Notes on Statistical Mechanics by MetroWind

They say this is a blank page; I say this is the tragic of life.

Notes on Statistical Mechanics Susan Gardner Mingyang Sun May 2, 2010

They say this is a blank page; I say this is the tragic of life.

Contents 1

Probability Theory

Entropy of Information

Kinetic Theory

Ensemble Theory 4.1 The micro canonical ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 The Canonical Ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 The Grand Canonical Ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Classical perfect gas in the gce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Slightly degenerated perfect gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Completely degenerated perfect Fermi gas (T <<<< TF = ÎľF /kB ) . . . . . . . . 4.7 Degenerated Fermi system (T TF but not T <<<< TF ) . . . . . . . . . . . . . . . 4.8 Degenerated Bose systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.9 Black-body radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Imperfect Gas

12 24 24 27 33 34 36 40 42 44 47 51

They say this is a blank page; I say this is the tragic of life.

1 Probability Theory How to approach equilibrium In equilibrium, if there are Γ0 possible states, then if each f them is equally likely, the probability of each state is Weq =

1 Γ0

The probability to find a system in state α is Wα =

1 Γα Γ0

in which Γα is the number of micro states associated with α. The average of a function f is f¯ =

Wα f α ,

Wα = 1

Consider two dies with 6 sides, Weq = 1/36. The probability to have a (6,6) is W66 = 1/36. The probability to have a (4,5) is W45 = 2/36, because the microstates associated with (4,5) are {(4,5), (5,4)}. P Probability function P , P [α] = x∈α P (x) is accepted if (1) P [α] ≥ 0 ∀ α ⊂ Ω, in which Ω is the set of all the states. (2) P [Ω] = 1 (3) P [α1 ∪ α2 ] = P (α1 )P (α2 ) Example: A set of N numbered balls. How many ways can they be seperated into 3 sets with N1 , N2 , N3 balls, N1 + N2 + N3 = N ? Number of ways to pick n of N : N (N − 1)· · ·(N − n + 1) = If order of pick up does not matter,

N! (N − n)!

N! = (N − n)! n!

N n

Therefore the number of way to separate them is N! N1 ! N2 ! N3 !

Example: What is the probability of picking n heads in a simultaneous toss of N coins? There are 2N possible outcomes, therefore 1 ∀x⊂Ω 2N 1 N ⇒ P [n heads] = N 2 n P (x) =

Define R(x) =

P [xN heads] P [ 12 N heads]

Then R(x) is a Gaussian.

The probability for a particle to be in dx dy is P (x,y) dx dy. Z Z P [A] = P (x,y) dx dy. P [Ω] = P (x,y) dx dy = 1 A

Conditional probability Suppose a statement q is true, what is the probability that q 0 is true? P [q 0 |q] =

P [q 0 &q] P [q]

If q and q 0 are independent, P [q 0 |q] =

P [q 0 ]P [q] = P [q 0 ] P [q]

P [q]P [q 0 |q] = P [q 0 &q] = P [q 0 ]P [q|q 0 ] 2

⇒ P [q 0 |q] =

P [q|q 0 ]P [q 0 ] P [q]

This is Bayes' Theorem.

Probability distribution function (pdf) Pdf p(x)dx. Define the moment as MN =

hxN i

+∞

p(x)xN dx

−∞

Generate moment using characteristic function * Z pe(k) =

he−ikx i

+∞

p(x)e−ikx dx

−∞

∞ X (−ik)N xN N!

N =0

∞ X (−ik)N = MN N! N =0

The cumulent generating function Define ∞ X (−ik)N N ln pe(k) = hx ic N! N =1

" ∞ # ∞ X X (−ik)M (−ik)N N hx i = exp hxM ic N! M!

N =0

M =1

Expand the exponential to equal power of k N hxi = hxic hx2 i = hx2 ic + hxi2c hx3 i = hx3 ic + 3hx2 ic hxic + hxi3c Example: Gaussian pdf

(x − λ)2 exp − 2σ 2 2πσ 2 Z +∞ k2 σ2 −ikx pe(x) = dx p(x)e = exp −ikλ − 2 −∞ p(x) = √

1 ⇒ ln pe(x) = −ikλ − k 2 σ 2 2 ⇒ hxic = λ,

hx2 ic = σ 2 ,

hxN ic = 0

Example: Binomial distribution Coin toss, pA + pB = 1. What is the possibility to get NA As? pN (NA ) = pe(k) = he

−ikNA

N NA

A N −NA pN A pB

N X N N −NB −ikNA A i= pN e = (e−ik pA + PB )N A pB NA NA =0

⇒ ln pe(k) = N ln(e−ik pA + pB )

When N → ∞, binomial distribution → Poission. Example: N particles move within a volume V . What is the probability of finding n in subvolume v? Any one particle in v is p = v/V . Therefore p[n in v] =

N n

pn (−p)N −n

Consider v V , n N . N! = N (N − 1)(N − 2)· · ·(N − n + 1) ∼ N n (N − n)! 1

(1 − p)N −n ≈ (1 − p)N = [(1 − p) p ]pN ∼ e−pN Therefore p[n in v] ≈ (N p)n e−pN

¯ n e−N¯ 1 N = n! n!

¯ := N p. Consider moments pn = which is a Poisson. N hn0 i =

∞ X

pn =

n=0

¯ ¯ n e− N N n!

∞ ¯ n −N ∞ ¯N ¯ X X N e N ¯ ¯ ¯ = e−N = e−N eN = 1 n! n! n=0 n=0

The central limit theorem pN (NA ) =

N NA

A N −NA pN , A pB

pA =

VA V

If ρA → 0 (V → ∞), but ρA N is finite (N → ∞).

NA in VA N in V

pN (NA ) → pN¯ (NA ) =

¯ NA e−N¯ N , NA !

¯ := pA N N

hNA0 i = 1 hNA i =

∞ X NA =0

NA p(NA ) =

∞ X

NA =1

∞ X ¯ NA e−N¯ ¯ NA −1 N N ¯ e−N¯ ¯ =N =N NA ! (NA − 1)! NA =1

Cumulents approach ln pe(k) = N ln(ρA e−ik + ρB ). ρA → 0 with ρA N being finite. Since x

ebx = lim (1 + ρb) ρ ρ→0

ln pe(k) = ln

1 1 + ρA (e−ik − 1) ρA

ρA N

= ln e(e

−ik

¯ ⇒ hNAN ic = ρA N = N ¯ → ∞, hNA i ∼ N , hN 2 i − hNA i2 ∼ N . If N A q hNA2 ic 1 ∼ √ →0 hNA i N So the Poisson pdf becomes sharp as N → ∞. 5

−1)ρA N

= ρA N (e−ik − 1)

Γ (z) =

∞

e−x xz−1 dx

is analytic except for z = 0, −1, −2, . . .. Suppose Z ∞ Z ∞ −x z e x dx = eg(x) dx, 0

g(x) = z ln x − x

g 0 (x) =

z − 1 = 0 iﬀ x = z, x

g 00 (z) = −

z 1 =− 2 z z

Therefore g(x) has maximum at x = z = xmax . Expand g(x) around xmax , 1 g(x) = g(xmax ) + g 00 (xmax )(x − xmax )2 + · · · 2 Z ∞ Z ∞ 1 ⇒ eg(x) dx = eg(xmax ) dx exp − |g 00 (xmax )| (x − xmax )2 2 0 0 If xmax ∈ [0, +∞), change the lower bound of the integral to −∞. Define ξ = x − xmax . Use saddle point approximation s Z ∞ 1 2π exp − |g 00 (xmax )|ξ 2 dξ = 00 2 |g (xmax )| 0 Z ⇒

∞

dxeg(x) ≈ eg(xmax )

2π

|g 00 (x

max )|

Let g(x) = z ln x − x, g(xmax ) = z ln z − z, Z ∞ √ dxe−x xz ∼ ez ln z−z 2πz, 0

Expand

√

z→∞

2πz to asymptotic series, √

2πz = 1 +

1 1 + 2 + ··· 12z z

If a partial sum SN (z), limz→∞ [f (z) − SN (z)] = 0, then the series for f (z) is asymptotic. 1

⇒ ln N ! ∼ N ln N − N + ln(2πN ) 2 + O(1/N ), 6

N →∞

Example: Poisson pdf when NA → ∞, ¯ NA ¯ NA e−N¯ ¯ NA e−N¯ N N N 1 ¯ pN¯ (NA ) = ∼ e−N +NA (2πNA )− 2 1 ∼ N A −N NA ! NA e A NA (2πNA ) 2 ln pN¯

¯ + NA ln ∼ NA − N

¯ N 1 − ln(2πNA ) 2 + · · · NA

¯ ¯ − NA N N log = ln 1 + NA NA As NA → ∞,

¯ −NA N NA

→ 0. Therefore ln

¯ ¯ ¯ − NA N N 1 N − NA 2 ≈ − + ··· NA NA 2 NA

¯ − NA )2 p 1 (N + · · · − ln 2πNA 2 NA 1 1 ¯ ⇒ pN¯ (NA ) ∼ √ exp − (N − NA )2 , NA → ∞ 2NA 2πNA

¯ +N ¯ − NA − ln pN¯ (NA ) ∼ NA − N

Also, Z

+∞

−∞

pN¯ (NA )dNA = 1,

+∞

−∞

¯, NA pN¯ (NA )dNA = N

+∞

−∞

¯2 +N ¯ dNA NA2 pN¯ (NA ) = N

¯ The Poisson √ pdf becomes Gaussian pdf as NA → ∞, steeply peaked about NA = N . Width of distribution ¯ σNA ∼ N , σNA 1 ∼ √ → 0, ¯ ¯ N N

¯ →∞ N

General form of central limit theorem Consider a sum of independent random variables {xi }, p(x,y,z, . . .) = p(x) p(y) p(z)· · · Then the distribution of x = (x1 + · · · + xN )/N becomes sharp and pxi (xi ) becomes a Gaussian for large N independent of pxi (xi ). 7

px (x) =

N 2πa2

1 2

N (x − x¯)2 exp − 2A2

The width is A ∆x = √ , N

A2 :=

1 X (∆xi )2 N i

Liouville's Theorem Blabla. . .

