PHY688, Statistical Mechanics

Transcription

1 Department of Physcs & Astronomy 449 ESS Bldg. Stony Brook Unversty January 31, 2017 Nuclear Astrophyscs

2 Thermodynamcs Internal Energy Densty and Frst Law: ε = E V = Ts P + n µ, dε = Tds + µ dn dp = sdt + n dµ Helmholtz Free Energy Densty: f = F V = ε Ts, Gbbs Free Energy Densty: g = G V = n µ, ( ) ε = T, s n ( ) f = s, T n ( ) P = s, T µ df = µ dn sdt dg = (µ dn + n dµ ) ( ) ε n ( ) f n ( ) P µ n j,t n j,t µ j,t = µ = µ = n

3 Classcal Statstcs A macrostate has N partcles arranged among m volumes, wth N ( = 1... m) partcles n the th volume. The total number of allowed mcrostates wth dstngushable partcles s W = N! m m N! ; ln W = ln N! ln N!. For a large number of partcles, use Strlng s formula ln N! = N ln N N. m ln W = N ln N N (N ln N N ). The optmum state s the macrostate wth the largest possble number of mcrostates, whch s found by maxmzng W, subject to the constrant that the total number of partcles N s fxed (δn = 0). In addton, we requre that the total energy be conserved. If w s the energy of the th state, ths s ( m ) m δ w N = w δn = 0.

4 Wth these constrants, the mnmzaton s [ ( )] m m δ ln W α N β w N = 0. m [ln N α βw ] δn = 0. N = αe βw = αe w /k B T, whch s the famlar Maxwell-Boltzmann, or classcal, dstrbuton functon.

5 Quantum Statstcs In the quantum mechancal vew, only wthn a certan phase space volume are partcles ndstngushable. The mnmum phase space s of order h 3. Denote the number of mcrostates per cell of phase space of volume h 3 as W. Then the number of mcrostates per macrostate s W = W. Note we have to consder both the partcles and the compartments nto whch they are placed. If the th cell has n compartments, there are n sequences of N + n 1 tems to be arranged. There are n(n + n 1)! ways to arrange the partcles and compartments, but we have overcounted because there are n! permutatons of compartments n a cell, and the order n whch partcles are added to the cell s also rrelevant (the factor N! we had n the classcal case). Thus W = n(n + n 1)! N!n! = (N + n 1)! N!(n 1)!.

6 Bosons Optmzng ths, we fnd δ ln W =δ [(n + N 1) ln(n + N 1) N ln N or (n 1) ln(n 1) ln αn βw N ] = [ ln n + N ] 1 ln α βw δn = 0, N ( ) 1 ( 1 N = (n 1) αe w /k B T 1 = (n 1) e (w µ)/k B T 1). Ths s the relevant expresson when there s no lmt to the number of partcles that can be put nto the compartment of sze h 3,.e., for bosons. We dentfed the chemcal potental µ = k B T ln α. The classcal lmt s α, µ, snce the 1 s then gnorable. In the boson case α 1 (µ 0) snce w > 0 and N > 0. Bosons become degenerate when α 1 or µ 0, whch s the stuaton for photons.

7 Fermons For fermons, only 2 partcles can be put nto a compartment, where 2 s the spn degeneracy. Thus, phase space s composed of 2n half-compartments, ether full or empty. There are no more than 2n thngs to be arranged and therefore no more than 2n! mcrostates. But agan, we overcounted. For N flled compartments, the number of ndstngushable permutatons s N!, and the number of ndstngushable permutatons of the 2n N empty compartments s (2n N )!. Thus Optmzng: or W = δ ln W =δ = [ ln 2n N N ( N = 2n αe w /k B T + 1 (2n)! N!(2n N )!. [2n ln(2n) (2n N ) ln(2n N ) ln αn βw N ] ln α βw ] δn = 0, ) 1 ( 1 = 2n e (w µ)/k B T + 1). In the fermon case, there s no restrcton on the value of µ, and fermons become degenerate when µ.

8 Statstcal Physcs of Fermons Perfect gas no ntractons sngle partcle energy ɛ 2 = m 2 c 4 + p 2 c 2 Occupaton ndex (T now has unts of energy, k B = 1) f = [ ( ɛ µ 1 + exp T )] 1 degenerate non-degenerate ɛ, µ, T scaled by mc 2 7ɛ

9 Fermon Thermodynamcs The number and energy denstes are n = g h 3 fd 3 p; ε = g h 3 ɛfd 3 p where g s the spn degeneracy. g = 2j + 1 for massve partcles wth spn j,.e., g = 2 for electrons, muons and nucleons, g = 1 for neutrnos. The entropy densty s ns = g h 3 [f ln f + (1 f ) ln(1 f )]d 3 p and the pressure s P = n 2 [ (ε/n) n ] = Tsn + µn ε = g s 3h 3 n = P ; ns = P. µ T T µ p ɛ p fd 3 p. Defne degeneracy parameters φ = µ/t and ψ = (µ mc 2 )/T : P = ɛ + n ɛ + T P P P ; = ns + nφ; = ns + nψ. n T T T T n φ ψ

