Classical Statistical Mechanics

A macrostate has N particles arranged among m volumes,with Ni(i = 1 . . .m) particles in the ith volume. The totalnumber of allowed microstates with distinguishable particles is

W =N !

∏mi Ni!

; lnW = lnN ! −m∑

i

lnNi! .

For a large number of particles, use Stirling’s formula

lnN ! = N lnN −N .

lnW = N lnN −N −m∑

i

(Ni lnNi −Ni) .

The optimum state is the macrostate with the largest possiblenumber of microstates, which is found by maximizing W , sub-ject to the constraint that the total number of particles N isfixed (δN = 0). In addition, we require that the total energybe conserved. If wi is the energy of the ith state, this is

δ

(

m∑

i

wiNi

)

=m∑

i

wiδNi = 0 .

With these constraints, the minimization is

δ

[

ln

(

W − α

m∑

i

Ni − β

m∑

i

wiNi

)]

= 0 .

m∑

i

[lnNi − α− βwi] δNi = 0 .

Ni = αeβwi = αe−wi/kT ,which is the familiar Maxwell-Boltzmann, or classical, distri-bution function.

Quantum Statistical Mechanics

In the quantum mechanical view, only within a certain phasespace volume are particles indistinguishable. The minimumphase space is of order h3. Now denote the number of mi-crostates per cell of phase space of volume h3 as Wi. Then thenumber of microstates per macrostate is

W =∏

i

Wi .

Note we have to consider both the particles and the compart-ments into which they are placed. If the ith cell has n com-partments, there are n sequences of Ni + n − 1 items to bearranged. There are n(Ni + n− 1)! ways to arrange the parti-cles and compartments, but we have overcounted because thereare n! permutations of compartments in a cell, and the orderin which particles are added to the cell is also irrelevant (thefactor Ni! we had in the classical case). Thus

W =∏

i

n (Ni + n− 1)!

Ni!n!=∏

i

(Ni + n− 1)!

Ni! (n− 1)!.

Optimizing this, we find

δ lnW =δ∑

i

[(n+Ni − 1) ln (n+Ni − 1) −Ni lnNi

− (n− 1) ln (n− 1) − lnαNi − βwiNi]

=∑

i

[

lnn+Ni − 1

Ni− lnα− βwi

]

δNi = 0 ,

or

Ni = (n− 1)(

αewi/kT − 1)−1

.

In fact, this is the relevant expression when there is no limitto the number of particles that can be put into the compart-ment of size h3, i.e., for bosons. Further, in the case when

bosons are photons, the condition δN does not apply, and thefactor α ≡ 1.

For fermions, only 2 particles can be put into a compart-ment, where 2 is the spin degeneracy. Thus, phase space iscomposed of 2n half-compartments, either full or empty. Thereare no more than 2n things to be arranged and therefore nomore than 2n! microstates. But again, we overcounted. For Nifilled compartments, the number of indistinguishable permuta-tions is Ni!, and the number of indistinguishable permutationsof the 2n−Ni empty compartments is (2n−Ni)!. In this case,we therefore have

W =∏

i

(2n)!

Ni! (2n−Ni)!.

As before, we optimize:

δ lnW =δ∑

i

[2n ln (2n) − (2n−Ni) ln (2n−Ni)

− lnαNi − βwiNi]

=∑

i

[

ln2n−NiNi

− lnα− βwi

]

δNi = 0 ,

or

Ni = 2n(

αewi/kT + 1)−1

.

The quantity lnα can be associated with the negative of thedegeneracy parameter µ/T , where µ is the chemical potential,of the system. The classical case is the limit of the fermionor boson case when α → ∞, since in this case the ±1 in thedenominator of the distribution function does not matter. Inthe boson case, also, α ≥ 1 since wi > 0 and Ni > 0. Bosonsbecome degenerate when α → 1. For photons, α = 1. Inthe fermion case, there is no restriction on the value of α, andfermions become degenerate when α→ −∞.

Thermodynamics

The internal energy U is

U = TS − PV +∑

i

µiNi

and the first law is

dU = TdS − PdV +∑

i

µidNi .