2 Entropy of Information Information content of a pdf Consider a random variable with M outcomes. Measure the random variable N times. Define “message” as the sequence of the results of the measurements. Binary bits needed to specify an outcome is log2 M . Binary bits needed to specify a message is N log2 M . If want probability pi associates with ith outcome, after N measurements, number of likely message is N! N1 ! N2 ! · · ·NM ! Actual number of bits to specify these messages is X N! log2 g = log2 QM = log2 N ! − log2 Ni ! i=1 Ni ! i=1 M

1 log(2πN ) + O(1/N ) 2 M X 1 1 ⇒ log g ≈ N log N − N + log(2πN ) − Ni log Ni − Ni + log(2πNi ) 2 2 log N ! ≈ N log N − N +

i=1

∞ X i=1

Ni log Ni = N

M X

pi log(N pi ) = N log N + N

i=1

∞ X i=1

Therefore log2 g ≈ −N

∞ X

pi log2 pi + · · ·,

N →∞

i=1

which is Shannon's theorem. Information per measurement is I = log2 M +

∞ X i=1

Another approach to Shannon's theorem

pi log2 pi

pi log pi

!N X

M Ni Y p i

i=1

{Ni }

Ni !

N → ∞, Ni = N pi

We can only look at the largest term to study a summation. S=

M X

eN φmax ∃ φmax 3 eN φmax eN φi

i=1

eN φmax ≤ S ≤ M eN φmax ⇒ N φmax ≤ log S ≤ log M + N φmax When N → ∞, N φmax

log M + N φmax ⇒ N φmax ⇒ eN φmax

log S

Therefore 1→g

M Y

pi pN , i

N →∞

i=1

Define entropy S=−

M X

p(i) log p(i)

i=1

having maximum when p(i) is unbiased. Example:

Suppose M = 2, S = −p log p − (1 − p) log(1 − p). dS 1−p 1 = log =0⇒p= dp p 2

Example: Loaded die 3 6 occurs twice as likely as 1. Calculate the unbiased probability to find each face. p6 = 2p1 ,

p 2 = p3 = p4 = p5 =

1 (1 − 3p1 ) 4

S = −p1 log p1 − 2p2 log(2p1 ) − (1 − 3p1 ) log[(1 − 3p1 )/4] dS (1 − 3p)2 1 = log 3 = 0 ⇒ p1 = 8 2 dp1 4 (2p1 ) p1 3 + 23 Information content is I = log2 6 +

6 X

pi log pi ≈ 0.03

Probability under constraints df Consider M possible xi , f ({xi }) = 0. Calculate {x∗i } 3 dx | ∗ = 0, in which x∗i is unbiased. i xi What if not all xi are independent? Find an extremum of a function f ({xi }) in which {xi } suﬀer holonomic constraints g({xi }) = 0. Use Lagrangian multipliers to find the maximum of

S=−

pi log pi = 0

while

i pi

= 1 holds. Maximize F =S−α

M X

pi − 1

i=1

dF = −1 − log pi − α = 0 ⇒ pi = e−(1+α) dpi

3 Kinetic Theory In equilibrium all allowed states are equally possible. How is this assumption true? •

Ergodic hypothesis. A system standing out in any allowed state will reach any other state allowed if one waits long enough. • Liouville's Theorem The density of points in phase space is conserved with time. − Ergodic approach Suppose Ergodic hypothesis is valid. The system has 3N degree of freedom qi ∈ {qi }. Z 1 T hf i = lim f (t)dt T →∞ T 0 dt → ρ({qi }, {pi })d{qi }d{pi } T →∞ T lim

in which T is the length of time the measurement takes place, and ρ is the pdf. Z ⇒ hf i = ρ({qi }, {pi })d{qi }d{pi }f ({qi }, {pi }) − Ensemble approach (Gibbs) {qi }, {pi } in 6N dimensional phase space N Y Γ = {qi ,pi } i=1

Picking a qi and pi for every i forms a representative point in Γ , which is called a micro state. Consider N copies of a particular macro state. Each copy has a diﬀerent micro state. Pick {pi }, {qi }, t so that dN ({pi }, {qi }, t) is the number of micro states in a volume dΓ =

6N Y i=1

dpi dqi

about {pi }, {qi }. The phase space density (pdf) is 1 dN ({pi }, {qi }, t) N →∞ N

ρ({pi }, {qi })d{pi }d{qi } = lim

The number of ensemble numbers of pi ,qi in dpi dqi about a choice {pi }{qi }. Therefore Z hf i = ρ({qi }, {pi })d{qi }d{pi }f ({qi }, {pi }) Proof of Liouville's theorem: Consider the volume element dΓ =

6N Y

dpi dqi

i=1

Show that the volume remains invariant so that dΓ = dΓ 0 ∀ time t0 Let dΓ 0 = J(t,t0 )dΓ in which J is the 3N × 3N Jacobian. J(t,t0 ) =

∂({p0k }, {qk0 }) ∂({pi }, {qi })

If t0 = t, ∂p0k ∂p0 = δik = k ⇒ J = 1 ∂pi ∂qi Let t0 = t + dt. Hamilton's equations are p˙k = − ⇒ p0k = pk −

∂H , ∂qk

q˙k =

∂H + O(dt2 ), ∂qk

∂H ∂pk

qk0 = qk +

∂p0k ∂ 2H = δik − dt + · · · ∂pi ∂pi ∂qk ∂qk0 ∂ 2H = dt ∂pi ∂pi ∂pk ∂p0k ∂ 2H =− dt ∂qi ∂qi ∂qk 13

∂H + ··· ∂pk

∂qk0 ∂ 2H = δik + dt ∂qi ∂qi ∂pk Y ∂ 2H ∂ 2H ⇒J = 1− dt 1+ dt ∂pk ∂qk ∂pk ∂qk k X ∂ 2H ∂ 2H ≈1+ − dt = 1 + O(dt2 ) ∂pk ∂qk ∂pk ∂qk k

For any t0 , dΓ 00 = J(t,t00 )dΓ ,

dΓ 0 = J(t00 ,t0 )dΓ 00 ,

dΓ → dΓ 0

dΓ 0 = J(t,t00 )J(t00 ,t0 )dΓ dΓ 0 = J(t,t0 )dΓ 00 0 ∂J(t, t0 ) 00 ) ∂J(t , t ) = J(t,t ∂t0 ∂t0

Choose t00 = t0 ,

∂J(t, t0 ) ∂J(t00 , t0 )

0 = J(t,t ) =0 ∂t0 ∂t0 t00 =t0 J(t,t) = 1,

J(t,t0 ) = 1

Therefore dΓ = dΓ 0 . Phase space fluids is incompressible dρ ∂ρ X = + dt ∂t N

i=1

∂ρ ∂ρ p˙i + q˙i ∂pi ∂qi

The phase space fluid & the bbgky hierarchy Apply Hamilton's equations to eq (3.1), ∂ρ = −{ρ,H} ∂t Behavior under time-reversal T t → −t, p → −p, q → q. 14

(3.1)

ρ(~p, ~q, t) = ρ(−~p, ~q, −t) Therefore ρ is T-invariant. Study dhOi/dt . . .in which O is an observable. hOi = hO(t)i = dhOi ⇒ = dt

dΓ ρ({pi }, {qi }, t)O({pi }, {qi })

∂ρ dΓ O = − ∂t

dΓ ρ{H,O} = h{O,H}i

In equilibrium, dhOi = 0, dt

dρ =0 dt

Structure of phase space fluid One-body density is defined as f1 (~p, ~q, t) =

*N X

δ 3 (~p − p~i )δ 3 (~q − ~qi )

i=1

f1 (~p, ~q, t) =

dΓ

N X

δ 3 (~q − ~qi )δ 3 (~p − p~i )ρ({~pi }, {~qi }, t)

i=1

dΓ δ 3 (~q − ~q1 )δ 3 (~p − p~1 )ρ({~p1 }, {~q1 }, t) Z Y N =N (d3~qi d3 p~i )ρ i=2

Two-body density is f2 (~p1 , ~q1 , p~2 , ~q2 , t) = N (N − 1)

dΓ 0 δ 3 (~q1 − ~q10 )δ 3 (~q2 − ~q20 )δ 3 (~p1 − p~01 )δ 3 (~p2 − p~02 )ρ

Z Y N = N (N − 1) (d3~qi d3 p~i )ρ i=3

S-particle density is N! fs (~p1 , ~q1 , . . ., p~s , ~qs ) = (N − s)!

