Instructor’s Manual Statistical Mechanics R. K. Pathria, Paul D. Beale, Exercises of Statistical mechanics

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Instructor’s Manual

Containing Solutions to

Over 280 Problems

Selected from

Statistical Mechanics

Third Edition

Chapter 1

1.1. (a) We expand the quantity ln Ω(0)(E 1 ) as a Taylor series in the variable (E 1 − E¯ 1 ) and get

ln Ω(0)(E 1 ) ≡ lnΩ 1 (E 1 ) + ln Ω 2 (E 2 ) (E 2 = E(0)^ − E 1 ) = {ln Ω 1 ( E¯ 1 ) + ln Ω 2 ( E¯ 2 )}+ { ∂ ln Ω 1 (E 1 ) ∂E 1

∂ ln Ω 2 (E 2 ) ∂E 2

∂E 2

∂E 1

E 1 = E¯ 1

(E 1 − E¯ 1 )+

∂^2 ln Ω 1 (E 1 ) ∂E 12

∂^2 ln Ω 2 (E 2 ) ∂E 22

∂E 2

∂E 1

E 1 = E¯ 1

(E 1 − E¯ 1 )^2 + · · ·.

The first term of this expansion is a constant, the second term van- ishes as a result of equilibrium (β 1 = β 2 ), while the third term may be written as

1 2

∂β 1 ∂E 1

∂B 2

∂E 2

eq.

E 1 − E¯ 1

kT 12 (Cv) 1

kT 22 (Cv) 2

(E 1 − E¯ 1 )^2 ,

with T 1 = T 2. Ignoring the subsequent terms (which is justified if the systems involved are large) and taking the exponentials, we readily see that the function Ω^0 (E 1 ) is a Gaussian in the variable (E 1 − E¯ 1 ), with variance kT 2 (Cv) 1 (Cv) 2 /{(Cv) 1 + (Cv) 2 }. Note that if (Cv) 2  (Cv) 1 — corresponding to system 1 being in thermal contact with a very large reservoir — then the variance becomes simply kT 2 (Cv) 1 , regardless of the nature of the reservoir; cf. eqn. (3.6.3). (b) If the systems involved are ideal classical gases, then (Cv) 1 = 32 N 1 k and (Cv) 2 = 32 N 2 k; the variance then becomes 32 k^2 T 2 · N 1 N 2 /(N 1 + N 2 ). Again, if N 2  N 1 , we obtain the simplified expression 32 N 1 k^2 T 2 ; cf. Problem 3.18.

1.2. Since S is additive and Ω multiplicative, the function f (Ω) must satisfy the condition f (Ω 1 Ω 2 ) = f (Ω 1 ) + f (Ω 2 ). (1)

Substituting V = L^3 and S = 6L^2 , we obtain eqns. (1.4.15 and 16). The expression for T now follows straightforwardly; we get 1 T

= k

∂ ln Ω ∂E

N

k hν

∂ ln Ω ∂R

N

k hν

ln

R + N

R

k hν

ln

Nhν E

so that T =

hν k

ln

Nhν E

For E  Nhν, we recover the classical result: T = E/Nk.

1.9. Since the function S(N,V,E) of a given thermodynamic system is an ex- tensive quantity, we may write

S(N, V, E) = Nf

V

N

E

N

= Nf (v, ε)

v =

V

N

, ε =

E

N

It follows that

N

∂S

∂N

V,E

= N

[

f + N

∂f ∂v

ε

−V

N 2

+ N

∂f ∂ε

v

−E

N 2

]

V

∂S

∂V

N,E

= VN

∂f ∂v

ε

N

and E

∂S.

∂E

N,V

= EN

∂f ∂ε

v

N

Adding these expressions, we obtain the desired result.

1.11. Clearly, the initial temperatures and the initial particle densities of the two gases (and hence of the mixture) are the same. The entropy of mixing may, therefore, be obtained from eqn. (1.5.4), with N 1 = 4NA and N 2 = NA. We get

(∆S)∗^ = k[4NA ln(5/4) + NA ln 5] = R[4 ln(5/4) + ln 5] = 2. 502 R,

which is equivalent to about 0.5 R per mole of the mixture.