This implies

V dP − SdT −∑

i

Nidµi = 0 .

The Helmholtz F and Gibbs G free energies are

F = U − TS ; G =∑

i

µiNi .

dF = −SdT−PdV +∑

i

µidNi ; dG = V dP−SdT+∑

i

dNi .

The thermodynamic potential Ω = −PV obeys

dΩ = −SdT − PdV −∑

i

Nidµi .

The following are useful thermodynamic relations:

∂U

∂S

∣

∣

∣

V,Ni= T

∂U

∂V

∣

∣

∣

S,Ni= −P ∂U

∂Ni

∣

∣

∣

S,V,Nj 6=i= µi

∂F

∂T

∣

∣

∣

V,Ni= −S ∂F

∂V

∣

∣

∣

T,Ni= −P ∂F

∂Ni

∣

∣

∣

T,V,Nj 6=i= µi

∂Ω

∂T

∣

∣

∣

V,µi= −S ∂Ω

∂V

∣

∣

∣

T,µi= −P ∂Ω

∂µi

∣

∣

∣

T,V,µj 6=i= −Ni

Then ∂P/∂T |V,µi = S/V and ∂P/∂µi|T,V,µj 6=i = Ni/V .

Statistical Physics of Perfect Gases—Fermions

The energy of a non-interacting particle is related to its restmass m and momentum p by the relativistic relation

E2 = m2c4 + p2c2. (1)

The occupation index is the probability that a given momen-tum state will be occupied:

f =

[

1 + exp

(

E − µ

T

)]−1

(2)

for fermions, where µ = ∂ε/∂n|s is the chemical potential andε is the energy density.

Figure 1: E, µ and T are scaled by mc2.

When the particles are interacting, E generally contains aneffective mass and a potential contribution. µ corresponds tothe energy change when 1 particle is added to or subtractedfrom the system. The entropy per particle is s. We will useunits such that kB=1; thus T = 1 MeV corresponds to T =1.16 × 1010 K. The number and internal energy densities aregiven, respectively, by

n =g

h3

∫

fd3p; ε =g

h3

∫

Efd3p (3)

where g is the spin degeneracy (g = 2j+1 for massive particles,where j is the spin of the particle, i.e., g = 2 for electrons,muons and nucleons, g = 1 for neutrinos). The entropy can beexpressed as

ns = − g

h3

∫

[f ln f + (1 − f) ln (1 − f)] d3p (4)

and the thermodynamic relations

P = n2∂ (ε/n)

∂n|s = Tsn+ µn− ε (5)

gives the pressure. Incidentally, the two expressions (Eqs. (4)and (5)) are generally valid for interacting gases, also. We alsonote, for future reference, that

P =g

3h3

∫

p∂E

∂pfd3p. (6)

Thermodynamics gives also that

n =∂P

∂µ

∣

∣

∣

T; ns =

∂P

∂T

∣

∣

∣

µ. (7)

Note that if we define degeneracy parameters φ = µ/T andψ = (µ−mc2)/T the following relations are valid:

P = −ε+n ∂ε∂n

∣

∣

∣

T+T

∂P

∂T|n;

∂P

∂T

∣

∣

∣

φ= ns+nφ;

∂P

∂T

∣

∣

∣

ψ= ns+nψ.

(8)

In many cases, one or the other of the following limits maybe realized: extremely degenerate (φ → +∞), nondegenerate(φ → −∞), extremely relativistic (p >> mc), non-relativistic(p << mc).

Non-relativistic

In this case, one expands Eq. (3) in the limit p << mc.Defining x = p2/(2mT ) and ψ = (µ−mc2)/T , one has

n =g (2mT )3/2

4π2h3

∫ ∞

0

x1/2dx

1 + ex−ψ≡ g (2mT )3/2

4π2h3F1/2 (ψ) (9)

ε = nmc2 +gT (2mT )3/2

4π2h3F3/2 (ψ) . (10)

Here, Fi is the usual Fermi integral which satisfies the recursion

dFi (ψ)

dψ= iFi−1 (ψ) . (11)

P =2

3

(

ε− nmc2)

; s =5F3/2 (ψ)

3F1/2 (ψ)− ψ. (12)

Fermi integrals for zero argument satisfy

Fi (0) =(

1 − 2−i)

Γ (i+ 1) ζ (i+ 1) , (13)

where ζ is the Riemann zeta function. Note that Fi(0) −−−→i→∞

i!.