N Y

(d3~qi d3 p~i )ρ

i=s+1

Assume N 2 N X p~i 1 X H= + U (~qi ) + V (~qi − ~qj ) 2N 2 i=1

i,j=1

Note that dV is a small volume in space while V (~r) is the potential. Rewrite H as H = Hs + HN −s + H 0 , s 2 s X pi 1X Hs = + U (~qi ) + V (~qi − ~qj ) 2N 2 i,j

i=1

HN −s =

N X

i=s+1

H0 =

s X

N p2i 1 X + U (~qi ) + V (~qi − ~qj ) 2N 2

N X

i,j=s+1

V (~qn − ~qi )

n=1 i=s+1

Z ⇒

N Y i=s+1

∂ρ ∂ρs dVi = = ∂t ∂t

N Y

dVi {ρ,H}

i=s+1

in which {ρ,H} = {ρ, Hs + HN −s + H 0 }. Look at the first term of eq (3.2) ( Z N ! ) Z Y N Y dVi {ρ,Hs } = dVi ρ , Hs = {ρs , Hs } i=s+1

Look at the second term of ?? Z Y Z N − Ni {ρ,HN −s } = i=s+1

i=s+1

N Y i=s+1

N X ∂ρ ∂HN −s ∂ρ ∂HN −s dVi − ∂~pj ∂~qj ∂~qj ∂~pj j=1

Assume that V (~qi − ~qj ) = V (~qj − ~qi ), 16

(3.2)

−

Z Y N

Ni {ρ,HN −s } =

Z Y N

i=s+1

dVi

i=s+1

N X j=s+1

∂ρ ∂~pj

N X ∂V (~qj − ~qk ) ∂U + ∂~qj ∂~qj

k=s+1

# ∂ρ pj − =0 ∂~qj bla. . .

Look at the third term of eq (3.2), Z Y N

N X ∂ρ ∂H 0 ∂ρ ∂H 0 dVi − , ∂~pj ∂~qj ∂~qj ∂~pj

i=s+1

j=1

Z Y N

dVi

i=s+1

∂H 0 X ∂V = , ∂~qj ∂~qj i=1

∂H 0 =0∀j ∂~pj

s N N s X X ∂ρ X ∂V (~qi − ~qj ) ∂ρ X ∂V (~qi − ~qj ) + ∂~pi ∂~qi ∂~pj ∂~qj i=1

j=s+1

i=1

The second term vanishes. ~qi is the variable integrated over. Assuming that there is no difference between particles, Z Y N

s N X ∂ρ X ∂V (~qi − ~qj ) ∂V (~qi − ~qs+1 ) dVi = (N − s) ∂~pi ∂~qi ∂~qi

i=s+1

i=1

j=s+1

Therefore eq (3.2) becomes (N − s)

s Z X i=1

∂V (~qi − ~qs+1 ) ∂ dVs+1 ∂~qi ∂~pi

Note that Z Y N

ρ dVi = ρs+1

i=s+2

Therefore

Z Y N i=s+2

ρ dVi

X ∂ρs − {Hs ,ρs } = (N − s) ∂t

dvs+1

i=1

X ∂fs ⇒ − {Hs ,fs } = (N − s) ∂t i=1

∂V (~qi − ~qi+1 ) ∂ρs+1 ∂~qi ∂~pi

dvs+1

∂V (~qi − ~qi+1 ) ∂fs+1 ∂~qi ∂~pi

This is the bbgky hierarchy. Look at the first two equations in the hierarchy. For s = 1, Z ∂ ∂U ∂ p~1 ∂ ∂V (~q1 − ~q2 ) ∂f2 − + f = dV2 ∂t ∂~q1 ∂~p1 M ∂~q1 ∂~q1 ∂~p1 For s = 2, Z ∂ ∂U ∂ ∂U ∂ p~1 ∂ p~2 ∂ ∂V (~q1 − ~q3 ) ∂ ∂V (~q2 − ~q3 ) ∂ − − + + f2 = dV3 + f3 ∂t ∂~q1 ∂~p1 ∂~q2 ∂~p2 M ∂~q1 M ∂~q2 ∂~q1 ∂~p1 ∂~q2 ∂~p2 To simplify these equations, we have to do some approximation. Consider extrinsic time scale 1 ∂U ∂ ∼ , τu ∂~q1 ∂~p1

τu ∼ 10−5 s

1 ∂V ∂ ∼ , τc ∂~q1 ∂~p1

τc ∼ 10−12 s

The collision time

Higher orders 1 ∼ τx

∂V ∂ fs+1 N d3 1 ∼ ∂~q ∂~p fs τc τc

Rewrite the first equation in the hierarchy as ∂ ∂U ∂ p~1 ∂ ∂f1 − + f1 = ∂t ∂~q1 ∂~p1 M ∂~q1 ∂t coll in which

∂f1 ∂t

coll

d3 p~1 d3~q

p~2 − p~1 ∂ f2 (~p1 , ~q1 , p~2 , ~q2 , t) M ∂~q

with 18

~q = ~q1 − ~q2 Decompose ~q into (a, ~b), in which a is the component along the direction of ~q, and ~b is a vector in the plane perpendicular to ~q. After the collision, a → ∞, therefore Z ∂f1 = d3 p~1 d2~b |~v2 − ~v1 | [f2 (~p1 , ~q, p~2 , +∞, t) − f2 (~p1 , ~q, p~2 , −∞, t)] ∂t coll p~1 + p~2 = p~01 + p~02 , Z ∂f1 = d3 p~2 d2~b|v2 − v1 |[f2 (~p01 , ~q10 , p~02 , −∞, t) − f2 (~p1 , ~q, p~2 , −∞, t)] ∂t coll Since Z

d2~b

dσ

dΩ

and f2 (~p01 , ~a, p~02 , −∞, t) ≈ f1 (~p1 , ~q, t)f1 (~p02 , ~q, t) ∂ ∂U ∂ p~1 ∂ ⇒ − + f1 ∂t ∂~q1 ∂~p1 M ∂~q1

dσ

|~v2 − ~v1 |[f1 (~p0 , ~q1 , t)f1 (~p0 , ~q1 , t) − f1 (~p1 , ~q, t)f1 (~p2 , ~q, t)] = d p~2 dΩ

1 2 dΩ

Another approach Suppose there is no collision, ~ δt, ~q1 + ~v δt, t + δt) d3~q0 d3 p~0 f1 (~p1 , ~q1 , t) d3 q~1 d3 p~1 = f1 (~p1 + F 1 1 According to Liouville's theorem, d3~q1 d3 p~1 = d3~q10 d3 p~01 , ~ δt, ~q1 + ~v δt, t + δt) ⇒ f1 (~p1 , ~q1 , t) = f1 (~p1 + F With collisions, ~ δt, ~q1 + ~v1 δt, t + δt) = f1 (~p1 , ~q1 , t) + f1 (~p1 + F

∂f1 ∂t

coll

δt

Expand the lhs about δt = 0, ∂f1 ∂f1 ~ ∂f1 + ~v · +F · = ∂t ∂~q1 ∂~p1

∂f1 ∂t

coll

in which ~v =

p~1 , M

~ = − ∂U F ∂~q1

Boltzmann's H-theorem ~ = 0. In equilibrium, f1 (~p1 , t), f eq (~p, t), f eq0 := f eq (~p0 , t). If no external force, F 1 1 1 ∂f1eq =0= ∂t

eq eq eq eq d3 p~2 dΩ

|~v2 − ~v1 |(f10 f20 − f1 f2 ) dΩ

If (f1eq0 f2eq0 − f1eq f2eq ) = 0, then ∂f1eq =0 ∂t The system is in equilibrium. Is it true vise-versa (suﬃciency)? Define Z H(t) = d3 p~f (~p, t) log f (~p, t) dH = dt If

∂f ∂t

d3 p~(1 + log f )

∂f ∂t

= 0, then dH =0 dt

Claim that if dH = 0 and f10 f2 − f1 f2 = 0 dt then f10 f20 − f1 f2 = 0 20

dH = dt

d3 p~1 d3 p~2 dΩ

dσ

0 0

dΩ |~v1 − ~v2 |(f1 f2 − f1 f2 )(1 + log f1 )

This is an Integral over all p, therefore index does not matter,

dH 1 dσ

|~v1 − ~v2 |(f 0 f 0 − f1 f2 )(2 + log f1 f2 ) ⇒ = d3 p~1 d3 p~2 dΩ

1 2 dt 2 dΩ

Z 1 = d3 p~1 d3 p~2 |Tfi |2 δ 4 (~p1 + p~2 − p~01 − p~02 )(f10 f20 − f1 f2 )(2 + log f1 f2 ) 2

dσ

dH 1 3 3

|~v1 − ~v2 |(f1 f2 − f 0 f 0 )(2 + log f1 f2 ) ⇒ = d p~1 d p~2 dΩ

1 2 dt 2 dΩ

Z dH 1 ⇒ = d3 p~1 d3 p~2 |Tfi |2 δ 4 (~p1 + p~2 − p~01 − p~02 )(f10 f20 − f1 f2 )(log f1 f2 − log f10 f20 ) dt 4 ⇒

dH ≤0 dt

dH = 0 iﬀ f2eq0 f1eq0 − f2 f1 = 0 dt

(3.3)

lim f (~p, t) = f eq (~p)

t→∞ dH dt

is not time-reversal invariant, because our assumption of molecular chaos f2 (~p1 , ~q1 , p~2 , ~b, −, t) = f1 (~p1 , ~q1 , t)f1 (~p2 , ~q1 , t)

and coarsegraining nd3 1 which avoids precise discreption at short distance. Take logarithm of eq (3.3), log f1eq + log f2eq = log f1eq0 + log f2eq0 Write log f1eq (~q, p~, t) as p~2 C(~q) − α ~ (~q) · p~ − β(~q) + U (~q) 2m

