Elements of Ensemble Theory - ikbooks.com

Chapter 2Elements of Ensemble Theory

In 1902 Gibbs, one of the founders of Statistical Mechanics, introduced the concept of ensemble(a French word meaning assembly of systems). An ensemble is a collection of essentially inde-pendent systems, which are macroscopically identical but microscopically different.

Three types of ensemble are important in statistical mechanics. The classification depends onthe matter in which the systems interact.

2.1 MICROCANONICAL ENSEMBLE

The microcanonical ensemble is a collection of independent systems, having the same numberof particles N, volume V and an energy between E and E + δE. So the individual systems ofa microcanonical ensemble are separated by rigid, impermeable and insulated walls, such that thevalues of E,V and N for a particular system are not affected by the presence of other systems. Thus,in microcanonical ensemble, neither energy nor matter is exchanged.

N,V, E N,V, E

N,V, E N,V, E

Fig. 2.1. Microcanonical ensemble.

57

60 • An Introduction to Equilibrium Statistical Mechanics

Now

E =1

2m

N

∑i=1

−→p 2i =

12m

N

∑i=1

3

∑α=1

p2iα

⇒ 2mE =N

∑i=1

3

∑α=1

p2iα (2.1.9)

where −→p 2i = p2

i1 + p2i2 + p2

i3, denoting x,y,z components by 1, 2, 3 respectively.Obviously, the sum contains 3N = f square terms ( f = degrees of freedom). For E = constant,

it describes a sphere of radius ς(E) = (2mE)12 in the f -dimensional space of the momentum coor-

dinates. Hence Ω(E) is proportional to the volume of phase space contained in the spherical shelllying between the sphere of radius ς(E) and that of slightly larger radius ς(E +δE).

We now calculate the volume of the N-dimensional sphere. From the dimensional analysis, thevolume of a N-dimensional sphere is

WN = CNςN (2.1.10)where CN is a constant and is given by

CN =π N

2

Γ(N

2 +1) (2.1.11)

(see Appendix II).Using (2.1.11) in (2.1.10), the volume of the N-dimensional sphere becomes

WN =π N

2

Γ(N

2 +1)ςN (2.1.12)

Hence, the volume of the 3N-dimensional sphere of radius√

2mE is

W3N (E) =π 3N

2

Γ( 3N

2 +1) (2mE)

3N2 (2.1.13)

∴ Ω(E) =V N

h3N0

∂WN (E)

∂EδE

⇒ Ω(E) =

(2mπh2

0

) 3N2 V N

Γ(3N

2)E

3N2 δE (2.1.14)

Neglecting 1 compared to N

Ω(E) = ΩEδE =

(2mπh2

0

) 3N2 V N

Γ( 3N

2)E

3N2 δE (2.1.15)

Since 3N = f , we see thatΩ(E) ∼ E f (2.1.16)

Elements of Ensemble Theory • 61

N is of the order of Avogadro’s number and so very large. Thus, Ω(E) is an extremely rapidlyincreasing function of the energy E of the system.

The entropy of the ideal gas is

S = k lnΩE

= Nk ln[

2mπEh2

0

] 32

+Nk lnV − k lnΓ(

3N2

)

(2.1.17)

Now Γ(

3N2

)

=

(3N2

−1)

! =3N2

!, since N 1.

Again by Stirling’s approximation

ln3N2

! =3N2

ln3N2

−3N2

= N ln(

3N2

) 32

−3N2

∴ lnΓ(

3N2

)

∼= N ln(

3N2

) 32

−3N2

(2.1.18)

Using (2.1.18) in (2.1.17)

S (E,V ) = Nk ln[

2mπEh2

0

] 32

+Nk lnV −Nk ln(

3N2

) 32

+32

Nk

⇒ S (E,V ) = Nk ln

[

V(

4mπ3h2

0

EN

) 32]

+32

Nk (2.1.19)

Solving for E in terms of S and V , we obtain the internal energy given by

E (S,V ) =

(3

4πh2

0m

)N

V 23

exp(

23

SNk

−1)

(2.1.20)

Again from thermodynamicsT dS = dE +PdV (2.1.21)

Then the temperature is given by

T =

(∂E∂S

)

V=

23

ENk

(2.1.22)

Hence, the specific heat at constant volume is given by

CV =

(∂E∂T

)

V=

32

Nk (2.1.23)

From (2.1.21), the pressure P is given by

P = −

(∂E∂V

)

S=

23

EV

(2.1.24)


Using (2.1.22) in (2.1.24), we finally get

P =NkT

V⇒ PV = NkT

This is the equation of state of the ideal gas.