Fi(ψ) may be expanded around ψ = 0 with

Fi (ψ) = Fi (0) + iFi−1 (0)ψ +i (i− 1)

2Fi−2 (0)ψ2 + · · · . (14)

Since F0(ψ) = ln(1 + eψ), Fermi integrals with integer indicesless than 0 do not exist. The recursion Eq. (11) can be em-ployed to define non-integer negative indices, however.

i Fi(0) i Fi(0)

-7/2 0.249109 3/2 1.152804

-5/2 0.2804865 2 1.803085

-3/2 -1.347436 5/2 3.082586

-1/2 1.07215 3 7π4/120 5.682197

0 ln(2) 0.693147 4 23.33087

1/2 .678094 5 31π6/252 118.2661

1 π2/12 0.822467

a. Non-degenerate and non-relativistic: In this limit,using the expansion

Fi (ψ) = Γ (i+ 1)∞∑

n=1

(−1)n+1 enψ

ni+1, ψ → −∞ (15)

we find

n = g

(

mT

2πh2

)3/2

eψ, P = nT, s = 5/2 − ψ. (16)

b. Degenerate, non-relativistic: In this limit, we usethe Sommerfeld expansion

Fi (ψ) =ψi+1

i+ 1

∞∑

n=0

(i+ 1)!

(i+ 1 − 2n)!

(

π

ψ

)2n

Cn, ψ → ∞ (17)

Some values for the constants Cn are C0 = 1, C1 = 1/6, C2 =7/360, and C3 = 31/15120. We find

n =g (2mψT )3/2

6π2h3

[

1 +1

8

(

π

ψ

)2

+ · · ·]

,

P =2nψT

5

[

1 +5

12

(

π

ψ

)2

+ · · ·]

,

s =π2

2ψ+ · · · .

(18)

Extremely relativistic

This case corresponds to setting the rest mass to zero. Eqs. (3)and (5) become

n =g

2π2

(

T

hc

)3

F2 (φ) ,

P =ε

3=gT

6π2

(

T

hc

)3

F3 (φ) ,

s =4F3 (φ)

3F2 (φ)− φ.

(19)

The above limiting expressions for the Fermi integrals may beused in these expressions.

a. Extremely relativistic and non-degenerate: Useof the expansion Eq. (15) results in

n =g

π2

(

T

hc

)3

eφ,

P =nT, s = 4 − ln

[

π2n

g

(

hc

T

)3]

= 4 − φ.

(20)

b. Extremely relativistic and extremely degener-

ate: The expansion Eq. (17) gives

n =g

6π2

( µ

hc

)3[

1 +

(

π

φ

)2

+ · · ·]

,

P =nµ

4

[

1 +

(

π

φ

)2

+ · · ·]

,

s =π2

φ+ · · ·

(21)

Extremely degenerate

This case corresponds to φ >> 0. It is useful to define theFermi momentum pf for which the occupation index f = 1/2,

i.e., where µ = Ef =√

m2c4 + p2fc2. In terms of the parameter

x = pf/mc, we have

µ = mc2√

1 + x2. (22)

In the case φ→ ∞, Eq. (2) becomes a step function, with f = 1for E ≤ µ; f = 0 for E > µ.

n =8A

mc2x3,

P = A[

x(

2x2 − 3)√

1 + x2 + 3 sinh−1 x]

,

ε− nmc2 = A[

3x(

2x2 + 1)√

1 + x2 − 8x3 − 3 sinh−1 x]

,

s = 0,(23)

where A = (gmc2/48π2)(mc/h)3.

Non-degenerate

This case corresonds to φ << 0. Because pair creation isoften important in this case, we delay detailed discussion oflimiting formulae for a later section. If pairs are neglected,results may be expressed in terms of Bessel functions:

n =(mc

h

)3 T

3mc2eφK2

(

mc2/T)

,

P = nT,

ε− nmc2 =(mc

h

)3 T

3eφ[−K1

(

mc2/T)

+(

3T/mc2 − 1)

K2

(

mc2/T)

],

s = 4 − mc2

T

K1(

mc2/T)

K2(

mc2/T) − φ.

