Classical Statistical Mechanics A macrostate has N particles arranged among m volumes, with Ni(i = 1 . . . m) particles in the ith volume. The total number of allowed microstates with distinguishable particles is N! ; W = Qm N ! i i

ln W = ln N ! −

m X

ln Ni ! .

i

For a large number of particles, use Stirling’s formula ln N ! = N ln N − N . m X ln W = N ln N − N − (Ni ln Ni − Ni) . i

The optimum state is the macrostate with the largest possible number of microstates, which is found by maximizing W , subject to the constraint that the total number of particles N is fixed (δN = 0). In addition, we require that the total energy be conserved. If wi is the energy of the ith state, this is ! m m X X δ w i Ni = wiδNi = 0 . i

i

With these constraints, the minimization is " !# m m X X δ ln W − α Ni − β w i Ni = 0. i

m X i

i

[ln Ni − α − βwi] δNi = 0 .

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

Quantum Statistical Mechanics In the quantum mechanical view, only within a certain phase space volume are particles indistinguishable. The minimum phase space is of order h3. Now denote the number of microstates per cell of phase space of volume h3 as Wi . Then the number of microstates per macrostate is Y W = Wi . i

Note we have to consider both the particles and the compartments into which they are placed. If the ith cell has n compartments, there are n sequences of Ni + n − 1 items to be arranged. There are n(Ni + n − 1)! ways to arrange the particles and compartments, but we have overcounted because there are n! permutations of compartments in a cell, and the order in which particles are added to the cell is also irrelevant (the factor Ni! we had in the classical case). Thus Y n (Ni + n − 1)! Y (Ni + n − 1)! = . W = Ni !n! Ni ! (n − 1)! i

i

Optimizing this, we find X δ ln W =δ [(n + Ni − 1) ln (n + Ni − 1) − Ni ln Ni i

− (n − 1) ln (n − 1) − ln αNi − βwiNi ]  X  n + Ni − 1 = ln − ln α − βwi δNi = 0 , Ni i

or



Ni = (n − 1) αewi/kT − 1

−1

.

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

bosons are photons, the condition δN does not apply, and the factor α ≡ 1. For fermions, only 2 particles can be put into a compartment, where 2 is the spin degeneracy. Thus, phase space is composed of 2n half-compartments, either full or empty. There are no more than 2n things to be arranged and therefore no more than 2n! microstates. But again, we overcounted. For Ni filled compartments, the number of indistinguishable permutations is Ni!, and the number of indistinguishable permutations of the 2n − Ni empty compartments is (2n − Ni)!. In this case, we therefore have Y (2n)! W = . Ni ! (2n − Ni)! i

As before, we optimize: X δ ln W =δ [2n ln (2n) − (2n − Ni) ln (2n − Ni ) i

or

− ln αNi − βwiNi ]  X  2n − Ni = ln − ln α − βwi δNi = 0 , Ni i

 −1 w /kT Ni = 2n αe i +1 .

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

Thermodynamics The internal energy U is X U = TS − PV + µi N i i

and the first law is

dU = T dS − P dV + This implies V dP − SdT −

X

X

µi dNi .

i

Ni dµi = 0 .

i

The Helmholtz F and Gibbs G free energies are X F = U − TS ; G= µi N i . i

dF = −SdT −P dV +

X i

µi dNi ;

dG = V dP −SdT +

The thermodynamic potential Ω = −P V obeys X dΩ = −SdT − P dV − Nidµi .

X

dNi .

i

i

The following are useful thermodynamic ∂U ∂U =T = −P ∂S V,Ni ∂V S,Ni ∂F ∂F = −S = −P ∂T V,Ni ∂V T,Ni ∂Ω ∂Ω = −S = −P ∂T V,µi ∂V T,µi

relations: ∂U = µi ∂Ni S,V,Nj6=i ∂F = µi ∂Ni T,V,Nj6=i ∂Ω = −Ni ∂µi T,V,µj6=i

Then ∂P/∂T |V,µi = S/V and ∂P/∂µi |T,V,µj6=i = Ni/V .

Statistical Physics of Perfect Gases—Fermions The energy of a non-interacting particle is related to its rest mass m and momentum p by the relativistic relation E 2 = m 2 c4 + p 2 c2 .

(1)

The occupation index is the probability that a given momentum state will be occupied:    E − µ −1 f = 1 + exp (2) T for fermions, where µ = ∂/∂n|s is the chemical potential and  is the energy density.

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

When the particles are interacting, E generally contains an effective mass and a potential contribution. µ corresponds to the energy change when 1 particle is added to or subtracted from the system. The entropy per particle is s. We will use units such that kB =1; thus T = 1 MeV corresponds to T = 1.16 × 1010 K. The number and internal energy densities are given, respectively, by Z Z g g  = 3 Ef d3p (3) n = 3 f d3p; h h 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 be expressed as Z g ns = − 3 [f ln f + (1 − f ) ln (1 − f )] d3p (4) h and the thermodynamic relations ∂ (/n) | = T sn + µn −  (5) P = n2 ∂n s gives the pressure. Incidentally, the two expressions (Eqs. (4) and (5)) are generally valid for interacting gases, also. We also note, for future reference, that Z ∂E 3 g P = 3 p f d p. (6) ∂p 3h Thermodynamics gives also that ∂P ∂P n= ns = (7) ; . ∂µ T ∂T µ Note that if we define degeneracy parameters φ = µ/T and ψ = (µ − mc2)/T the following relations are valid: ∂P ∂P ∂P ∂ | ; P = −+n +T = ns+nφ; = ns+nψ. ∂n T ∂T n ∂T φ ∂T ψ (8)

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