phy688, statistical mechanics

PHY688, Statistical Mechanics

James Lattimer

Department of Physics & Astronomy449 ESS Bldg.

Stony Brook University

January 31, 2017

Nuclear Astrophysics [email protected]

James Lattimer PHY688, Statistical Mechanics

Thermodynamics

Internal Energy Density and First Law:

ε =E

V= Ts − P +

∑i

niµi , dε = Tds +∑

i

µidni

dP = sdT +∑

nidµi

Helmholtz Free Energy Density:

f =F

V= ε− Ts, df =

∑µidni − sdT

Gibbs Free Energy Density:

g =G

V=∑

i

niµi , dg =∑

i

(µidni + nidµi )(∂ε

∂s

)ni

= T ,

(∂ε

∂ni

)nj 6=i ,T

= µi(∂f

∂T

)ni

= s,

(∂f

∂ni

)nj 6=i ,T

= µi(∂P

∂T

)µi

= s,

(∂P

∂µi

)µj 6=i ,T

= ni


Classical Statistics

A macrostate has N particles arranged among m volumes, withNi (i = 1 . . .m) particles in the ith volume. The total number of allowedmicrostates with distinguishable particles is

W =N!∏mi Ni !

; ln W = ln N!−m∑i

ln Ni ! .

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

ln N! = N ln N − N .

ln W = N ln N − N −m∑i

(Ni ln Ni − Ni ) .

The optimum state is the macrostate with the largest possible number ofmicrostates, which is found by maximizing W , subject to the constraintthat the total number of particles N is fixed (δN = 0). In addition, werequire that the total energy be conserved. If wi is the energy of the ithstate, this is

δ

(m∑i

wiNi

)=

m∑i

wiδNi = 0 .


With these constraints, the minimization is

δ

[ln

(W − α

m∑i

Ni − βm∑i

wiNi

)]= 0 .

m∑i

[ln Ni − α− βwi ] δNi = 0 .

Ni = αeβwi = αe−wi/kBT ,

which is the familiar Maxwell-Boltzmann, or classical, distributionfunction.


Quantum Statistics

In the quantum mechanical view, only within a certain phase spacevolume are particles indistinguishable. The minimum phase space is oforder h3. Denote the number of microstates per cell of phase space ofvolume h3 as Wi . Then the number of microstates per macrostate is

W =∏

i

Wi .

Note we have to consider both the particles and the compartments intowhich they are placed. If the ith cell has n compartments, there are nsequences of Ni + n − 1 items to be arranged. There are n(Ni + n − 1)!ways to arrange the particles and compartments, but we haveovercounted because there are n! permutations of compartments in acell, and the order in which particles are added to the cell is alsoirrelevant (the factor Ni ! we had in the classical case). Thus

W =∏

i

n(Ni + n − 1)!

Ni !n!=∏

i

(Ni + n − 1)!

Ni !(n − 1)!.


Bosons

Optimizing this, we find

δ ln W =δ∑

i

[(n + Ni − 1) ln(n + Ni − 1)− Ni ln Ni

− (n − 1) ln(n − 1)− lnαNi − βwiNi ]

=∑

i

[ln

n + Ni − 1

Ni− lnα− βwi

]δNi = 0 ,

or

Ni = (n − 1)(αewi/kBT − 1

)−1

= (n − 1)(

e(wi−µ)/kBT − 1)−1

.

This is the relevant expression when there is no limit to the number ofparticles that can be put into the compartment of size h3, i.e., forbosons. We identified the chemical potential µ = −kBT lnα.

The classical limit is α→∞, µ→ −∞, since the −1 is then ignorable.

In the boson case α ≥ 1 (µ ≤ 0) since wi > 0 and Ni > 0.

Bosons become degenerate when α→ 1 or µ→ 0, which is the situationfor photons.