PROBLEM 2.1: Consider an isolated system consisting of a large number N of weakly interactinglocalized particles of spin 1/2 in an external magnetic fieldH. Each particle has a magnetic momentµ, which can point either parallel or anti-parallel to the field H. Consider the energy range betweenE and E + δE where δE is very small compared to E but microscopically large so that δE µH.What is the total number of states Ω(E) lying in this energy range? Hence, find the temperature T .SOLUTION

Let n be the number of spins aligned parallel to −→H . So (N −n) is the number of spins alignedanti-parallel to −→H .∴ The total energy of the system is

E = −nµH +(N −n)µH

= NµH −2nµH

⇒ n =N2−

E2µH

(i)

Then, if the energy of the system is exactly E, the number of accessible microstates is

ΩE = NCn =N!

n!(N −n)!=

N!(

N2−

E2µH

)

!(

N2

+E

2µH

)

!(ii)

where we use equation (i).

From equation (i), δn =δE

2µH(considering magnitude only).

Hence, the total number of states in the range E to E +δE is

Ω(E) = ΩEδn =N!

(N2−

E2µH

)

!(

N2

+E

2µH

)

!·

δE2µH

(iii)

From equation (1.22.10) the temperature T is defined through the relation1

kT=

∂ lnΩ∂E

(iv)

Taking logarithms on both sides of equation (ii)

lnΩ = lnN!− ln(

N2−

E2µH

)

!− ln(

N2

+E

2µH

)

!. (v)

Using Stirling’s approximation

lnΩ = N lnN −N −

(N2−

E2µH

)

ln(

N2−

E2µH

)

+

(N2−

E2µH

)

−

(N2

+E

2µH

)

l1n(

N2

+E

2µH

)

+

(N2

+E

2µH

)


= N lnN −

(N2−

E2µH

)

ln(

N2−

E2µH

)

−

(N2

+E

2µH

)

ln(

N2

+E

2µH

)

∴

∂ lnΩ∂E

=1

2µHln

(N2−

E2µH

)

+1

2µH−

12µH

ln(

N2

+E

2µH

)

−1

2µH

=1

2µHln

N2−

E2µH

N2

+E

2µH

1kT

=1

2µHln

N2−

E2µH

N2

+E

2µH

⇒

N2−

E2µH

N2

+E

2µH

= e2µHkT

⇒N

−E

µH

=e

2µHkT +1

e2µHkT −1

⇒ −NµH

E=

eµHkT +e−

µHkT

eµHkT −e−

µHkT

= coth(

µHkT

)

∴ E = −NµH tanh(

µHkT

)

This is the relation between energy E and temperature T .

PROBLEM 2.2: Consider an extreme relativistic gas, characterized by the single particle energy

states ε(nx,ny,nz) =hc2l

(n2

x +n2y +n2

z) 1

2 . Calculate Ω and hence show that the ratio γ =Cp

Cv=

43

.

SOLUTIONThe single particle energy states are given by

ε(nx,ny,nz) =hc2l

(n2

x +n2y +n2

z) 1

2 (i)

∴ n2x +n2

y +n2z =

4ε2L2

h2c2 =4ε2V 2

3

h2c2 (ii)

Then the total energy of the system is given by

E = ∑r

εr =hc2L

N

∑r=1

(n2

rx +n2ry +n2

rz) 1

2

⇒ n21x +n2

1y +n21z +n2

2x +n22y +n2

2z + · · ·+n2Nx +n2

Ny +n2Nz =

4L2E2

h2c2 = R2 (say)

(iii)


This is the equation of a sphere of radius R in 3N-dimensional space; the surface of the spherecorresponds to a constant energy E. The number of microstates Ω(N,V,E) is equal to the numberof lattice points on the positive segment of this spherical surface. If the energy of the system isspecified in between E and E + δE, the number of microstates Ω(N,V,E → E +δE) within thisrange is equal to the number of lattice points lying within the positive part of the spherical shellformed between the surfaces of radii R and R+δR in the 3N-dimensional space.