(24)

General Comments About Fermions

Convenient scalings for electrons are achieved using

nc =( g

2π2

)(mec

h

)3= 1.76 × 10−9 fm−3

mec2 =0.511 MeV

Fermions become relativistic under non-degenerate condi-tions when T > mc2 (T > 5 × 109 K for electrons) for anydensity, and, under degenerate conditions, when pfc > mc2

(ρYe > 2 × 106 g cm−3 for electrons) for any temperature.Here, ρ is the baryon density, and the number of electrons perbaryon is Ye. n(≡ ne) = ρNoYe. No is Avogadro’s number.

ψ ' 0 demarks the degenerate and non-degenerate regionsunder all relativity conditions.

n = g(2mT )3/2

4π2h3F1/2 (0) ;

ρYe ' 2 × 106(

T

5 × 109 K

)3/2

g cm−3 non − relativistic;

n =g

2π2

(

T

hc

)3

F2 (0) ;

ρYe ' 2 × 106(

T

5 × 109 K

)3

g cm−3 relativistic

separate the degenerate from the non-degenerate regions.Interacting baryons are far more complicated. At subnu-

clear densities (ρ < ρo ≡ 2.7 × 1014 g cm−3) they cluster intonuclei with internal densities near ρo. The nuclei themselvesare dilute, comprising a non-degenerate, non-relativistic gas,but with a strong Coulombic (lattice) interaction. At veryhigh temperatures, the nuclei dissociate. Above ρo, nuclear in-teractions and degeneracy effects dominate. Baryons becomerelativistic at a density (mbaryon/melectron)

3 times higher than

the electrons, or about 1016 g cm−3. This is above the tran-sition density to quark matter. At these densities, quarks canbe approximated as a perfect gas due to asymptotic freedom.

Fermion–Antifermion particle pairs

Under conditions found in the evolution of very massivestars, the temperature may be high enough to produce electron-positron pairs, while the electrons are non-relativistic. Duringgravitational collapse a degenerate neutrino-antineutrino gasforms when densities large enough to trap neutrinos on dy-namical time scales are reached (ρ > 1012 g/cm3). For particle-antiparticle pairs in equilibrium, µ+ = −µ−. The net difference

of particles and anti-particles and the total pressure are

n =n+ − n− =4πg

h3

∫ ∞

0p2[

1

1 + e(E−µ)/T− 1

1 + e(E+µ)/T

]

dp,

P =P+ + P− =4πg

3h3

∫

p3∂E

∂p

[

1

1 + e(E−µ)/T+

1

1 + e(E+µ)/T

]

dp.

(25)

Thus, when pairs are included, and n is positive, µ ≡ µ+must be positive, i.e., there will not be cases involving extremenon-degeneracy. However, pairs will never be important when-ever µ/T >> 0, that is, under extremely degenerate conditions.With the substitutions x = pc/T , z = mc2/T , we may write

n =g

2π2

(

T

hc

)3

sinhφ

∫ ∞

0

x2

coshφ+ cosh√z2 + x2

dx,

P =gT

6π2

(

T

hc

)3 ∫ ∞

0

x4√z2 + x2

coshφ+ e−√z2+x2

coshφ+ cosh√z2 + x2

dx.

(26)

a. Extremely relativistic case: µ >> mc2 or T >> mc2.This applies to neutrinos. With µ = µ+ = −µ−, i.e., z → 0,

n = n+ − n− =g

2π2

(

T

hc

)3 [

F2

(µ

T

)

− F2

(

−µT

)]

=g

6π2

( µ

hc

)3[

1 +

(

πT

µ

)2]

;

ε/3 = P = P+ + P− =gT

6π2

(

T

hc

)3 [

F3

(µ

T

)

+ F3

(

−µT

)]

=gµ

24π2

( µ

hc

)3[

1 + 2

(

πT

µ

)2

+7

15

(

πT

µ

)4]

;

s =gTµ2

6n (hc)3

[

1 +7

15

(

πT

µ

)2]

.