1.12. (a) The expression in question is given by eqn. (1.5.3a). Without loss of generality, we may keep N 1 , N 2 and V 1 fixed and vary only V 2. The first and second derivatives of this expression are then given by

k

[

N 1 + N 2

V 1 + V 2

N 2

V 2

]

and k

[

N 1 + N 2

(V 1 + V 2 )^2

N 2

V 22

]

(1a,b)

respectively. Equating (1a) to zero gives the desired condition, viz. N 1 V 2 = N 2 V 1 , i.e. N 1 /V 1 = N 2 /V 2 = n, say. Expression (1b) then reduces to

k

[

n V 1 + V 2

n V 2

]

knV 1 V 2 (V 1 + V 2 )

Clearly, (∆S) 1 ≡ 2 is at its minimum when N 1 /V 1 = N 2 /V 2 , and it is straightforward to check that the value at the minimum is zero.

(b) The expression now in question is given by eqn. (1.5.4). With N 1 = αN and N 2 = (1 − α)N , where N = N 1 + N 2 (which is fixed), the expression for (∆S)∗/k takes the form

−αN ln α − (1 − α)N ln (1 − α).

The first and second derivatives of this expression with respect to α are

[−N ln α + N ln(1 − α)] and

[

N

α

N

1 − α

]

(2a,b)

respectively. Equating (2a) to zero gives the condition α = 1/2, which reduces (2b) to − 4 N. Clearly, (∆S)∗/k is at its maximum when N 1 = N 2 = (1/2)N , and it is straightforward to check that the value at the maximum is N ln 2.

1.13. Proceeding with eqn. (1.5.1), with T replaced by Ti, it is straightforward to see that the extra contribution to ∆S, owing to the fact that T 1 6 = T 2 , is given by the expression

3 2

N 1 k ln (Tf /T 1 ) +

N 2 k ln(Tf /T 2 ),

where Tf = (N 1 T 1 + N 2 T 2 )/(N 1 + N 2 ). It is worth checking that this expression is always greater than or equal to zero, the equality holding if and only if T 1 = T 2. Furthermore, the result quoted here does not depend on whether the two gases were different or identical.

1.14. By eqn. (1.5.1a), given on page 19 of the text, we get

(∆S)v =

Nk ln(Tf /Ti).

Now, since PV = NkT , the same equation may also be written as

S = Nk ln

kT P

Nk

• ln

2 πmkT h^2

. (1b)

It follows that

(∆S)P =

Nk ln(Tf / Ti) =

(∆S)V.

A numerical verification of this result is straightforward. It should be noted that, quite generally,

(∆S)P (∆S)V

T (∂S / ∂T )P

T (∂S / ∂T )V

CP

CV

= γ

which, in the present case, happens to be 5/3.

Chapter 2

2.3. The rotator in this problem may be regarded as confined to the (z = 0)- plane and its position at time t may be denoted by the azimuthal angle ϕ. The conjugate variable pϕ is then mρ^2 ϕ˙, where the various symbols have their usual meanings. The energy of rotation is given by

E =

m(ρ ϕ˙)^2 = p^2 ϕ / 2 mρ^2.

Lines of constant energy in the (ϕ, pϕ)-plane are “straight lines, running parallel to the ϕ-axis from ϕ = 0 to ϕ = 2π”. The basic cell of area h in this plane is a “rectangle with sides ∆ϕ = 2π and ∆pϕ = h/ 2 π”. Clearly, the eigenvalues of pϕ, starting with pϕ = 0, are nℏ and those of E are n^2 ℏ^2 / 2 I, where I = mρ^2 and n = 0, ± 1 , ± 2 ,... The eigenvalues of E obtained here are precisely the ones given by quan- tum mechanics for the energy “associated with the z-component of the rotational motion”.

2.4. The rigid rotator is a model for a diatomic molecule whose internuclear distance r may be regarded as fixed. The orientation of the molecule in

space may be denoted by the angles θ and ϕ, the conjugate variables being pθ = mr 2 θ˙ and pϕ = mr 2 sin^2 θ ϕ˙. The energy of rotation is given by

E =

m(r θ˙)^2 +

m(r sin θ ϕ˙)^2 =

p^2 θ 2 mr 2

p^2 ϕ 2 mr 2 sin^2 θ

M 2

2 I

where I = mr 2 and M 2 = p^2 θ +

p^2 ϕ/ sin^2 θ

The “volume” of the relevant region of the phase space is given by the integral

dpθ dpϕdθ dϕ, where the region of integration is constrained by the value of M. A little reflection shows that in the subspace of pθ and pϕ we are restricted by an elliptical boundary with semi-axes M and M sin θ, the enclosed area being πM 2 sin θ. The “volume” of the relevant region, therefore, is

∫^ π

θ=

∫^2 π

ϕ=

(πM 2 sin θ)dθ dϕ = 4π^2 M 2.