If Φ(N,V,E) is the volume of the positive part of the 3N-dimensional sphere of radius R, then

Ω(N,V,E → E +δE) = Ω = Φ(N,V,E +δE)−Φ(N,V,E)

By Taylor’s expansion,

Φ(N,V,E +δE) = Φ(N,V,E)+∂Φ∂E

δE +12!

∂2Φ∂E2 (δE)2 + . . .

∼= Φ(N,V,E)+∂Φ∂E

δE

∴ Ω = Φ(N,V,E +δE)−Φ(N,V,E) ∼=∂Φ∂E

δE (iv)

Now from appendix II, the volume of the 3N-dimensional sphere of radius R is

V3N (R) =π 3N

2(

3N2

)

!R3N (v)

Hence,

Φ(N,V,E) =

(12

)3N π 3N2

3N2

!

(2LEhc

)3N=

(√πL

hc

)3NE3N

(3N2

)

!(vi)

∴

∂Φ∂E

=3N

(√πL

hc

)3NE3N−1

(3N2

)

!=

3NE

Φ (vii)

Thus,

Ω =∂Φ∂E

δE = 3NΦ(N,V,E)δEE

(viii)

Taking log on both sides

lnΩ = lnΦ+ ln3N + lnδEE

(ix)

But from (vi), we have

lnΦ = N ln[√

πLEhc

]

− ln(

3N2

)

! ∼= N ln[√

πLEhc

]

−3N2

ln(

3N2

)

+3N2

(x)


where we use Stirling’s approximation.

∴ lnΩ = N ln[√

πLEhc

]

−3N2

ln(

3N2

)

+3N2

+ ln3N + lnδEE

=32

N ln

[

23N

(√πLEhc

) 23]

+3N2

+ ln3N + lnδEE

For macroscopic systems, N 1 and so ln(3N) 3N. Again since δE E, ln(

δEE

)

is negligible

and so

lnΩ ∼=32

N ln

[

23N

(√πLEhc

)2/3]

+3N2

(xi)

Then the entropy of the system is

S = k lnΩ =32

Nk ln

[

23N

(√πLEhc

)2/3]

+3Nk

2(xii)

Solving for E,

E =hc√

πL

(3N2

) 32

e(s

Nk32) (xiii)

The temperature of the gas is given by1T

=

(∂S∂E

)

N,V=

NkE

⇒ E = NkT (xiv)

The specific heat at constant volume is

CV =

(∂E∂T

)

N,S= Nk (xv)

For the pressure of the system, we obtain

P = −

(∂E∂V

)

N,S=

E3V

(xvi)

The specific heat at constant pressure is

CP =

(∂E∂T

)

N,P+P

(∂V∂T

)

N,P=

(∂E∂T

)

N,P+

13

(∂E∂T

)

N,P=

43

(∂E∂T

)

N,P=

43

Nk (xvii)

∴ γ =CP

V=

43

(Proved) (xviii)

PROBLEM 2.3: The quantity Ω(N,V,E) is called the microcanonical partition function. Show thatlnΩ(N,V,E) = ln(Ω1 +Ω2 + · · ·) ∼= lnΩmax, where Ωmax is the largest Ωi in the series.


SOLUTIONLet us assume that there are as many N systems with Ωi’s comparable to Ωmax, that is, as many

such systems as there are particles in any one system. Thus,Ω ∼= N ×Ωmax ⇒ lnΩ ∼= lnΩmax + lnN (i)

But lnN N.∴ lnΩ(N,V,E) ∼= lnΩmax (Proved) (ii)

PROBLEM 2.4: A one-dimensional chain is made up of N identical elements of length l. The anglebetween successive elements can be either 0 or 180, but there is no difference in internal energybetween these two possibilities. For the sake of counting, one can think of each element as eitherpointing to the right (+) or to the left (−). Then one has

N = n+ +n−L = l (n+−n−) = l (2n+−N) .

(a) Use the microcanonical ensemble to find the entropy as a function of N and n+, S (N,n+).(b) Find an expression for the tension in the chain as a function of T , N and n+, ℑ(T,N,n+).(c) Rearrange (b) to give the length as a function of N, T and ℑ.

SOLUTION

(a) The number of ways of choosing n+ elements from a total of N isN!

(N −n+)!n+!. It follows

thatΩ(N,n+) =

N!(N −n+)!n+!

(i)

The entropy of the system isS (N,n+) = k lnΩ = k [lnN!− ln(N −n+)!− lnn+!]