10 φ = µ/t, ψ = (µ mc 2 )/T mc 2 = MeV n c = (g/2π 2 )(mc/ ) 3 = fm 3

11 Non-Relatvstc Fermons Assume p << mc. Defne x = p 2 /(2mT ) and ψ = (µ mc 2 )/T. n = g(2mt )3/2 4π ε = nmc 2 + P = 2 3 (ε nmc2 ) = x 1/2 dx g(2mt )3/2 1 + ex ψ 4π 2 3 F 1/2 (ψ) gt (2mT )3/2 4π 2 3 F 3/2 (ψ). gt (2mT )3/2 6π 2 3 F 3/2 (ψ). s = 5F 3/2(ψ) 3F 1/2 (ψ) ψ. F (ψ) s the non-relatvstc Ferm ntegral, satsfyng df (ψ) dψ = F 1(ψ).

12 Ferm Integrals F 0 (ψ) = ln(1 + e ψ ) For zero argument and > 0 (ζ s the Remann zeta functon): F (0) = (1 2 )Γ( + 1)ζ( + 1) F (0) =! ( 1) F (ψ) F (0) + F 1 (0)ψ + F 2 (0)ψ The recurson df /dψ = F 1 can be used to defne non-nteger negatve ndces, but F wth negatve nteger does not exst. F (0) F (0) -7/ / / / / / π 4 /120 = ln(2) = / / π 2 /12 = π 6 /252 =

13 Non-Relatvstc Lmtng Expressons Non-degenerate and non-relatvstc: In ths lmt, use the expanson F (ψ) = Γ( + 1) ( 1) n+1 e nψ n (+1), ψ n=1 ( ) 3/2 mt n = g 2π 2 e ψ, P = nt, s = 5 2 ψ. Degenerate and non-relatvstc: In ths lmt, use the Sommerfeld expanson: ( ) 2n F (ψ) = ψ+1 ( + 1)! π + 1 ( + 1 2n)! C n, ψ ψ n=0 The constants C n are C 0 = 1, C 1 = 1/6, C 2 = 7/360, C 3 = 31/ [ g(2mψt )3/2 n = 6π ( ) ] 2 π +, s = π2 8 ψ 2ψ + [ gψt (2mψT )3/2 P = 15π ( ) ] 2 π + = 1 ( 6π 2 3 ) 2/3 n 5/ ψ 5m g

14 Relatvstc Fermons Ths case corresponds to settng the rest mass to zero. φ = µ/t n = g ( ) 3 T 2π 2 F 2 (φ), P = ε c 3 = gt ( ) 3 T 6π 2 F 3 (φ), s = 4F 3(φ) c 3F 2 (φ) φ. Extremely relatvstc and non-degenerate: Expand n powers of e φ n = g π 2 ( T c ) 3 e φ, P = ε 3 =nt, s = 4 ln [ π 2 n g ( ) ] 3 c = 4 φ. T Extremely relatvstc and extremely degenerate: Use the Sommerfeld expanson [ n = g ( µ ) ( ) ] 3 2 π 6π , s = π2 c φ φ + [ P = ε 3 = gµ ( µ ) ( ) ] 3 2 π 24π = c ( ) 6π 2 1/3 n 4/3 +, c φ 4 g

15 Degenerate Fermons Ths case corresponds to φ >> 0. For extreme degeneracy, f approaches a step functon at ɛ = ɛ F, the Ferm energy. For any degeneracy, t s useful to defne the Ferm momentum ɛ F = m 2 c 4 + p 2 F c2. Defne x = p F /mc, so that µ = mc x 2 n extreme degeneracy. In the case µ/t, the occupaton f becomes a step functon, wth f = 1 for ɛ µ and f = 0 for ɛ > µ. Then one obtans n = g ( mc ) 3 x 3 6π 2 = g ( pf ) 3, 6π 2 ( P = gmc2 mc ) 3 [x(2x 2 48π 2 3) 1 + x snh 1 x], ( ε nmc 2 = gmc2 mc ) 3 [3x(2x 2 48π 2 + 1) 1 + x 2 8x 3 3 snh 1 x], s = 0. One can recover the extreme degenerate non-relatvstc and relatvstc expressons derved earler usng the respectve lmts x 0 and x. Note that p F s defned by n = g/(6π 2 )(p F / ) 3, and holds for any degeneracy. If degeneracy s not extreme, µ/ɛ F = 1 (πt /ɛ F ) 2 /12 +.