(27)These expressions are exact . The exponential terms ignoredin the Sommerfeld expansion of the +µ/T Fermi integral areexactly canceled by those of the −µ/T Fermi integral. The pairFermi integral

Gi (η) ≡ Fi (η) + (−1)i+1 Fi (−η) i ≥ 0

obeys the same recursion formula as Fi(η) for i ≥ 1.n(µ) is a cubic in µ, which can be inverted:

µ = r − q/r, r =

[

(

q3 + t2)1/2

+ t

]1/3

, (28)

where t = 3π2(hc)3n/g and q = (πT )2/3. For T → ∞, one hasµ→ 6n(hc)3/gT 2 → 0+. For all µ and T the adiabatic index

Γ1 =d lnP

d lnn

∣

∣

∣

s=d lnP

d lnn

∣

∣

∣

T+T

P

(

dP

dT

)2

n

(

dε

dT

)−1

n= 4/3. (29)

One may include the lowest order corrections for finite restmass by expanding the integrands of Eq. (3) and using therecursion relations for the Fermi integrals:

n =g

6π2

( µ

hc

)3[

1 + µ−2(

π2T 2 − 3

2m2c4

)]

,

P =gµ

24π2

( µ

hc

)3[

1 + µ−2(

2π2T 2 − 3m2c4)

+π2T 2

µ4

(

7

15π2T 2 −m2c4

)]

,

ε =gµ

8π2

( µ

hc

)3[

1 + µ−2(

2π2T 2 −m2c4)

+π2T 2

µ4

(

7

15π2T 2 − 1

3m2c4

)]

,

s =gTµ2

6n (hc)3

[

1 + µ−2(

7

15π2T 2 − 1

2m2c4

)]

.

(30)The relativistic relationship ε = 3P no longer holds. Interest-ingly, the cubic relationship between µ and n is preserved inthis approximation, and the solution is still given by Eq. (28) ifwe simply redefine q = (πT )2/3−m2c4/2. Including the finiterest mass terms lowers Γ1 below 4/3:

Γ1 =4

3

(

1 − 5

11

(

mc2

πT

)2)

(31)

when photons (see below) are also included.

b. Non-relativistic case: µ << mc2 and T << mc2.

In the degenerate case, T → 0, µ→ (mc2)+ and pairs are ofnegligible importance. We can use the non-relativistic, degen-erate formulas already obtained for particles alone. At highertemperatures, µ reaches a maximum, and then decreases, even-tually becoming less than mc2, so that µ′ < 0. The gas is thusat most only partially degenerate when pairs are present and

n = n+ − n− > 0. The non-degenerate expansion yields

n± ' g

(

mT

2πh2

)3/2

e[±µ−mc2]/T . (32)

Noting that n = n+ − n− and

n+n− = g2(

mT

2πh2

)3

e−2mc2/T ≡ n21, (33)

we can instead write

n± = ∓n2

+

[

(n

2

)2+ n2

1

]1/2

. (34)

P = (n+ + n−)T =(

n2 + 4n21

)1/2T, (35a)

ε = (n+ + n−)

(

mc2 +3

2T

)

, (35b)

s =

(

5

2+mc2

T

)

(n+ + n−)

n− µ

T, (35c)

µ = T ln

n

2n1+

(

n2

4n21

+ 1

)1/2

. (35d)

Pairs are important in the non-relativistic case when n ≤ n1.Including photon pressure (see below), in the case when n <<n1, one has

Γ1 ' 4

3

[

1 − 15

32

(

2mc2

πT

)7/2

e−mc2/T

]

. (36)

Thus Γ1 reaches a minimum value (1.02) when T = 27mc

2, and

is always less than 43. The creation of a pair costs an energy of

2mc2 which is non-negligible in the non-relativistic case.

c. Non-degenerate case: ψ < 0Consider the region for which coshφ − 1 << cosh z. Since

µ = Tφ cannot be negative when pairs are included, the gasis at most partially degenerate in this non-degenerate limit.Expanding the coshφ in the denominator terms to lowest orderin coshφ− cosh z,

n = ncz−3 sinhφ

∫ ∞

0

x2dx

1 + cosh√z2 + x2

,

P = ncTz−3[(coshφ− 1)