Using Stirling’s approximation, the above equation becomesS (N,n+) ∼= k [N lnN −N − (N −n+) ln(N −n+)−n+ lnn+ +n+]

= k [N lnN − (N −n+) ln(N −n+)−n+ lnn+] (ii)(b)

ℑT

= −

(∂S∂L

)

N,E= −

∂S∂n+

∂n+

∂L= −

∂S∂n+

×12l

= −k2l

[N −n+

N −n++ ln(N −n+)−

n+

n+− lnn+

]

= −k2l

ln(

N −n+

n+

)

∴ (N,T,n+) = −kT2l

ln(

N −n+

n+

)

(iii)


(c) From (iii), we haven+ = N

1

1+ exp(

−2lℑkT

) (iv)

We now use this expression for n+ to find L:

L= l (2n+−N)=Nl

2

1+ exp(

−2/ℑkT

) −1

=Nl

2

1+ exp(

−2/ℑkT

) −

1+ exp(

−2/ℑkT

)

1+ exp(

−2/ℑkT

)

= Nl1− exp

(

−2/ℑkT

)

1+ exp(

−2/ℑkT

) = Nlexp

(

−1/ℑkT

)

− exp(

−1/ℑkT

)

exp(

−1/ℑkT

)

+ exp(

−1/ℑkT

)

∴ L = Nl tanh(

lℑkT

)

(v)

For high temperatures, where kT lℑ, tanhx −→ x for x, so

L =Nl2

kTℑ (vi)

The fact that the length is proportional to the tension ℑ shows that Hooke’s law applies to thissystem, at least for high temperatures.

2.1.3 A two-level systemConsider a system of N (non-interacting and distinguishable) particles. Let the only accessiblestates for the system be the ground state of zero energy and an excited state of energy ε.

ε

Fig. 2.2. Two non-degenerate energy levels separated by energy.

Let the number of particles in the ground state of zero energy be N0 and that in the excited statewith energy ε be Nε.∴ N0 +Nε = N (2.1.25)The total energy of the system is

E = (N0 ×0)+(Nε × ε) ⇒ Nε =Eε

(2.1.26)


If the particles are distinguishable, the number of microstates accessible to the system is equal tothe number of ways choosing Nε particles from N particles and is given by

Ω = NCNε =N!

Nε!(N −Nε)!=

N!Eε

!(

N −Eε

)

!(2.1.27)

Assuming that N, Nε and (N −Nε) are all large numbers compared to 1, we use Stirling’s approxi-mation for the factorials of large numbers and then the entropy is

S = k lnΩ = k [N lnN −Nε lnNε − (N −Nε) ln(N −Nε)]

= k[(

Eε−N

)

ln(

1−ENε

)

−Eε

lnENε

]

(2.1.28)

The temperature of the system is given by1T

=

(∂S∂E

)

N,V=

kε

ln(

NεE

−1)

(2.1.29)

from which the energy E of the system is given by

E =Nε

1+ e εkT

(2.1.30)

∴ Nε =Eε

=N

1+ e εkT

(2.1.31)

Now for T → 0, i.e., εkT 1,Nε → 0, that is, all particles are frozen in the ground state. As T

increases, both E and Nε increase and when T → ∞, i.e., εkT 1,Nε; N

2 , that is, half the particles arein the ground state and half in the excited state and the energy of the system attains its maximumvalue of E ∼= Nε

2 . The specific heat at constant volume is given by

CV =

(∂E∂T

)

V= Nk

( εkT

)2 e εkT

(1+ e εkT )2

(2.1.32)

kTε

CV

0.42

Fig. 2.3. The specific heat at constant volume of a two-state system, showing Schottky hump.


The specific heat CV is zero at both very low and very high temperatures and is maximum at T ∼=0.42 ε

k . The hump, observed in the CV vs T plot, is called Schottky anomaly. Such a peak, whenobserved experimentally, is therefore an indication of a gap in its energy states.

2.2 CANONICAL ENSEMBLE

This ensemble is a collection of independent systems having same temperature T , volume V andnumber N of particles. So the systems of this ensemble are separated by rigid, impermeable, butconducting walls. Since the systems are separated by conducting walls, heat can be exchangedbetween the systems till they reach a common temperature T . Thus, in canonical ensemble, thesystems can exchange energy but not particles.

N,V,T N,V,T

N,V,T N,V,T

Fig. 2.4. Canonical ensemble.