The number of microstates available to the rotator is then given by 4π^2 M 2 /h^2 , which is precisely (M/ℏ)^2. At the same time, the number of microstates associated with the quantized value M (^) j^2 = j(j + 1)ℏ^2 may be estimated as

1 ℏ^2

[

M (^) j^2 + 1 2 − M (^) j^2 − 1 2

]

j +

j +

j −

j +

= 2j + 1.

This is precisely the degeneracy arising from the eigenvalues that the az- imuthal quantum number m has, viz. j, j − 1 ,... , −j + 1, −j.

2.6. In terms of the variables θ and L(= m`^2 θ), the state of the simple pendu- lum is given by, see eqns. (2.4.9),

θ = (A/) cos(ωt + ϕ), L = −mωA sin(ωt + ϕ), with E = 12 mω^2 A^2 and τ = 2π/ω. The trajectory in the (θ, L)-plane is given by the equation

θ^2 (A/`)^2

L^2

(m` ωA)^2

which is an ellipse — just like in Fig. 2.2. The enclosed area turns out to be πmωA^2 , which is precisely equal to the product Eτ.

2.7. Following the argument developed on page 68–69 of the text, the number of microstates for a given energy E turns out to be

Ω(E) = (R + N − 1)!/ R!(N − 1)!, R =

E −

N ℏω

/ℏω. (1)

so that Σ(n, V, E) = V N^ (8π E^3 / h^3 c^3 )N^ / (3N )!,

which is a function of N and VE 3. An isentropic process then implies that VE 3 = const.

The temperature of the system is given by

1 T

∂(k ln Σ) ∂E

N,V

3 Nk E

, i.e. E = 3NkT.

The equation for the isentropic process then becomes VT 3 = const., i.e. T ∝ V −^1 /^3 ; this implies that γ = 4/3. The rest of the thermodynamics follows straightforwardly. See also Problems 1.7 and 3.15.

Chapter 3

3.4. For the first part, we use eqn. (3.2.31) with all ωr = 1. We get

k N ln Γ = k ln

r

e−βEr

• kβU,

which is indeed equal to −(A/T ) + (U/T ) = S. For the second part, we use eqn. (3.2.5), with the result that

k N

ln W {n∗ r } = k N

[

N ln N −

r

n∗ r ln n∗ r

]

= −k

r

n∗ r N

ln

n∗ r N

= −k

ln

n∗ r N

Substituting for n∗ r from eqn. (3.2.10), we get

k N

ln W {n∗ r } = kβ〈Er 〉 + k ln

r

e−βEr

which is precisely the result obtained in the first part.

3.5. Since the function A(N, V, T ) of a given thermodynamic system is an extensive quantity, we may write

A(N, V, T ) = Nf (v, T ) (v = V / N ).

It follows that

N

∂A

∂N

V,T

= N

[

f + N

∂f ∂v

T

−V

N 2

]

, and V

∂A

∂V

N,T

= VN

∂f ∂v

T

N

Adding these expressions, we obtain the desired result.

3.6. Let’s go to part (c) right away. Our problem here is to maximize the expression S/k = −

r,s

Pr,s ln Pr,s, subject to the constraints

r,s

Pr,s =

It follows that T (∂ ln Q 1 /∂T )P = 5/2; the expression on the right-hand side of the given equation then is

ln

V

N

(2πmkT )^3 /^2 h^3

which, by eqn. (3.5.13), is indeed equal to the quantity S/Nk.

3.12. We start with eqn. (3.5.5), substitute H(q,p) =

i

p^2 i / 2 m

• U (q) and integrate over the pi′s, to get

QN (V, T ) =

N!

2 πmkT h^2

) 3 N/ 2

ZN (V, T ), where ZN (V, T ) =

e−U^ (q)/kT^ d^3 N^ q.

It follows that, for N  1,

A = NkT

[

ln

N

h^2 2 πmkT

]

− kT ln Z, whence

S = Nk

[

ln

N

2 πmkT h^2

]

• k ln Z + kT

∂ ln Z ∂T

N,V

Now

kT

∂ ln Z ∂T

N,V

kT

e−U/kT^ (U/kT 2 )d^3 N^ q ∫ e−U/kT^ d^3 N^ q

U¯

T

, while

k ln Z = k ln

V¯ N^ e−^ U /¯kT^

= Nk ln V¯ −

U¯

T

Substituting these results into the above expression for S, we obtain the desired result for S. In passing, we note that 〈H〉 ≡ A + TS = 32 NkT + U¯. For the second part of the question, we write U (q) =

i<j

u(rij ), so that

e−βU^ (q)^ =

i<j

e−βu(rij^ )^ =

i<j

(1 + fij ) ,

and follow Problems 3.23 and 1.4. The quantity V¯ then appears to be in the nature of a “free volume” for the molecules of the system.