∫ ∞

0

x2dx

1 + cosh√z2 + x2

+2

3

∫ ∞

0

x4dx√x2 + z2

1

1 + e√x2+z2

],

ε = ncTz−3[(coshφ− 1)

∫ ∞

0

x(

x2 + z2)

dx

1 + cosh√z2 + x2

+ 2

∫ ∞

0

x2√z2 + x2

1 + e√z2+x2

dx],

(37)

where nc = (g/2π2)(mc/h)3 = 6×106 g cm−3 for g = 2. This isan interesting approximation because, given n and T , one canimmediately evaluate µ or φ because they no longer appearwithin the integrals. The integrals can be easily evaluated byquadrature, with relatively few points, using Gauss-Laguerrefor z < 30 and Gauss-Hermite for z > 30.

When are pairs important?

In the relativistic case, n− = 0.1n+ is equivalent to F2(φ) '10F2(−φ) or φ ' 0.9. In the non-relativistic case, we findφ ' ln

√10 or φ ' 1.15. φ ' 1 is the effective boundary.

The intrusion of this boundary into the NDNR region meansthat there are actually five limiting cases when pairs are con-sidered, as opposed to four when pairs are ignored. This is anunfortunate complication.

Generalized Approximation

We explore here a technique invented by Eggleton, Faulknerand Flannery (A& A 23, 325 [1973]) to bridge the limitingregions for a fermion gas. It is essential to maintain thermody-namic consistency in this approximation. To include pairs wesimply apply the scheme separately to electrons and positrons.The scheme establishes an analytic formula for the thermody-namic potential (or pressure) as an explicit function of chem-ical potential and temperature. Then n = T−1∂P/∂ψ;ns =

∂P/∂T − nψ; ε = T (∂P/∂T ) − P + nmc2. Density and tem-perature are inputs, so iteration is necessary to determine thechemical potential. Johns, Ellis & Lattimer (ApJ 473, 1020[1996]) improved the accuracy of the scheme and corrected thebehavior of the entropy in the degenerate limit.

The four limiting cases we have discussed are:

P

ncmc2=

(ψT )4∑∑

amnψ−2m (ψT )−n ER,ED : ψT >> mc2, ψ >> 1

(ψT )5/2∑∑

bmnψ−2m (ψT )n NR,ED : ψT << mc2, ψ >> 1

T 4eψ∑∑

cmnemψT−n ER,ND : ψ << −1, T >> mc2

T 5/2eψ∑∑

dmnemψTn NR,ND : ψ << −1, T << mc2

(38)

where nc = (g/2π2)(mc/h)3. The coefficients amn, bmn, cmnand dmn (m,n ∈ 0 . . .∞) can be determined from the limits.The key is to find functions f(ψ), g(ψ, T ) such that Eq. (38)can be rewritten as

P

ncmc2=

g4∑∑

a′mnf−mg−n ER,ED : g >> 1, f >> 1

g5/2∑∑

b′mnf−mgn NR,ED : g << 1, f >> 1

fg4∑∑

c′mnfmg−n ER,ND : g >> 1, f << 1

fg5/2∑∑

d′mnfmgn NR, ND : g << 1, f << 1.

(39)This is possible provided that

f (ψ) =

ψ2∑

rmψ−2m ED : ψ >> 1

eψ∑

smemψ ND : ψ << −1

(40)

and

g (ψ, T ) =

ψT∑

tmψ−2m ED : ψ >> 1

T∑

umemψ. ND : ψ << −1

(41)

rmn, smn, tmn and umn are additional coefficients. Then

P

ncmc2=

f

1 + fg5/2 (1 + g)3/2

∑M0∑N

0 Pmnfmgn

(1 + f)M (1 + g)N(42)

has the proper limits for any M,N ≥ 1 when f and g are eitherlarge or small. Pmn are coefficients which are least squares fitto a numerical evaluation of P . From Eq. (41), g ∝ T , and

g =(

T/mc2)