2.2.1 Probability distribution and canonical partition functionConsider a small system A in thermal interaction with a heat reservoir A′.

What is the probability Pr of finding the system A in any one particular microstate r of energyEr?

A A′

Fig. 2.5. A small system ‘A’ in thermal interaction with a heat reservoir ‘A′’.

We assume weak interaction between A and A′, so that their energies are additive. The energyof A is, of course, not fixed. Only the total energy of the combined system, A0 = A + A′ (which isisolated) has a constant value in some range between E0 and E0 +δE:

Er +E ′ = E0 = constant (2.2.1)


where E ′ denotes the energy of the reservoir A′. Thus, when A has an energy Er, the reservoir A′

must have an energy E ′ = E0 −Er.Let Ω′ (E ′) denote the number of accessible microstates of A′, when it has an energy in the

range near E ′. If the system A is in the particular state r, the number of states accessible to thecombined system A0 is 1×Ω′ (E ′) = Ω′

(E0 −Er

).

Then the probability of finding the system A in this state is

Pr = Pr (Er) =Ω′ (E ′)

Ω0total

=Ω′

(E0−Er

)

Ω0total

,

where Ω0total denotes the total number of states accessible to the combined system A0.

But Ω0total = constant = c0 (say).

∴ Pr = c0Ω′(E0 −Er

)(2.2.2)

⇒ lnPr = lnc0 + lnΩ′ (E0 −Er) (2.2.3)

Since A is very small compared to A′,Er E0. Expanding lnΩ′ about E ′ = E0,

lnΩ′(E0 −Er

)= lnΩ′

(E0)−

[∂ lnΩ′

∂E ′

]0

EEr − . . .

Since A′ acts as a reservoir, Er E0 and so we can neglect higher order terms.

∴ lnΩ′(E0 −Er

)∼= lnΩ′

(E0)−

[∂ lnΩ′

∂E ′

]

E0Er (2.2.4)

The derivative[

∂ lnΩ′

∂E ′

]

E0is evaluated at the fixed energy E ′ = E0 and is thus a constant, indepen-

dent of the energy E ′ and[

∂ lnΩ′

∂E ′

]

E0= β =

1kT

, where T is the temperature of the reservoir.The equation (2.2.4) then becomes

lnΩ′(E0 −Er

)= lnΩ′

(E0)−βEr

⇒ Ω′(E0 −Er

)= Ω′

(E0)e−βEr

Since Ω′(E0) is just a constant, independent of r, equation (2.2.2) becomes

Pr = ce−βEr (2.2.5)

where c is a constant, independent of r.The probability (2.2.5) is a very general result and is of fundamental importance in statisti-

cal mechanics. The exponential factor e−βEr is called the ‘Boltzmann factor’; the correspondingprobability distribution (2.2.5) is known as the ‘canonical distribution’.

Now according to the normalization conditions for probabilities,

∑r

Pr = 1 ⇒ c−1 = ∑r

e−βEr


∴ Pr =e−βEr

∑r

e−βEr=

e−βEr

Z(2.2.6)

whereZ = ∑

re−βEr (2.2.6a)

‘Z’ is called the canonical partition function. (The letter Z is used because the German name isZustandsumme.) It is the sum over all accessible microstates of the system A.

It should be noted that the expression (2.2.6a) is correct, if the energy levels are discrete andnon-degenerate. In the general case, an energy level Er consists of a group of states gr in number allof the same energy Er and of equal probability. Thus in summing over all accessible microstates, itis necessary to repeat equal terms (involving Er) gr times. Instead, it is simpler to use a modifiedform of the partition function

Z = ∑r

gre−βEr (2.2.6b)

where the label r goes over all energy levels. If the level is non-degenerate, then gr = 1. Equation(2.2.6b) is the general form of the partition function.