Fermions

For fermions, only 2 particles can be put into a compartment, where 2 isthe spin degeneracy. Thus, phase space is composed of 2nhalf-compartments, either full or empty. There are no more than 2nthings to be arranged and therefore no more than 2n! microstates. Butagain, we overcounted. For Ni filled compartments, the number ofindistinguishable permutations is Ni !, and the number of indistinguishablepermutations of the 2n − Ni empty compartments is (2n − Ni )!. Thus

W =∏

i

(2n)!

Ni !(2n − Ni )!.

Optimiziing:

δ ln W =δ∑

i

[2n ln(2n)− (2n − Ni ) ln(2n − Ni )− lnαNi − βwiNi ]

=∑

i

[ln

2n − Ni

Ni− lnα− βwi

]δNi = 0 ,

or

Ni = 2n(αewi/kBT + 1

)−1

= 2n(

e(wi−µ)/kBT + 1)−1

.

In the fermion case, there is no restriction on the value of µ, andfermions become degenerate when µ→∞.


Statistical Physics of FermionsPerfect gas – no intractions – single particle energy

ε2 = m2c4 + p2c2

Occupation index (T now has units of energy, kB = 1)

f =

[1 + exp

(ε− µ

T

)]−1

ε, µ, T scaled by mc2

degenerate

non-degenerate

~εJames Lattimer PHY688, Statistical Mechanics

Fermion Thermodynamics

The number and energy densities are

n =g

h3

∫fd3p; ε =

g

h3

∫εfd3p

where g is the spin degeneracy. g = 2j + 1 for massive particles with spinj , i.e., g = 2 for electrons, muons and nucleons, g = 1 for neutrinos.

The entropy density is

ns = − g

h3

∫[f ln f + (1− f ) ln(1− f )]d3p

and the pressure is

P = n2

[∂(ε/n)

∂n

]s

= Tsn + µn − ε =g

3h3

∫p∂ε

∂pfd3p.

n =∂P

∂µ

∣∣∣∣∣T

; ns =∂P

∂T

∣∣∣∣∣µ

.

Define degeneracy parameters φ = µ/T and ψ = (µ−mc2)/T :

P = −ε+ n∂ε

∂n

∣∣∣∣∣T

+ T∂P

∂T

∣∣∣∣∣n

;∂P

∂T

∣∣∣∣∣φ

= ns + nφ;∂P

∂T

∣∣∣∣∣ψ

= ns + nψ.


φ = µ/T , ψ = (µ−mc2)/T

nc = (g/2π2)(mc/~)3 = 1.76× 10−9 fm−3mc2 = 0.511 MeV


Non-Relativistic Fermions

Assume p << mc . Define x = p2/(2mT ) and ψ = (µ−mc2)/T .

n =g(2mT )3/2

4π2~3

∫ ∞0

x1/2dx

1 + ex−ψ ≡g(2mT )3/2

4π2~3F1/2(ψ)

ε = nmc2 +gT (2mT )3/2

4π2~3F3/2(ψ).

P =2

3(ε− nmc2) =

gT (2mT )3/2

6π2~3F3/2(ψ).

s =5F3/2(ψ)

3F1/2(ψ)− ψ.

Fi (ψ) is the non-relativistic Fermi integral, satisfying

dFi (ψ)

dψ= iFi−1(ψ).


Fermi Integrals

F0(ψ) = ln(1 + eψ)

For zero argument and i > 0 (ζ is the Riemann zeta function):

Fi (0) = (1− 2−i )Γ(i + 1)ζ(i + 1)

Fi→∞(0) = i !

Fi (ψ) ' Fi (0) + iFi−1(0)ψ +i(i − 1)

2Fi−2(0)ψ2 + · · ·

The recursion dFi/dψ = iFi−1 can be used to define non-integer negativeindices, but Fi with negative integer i does not exist.

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 7/2 11.183721/2 0.678094 4 23.33087

1 π2/12 = 0.822467 5 31π6/252 = 118.2661


Non-Relativistic Limiting Expressions

Non-degenerate and non-relativistic: In this limit, use the expansion

Fi (ψ) = Γ(i + 1)∞∑

n=1

(−1)n+1enψn−(i+1), ψ → −∞

n = g

(mT

2π~2

)3/2

eψ, P = nT , s =5

2− ψ.