Add constraint f1eq = f1eq in order to have {f1eq ,H} = 0, so that streaming term do not generate change in f1eq . Then α ~ (~q) = 0, C(~q) = C, β(~q) = β. Write f1eq as 21

2 p ~ e exp −β =N + U (~q) 2m

f1eq

e is normalization constant. To determine N e , use in which N Z d3 p~ d3 p~ f1eq = N Example:

Particle in a box in which U (~q) = 0. eV N =N

3 Z +∞ 3 3 2 2 p~2 2m 2 e 2 e V 2πm d3 p~ exp −β = NV dU e−U =N 2m β β −∞ ⇒

f1eq (~ p)

=ρ

β 2πm

3 2

p~2 exp −β 2m

in which ρ = N /V . To find β, calculate another observable (assemble average energy)

p~2 2m

p ~ d3 p~f1eq (~ p) 2m R = d3 p~f1eq (~ p)

β 2πm

3 2

1 2m

p ~2

d3 p~ p~2 e−β 2m =

3 β 2

with notice that Z

d3 p~ f1eq (~ p) =

N V

Therefore β= H eq = V

ρ 3

(2πmkB T ) 2

d3 p~ e

= N log

1 kB T 2

p ~ − 2mk

p~2 log 3 − 2mkB T (2πmkB T ) 2 "

3 − N 2 (2πmkB T ) ρ

3 2

We expect H eq ∼ S. From ideal gas,

∂E ∂T

∂H ∂T

N ,V

=T V ,N

∂S ∂T

= V ,N

3 N kB 2

∂ 3 3 =T − N log (2πmkB T ) = − N ∂T 2 2

⇒ S = −kB H Check with T dS = dE + P dV P+

∂E ∂V

=T T

∂S ∂V

= T ,N

N kB T V

4 Ensemble Theory 4.1 The micro canonical ensemble For a mechanically and adiabatically isolated system, internal energy and generalized coordinates are fixed, what is the set of microstates consistent with them? A surface (sphere) in phase space on which E and ~x are fixed. Define Ω(E, ~x) as the area of the surface. According to the equal-probability assumption, the joint pdf for one microstate µ is 1 1 for E − ∆E ≤ H(µ) ≤ E + ∆E pE,~x (µ) = Ω(E, ~x) 0 otherwise in which the ∆E is for quantum consideration. Example: Particle in a box. H= 1 p(µ) = Ω(E, V , N )

1 0

N X p2i + U (~ qi ) 2m i=1

E − ∆E =∕ otherwise

p2i i 2m

≤ E + ∆E

Ω(E,V ,N ) is a thin spherical shell in the phase space. Ω(E,V ,N ) = V N A3N ∆R in which A3N is the area of the sphere, and ∆R is the thickness. To determine ∆R , N X Ri2 = E + ∆E 2m i

⇒ R+ =

p 2m(E + ∆E),

R− =

2m(E − ∆E)

Expand R+ around E = E + ∆E. R+ =

√

2mE +

1 2m √ ∆E + · · · 2 2mE

⇒ ∆R =

2m ∆E E

To find the area of a hypersphere Ad , Ad = SA Rd−1 in which SA is the solid angle. To find Sd , consider integral Id =

+∞

dxe−x

= π2

−∞

Change to spherical coordinates Id =

d +∞ Y

−∞ i=1

dxi exp(−x2i ) =

+∞

dR Sd Rd−1 e−R = π 2

2π 2 Γ d2 d

⇒ Sd = Therefore

Ω(E,V ,N ) = V N

2π 2 N 3N −1 (2mE) 2 ∆R Γ 3N 2

3N 2∆R 3N 2 S = kB log Ω = kB N log V − log Γ + log(2πmE) + log √ 2 2mE When N → ∞ log Γ (z + 1) ∼ z log z − z + O(log z) 2∆R 2∆E log √ = log ∼ log N E 2mE and can be ignored. log Γ

3N 2

≈

3N 3N 3N − 1 log −1 ≈ 2 2 2

Therefore " 3 # 4πemE 2 S = N kB log V 3N Also dE = T dS − P dV + µdN

V and N are fixed. Therefore ∂S 1 3 N kB = ∼ ∂E T 2 E ⇒E∼

3 N kB T 2

From the kinetic theory, H

" # S N 3 =− = N log − N 3 kB 2 V (2πmkB T ) 2

The diﬀerence N kB log N comes from the indistinguishability of particles.

The Maxwell-Boltzmann distribution is 3 mβ 2 1 2 p(~v ) = exp − mv β 2π 2 Proof: µ → {~qi }{~pi } p(~p1 ) =

d3~qi

N Z Y

d3~qi d3 p~i P ({~qi }, {~pi })

i=2 3

Ω(E,V ,N ) =

2π 2 N 1 (2mE) 2 (3N −1) ∆R 3 Γ (2N)

p21 1 P (~p1 ) − Ω(E − , V , N − 1) Ω(E, V , N ) 2m in which the second Ω is the Ω for the system without the first particle. 3

( 32 N − 1)! V N π 2 (N −1) (2mE − p~21 ) 2 (3N −4) P (~p1 ) = 3 1 [3(N − 1) 12 − 1]! V N π 2 N (2mE) 2 (3N −1) =

p~2 1− 1 2mE

3 N −2

( 32 N − 1)!

1 (2mE) 2 [3(N − 1) 2 − 1]! p21 3N 3N p21 1 = exp − = − 3 exp 4πmE 2 2mE 2mkB T (2πmkB T ) 2 3

4.2 The Canonical Ensemble Fix T , V , N . Call the set of system with fixed T , V , N and diﬀerent microstates a bath. The total number of microstates is Ωtot = Ωsys Ωreservoir in which Ωsys is the number of microstates for one system, and Ωreservoir is the number of microstates of the bath. Suppose the system is in one microstates µ, Ωtot = Ωµ0 in which Ωµ0 is the number of states for the bath if the system has states µ. Stot = kB log Ωtot , Stot − Sµ0

Sµ0 = kB log Ωµ0

Ωµ0 = −kB log Ωtot

The probability the system has microstate µ is Pµ =

Ωµ0 Ωtot

The entropy is S=

(Stot − Sµ0 )Pµ =

Sµ0

Pµ (−kB log Pµ ) = −kB

S 0 (E

tot

Pµ log Pµ

− Eµ , Ntot − Nµ , V ),

Stot − Sµ0 Pµ = exp − kB

Eµ Etot , then Sµ0

≈

S 0 (E

0 ∂S 0 ∂S Eµ − Nµ tot ,N ,V ) − 0 ∂E V ,N ∂N 0 V 0 ,E 0 0 ∂S 1 = 0 ∂E V ,N T

⇒ Pµ = C exp(−βEµ ),

β=

1 kB T

Normalization: X µ

exp(−βEµ ) Pµ = 1 ⇒ P µ = P µ exp(−βEµ )

S = −kB

Pµ (log C − βEµ )

Set

µ Eµ Pµ

= E, S = −kB log C + kB βE

⇒ kB T log C = E − T S

Define partition function Z(T ,V ,N ) =

exp(−βEj )

−kB T log Z = E − T S = F Although the canonical ensemble has systems with all E, the majority have the same E in the thermodynamic limit. ¯ 2i = h(δE)2 i = h(E − E)

Pµ (∆E)2

h∆E 2 i 1 ∼ √ as N → ∞ E N

For a single particle of ideal gas, exp(−βE) p2 , E = 2m d3 p exp(−βE) p2 1 exp − 2mkB T 3 3 P (p,q)d3 p~ d3~q = d p~ d ~q V (2πmkB T ) 32 P (E) = R

P (p)d3 p~ =

p2 exp − 2mk BT (2πmkB T )

3 2

d3 p~

d3 p~ = m3 d3~v = m3 4πv 2 dv 4πm3 v 2 mv 2 ⇒ P (v)dv = − dv 3 exp 2kB T (2πmkB T ) 2 Equipartition Theorem Every degree of freedom which contributes a square term in either coordinate or momentum to the total energy has a mean energy of 12 kB T in that dof. Proof: Define xi to be either qi or pi , i ∈ [1, 3N ]. Z ∂H 1 ∂H xi = {dpi }{dqi }xi ∂xj Ω(E) E<H<E+∆ ∂xj P X X ∂ (E) Ω(E) = (E + ∆) − (E) = ∆ ∂E P in which Ω(E) is the number of states with energy E, (E + ∆) is the number P of states with P ∂ (E) energy up to E + ∆, (E) is the number of states with energy up to E. ∂E = P (E). Z ∂H ∆ ∂ ∂H xi = {dpi }{dqi }xi ∂xj Ω(E) ∂E E<H<E+∆ ∂xj xi

∂H ∂ ∂ = xi (H − E) = [xi (H − E)] − δij (H − E) ∂xj ∂xj ∂xj

The integral evaluate at the surface of H < E, which is H = E. Z Z δij ∂ δij ∂H xi = {dpi }{dqi }(E − H) = {dpi }{dqi } ∂xj P (E) ∂E H<E P (E) H<E Since P (E) =

∂Σ ∂E ,

δij X (E) = δij P (E)

1 ∂Σ Σ ∂E

= δij

∂ log Σ ∂E

−1

1 ∂S 1 = kB ∂E kB T Therefore δij = δij kB T P (E) ∂H ∂H ⇒ pi = qi = kB T ∂pi ∂qi H=