Accordingly,

Pr =gre−βEr

∑r

gre−βEr(2.2.6c)

Once the probability distribution is known, various mean values can be computed. For example, letx be any quantity assuming the value xr in state r of the system A. Then the mean value of x is

x =∑r

xrPr

∑r

Pr= ∑

rxrPr =

∑r

xre−βEr

∑r

e−βEr(2.2.7)

2.2.2 Energy fluctuations in the canonical ensemble: correspondence withthe microcanonical ensemble

According to equation (2.2.7), the mean energy value is

E =∑r

Ere−βEr

∑r

e−βEr(2.2.8)

Now the canonical partition function is

Z = ∑r

e−βEr ⇒−∂Z∂β

= ∑r

Ere−βEr (2.2.9)

Using (2.2.9) in (2.2.8)

E = −1Z

∂Z∂β

= −∂ lnZ

∂β. (2.2.10)


2.2.3 To calculate the canonical partition function

1. Classical ideal gas

We consider a classical monatomic gas at an absolute temperature T , contained in a volume V . Letus denote the position vector of the ith molecule by −→r i, its momentum by −→p i. Then the total energyof the gas is given by

E =N

∑i=1

p2i

2m+U (−→r1 ,−→r2 . . .−→rN) (2.2.17)

The first term on the right-hand side of the above equation represents the total kinetic energy ofall the molecules and the second term represents the potential energy of interaction between themolecules.

The classical canonical partition is given by

Z =

∫

exp[

−β

12m

(p2

1 + · · ·+ p2N)+U (−→r 1, . . . ,

−→r N)

]d3r1 . . .d3rNd3 p1 . . .d3 pN

h3N0

=1

h3N0

∫

e−βp2

12m d3 p1 . . .

∫

e−βp2

N2m d3 pN

∫

e−βU(−→r 1...−→r N)d3r1 . . .d3rN

Since the kinetic energy is a sum of terms, one for each molecule, the corresponding part of the par-tition function breaks into a product of N integrals, each identical except for the irrelevant variableof integration and equal to

∞∫

−∞e−

βp22m d3 p.

Since U (−→r 1, . . . ,−→r N) is not in the form of a simple sum of terms for individual molecules, the

integration over the coordinates −→r 1, . . . ,−→r N is very difficult to carry out. This is why the treatment

of non-ideal gases is complicated (Chapter 7).For an ideal gas, we can set U = 0 and so the integral becomes trivial:

∫

d3r1 . . .d3rN = V N (2.2.18)

since each integration extends over the volume V of the container.

∴ Z = ξN (2.2.19)

where

ξ =Vh3

0

∫ ∞

−∞e−

βp22m d3 p (2.2.20)

is the partition function for a single particle.Equation (2.2.19) is the relation between the single particle canonical partition function and that

for the whole system, if the particles of the system are identical, non-interacting and distinguishable.


Now∞∫

−∞

e−βp22m d3 p =

∞∫

−∞

e−β

2m

(

p2x + p2

y + p2z)

d pxd pyd pz

=

∞∫

−∞

e−βp2x2m d px

∞∫

−∞

e−βp2y2m d py

∞∫

−∞

e−βp2z2m d pz =

[√

2πmβ

]3

,

where we use the standard integral∞∫

−∞

e−αx2dx =

√πα

∴ ξ =Vh3

0

(2πm

β

) 32

=Vh3

0(2πmkT )

32 (2.2.21)

Using (2.2.21) in (2.2.19)

Z (N,V,T ) =V N

h3N0

(2πmkT )3N2 (2.2.22)

This is the canonical partition function of a classical ideal gas.What is going to happen, if we would use the same formula, i.e., equation (2.2.19) to calculate

the partition function for a system of N indistinguishable, identical, non-interacting particles. Toillustrate that this is wrong, we consider two indistinguishable particles:

Z (N = 2,V,T ) = ξ2 = ∑r1

e−βεr1 ×∑r2

e−βεr2

= ∑r=r1=r2

e−2βεr

︸︷︷︸particles both

in the same state

+ ∑r1 =r2

∑e−β(εr1+εr2)

︸︷︷︸particles in different

states

However, the particles in the second term are counted twice, but since the particles are indistin-guishable this does not generate a new state, because

e−β(εr1+εr2) = e−β(εr2+εr1).

Hence, Z (N = 2,V,T ) = ξ2 counts the particles in different states falsely twice!We can correct for that by

Z (N = 2,V,T ) = ∑r=r1=r2

e−2βεr +12 ∑

r1 =r2

∑e−β(εr1+εr2)

= ∑r=r1=r2

e−2βεr +12! ∑

r1 =r2

∑e−β(εr1+εr2).