Degenerate and non-relativistic: In this limit, use the Sommerfeldexpansion:

Fi (ψ) =ψi+1

i + 1

∞∑n=0

(i + 1)!

(i + 1− 2n)!Cn

(π

ψ

)2n

, ψ →∞

The constants Cn are C0 = 1,C1 = 1/6,C2 = 7/360, C3 = 31/15120.

n =g(2mψT )3/2

6π2~3

[1 +

1

8

(π

ψ

)2

+ · · ·

], s =

π2

2ψ+ · · ·

P =gψT (2mψT )3/2

15π2~3

[1 +

5

8

(π

ψ

)2

+ · · ·

]=

1

5m

(6π2~3

g

)2/3

n5/3 + · · · .


Relativistic Fermions

This case corresponds to setting the rest mass to zero. φ = µ/T

n =g

2π2

(T

~c

)3

F2(φ), P =ε

3=

gT

6π2

(T

~c

)3

F3(φ), s =4F3(φ)

3F2(φ)− φ.

Extremely relativistic and non-degenerate: Expand in powers of eφ

n =g

π2

(T

~c

)3

eφ,

P =ε

3=nT , s = 4− ln

[π2n

g

(~c

T

)3]

= 4− φ.

Extremely relativistic and extremely degenerate: Use theSommerfeld expansion

n =g

6π2

( µ~c

)3[

1 +

(π

φ

)2

+ · · ·

], s =

π2

φ+ · · ·

P =ε

3=

gµ

24π2

( µ~c

)3[

1 + 2

(π

φ

)2

+ · · ·

]=

~c

4

(6π2

g

)1/3

n4/3 + · · · ,


Degenerate Fermions

This case corresponds to φ >> 0. For extreme degeneracy, f approachesa step function at ε = εF , the Fermi energy. For any degeneracy, it isuseful to define the Fermi momentum εF =

√m2c4 + p2

F c2. Define

x = pF/mc , so that µ = mc2√

1 + x2 in extreme degeneracy.

In the case µ/T →∞, the occupation f becomes a step function, withf = 1 for ε ≤ µ and f = 0 for ε > µ. Then one obtains

n =g

6π2

(mc

~

)3

x3 =g

6π2

(pF

~

)3

,

P =gmc2

48π2

(mc

~

)3

[x(2x2 − 3)√

1 + x2 + 3 sinh−1 x ],

ε− nmc2 =gmc2

48π2

(mc

~

)3

[3x(2x2 + 1)√

1 + x2 − 8x3 − 3 sinh−1 x ],

s = 0.

One can recover the extreme degenerate non-relativistic and relativisticexpressions derived earlier using the respective limits x → 0 and x →∞.Note that pF is defined by n = g/(6π2)(pF/~)3, and holds for anydegeneracy. If degeneracy is not extreme, µ/εF = 1− (πT/εF )2/12 + · · · .


Fermions With Pair Equilibrium

Under many conditions, temperatures become large enough that e−e+

pairs are formed in abundance, even when T < mec2.For particle (+)-antiparticle(−) pairs in equilibrium, µ+ = −µ−. The netdifference of particles and anti-particles and the total pressure are

n =n+ − n− =4πg

h3

∫ ∞0

p2

[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

Defining x = pc/T and z = mc2/T ,

n =g

2π2

(T

~c

)3

sinhφ

∫ ∞0

x2

coshφ+ cosh√

z2 + x2dx ,

P =gT

6π2

(T

~c

)3 ∫ ∞0

x4

√z2 + x2

[coshφ+ e−

√z2+x2

coshφ+ cosh√

z2 + x2

]dx .

For n > 0, µ ≡ µ+ > 0, so this gas is never extremely non-degenerate;however, pairs won’t be important under extremely degenerate conditionswhen µ/T >> 0 either.