√

1 + f (43)

guarantees the right limiting behavior in Eq. (41).On the other hand, it is also clear that ∂f/∂ψ is either

√f

(f → ∞) or f (f → 0). We choose

∂f/∂ψ = f/√

1 + f/a, (44)

where a is an adjustable parameter introduced by Johns, Ellis& Lattimer (1996) which greatly improves the accuracy of thescheme. Integrating this relation, we find the required relationbetween ψ and f :

ψ = 2√

1 + f/a+ ln

√

1 + f/a− 1√

1 + f/a+ 1. (45)

Explicitly evaluating the density, we find

n =1

T

∂P

∂ψ

∣

∣

∣

T=

1

T

∂f

∂ψ

(

∂P

∂f

∣

∣

∣

g+∂g

∂f

∣

∣

∣

T

∂P

∂g

∣

∣

∣

f

)

; (46)

n

nc=

fg3/2∑M0

∑N0 Pmnf

mgn√

1 + f/a (1 + f)M+1/2 (1 + g)N−3/2×

[

1 +m+f

1 + f

(

1

4+n

2−M

)

+fg

(1 + f) (1 + g)

(

3

4− N

2

)]

.

(47)

From P +U = T (∂P/∂T )ψ, where U = ε−nmc2 is the internalenergy density,

U

ncmc2= fg5/2 (1 + g)3/2

∑M0

∑N0 Pmnf

mgn

(1 + f)M+1 (1 + g)N×

[

3

2+ n+

g

1 + g

(

3

2−N

)]

.

(48)

Given n and T , we invert Eq. (47) to determine f ; g is triviallyfound. The pressure is given by Eq. (42), the chemical potentialby Eq. (45), and the energy density by Eq. (48).

The entropy is found from s = n−1(∂P/∂T )ψ − ψ. A draw-back of the Eggleton et al. scheme was that in the degeneratelimit, although the entropy per particle has the correct asymp-totic dependence 1/

√f , the coefficient is not exact:

s =

√

a

f

(

2 +1

a− M

a+PM−1,N

aPM,N

)

−−−−−−→M,N→∞

π2

2

√

a

fED,ER

8

5

√

a

f

(

5

4+

1

4a−M

a+PM−1,0

aPM,0

)

−−−−−−→M,N→∞

π2

4

√

a

f. ED,NR

(49)For M,N = 2(3), Eqs. (49) have errors of 1.35 (0.0165)% and0.254 (0.0637)%, respectively, for the ER and NR cases. Theoriginal Eggleton et al. errors for these cases (they assumeda = 1) are (2.4 (1.40)%, 1.0 (0.563)%), respectively.

From Eqs. (49), the corner values of Pmn should be:

Pmn n = 0 n = N

m = 0 e2√

π/32/a e2/(2a)

m = M − 15π2−40+(32M−8)/a

15a1/42π2−8+4(M−1)/a

3a

m = M 32/(15a5/4) 4/(3a2)

Johns et al. constrained the fit so that these corner valuesand the ED entropies in Eq. (49) are exactly fulfilled. ForM = N = 3 we find an optimum fit for a = 0.433 with aroot-mean-square error of 8.1 · 10−5 and a maximum error of3.0 · 10−4 at the fitting points. The coefficients Pmn are:

Pmn n = 0 n = 1 n = 2 n = 3

m = 0 5.34689 18.0517 21.3422 8.53240

m = 1 16.8441 55.7051 63.6901 24.6213

m = 2 17.4708 56.3902 62.1319 23.2602

m = 3 6.07364 18.9992 20.0285 7.11153

To extend this scheme to include pairs, let the subscript +refer to particles and − refer to antiparticles. Then

n = n+ (f+, g+) − n− (f−, g−) (50)

where g± = (T/mc2)√

1 + f± and

ψ± = 2√

1 + f±/a+ ln

√

1 + f±/a− 1√

1 + f±/a+ 1. (51)

One solves the simultaneous equations

A = n− n+ (f+, T ) + n− (f−, T ) = 0;

B = ψ+ (f+) + ψ− (f−) + 2mc2/T = 0,(52)

where the second follows from µ− = −µ+. This is readilyhandled, since the derivatives are analytic:

∂A/∂f± = ∓∂n±/∂f±; ∂B/∂f± =√

1 + f±/a/f±. (53)

Boson Gas

The boson pressure and energy density are obtained by em-ploying the same equations as for fermions, but using the Bosedistribution function

fB =

[

exp

(

E − µ

T

)

− 1

]−1

, (54)

and a slightly different entropy formula

ns = − g

h3

∫

[fB ln fB − (1 + fB) ln (1 + fB)] d3p. (55)

Since the occupation index cannot be negative, a free (non-interacting) Bose gas µ ≤ mc2. If µ → mc2, a “Bose conden-sate” appears and there will be a finite number of particles ina zero-momentum state. Some limiting cases:

a. Extremely Non-degenerate:

In the non-degenerate limit, µ/T → −∞, the Bose andFermion distributions become indistinguishable, so the lim-its for thermodynamic quantities evaluated previously for theFermi gas apply.

b. Extremely Degenerate

For bosons, the “degenerate” limit is µ = mc2 or ψ = 0; thenumber density is

n =g

2π2h3

∫ ∞

0

p2

e(E−mc2)/T − 1dp. (56)

The integrals in this case can be written simply in terms ofzero argument Fermi integrals:∫ ∞

0

xi

ex − 1dx =

(

1 − 2−i)−1

Fi (0) = Γ (i+ 1) ζ (i+ 1) .

(57)

Therefore we have the following additional limits:i. Relativistic (T >> mc2, ψ = 0)

n =4g

6π2

(

T

hc

)3

F2 (0) , ε = 3P ;

P =4gT

21π2

(

T

hc

)3

F3 (0) =gπ2T

90

(

T

hc

)3

, s =π4

15F2 (0)' 3.601571.

ii. Non-relativistic (T << mc2, ψ = 0)

n =g

π2

(mT )3/2

h3

F1/2 (0)√

2 − 1, ε = nmc2 +

3

2P ;

P =4gT

3π2

(mT )3/2

h3

F3/2 (0)

23/2 − 1, s =

10

3

F3/2 (0)

F1/2 (0)

21/2 − 1

23/2 − 1' 1.283781.

Note that in these limits the entropy per boson is constant. Thelocation of the ψ = 0 trajectory in a boson density-temperatureplot is not far from the same curve for fermions.

c. Extremely Relativistic

In this case, we take m → 0, and we arrive at the simplestbose gas, the photon gas, for which µγ = 0. With gγ = 2, oneobtains

εγ = 3Pγ =3

4TSγ =

π2T

15

(

T

hc

)3

, (58)

which is (8/7g) times the value for a relativistic fermion gas.Here S is the entropy density. In any regime where electron-positron pairs are important, the photon pressure is also im-portant. In the non-degenerate, relativistic domain, the totalpressure from photons and electron-positron pairs is therefore11Pγ/4, Under situations when neutrino pairs of all three fla-vors are trapped in the matter, the total pressure increases to43Pγ/8. In the regime where electrons are degenerate, however,photon pressure is negligible.

In the regime where the electrons are non-degenerate andpairs are not important, the non-degenerate gas pressure ofnuclei must be included, and photon pressure may or not beimportant. The pressure due to photons is important at lowertemperatures than pair pressure, owing to the expense of creat-ing electron-positron pairs. Since the photon pressure is (8/7g)times the relativistic non-degenerate pair pressure, the bound-ary to the region in which photon pressure is significant is sim-ply obtained by a continuation of the straight line relativisticboundary ρ ∝ T 3 to low densities and is akin to the line φ = 1in the fermion-antifermion pair case.

Even excluding the contribution from electron-positron pairs,the adiabatic index of a non-relativistic gas changes from 5/3to 4/3 as the temperature is increased and the contribution ofradiation pressure increases. Denoting the fraction of the totalpressure due to gas pressure (assuming complete ionization) byβ,

Γ1 =32 − 24β − 3β2

24 − 21β. (59)

This ultimately sets an upper limit to the mass of main se-quence stars.

Johns, Ellis & Lattimer (1996) extended the Eggleton et al.scheme to handle a boson gas.