Similarly, for a system of 3 particles

Z (N = 3,V,T ) = ∑r=r1=r2=r3

particles in same state

e−3βεr + · · ·some particlesin same state

+16 ∑

r1 =r2 =r3all particles indifferent states

e−β(εr1+εr2

+εr3)

= ∑r=r1=r2=r3

particles in same state

e−3βεr + · · ·some particlesin same state

+13! ∑

r1 =r2 =r3all particles indifferent states

e−β(εr1+εr2

+εr3)

In general, for a system of N indistinguishable particles,

Z (N,V,T ) = ∑r=r1=r2=r3=···=rN

all particles insame state

e−Nβεr + · · ·some particlesin same state

+1

N! ∑r1 =r2 =r3

e−β(εr1

+εr2+εr3+ ···+εrN )

all particles indifferent states

(2.2.23)

We now use semi-classical argument according to which1. gas particles are identical2. only very weak or no interaction between the particles exists3. there are much more available energy levels than particles (classical continuum of quantum

mechanics).Thus, the probability that any single particle state is occupied by more than one particle is verysmall. This argument greatly simplifies (2.2.23) because only the last term counts.

Thus, for a system of N indistinguishable, identical, non-interacting particles

Z (N,V,T ) =ξN

N!(2.2.23a)

2. Classical non-ideal gasFor a non-ideal monatomic gas, the interaction term U = 0. Therefore, the canonical partitionfunction for a classical non-ideal gas is given by

Z (N,V,T ) =∫

exp[

−β

12m

(p2

1 + · · ·+ p2N)+U (−→r 1, . . . ,

−→r N)

]d3r1 . . .d3rNd3 p1 . . .d3 pN

h3N0

=1

h3N0

∫

e−βp2

12m d3 p1 . . .

∫

e−βp2

N2m d3 pN

∫

e−βU(−→r 1...−→r N)d3r1 . . .d3rN

∴ Z (N,V,T ) =1

h3N0

(2πm

β

) 3N2

Q (2.2.24)

whereQ =

∫

e−βU(−→r 1...−→r N)d3r1 . . .d3rN

known as configurational partition function (chapter 7). The calculation of Q is quite difficult. Forthe sake of simplicity, we apply mean field approximation. Then equation (2.2.24) becomes

Z (N,V,T ) =1

h3N0

(2πm

β

) 3N2

[∫ V

0e−UMβd3r

]N


where UM is the mean field. We take it as follows:

UM =

∞,r < r0

−U ,r ≥ r0

UM

r0r

U

0

Fig. 2.6. UM vs. r plot.

Therefore,

Z (N,V,T ) =1

h3N0

(2πm

β

) 3N2

V0∫

0

e−UMβd3r +

V∫

V0

e−UMβd3r

N

(2.2.25)

where V0 is the volume corresponding to r = r0. Now in the range 0 to V0,UM → ∞ and in the rangeV0 to V , UM = −U . Therefore, equation (2.2.25) becomes

Z (N,V,T ) =1

h3N0

(2πm

β

) 3N2

[∫ V

V0eUβd3r

]N

⇒ Z (N,V,T ) =1

h3N0

(2πmkT )3N2

[

(V −V0)eUkT

]N(2.2.26)

This is the canonical partition function for a classical non-ideal gas in the mean field approximation.This topic will be further discussed in Chapter 7.

2.2.4 Thermodynamic quantities in terms of partition functionAll the physical quantities can be expressed completely in terms of the partition function Z. Someof them are shown below.


1. EntropyAccording to Boltzmann’s entropy relation, the entropy S is related to the thermodynamic proba-bility Ω through the relation

S = k lnΩ (2.2.27)For a classical system, the total number of molecules is

N = ∑i

ni (2.2.28)

andΩ = N!∏

i

gnii

ni!(2.2.29)

Taking logarithm and then using Stirling’s approximationlnΩ = N lnN −N +∑(ni lngi −ni lnni +ni) (2.2.30)

According to classical statistics, for equilibrium state [see equation (3.1.7)]ni = gie−αe−βEi = giAe−βEi (2.2.31)

where A is a new constant. Using (2.2.31) in (2.2.30)

lnΩ = N lnN −N +∑i

(

ni lngi −ni lngiAe−βEi +ni

)

= N lnN −N +∑i

ni lngi −∑i

ni lngi − lnA∑i

ni +β∑i

niEi +∑i

ni

= N lnN −N −N lnA+βE +N= N lnN −N lnA+βE

∴ lnΩ = N lnNA

+βE (2.2.32)

where we put ∑i

ni = N = total number of particles and ∑i

niEi = E = total energy of the system.

