/2 e µ/kt e x2 x 2 dx e 2µ/kT e 2x2 x 2 dx h3 0 0 e βx 2 dx = e βx2 dx = 1 e 0 β 0 β 4 βr2 2πrdr 0 = π π e β 4β u du = 1 π β 3/2 = 4 2 n e = 0 2(2πmkT )3/2 h 3 e µ/kt 1 eµ/kt 2 3/2 4β 3/2

12 NR Gas, Weak Degeneracy n e = 2(2πmkT )3/2 h 3 e µ/kt 1 eµ/kt 2 3/2 For a weakly degenerate gas, µ/kt is large and negative, so the second term in parentheses is small compared with 1, and we can approximate by setting µ = µ 0 (the chemical potential for a nondegenerate gas). This will allow us to invert the expression and solve for µ n e 2(2πmkT )3/2 h 3 e µ/kt 1 eµ 0/kT where µ 0 = kt ln(t 3/2 /n) 3 2 ln 2πmk h 2 2 3/2 kt ln 2 Substituting, and solving for µ we obtain (after a fair bit of algebra): 2πmk µ weak degen = kt ln h 2 3/2 kt ln(2t 3/2 /n e ) n e 2 1/2 h 2 2πmkT 3/2

13 NR Gas, Weak Degeneracy Armed with the chemical potential: 2πmk µ weak degen = kt ln and the occupation probability h 2 3/2 kt ln(2t 3/2 /n e ) n e 2 1/2 f e (E µ)/kt (1 e (E µ)/kt ) h 2 2πmkT we can solve for thermodynamic quantities like pressure and internal energy by performing the appropriate moment integrals of the probability density functions. for internal energy we can calculate U = 8π h 3 where E is the kinetic energy, for a NR gas E = p 2 /2m and for a relativistic gas E = E m 0 c 2 0 Ef(p)p 2 dp Substituting for the chemical potential, and doing lots of math, we obtain U e 3 2 n ekt 1+ n e h 2 3/2 2 7/2 2πmkT U And the heat capacity per unit volume is c v = = 3 T 2 n ek 1 n e 2 9/2 V 3/2 h 2 2πmkT 3/2

14 Relativistic Gas, Weak Degen. For an ultrarelativistic gas, with weak degeneracy, we follow the same procedure, but set E = pc and introduce the variable x = pc/kt in performing the integrals over the probability distribution function f. We start by calculating the chemical potential for an ultrarelativistic nondegenerate gas µ r,nd n e = 8π (hc) 3 eµ r,nd/kt e pc/kt c 3 p 2 dp Evaluating the intergral and solving for µ gives µ r,nd = kt ln(n e /16π) 3kT ln(kt/hc) For an ultrarelativistic, but weakly degenerate gas, the total number density is given by: 3 kt n e 16π e µ/kt 1 e µ r,nd /kt 2 hc 3 Solving for µ (which we will call µ r,d ) we obtain 3 16π kt µ r,d = kt ln + n 3 e hc n e hc 2 7 π kt U e 3n e kt 1+ n 3 e hc 256π kt c V 3n e k 0 1 n e 128π hc kt 3

15 Atoms If there are additional internal degrees of freedom in our particles (e.g., atomic energy levels, vibrational states in molecules, etc.) we start with a modified formula for the internal energy per particle: u p =3/2kT + n g ne n e E n/kt n g ne E n/kt this is more compactly written in terms of the partition function, which we can define as Z n u p = 3 2 g n e E n/kt kt + kt 2 ln Z T Following the usual steps, one can derive the chemical potential, specific heat and entropy. Of greatest importance is the chemical potential: µ A = kt ln(t 3/2 /n A ) 3 2 kt 2πmA k h 2 kt ln g n e E n/kt n

16 Reaction Equilibrium The Gibbs free energy for a system with n different species, of number densities n i is given by n dg = SdT + V dp + µ i dn i if changes occur under constant P and T then (dg) P,T = n µ i dn i i=1 And integrating gives the result (derived previously by substituting for U in the expression G = U + PV TS) n G = µ i n i i=1 Say that we have reversible reactions that convert particles of type ν i particles of type χ i into other species, e.g.: ν 1 χ 1 + ν 2 χ 2 ν 3 χ 3 + ν 4 χ 4 then changes in the number density of the constituents are constrained such that dn 1 = dn 2 = dn 3 = dn 4 ν 1 ν 2 ν 3 ν 4 i=1

17 Reaction Equilibrium (dg) P,T = 4 i=1 µ i dn i = dn 1 µ 1 + ν 2 µ 2 ν 3 µ 3 ν 4 µ 4 ν 1 ν 1 ν 1 ν1 or (dg) P,T = dn 2 µ 1 + µ 2 ν 3 µ 3 ν 4 µ 4, etc. ν 2 ν 2 ν 2 In equilibrium dg = 0 and, regardless of how we write dg we get the same condition for equilibrium: For other reactions, e.g., ν 1 µ 1 + ν 2 µ 2 = ν 3 µ 3 + ν 4 µ 4 ν 1 χ 1 ν 2 χ 2 + ν 3 χ 3 we get similar equations ν 1 µ 1 = ν 2 µ 2 + ν 3 µ 3 and the prescription for going from any arbitrary reactions (in equilibrium with the reverse reaction) is straightforward.

18 Ionization Equilibrium H e + p µ H = µ e + µ p µ e = kt ln(t 3/2 /n e ) 3 2 kt ln 2πme k h 2 µ p = kt ln(t 3/2 /n p ) 3 2 kt ln 2πmp k h 2 kt ln 2 kt ln 2 µ H = kt ln(t 3/2 /n H ) 3 2 kt 2πmH k h 2 kt ln n g n e E n/kt assuming m p m H n H n e n p = gn e E n/kt 4 2πmkT h 2 3/2 Giving the Saha equation (after a bit of math!)