3.14. a) The Lagrangian is given by

L = K − V =

iα

m r˙^2 iα −

i<j

u(rij ) −

iα

[uw (riα) + uw (L − riα)],

where i = 1, · · · , N denotes the particle number, α = x, y, z denotes the cartesian directions, and r^2 ij =

α (riα^ −^ rjα)

(^2). The canonical momenta are

piα =

∂L

∂ r˙iα

= m r˙iα.

The Hamiltonian is given by

H =

iα

piα r˙iα − L

iα

p^2 iα 2 m

i<j

u(rij ) +

iα

[uw (riα) + uw (L − riα)].

The canonical pressure can be written

P = −

∂H

∂V

3 L^2

∂H

∂L

3 L^2

iα

u′ w(L − riα) =

3 L^2

(Fx + Fy + Fz ).

This is clearly the instantaneous force per unit area on the right, back, and top walls.

b) The cartesian coordinates for the scaled position inside the box are siα = riα/L so the Lagrangian becomes

L =

iα

mL^2 s˙^2 iα −

i<j

u(Lsij ) −

iα

[uw (Lsiα) + uw (L − Lsiα)].

In this case the canonical momenta are

˜piα =

∂L

∂ s˙iα

= mL^2 s˙iα.

This leads to a Hamiltonian of the form

H =

iα

p ˜^2 iα 2 mL^2

i<j

u(Lsij ) +

iα

[uw (Lsiα) + uw (L − Lsiα)],

with canonical pressure is

P = +

3 L^2

iα

p ˜^2 iα mL^3

−

3 L^2

i<j

u′(Lsij )sij

3 L^2

iα

[u′ w(Lsiα)siα + u′ w(L − Lsiα)(1 − siα)].

Converting back to normal cartesian coordinates and momenta gives

P = +

3 V

iα

p^2 iα 2 m

−

3 V

i<j

u′(rij )rij

3 V

iα

[u′ w(riα)riα + u′ w(L − riα)(L − riα)].

3.18. We start with eqn. (3.6.2), viz.

∂U

∂β

r

E r^2 e−βEr ∑ r

e−βEr^

+ U 2 , (1)

and differentiate it with respect to β, keeping the Er fixed. We get

∂^2 U ∂β^2

= 〈E^3 〉 − 〈E^2 〉〈E〉 + 2U

∂U

∂β

Substituting for (∂U/∂β) from eqn. (1), we get

∂^2 U ∂β^2

= 〈E^3 〉 − 3 〈E^2 〉U + 2U 3 ,

which is precisely equal to 〈(E −U )^3 〉. As for ∂^2 U/∂β^2 , we note that, since ( ∂U ∂β

Er

= −kT 2

∂U

∂T

V

= −kT 2 CV, ( ∂^2 U ∂β^2

Er

= −kT 2

[

∂T

−kT 2 CV

]

V

= k^2 T 2

[

2 TC V + T 2

∂CV

∂T

V

]

Hence the desired result. For the ideal classical gas, U = 32 NkT and CV = 32 Nk , which readily yield the stated results.

3.19. Since G =

i

qipi, G˙ =

i

( ˙qipi + qi p˙i). Averaging over a time interval τ , we get

τ

∫^ t+τ

t

i

( ˙qipi + qi p˙i)dt =

τ

∫^ t+τ

t

G^ ˙dt = G(t^ +^ τ^ )^ −^ G(t) τ

For a finite V and finite E, the quantity G is bounded ; therefore, in the limit τ → ∞, the right-hand side of (1) vanishes. The left-hand side then gives (^) 〈 ∑

i

( ˙qipi + qi p˙i)

which leads to the desired result.

3.20. The virial of the noninteracting system, by eqn. (3.7.12), is − 3 PV. The contribution from interparticle interactions, by eqn. (3.7.15), is given by the “expectation value of the sum of the quantity −r(∂u/∂r) over all pairs of particles in the system”. If u(r) is a homogeneous function (of degree n) of the particle coordinates, this contribution will be −nU , where U is

the mean potential energy (not the internal energy) of the system. The total virial is then given by

V = − 3 PV − nU.

The relation K = − 12 V still holds, and the rest of the results follow straightforwardly.

3.21. All systems considered here are localized. The pressure term, therefore, drops out, and we are left with the result

K =

n 2

U =

n n + 2

E.

Example (a) pertains to n = 2, while examples (b) and (c) pertain to n = −1. In the former case, K = U = 12 E; in the latter, K = − 12 U = −E. The next problem pertains to n = 4.