φ = µ/T , ψ = (µ−mc2)/T

nc = (g/2π2)(mc/~)3 = 1.76× 10−9 fm−3mc2 = 0.511 MeV


Relativistic Pairs

µ >> mc2 or T >> mc2

n = n+ − n− =g

2π2

(T

~c

)3 [F2

( µT

)− F2

(− µ

T

)]=

g

6π2

( µ~c

)3[

1 +

(πT

µ

)2]

;

ε

3= P = P+ + P− =

gT

6π2

(T

~c

)3 [F3

( µT

)+ F3

(− µ

T

)]=

gµ

24π2

( µ~c

)3[

1 + 2

(πT

µ

)2

+7

15

(πT

µ

)4]

;

s =gTµ2

6n(~c)3

[1 +

7

15

(πT

µ

)2].

These expressions are exact. Note the inversion to obtain µ(n,T ):

µ = r − q/r , r =[(

q3 + t2)1/2

+ t]1/3

where t = 3π2(~c)3n/g and q = (πT )2/3.For T →∞, µ→ 6π2(~c)3n/(gT 2)→ 0+.


Non-relativistic Pairs

When T → 0, µ→ (mc2)+ and pairs are negligible.With increasing temperature, µ increases to a maximum, then decreasesbelow mc2. The non-degenerate expansion gives

n± ' g

(mT

2π~2

)3/2

e(±µ−mc2)/T , n+n− ≡ n21 = g2

(mT

2π~2

)3

e−2mc2/T

n± = ±n

2+

[(n

2

)2

+ n21

]1/2

P = (n+ + n−)T =(n2 + 4n2

1

)1/2T ,

ε = (n+ + n−)

(mc2 +

3

2T

),

s =

(5

2+

mc2

T

)n+ + n−

n− µ

T,

µ = T ln

[n

2n1+

(n2

4n21

+ 1

)1/2].


Bosons

f =[e(ε−µ)/T − 1

]−1

ns = − g

h3

∫[f (1− f )− (1− f ) ln(1− f )] d3p

f ≥ 0 and ε ≥ mc2 imply µ ≤ mc2. In the limit µ→ mc2, a Bosoncondensate appears and an unlimited number of particles can exist in azero-momentum state.

Extremely Non-Degenerate: µ/T → −∞, so the fermion and bosondistributions are indistinguishable from the classical Maxwell-Boltzmanndistributions.

Extremely Degenerate: In this limit, µ = mc2 or ψ = 0.

n =g

2π2~3

∫ ∞0

p2

e(ε−mc2)/T − 1dp, P =

2g

3π2~3

∫ ∞0

p3

e(ε−mc2)/T − 1dp.

One can show that∫ ∞0

x i

ex − 1dx =

(1− 2−i

)−1Fi (0) = Γ(i + 1)ζ(i + 1).


Extremely Degenerate and Relativistic: (T >> mc2, ψ = 0)

n =2g

3π2

(T

~c

)2

F2(0), ε = 3P,

P =4gT

21π2

(T

~c

)3

F3(0) =gπ2T

90

(T

~c

)3

, s =π4

15F2(0)' 3.601571.

Extremely Degenerate and Non-Relativistic: (T << mc2, ψ = 0)

n =g

π2

(mT )3/2

~3

F1/2(0)√

2− 1, ε = nmc2 +

3

2P,

P =4gT

3π2

F3/2(0)

23/2 − 1, s =

10

3

F3/2(0)

F1/2(0)

21/2 − 1

23/2 − 1' 1.283781.


Extremely Relativistic: (mc2 → 0, µ→ 0)

This case includes the photon gas. With gr = 2,

εr = 3Pr =3

4TSr =

π2T

15

(T

~c

)3

= aT 4.

S is the entropy density and a is the radiation constant. Theseexpressions are 8/(7g) times the value for relativtic fermions.

Pairs are always important if e− − e+ pairs are important, in which casethe combined pressure is (11/4)Pr .

If neutrinos of all 3 flavors are trapped in matter, the total pressureincreases to (43/8)Pr .

If electrons are degenerate, the photon pressure is negligible.


phy688, statistical mechanics

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