3N X

(ap2i + bqi2 )

i=1

1 ⇒ hap2i i = hbqi2 i = kB T 2 Equipartition theorem for quantum systems Outline: Show that Gibbs distribution is compatible with quantum mechanics (method of most probable distribution). Define a microstate as a specific set of occupation numbers {Ni }. N1 is the number of particles in quantum state 1, N2 is the number of particles in quantum state 2, etc.. Fix total energy, work in under microcanonical ensemble. The statistical weight is degeneracy W ({Ni }) =

N! N1 ! N2 ! · · ·

¯ j /N is the Nj /N is the fraction of particles in quantum state j (associated with Ej ). Pj = N ensemble average fraction in jth quantum state. P 1 {Ni } W ({Ni })Nj P Pj = N {Ni } W ({Ni }) P in which {Ni } W ({Ni }) is the number of ensemble members. Let N and Ni go to infinity. The most probable set of Ni , {Ni∗ } is the one that maximize W ({Ni }). Then ∗ ∗ ¯i Nj∗ N 1 W ({Ni })Nj Pj = ≈ = N N W ({Ni∗ }) N

P P Find {Ni∗ } which maximize W subject to constraints j Nj = N and j Nj Ej = E. " ! !# X X ∂ log W ({Ni }) − α Nj − N − β Nj Ej − E =0 ∂Nj j

is satisfied at {Nj∗ }. ⇒ − log Nj∗ − α − βEj = 0 ⇒ Nj∗ = exp(−α − βEj ) Normalization: X

Nj∗ = N =

exp(−α − βEj )

N −βEj je

⇒ e−α = P ⇒

Nj∗

e−βEj = P −βE = Pj j N je

Example: One-dimensional harmonic solid. Single atom Hamiltonian is p2 mω 2 2 + q 2m 2 1 ε(n) = n + ~ω 2 exp − n + 12 ~ωβ P (n) = P∞ 1 n=0 exp − n + 2 ~ωβ H=

Define f=

1 exp − n + ~ωβ 2 n=0 ∞ X

Then hεi =

1 1 ∂f P (n) n + ~ω = − 2 f ∂β n=0 ∞ X

X ∞ 1 1 1 n f = exp − ~ωβ [exp(−~ωβ)] = exp − ~ωβ 2 2 1 − exp(−~ωβ) n=0

1 ~ωβ 2 cosh 12 ~ωβ 1 1 1 1 ⇒ hεi = ~ω = ~ω coth ~ωβ 2 2 sinh2 12 ~ωβ csch 12 ~ωβ 2 =

1 csch 2

But the equipartition theorem says hεi = kB T ??!! Wtf?? Expand hεi at β = 0 coth

1 ~ωβ 2

= ∼

dhεi CV = = kB dT

2 12 β~ω

+ ···

2kB T ~ω

~ωβ 2 sinh 12 ~ωβ

State counting If the energy level splitting is small comparing to kB T (large kB T means large occupation number), Z X exp(−βEj ) → dερ(ε) j

For a single spinless free particle in three-dimensional space, d3 p~ d3~q V 4πp2 dp → (2π~)3 (2π~)3 4πV p2 dp ρ(ε)dε = dε (2π~)3 dε Since p =

√

2mε, 1

ρ(ε)dε ∼ ε 2 What is the partition function of a many-particle system in canonical ensemble? 32

Etot = EA + EB ,

Ztot = ZA ZB

If the particles are distinguishable, Ztot = Z1N . If not, Method 1. Enumerate the quantum numbers of each particle and totally the combination. Method 2. Give the number of particles in each single-particle state and then count all the single particle states. The weak-degenerate limit: large T , small ρ. Over count N ! times, Ztot =

Z1N Z!

4.3 The Grand Canonical Ensemble Fix T , V , µ. S0 − Sα0 = −kB log Pα in which S0 is the entropy of the system and the bath. and Sα0 is the entropy of the bath if the system is in state α. 0 0 Sα −(S −S )/k ⇒ Pα = e 0 α B = A exp kB Sα0 S 0 (E0 − Eα ,N0 − Nα ) Expand around E0 and N0 ,

∂S ∂S

Sα0 ≈ S 0 (E0 ,N0 ) − Eα − Nα ∂E 0 V0 ,N0 ∂N 0 V0 ,E0

∂S 0

∂(S 0 , E 0 )

∂(E 0 , S 0 ) ∂(N 0 , S 0 )

= =− ∂N 0 V 0 ,E 0 ∂(N 0 , E 0 ) V 0 ∂(N 0 , S 0 ) ∂(N 0 , E 0 ) V 0

∂E 0

∂S 0

µ =− =

0 0 ∂N S 0 ,V 0 ∂E N 0 ,V 0 T

⇒

Sα0

Eα µNα = const − + , T T

Eα µNα Pα = C exp − + kB T kB T

Normalization: exp[−β(Eα − µNα )] Pα = P α exp[−β(Eα − µNα )] X X S = −kB Pα log Pα = −kB Pα log{C exp[−β(Eα − µNα )]} α

= −kB log C + kB β

Pα Eα − kB βµ

L(T ,V ,µ) =

Nα Pα

exp[−β(Eα − µNα )]

We can connect gce results to ce ones. Let Nα = N for any α, " # X Ω = −kB T log Z = −kB T log e−β(Eα 0µNα ) α

= −kB T log eβµN

!# X

e−βEα

= −kB T log[eβµN Z(T ,V ,N )]

Ω = F − µN = −µN − kB T log Z

4.4 Classical perfect gas in the gce Z=

exp[−β(Eα − µNα )] =

exp[−Nq β(εq − µ)]

In the single particle state, Eα = Nq εq , εq =

~2 q 2 2m

   Y X X Z= Z~q =  e−Nq1 β(εq −µ)   e−Nq2 β(εq −µ)  ~q

Nq1

Nq2

â&#x201E;Ś=

â&#x201E;Ś~q = â&#x2C6;&#x2019;kB T log

â&#x201E;Ś~q ,

exp[â&#x2C6;&#x2019;Nq Î˛(Îľq â&#x2C6;&#x2019; Âľ)]

For a classical perfect gas, there are huge number of available states. Therefore for each state ~q, PNq =0 â&#x2030;&#x2C6; 1, PNq =1 â&#x2C6;ź , PNq =2 â&#x2C6;ź 2 . . . P exp[â&#x2C6;&#x2019;Î˛(Îľq â&#x2C6;&#x2019; Âľ)] Nq Nq exp[â&#x2C6;&#x2019;Nq (Îľq â&#x2C6;&#x2019; Âľ)Î˛] ÂŻq = P N = â&#x2030;&#x2C6; exp[â&#x2C6;&#x2019;Î˛(Îľq â&#x2C6;&#x2019; Âľ)] 1 1 + exp[â&#x2C6;&#x2019;Î˛(Îľq â&#x2C6;&#x2019; Âľ)] Nq exp[â&#x2C6;&#x2019;Nq (Îľq â&#x2C6;&#x2019; Âľ)Î˛] ÂŻ q ) â&#x2030;&#x2C6; â&#x2C6;&#x2019;kB T N ÂŻq â&#x2021;&#x2019; â&#x201E;Ś~q = â&#x2C6;&#x2019;kB T log(1 + N X X ÂŻ q = â&#x2C6;&#x2019;kB T N â&#x2021;&#x2019; P V = N kB T â&#x201E;Ś = â&#x2C6;&#x2019;P V = â&#x201E;Ś~q = â&#x2C6;&#x2019;kB T N q

exp[â&#x2C6;&#x2019;Î˛(Îľq â&#x2C6;&#x2019; Âľ)] =

+â&#x2C6;&#x17E;

Ď (Îľ)dÎľ exp[â&#x2C6;&#x2019;Î˛(ÎľÎ¸ â&#x2C6;&#x2019; Âľ)]

For three-dimensional space, â&#x2C6;&#x161; 3 1 4Ď&#x20AC; 2V m 2 Îľ 2 Ď (Îľ) = (2Ď&#x20AC;~)3 â&#x2021;&#x2019;N =

eÎ˛Âľ

Z 0

+â&#x2C6;&#x17E;

Ď (Îľ)dÎľeâ&#x2C6;&#x2019;Î˛Îľ

= exp

Âľ kB T

Define Î&#x203A;= â&#x2C6;&#x161;

2Ď&#x20AC;~ 2Ď&#x20AC;mkB T

Then N = eÎ˛Âľ

V Î&#x203A;3

To guarantee exp[â&#x2C6;&#x2019;Î˛(Îľq â&#x2C6;&#x2019; Âľ)] 1, exp(Î˛Âľ) must be much smaller than 1. N Î&#x203A;3 1 V Therefore the pressure is small or temperature is high.

Ď&#x20AC; 2 m 2 (kB T ) 3 V2 (2Ď&#x20AC;~)3 3 2

P Λ3 = eβµ kB T

⇒ µ = −kB T log

kB T P Λ3

With eβµ 1, kB T 1 P Λ3

kB T ⇒ P

√

2πkB T m 2π~

⇒P 1

kB T ∂G G = µN = −N kB T log , = −S P Λ3 ∂T P ,N " # √ kB T P Λ3 kB kB T 2πkB m 3 3 1 ⇒ S = N kB log + N kB T + T2 P Λ3 kB T P Λ3 P 2π~ 2 ⇒ S = N kB log

kB T 5kB + N P Λ3 3 PΛ 2P Λ3

Classical ideal monatomic gas Sucks . . . Classical ideal diatomic gas Sucks . . .