16 Fermons Wth Par Equlbrum Under many condtons, temperatures become large enough that e e + pars are formed n abundance, even when T < m e c 2. For partcle (+)-antpartcle( ) pars n equlbrum, µ + = µ. The net dfference of partcles and ant-partcles and the total pressure are n =n + n = 4πg h 3 P =P + + P = 4πg 3h 3 Defnng x = pc/t and z = mc 2 /T, n = g ( ) 3 T 2π 2 snh φ c 0 P = gt ( ) 3 T 6π 2 c 0 0 [ ] p e 1 dp, (E µ)/t 1 + e (E+µ)/T p 3 E [ ] 1 p 1 + e + 1 dp (E µ)/t 1 + e (E+µ)/T x 2 cosh φ + cosh z 2 + x 2 dx, [ x 4 z2 + x 2 cosh φ + e z 2 +x 2 cosh φ + cosh z 2 + x 2 For n > 0, µ µ + > 0, so ths gas s never extremely non-degenerate; however, pars won t be mportant under extremely degenerate condtons when µ/t >> 0 ether. ] dx.

17 φ = µ/t, ψ = (µ mc 2 )/T mc2 = MeV n c = (g/2π 2 )(mc/ ) 3 = fm 3

18 Relatvstc Pars µ >> mc 2 or T >> mc 2 n = n + n = g ( ) 3 T [ ( µ ) ( 2π 2 F 2 F 2 µ )] c T T [ = g ( µ ) ( ) ] 3 2 πt 6π ; c µ ε 3 = P = P + + P = gt ( ) 3 T [ ( µ ) ( 6π 2 F 3 + F 3 µ )] c T T [ = gµ ( µ ) ( ) 3 2 πt 24π ( πt c µ 15 µ [ gt µ2 s = 6n( c) ( ) ] 2 πt. 15 µ These expressons are exact. Note the nverson to obtan µ(n, T ): [ (q µ = r q/r, r = 3 + t 2) ] 1/2 1/3 + t where t = 3π 2 ( c) 3 n/g and q = (πt ) 2 /3. For T, µ 6π 2 ( c) 3 n/(gt 2 ) 0 +. ) 4 ] ;

19 Non-relatvstc Pars When T 0, µ (mc 2 ) + and pars are neglgble. Wth ncreasng temperature, µ ncreases to a maxmum, then decreases below mc 2. The non-degenerate expanson gves ( ) 3/2 ( ) 3 mt mt n ± g 2π 2 e (±µ mc2)/t, n + n n1 2 = g 2 2π 2 e 2mc2 /T n ± = ± n 2 + [ (n 2 ) 2 + n 2 1 ] 1/2 P = (n + + n )T = ( n 2 + 4n1) 2 1/2 T, ε = (n + + n ) (mc ) T, ( ) 5 s = 2 + mc2 n+ + n µ T n T, [ ( ) n n 2 1/2 ] µ = T ln n 1 4n 2 1

20 Bosons [ ] 1 f = e (ɛ µ)/t 1 ns = g h 3 [f (1 f ) (1 f ) ln(1 f )] d 3 p f 0 and ɛ mc 2 mply µ mc 2. In the lmt µ mc 2, a Boson condensate appears and an unlmted number of partcles can exst n a zero-momentum state. Extremely Non-Degenerate: µ/t, so the fermon and boson dstrbutons are ndstngushable from the classcal Maxwell-Boltzmann dstrbutons. Extremely Degenerate: In ths lmt, µ = mc 2 or ψ = 0. n = g 2π p 2 2g dp, P = e (ɛ mc2 )/T 1 3π 2 3 One can show that x e x 1 dx = ( 1 2 ) 1 F (0) = Γ( + 1)ζ( + 1). 0 0 p 3 e (ɛ mc2 )/T 1 dp.

21 Extremely Degenerate and Relatvstc: (T >> mc 2, ψ = 0) n = 2g ( T 3π 2 c P = 4gT 21π 2 ( T c ) 2 F 2 (0), ε = 3P, ) 3 F 3 (0) = gπ2 T 90 ( ) 3 T, s = π4 c 15F 2 (0) Extremely Degenerate and Non-Relatvstc: (T << mc 2, ψ = 0) n = g π 2 (mt ) 3/2 3 F 1/2 (0) 2 1, ε = nmc P, P = 4gT F 3/2 (0) 3π 2 2 3/2 1, s = 10 3 F 3/2 (0) 2 1/2 1 F 1/2 (0) 2 3/

22 Extremely Relatvstc: (mc 2 0, µ 0) Ths case ncludes the photon gas. Wth g r = 2, ε r = 3P r = 3 4 TS r = π2 T 15 ( ) 3 T = at 4. c S s the entropy densty and a s the radaton constant. These expressons are 8/(7g) tmes the value for relatvtc fermons. Pars are always mportant f e e + pars are mportant, n whch case the combned pressure s (11/4)P r. If neutrnos of all 3 flavors are trapped n matter, the total pressure ncreases to (43/8)P r. If electrons are degenerate, the photon pressure s neglgble.