3.22. Note that a force proportional to q^3 implies a potential energy proportional to q^4. Thus

H =

2 m

p^2 + cq^4 (c > 0).

It follows that

2 m

p^2

∫^ ∞

−∞

e−βp

(^2) / 2 m (p^2 / 2 m)dp ∞∫

−∞

e−βp^2 /^2 m^ dp

2 β

for the values of these integrals, see eqns. (13a) of Appendix B. Next,

〈cq^4 〉 =

∫^ ∞

−∞

e−βcq 4 (cq^4 )dq

∫^ ∞ −∞

e−βcq^4 dq

∂β

ln I(B),

where I(β) denotes the integral in the denominator. It is straightforward to see that I(β) is proportional to β−^1 /^4 , whence 〈cq^4 〉 = 1/ 4 β, which proves the desired result.

3.23. The key to this derivation is writing the partition function in terms of position integrals over scaled coordinates. Assume a cubic box of size L and volume V = L^3. The scaled position for particle i is si = ri/L. The

Calculation of the various thermodynamic quantities is now straightfor- ward. The results are found to be essentially the same as for a system of sN one-dimensional oscillators. However, since

Q( Ns) (β) = Q(1) Ns (β),

the chemical potential μs will turn out to be s times μ 1.

3.28. (a) When one of the oscillators is in the quantum state n, the energy left for the remaining (N − 1) oscillators is E −

n + (^12)

ℏω; the corre- sponding number of quanta to be distributed among these oscillators is R−n; see eqn. (3.8.24). The relevant number of microstates is then given by the expression (R − n + N − 2)!/(R − n)!(N − 2)!. Combined with expression (3.8.25), this gives

pn =

(R − n + N − 2)! (R − n)!(N − 2)!

÷

(R + N − 1)!

R!(N − 1)!

It follows that pn+ pn

R − n R − n + N − 2

R

R + N

n¯ ¯n + 1

By iteration, pn = p 0 {¯n/(¯n + 1)}n. Going back to eqn. (1), we note that

p 0 =

N − 1

R + N − 1

N

R + N

¯n + 1

which completes the desired calculation. (b) The probability in question is proportional to gN − 1 (E − ε), i.e. to (E − ε) (^32) (N −1)− 1

. For 1  N , this is essentially proportional to (1 − ε/E) (^32) N and, for ε  −E, to e−^3 N ε/^2 E^.

3.29. The partition function of the anharmonic oscillator is given by

Q 1 (β) =

h

∫^ ∞

−∞

−∞

e−βH^ dq dq

H =

p^2 2 m

• cq^2 − gq^3 − fq^4

The integration over p gives a factor of

2 πm/β. For integration over q, we write

e−βcq

2 eβ(gq

(^3) +fq (^4) ) = e−βcq

2

[

1 + β(gq^3 + fq^4 ) +

β^2 (gq^3 + fq^4 )^2 +...

]

the integration then gives √ π βc

• βf ·

π β^5 c^5

β^2 g^2 ·

π β^7 c^7

It follows that

Q 1 (β) = π βh

2 m c

[

3 f 4 βc^2

15 g^2 16 βc^3

]

so that ln Q 1 (β) = const. − ln β + 3 f 4 βc^2

15 g^2 16 βc^3

whence U (β) =

β

3 f 4 β^2 c^2

15 g^2 16 β^2 c^3

and C(β) = k + 3 f k^2 T 2 c^2

15 g^2 k^2 T 8 c^3

Next, the mean value of the displacement q is given by

〈q〉 =

−∞

−∞

exp(−βH)q dp dq /

−∞

−∞

exp(−βH)dpdq.

In the desired approximation, we get

〈q〉 ' βg

∫^ ∞

−∞

e−βcq

2 q^4 dq

−∞

e−βcq

2 dq

= βg ·

π β^5 c^5

π βc

3 g 4 βc^2

3.30. The single-oscillator partition function is now given by

Q 1 (β) =

∑^ ∞

n=

e−β(n+^

(^12) )ℏω+βx(n+ (^12) )^2 ℏω .

For x  1, we may write

Q 1 (β) =

∑^ ∞

n=

e−β(n+^

(^12) )ℏω

[

1 + βx

n +

ℏω +...

]

With u = βℏω, the sums involved are

S 1 (u) =

∑^ ∞

n=

e−u(n+^

[

2 sinh

u

)]− 1

, and

S 2 (u) =

∑^ ∞

n=

e−u(n+^

n +

d^2 du^2

S 1 (u) =

[

4 sinh

u

)]− 1 {

coth^2

u