4.5 Slightly degenerated perfect gas Ω = −kB T log Z,

exp[−β(Eα − µNα )],

Eα = Nq εq

⇒Z=

exp[−βNq (εq − µ)]

¯ q exactly. For Fermi gas, Nq = 0 or 1. For Bose In the classical case, Nq = 1. Now treat N gas, there is no restriction on Nq . Define Zq =

exp[−βNq (εq − µ)]

Ωq = −kB T log Zq = −kB T log{1 + exp[−β(εq − µ)]} −

∂Ωq ¯q =N ∂µ

For fermions, 1 ¯ q − kB T β exp[β(µ − εq )] = N ≤1 1 + exp[β(µ − εq )] exp[β(εq − µ)] + 1 For bosons, Ωq = −kB T log

∞ X

exp[−βNq (εq − µ)] = −kB T log{exp[−β(εq − µ)]}N

Nq =0

¯q = − N ⇒ Ω = ∓kB T

∂Ω ∂µ

= T ,N

exp[β(µ − εq )] 1 + exp[β(µ − εq )]

log{1 ± exp[−β(εq − µ)]} = N1

X q

1 exp[β(εq − µ)] ± 1

in which fermions use the upper sign, and bosons use the lower sign. To reduce to classical case, let exp[−β(εq − µ)] 1. Then log(1 + x) = x + O(x2 ). Now do the quantum correction. 1 log(1 + x) = x − x2 + O(x3 ) 2 X X 1 ⇒ Ω = −kB T exp[−β(εq − µ)] ± kB T exp[−2β(εq − µ)] + · · · 2 q q in which the first term is just Ωclass . Assume kB T energy splitting, and consider spin degeneracy gs = 2s + 1. √ 3 1 4π 2gs V m 2 ε 2 ρ = gs ρold (ε) = (2π~)3 Z +∞ X exp(−2βεq ) = ρ(ε) exp(−2βε)dε q

kB T 2βµ 1 ⇒ Ω = Ωclass ± e 3 2 22 37

Z 0

+∞

ρ(ε) exp(−βε)dε

= Ωclass 1 ∓

eβµ 5

⇒ P V = N kB T

1∓

eβµ 5

+ ···

Correct µ: µclass = kB T log ⇒ Ω = ∓kB T

N Λ3 , V

µ = kB T log

log{1±exp[−β(εq −µ)]} = ∓kB T

X q

exp[−β(εq − µ)] ± 1

+∞

ρ(ε)dε

ρ(ε)dε log{1±exp[−β(ε−µ)]}

N Λ3 V gs

exp[−β(ε − µ)] 1 ± exp[−β(ε − µ)]

Define the fugacity as λ = eβµ , Ω = ∓kB T N=

Z 0

+∞

1 2 3 ρ(ε)dε ±λ exp(−βε) − λ exp(−2βε) + O(λ ) 2

+∞

ρ(ε)dε λ exp(−βε)[1 ∓ λ exp(−βε) + O(λ2 )]

Let ρ(ε) = ξV ε 2 . then Z +∞ Z 1 2 N = V ξλ ε 2 dε exp(−βε) ∓ λ V ε 0

+∞ 1

ε 2 dε exp(−2βε) + O(λ3 )

3 3 3 3 − − 2 2 = N V ξλβ Γ ∓ (2β) Γ 2 2 3 √ 1√ (kB T ) 2 π 2 ⇒ P = (kB T ) ξλ π ∓ ξλ + O(λ3 ) = ρ(λ) 3 2 2 2 2 3 2

We need to find λ(ρ) in the form of a1 ρ + a2 ρ2 + · · ·. √ h i π 3 3 ρ = (kB T ) 2 ξ (a1 ρ + a2 ρ2 + · · ·) ∓ 2− 2 (a1 ρ + a2 ρ2 + · · ·)2 2 Compare the coeﬃcients, √

π ⇒ 1 = (kB T ) ξ a1 , 2 3 2

3 2

0 = (kB T ) ξ

√

π 3 (a2 ∓ 2− 2 a21 ) 2

2 1 1 ⇒λ= √ 3 ξρ ± 3 π (kB T ) 2 22

2 1 √ π (kB T ) 32 ξ

ρ2 + O(ρ3 )

1 3 1 Λ ρ + 3 (Λ3 ρ)2 + O(ρ3 ) gs 2 2 gs2 Z +∞ Z kB T 2 +∞ Ω = −kB T λ ρ(ε)dε exp(−βε) ± λ ρ(ε)dε exp(−2βε) + O(λ3 ) 2 0 0 1 λ2 V gs = −kB T λV gs 3 ± kB + 3 + O(λ3 ) 3 Λ Λ 2 2 2 V gs λ 3 = −kB T 3 λ ∓ 3 + O(λ ) Λ 22 ! V gs 1 3 Λ3 ρ = −kB T 3 Λ ρ 1 ± 5 + O(ρ3 ) Λ gs 2 2 gs =

= −P V ⇒ P = ρkB T

1±

Λ3 ρ

2 2 gs

+ O(ρ3 )

Another way: perturbation. λ 2 Ω = Ωclass 1 ± 5 + O(λ ) , 22 Ω − Ωclass = δΩ,

λ = eβµ=

F − Fclass = δF ,

Λ3 ρ gs

...

Hold all nature variables fixed, then (δΩ)V ,T ,N = (δF )V ,T ,N = (δG)P ,T ,N = · · · = ∓ P =−

∂F ∂V

T ,N

∂ = Pclass ∓ ∂V

λ 2

5 2

Ωclass = ±

N kB T Λ3 N

= ρkB T ± 5

2 2 gs V T ,N

The totally QM way

N kB T 5 2

2 gs

Λ3 ρ

N 2 kB T Λ3 5

2 2 gs V 2

e−βεα Pα = P −βεα , {~qi }, {~pi } → |ψi αe X X X hOi = Pα hψα |O|ψα i = Pα hψα |mihn|ψα ihm|O|ni = hn|ρ|mihm|O|ni α

α,m,n

m,n

Define the density matrix to be hn|ρ(t)|mi =

hn|ψα (t)iPα hψα |mi

⇒ ρ(t) =

Pα |ψα (t)ihψα (t)|

Property of density matrix: 1. Normalization tr{ρ} = 1 2. Hermiltian In the micro canonical ensemble, ρ(E) = ⇒ hn|ρ|mi =

δ(H − E) Ω(E)

|hn|ψi|2 = 1/Ω 0

if εn = E, n = m otherwise

In the canonical ensemble, ρ(β) =

exp(−βH) , Z(β)

Z(ρ) = tr(e−βH )

4.6 Completely degenerated perfect Fermi gas (T <<<< TF = εF /kB ) At β → ∞, N=

X q

1 exp[β(εq − 1)] + 1

 ∞ exp[β(εq − µ)] = 0  1 40

εq > µ εq < µ εq = µ

Number of states between p and p + dp is gs

4πp2 dpV (2π~)3

Number of states is N=

gs 4πp2 dpV gs 4πV 3 = p 3 (2π~) 3(2π~)3 F

Z 0

1 3 3 ⇒ pF = 2π~ 4πgs If non-relativistic, 2 p2F ~2 6π 2 3 2 εF = = ρ3 , 2m 2m gs E=

µ = εF

5 2 4πp2 dp p2 gs V 4π 1 p5F V = ∝ V ρ3 ∝ Nρ3 3 3 (2π~) 2m (2π~) 2m 5

Or use N= to yield the same results.q If relativistic, εF =

εF

Z 0

ρ(ε)dε

p2F + m2 with c = 1. If extremely relativistic, εF = cpF .

Hydrostatic equilibrium Internal pressure balance the gravity. Fin =

GM (R)dm , R2

Fout = −[P (R + dR) − P (R)]da = − ⇒

M (R) =

Z 0

dP GM (R)ρ =− dR R2

ρ(R)4πR2 dR ⇒

dM (R) = ρ(R)4πR2 dR

dP da dR

d ⇒ dR

R2 dρ ρ dR

=−

GdM (R) = −Gρ(R)4πR2 dR

For white dwarf, the internal pressure comes from the degenerated electron gas. Use the polytropic equation of states P = Kργ . Let ρ = ρc θn , γ = 1 + n1 . Define ξ as R = Aξ. θ and ξ are dimensionless. 1 12 n −1 K(n + 1)ρc 1 d dθ 2 A= ⇒ 2 ξ = −θn 4πG ξ dξ dξ

This is the Lant-Emden equation for polytropic index n. The boundary conditions are θ(ξ = 0) = 1 and dρ → 0 (R → 0) ⇒ θ0 (ξ = 0) = 0 dR For n < 5, θ(ξ) decrease with ξ. M=

4πA3 ρ

Z c

ξ1

ξ 2 θn dξ =

−4πA3 ρ

ξ12 |θ0 (ξ1 )|4π

dθ ξ2

ξ1

= 4πA3 ρc ξ 2 |θ0 (ξ1 )| 1 dξ

3 (n + 1)K 2 3−n ρc2n 4πG

For non-relativistic electron gas, 5 2 γ= , N= , 3 3

− 12

M ∼ ρc2 , V ∼ ρc

When pF ≥ me c2 , the gas becomes extremely relativistic, 4 γ = , N = 3, 3

− 13

M ∼ ρ0c , R ∼ ρc

2 2 M = 1.457 M

µe

4.7 Degenerated Fermi system (T TF but not T <<<< TF ) In gce, 42

ÂŻq = N

X q

1 exp[Î˛(Îľq â&#x2C6;&#x2019; Âľ)] Âą 1

Assume the system is non-relativistic. P V = 23 E, â&#x2C6;&#x161; 3 4Ď&#x20AC; 2gs M 2 = (2Ď&#x20AC;~)3

1 2

Ď (Îľ) = ÎžV Îľ , 2 â&#x201E;Ś = â&#x2C6;&#x2019;ÎžV 3

+â&#x2C6;&#x17E;

Îľ 2 dÎľ , exp[Î˛(Îľ â&#x2C6;&#x2019; Âľ)] Âą 1

N = ÎžV

+â&#x2C6;&#x17E;

Îľ 2 dÎľ exp[Î˛(Îľ â&#x2C6;&#x2019; Âľ)] Âą 1

Generally, I=

+â&#x2C6;&#x17E;

f (Îľ)dÎľ exp[Î˛(Îľ â&#x2C6;&#x2019; Âľ)] Âą 1

Consider fermions. Define z = Î˛(Îľ â&#x2C6;&#x2019; Âľ), Z +â&#x2C6;&#x17E; f Âľ + z dz Z Î˛Âľ f Âľ â&#x2C6;&#x2019; z dz Z +â&#x2C6;&#x17E; f Âľ + z dz Î˛ Î˛ Î˛ I= = + z â&#x2C6;&#x2019;z z (e + 1)Î˛ (e + 1)Î˛ (e + 1)Î˛ â&#x2C6;&#x2019;Î˛Âľ 0 0 Z Î˛Âľ Z Î˛Âľ f Âľ + z â&#x2C6;&#x2019; f Âľ â&#x2C6;&#x2019; z Z +â&#x2C6;&#x17E; Âľ + z Î˛ Î˛ dz Î˛ z dz = f Âľâ&#x2C6;&#x2019; + + dz Î˛ Î˛ ez + 1 Î˛ (ez + 1)Î˛ 0 0 ÂľÎ˛ Let the upper limit of the 2nd integral go to +â&#x2C6;&#x17E;, then the 3rd integral vanishes. The corrections are exponentially suppressed. Expand f Âľ + f

z Âľ+ Î˛

â&#x2021;&#x2019;I=

â&#x2C6;&#x2019;f Z

Âľ

Z 0

z Âľâ&#x2C6;&#x2019; Î˛

f (Îľ)dÎľ + +â&#x2C6;&#x17E;

Z 0

z Î˛

â&#x2C6;&#x2019;f Âľâ&#x2C6;&#x2019;

about z/Î˛ = 0.

3 2 000 z + f (Âľ) + O(z 5 ) Î˛ 3! Î˛ dz 2z f 0 (Âľ) + Âˇ Âˇ Âˇ Î˛(ez + 1) Î˛

2z f 0 (Âľ)

+â&#x2C6;&#x17E;

z Î˛

z 2nâ&#x2C6;&#x2019;1 dz 22nâ&#x2C6;&#x2019;1 â&#x2C6;&#x2019; 1 = + Ď&#x20AC; 2n Bn ez + 1 2n

in which Bn is the Bernoulli numbers

+∞

Z 0

z x−1 dz = (1 − 21−x )Γ (x)ζ(x) x > 0, x =∕ 1 ez + 1

in which ζ is the Riemann Zeta function. Z µ 2 1 I= f (ε)dε + f 0 (µ) 2 π 2 B1 + O(β −4 ) β 2 0 3

Note that f (ε) = ε 2 , and µ = εF . √ 3 2 4π 2gs M 2 3 1 π 2 Ω = Ω|T =0 − V µ2 + O(β −4 ) 3 (2π~)3 2 4β 2 in which Ω|T =0 is the Ω for completely degenerated case. √

(δΩ)T ,V ,µ = (δF )T ,V ,N

3 2

V gs 2m ⇒ F = F |T =0 − 2 6~3 β 2

= F |T =0 − N ρ

− 23

3π 2 2 3 gs

2 3

#1 2

" 2 # 12 1 ~2 3π 2 2 3 2 ρ3 + O(β −4 ) 2 β 2m gs

∂β 1 β =− =− 2 ∂T kB T T V ,N 1 −2 2 β 2 T N ρ− 23 η ⇒ S = Nρ 3 η − 3 − = kB 2 β T ∂S 2 N ρ− 23 N CV = T = T kB ∂T V ,N S=−

∂F ∂T

T CP − CV = V kB

∂V ∂T

∼ T 3,

T →0

4.8 Degenerated Bose systems For a perfect Bose gas, Ωq = −kB T log

∞ X

exp[−βNq (εq − µ)] = kB T log{1 − exp[β(µ − εq )]}

Nq =0

1 exp[β(εq − µ)] − 1 X X Ω= Ωq , N = Nq Nq =

Let N → ∞, Ω = kB T

+∞

ρ(ε)dε log{1 − exp[−β(ε − µ)]}

Z 0

+∞

ρ(ε)dε exp[β(ε − µ)] − 1

Let ρ(ε) = ξV ε 2 , ⇒ N = ξV

Z 0

+∞

ε 2 dε 3 = ξV β − 2 exp[β(ε − µ)] − 1

+∞

z 2 dz exp(z − βµ) − 1

Fix N and V , lower T , then µ must go less negative. At finite Tc , µ may go to 0, which implies macroscopic occupation of ground state. N 3 = ξ(kB Tc ) 2 V

Z 0

+∞ z 12 dz

3 3 = ξ(kB Tc ) Γ ζ z e −1 2 2 3 2

√ 3 4π 2gs m 2 3 1 √ 2.6 = (kB Tc ) 2 π 3 (2π~) 2 2 2

⇒ Tc = 3.31

~2 ρ 3

kB mgs3 √ N 2πmkB Tc 3 2.612 gs = 2.612 gs = V 2π~ (Λ|t=Tc )3 This is the critical density which happens when the wave packets begin to overlap (in the same order). Go beyond Tc

Nq =

1 exp[β(εq − µ)] − 1

For ground state, εq = 0. ⇒ N0 =

1 e−βµ

−1

∼−

kB T as T → 0 µ 1

When T goes up, N 0 increases. This is inconsistent with ρ(ε) ∼ ε 2 . However ρ(ε) is still usable. Z +∞ 1 z 2 dz 3 2 Nε>0 = ξV (kB T ) ez − 1 0 " 3 # T 2 N0 = N − Nε>0 = N 1 − Tc The chemical potential remains zero when T < Tc . E=

+∞

εdNε>0 = ξV (kB T ) CV =

∂E ∂T

5 2

+∞

= V

z 2 dz 5 ∼T2 z e −1

5E 2T

When T < Tc , Ω = F − G = F . ∂S ∂S 5 3 CV = T ⇒ ∝ T2 ∂T V ∂T V 2

5 3 5E ⇒S ∝ T2 = 3 3T

2 F = E − T S = − E = Ω = −P V 3 2 5 5 5 ⇒ P = ξ(kB T ) 2 Γ ζ 3 2 2 which is independent of V . For T > Tc , N = ξV (kB T )

3 2

Z 0

+∞

z 2 dz ez−βµ − 1

The derivative of CV discontinues at T = Tc .

4.9 Black-body radiation Perfect gas of photons. N is not constant. The reaction N γ + A ←→ M γ + A exists. Therefore (N − M )µγ = 0 ⇒ µγ = 0 Photons are spin 1, c is fixed. Therefore there are only two spin states. ω = ck, gs = 2. The number of modes is V d3~k (2π)3 ρ(ω)dω = Nq =

V ω 2 dω π 2 c3

1 exp[β(εq − µ)] − 1

⇒ dNω = dEω =

V 1 ω 2 dω β~ω 2 3 π c e −1 V dω ~ω 3 β~ω−1 2 3 π c e

If β~ω 1, dEω =

V ~3 dω 1 V ω 2 dω = k T B π 2 c3 β~ω π 2 c3

If β~ω 1, V ~ω 3 dωe−β~ω π 2 c3 Z +∞ X V ω 2 dω −βε q Ω= kB T log(1 − e )⇒F = kB T log(1 − e−β~ω ) 2 c3 π 0 q dEω =

V = 2 3 kB T (β~)−3 π c

Z 0

+∞

x2 dx log(1−e−x )

V = − 2 3 3 (kB T )4 π c ~

+∞

x3 dx e−x 4σV T 4 = − 3 1 − e−x 3c

in which 4 π 2 kB 60~3 c2 ∂F ∂F − = P, − =S ∂V T ∂T V

σ=

⇒P =

4 σT 4 , 3c

16 σV T 3 3c

Since µ = 0, E = T S − P V + µN =

4σV T 3 = 3P V c

This is consistent with the case of relativistic perfect gas. Z +∞ V ω 2 dω 1 V (β~)−3 N= = Γ (3)ζ(3) π 2 c3 eβ~ω − 1 π 2 c3 0 For isotropic process (fix S ∼ V T 3 ), 1 4σ PV = E = V T 3T 3 3c 1

Since V T 3 ∝ N , T ∼ (N /V ) 3 , P ∝ ρ 3 . Emissivity of a body dEω =

V ~ω 3 dω 1 2 3 β~ω π c e −1

The spectral energy density per unit solid angle per volume is e0 (ω) =

1 dEω 4πV dω

Then the energy flux in dΩdω is ce0 (ω). Only the component perpendicular to the surface can be absorbed. Therefore the radiation absorbed in dΩdω is ce0 (ω)A(ω,θ) cos θ, in which 48

A is the absorption factor. Define the emitted energy flux in dΩdω to be J(ω,θ). Then in equilibrium, Z Z J(ω,θ)dΩdω = ce0 (ω)A(ω,θ) cos θ dΩ dω For a block body, A = 1. Therefore J(ω,θ) = ce0 (ω). The total intensity of radiation is J0 =

Z 0

+∞

dω

Z 0

sin θ dθ 2πJ(ω,θ) = cπ =

Z 0

+∞

e0 ω dω = cπ

πω 3 dω 4π 3 c3 (eβ~ω − 1)

+∞

cE c 4σV T 4 = = σT 4 4V 4V c

Einstein model revisited In Einstein model, Z=

Z13N ,

Z1 =

∞ X n=0

1 exp − n + ~ωβ 2

But experimentally, CV ∼ T 3 . Z does not depend on T . Einstein lacks the consideration of collective behavior, and misses modes. For low energy, the main contribution comes from phonons. ω = cg, in which c is the speed of sound. The excitations are the same as a particle in a box qx =

3πnx , L

d3 n =

V 4πV 2 d3~q = q dq 3 (2π) (2π)2

In each direction, 4πV 2 4πV q (ω)dq(ω) = ω 2 dω 3 (2π) (2π)3 c3i ! 4πV 1 2 ρ(ω)dω = + 3 ω 2 dω (2π)3 c3l ct

ρi (ω)dω =

in which cl is speed of longitudinal wave, and ct is the speed of translational wave. Define

3 1 2 = 3+ 3 3 c¯ cl ct ρ(ω)dω = E=

+∞

12πV 2 ω dω (2π¯ c)3

Z Z +∞ 3 12πV ~ +∞ ω 3 dω x dx −4 N (ω)ρ(ω)~ω dω = ∼ (β~) ∝ T4 x β~ω (2π¯ c)3 0 e − 1 e −1 0 Z ω Z ω D D 12πV 4πV 3 3N = ρ(ω)dω = ω 2 dω = ω 3 (2π¯ c) 0 (2π¯ c)3 D 0

in which ωD is the Debye frequency. 3N ⇒ ωD = 2π¯ c 4πV

CV =

∂E ∂T

V ,N

9N = 3 ωD

1 3

ρ(ω) =

9N ω 2 3 ωD

~ω 3 dω β~ω ~ω = 9N kB e 3 β3 kB T 2 (e~ωβ − 1)2 ~3 ωD

ωD

Z 0

θD T

x4 dx (ex − 1)2

in which kB θD = ~ωD . As T → 0, CV → T 3 . As T → ∞, consider only small x, Z 0

θD T

x2 dx

1 ∼ 3

θD T

5 Imperfect Gas Discuss under classical limit only. P = ρkB T [1 + ρB2 (T ) + ρ2 B3 (T ) + · · ·] This is the Virial expansion. For classical imperfect gas in canonical ensemble, N indistinguishable particles, Z 1 Z= {d3 p~i }{d3~qi }e−βε (2π~)3N N ! ∞ X p2i ε({~pi }, {~qi }) = + u({~ri }) 2m i=1

Z Z p2i 1 3 ⇒Z= exp −β {d p~i } e−βu {d3~ri } 2m (2π~)3N N ! Z N 1 1 −βV 2m 3 3 = e d p~d r QN N! VN in which QN is the configuration integral. N 1 V Zideal = , N ! Λ3

Z = ZN = Zideal

QN VN

In grand canonical ensemble Z=

exp[−β(Eα − µNα )] =

X N

in which X

e−βEN i = ZN

Define z = eβµ . Then Z=

X z n QN n

n!Λ3n

eβµN

X i

e−βEN i

QN =

exp[−βu(~ri )]{d3~ri }

{d3~ri } = d3~ri d3 (~r2 − ~r1 ) · · · d3 (~rN − ~ri ) Define d{N − 1} = d3 (~r2 − ~r1 ) · · · d3 (~rN − ~ri ) Z ⇒ QN = V exp[−βu({N })]d{N − 1} Consider δ(~r + ~ri − ~r2 ). At fixed N , Z V hδ(~r + ~ri − ~r2 )i = δ(~r + ~ri − ~r2 ) exp[−βu({N })]d{N − 1} QN Z V = exp[−βu({N })]d{N − 2} QN Suppose the gas is dilute, ρB2 , ρ2 B3 , · · · 1. Assume only pair-wise interaction. u({ri }) =

X 1X u(rij ) = u(rij ) 2 i=∕j

exp −

i<j

βu(rij ) =

i<j

exp[−βu(rij )]

i<j

fij = exp[−βu(rij )] − 1 If u → 0, then fij → 0. ⇒ QN =

exp −

Z Y βu(rij ) d{N } = (1 + fij )d{N }

i<j

N Y (1 + fij ) = 1 + f12 + f13 + f23 + · · · + f12 f34 + f12 f23 + · · · i<j

If the gas the dilute, f is small.

⇒ QN =

! X

fij

d{N } =

+ V N −2

i<j

fij d3~ri d3~rj

i<j

N (N − 1) = V N + V N −2 V d3~r12 fij (~r12 ) 2 Z N (N − 1) =VN 1+ f12 d3~r12 2N Z QN N (N − 1) ⇒ Z = Zideal N = Zideal 1 + f12 d3~r12 2N V Z Z N (N − 1) N (N − 1) 3 F = Fideal − kB T log 1 + f12 d ~r12 = Fideal − kB T d3~r12 f12 2N 2V Z ∂F N (N − 1) P =− = ρkB T − kB T d3~r12 f12 ∂V T ,N 2V 2 Z 1 B2 (T ) = − d3~r12 f12 2 Z

Lennard-Jones potential u(r) = ε0

σ 12 r

−

σ 6 r

Approximately, for r < σ, the potential is a infinitely-strong “hardcore”. Outside r = σ, the potential is attractive. ∞ r≤σ Vhardcore = 0 r>σ Z 1 B2 (T ) = − d3~r(e−βu(r) − 1) 2 If r > σ, e−βu − 1 ≈ −βu(r). 1 ⇒ B2 (T ) ≈ 2

Z 0

d3~r

1 + 2kB T

in which

+∞

4πu(r)r2 dr = b0 −

A kB T

b0 =

1 4πσ 3 2 3

B2 (T ) → b2 > 0 (T → ∞),

B2 (T ) < 0 (T → 0)

N kB T N kB T P + B2 (T ) + · · · P P kB T N kB T = + N B2 + · · · P N dB2 µJT = T − B2 CP dT

V =

About Bn (T ) QN

Z Y Z = (1 + fij ) d{N } = d{N }[1 + f12 + · · · + f12 f13 + · · ·] i<j

This equals the sum of all the distinct ways to connect N particles (N -particle graphs). Define irreducible l-cluster integral bl (V ,T ) =

l!Λ3l−3 V

[sum over all possible connected clusters]

A general N -particle graph in QN . Define ml to be the number of connected l-clusters, then N X

lml = N

l=1

For clusters 

  5



2 7

4 8 ,

 N = 2(l = 2) + 2(l = 1) + 1(l = 4) = 10 7

3, 2 4, 5 6, 9 2, 3 4, 5 6, 9 For fixed {ml }, [ 1 10 ] and [ 1 10 ] are two contributors with the same {ml } to QN . The sum of all distince graphs with fixed {ml } is

S({ml }) =

(·)m1 (

)m2 (

)m3 · · ·

N Y 1 =N (V Λ3l−3 bl )ml ml ! l=1 ml N Y 1 V 3N = N !Λ bl ml ! Λ3 l=1

ml N X XY 1 QN 1 1 V = = S({ml }) = bl N ! Λ3N ml ! Λ3 N !Λ3N {ml }

{ml } l=1

In the grand canonical ensemble, Z=

∞ X z n Qn , n!Λ3n

z = eβµ

n=1

⇒Z=

∞ X ∞ X m1 =0 m2 =0

m2 X 1 V zb1 m1 1 V 2 ··· z b2 ··· m1 ! Λ3 m2 ! Λ3

V zb1 V z 2 b2 exp ··· Λ3 Λ3 X V = exp z l bl 3 Λ = exp

Ω = −P V = −kB T log Z ⇒ P =

∞ kB T X l bl z Λ3 l=1

−N =

∂Ω ∂µ

⇒N =z T ,V

∞ ∂ log Z zV X = 3 bl lz l−1 ∂z Λ l=1

⇒ρ=

∞ 1 X bl lz l Λ3 l=1

Now let V → ∞, gce → ce. Define ˜bl (T ) = lim bl (V T ) V →∞

P = kB T

∞ 1 X˜ l bl z , Λ3 l=1

ρ=

∞ 1 X ˜ l l bl z Λ3 l=1

P∞ ˜ l ∞ X P l=1 bl z = Bl (T )ρl−1 = P∞ ˜ l ρkB T l=1 lbl z l=1

Define el (T ) = Bl (T ) B Λ3(l−1) e1 (T ) + B e2 (T ) B

∞ X

l˜bl

P∞ ˜ l l=1 bl z + · · · = P∞ l˜bl z l l=1

l=1

Compare both sides, e1 = 1, ⇒B

e2 = −˜b2 , B

e3 = 4˜b2 − 2˜b3 , · · · B 2

Consider an eos which effectively includes interactions. Most interaction potentials have a repulsive core. For example, Van der Waals gas, Veff = V − bN 2 N P = Pkinetic − A , Pkinetic Veff = N kB T V 2 N ⇒ P +a (V − N b) = N kB T V aρ ⇒ P = ρkB T + bP ρ − aρ2 + abρ3 = ρkB T 1 + bρ + + O(ρ3 ) kB T Z Z a 1 σ 3 2πσ 3 1 +∞ ⇒ B2 = b − , b= d ~r = , −a = u(r)d3~r kB T 2 0 3 2 σ