Again using (2.2.28) in (2.2.31)

∑i

ni = A∑i

gie−βEi ⇒ N = AZ ⇒ Z =NA

(2.2.33)

where Z = ∑i

gie−βEi −→ partition function.

After using (2.2.33), equation (2.2.32) becomeslnΩ = N lnZ +βE (2.2.34)

Finally, using (2.2.34) in (2.2.27), the expression for entropy becomesS = Nk lnZ + kβE

= Nk lnZ + k(

1kT

)

E

∴ S = Nk lnZ +ET

(2.2.35)


Taking logarithms on both sides

lnZ = N lnV +3N2

[

ln(

2πmkh2

0

)

+ lnT]

.

Differentiating partially with respect to V∂ lnZ∂V

=NV

(2.2.49)

Using (2.2.49) in (2.2.48)pV = NkT (2.2.50)

This is equation of state for an ideal gas.

2. Reproduction of van der Waals’ equation of stateFrom equation (2.2.26), the canonical partition function for a non-ideal gas in the mean field ap-proximation is

Z (N,V,T ) =1

h3N0

(2πmkT )3N/2[

(V −V0)eUkT

]N


lnZ =3N2

ln(

2πmkTh2

0

)

+N ln(V −V0)+NUkT

(2.2.51)

Now, V0 ∝ N ⇒V0 = bN, where b is a constant and U ∝NV

⇒U =aNV

, where a is also a constant.Then equation (2.2.51) becomes

lnZ =3N2

ln(

2πmkTh2

0

)

+N ln(V −bN)+aN2

kTVDifferentiating partially with respect to V

∂ lnZ∂V

= −akT(

NV

)2+

NV −bN

(2.2.52)

Using (2.2.52) in (2.2.48)

p = −a(

NV

)2+

kT NV −bN

⇒ p = −a

(VN)2 +

kTVN −b

⇒

[

p+a

(VN)2

][VN−b

]

= kT

LetVN

= v, known as specific volume.

∴

(

p+av2

)

(v−b) = kT (2.2.53)

This is the van der Waals equation of state for a real gas.


2.2.6 Generalized equipartition theoremLet qi be the generalized co-ordinate. We want to calculate the ensemble average of the quantity

qi

(∂H∂q j

)

, where H is the Hamiltonian of the system under consideration. Using dτ ≡ d3Nqd3N p,

we can write in the canonical ensemble⟨

qi∂H∂q j

⟩

=

∫(

qi∂H∂q j

)

e−βHdτ∫

e−βHdτ(2.2.54)

Let us now consider the integral in the numerator. Integrating over q j by parts, it becomes∫ [

−1β

qie−βH (q j)2(q j)1

+1β

∫ (∂qi

∂q j

)

e−βHdq j

]

dτ( j) (2.2.55)

where (q j)1 and (q j)2 are the extreme values of the coordinate q j. If q j is a space coordinate, thenits extreme values will correspond to positions of the walls of the container and so the potentialenegy would be infinite The infinite potential energy immediately implies that the Hamiltonian H

of the system would be infinite. Therefore, the integrated part, i.e.,[

−1β

qie−βH (q j)2(q j)1

]

vanishes.

The integral part contains the term (∂qi/∂q j), which is equal to δi j. Finally, (2.2.55) becomes1β

δi j

∫

e−βHdτ (2.2.55a)

Substituting this in (2.2.54)⟨

qi∂H∂q j

⟩

= δi jkT (2.2.56)

independent of H.Similarly

⟨

pi∂H∂p j

⟩

= δi jkT (2.2.57)

which is independent of the form of the Hamiltonian H.Equations (2.2.55) and (2.2.56) are known as generalized equipartition theorem.This topic will be further discussed in Chapter 3.

2.2.7 Virial theoremAccording to the Hamilton’s equations of motion,

pi = −∂H∂qi

(2.2.58)

Using this in (2.2.57)〈qi pi〉 = −kT (2.2.59)

Adding over all i from i = 1 to 3N, we obtain⟨

3N

∑i=1

qi pi

⟩

= −3NkT (2.2.60)


This is known as the virial theorem, because ∑i

qi pi is known as virial in classical mechanics.

PROBLEM 2.5: Find the classical canonical partition function of a one-dimensional linear har-monic oscillator. Also find its average energy 〈E〉.

SOLUTIONSince this is a one-dimensional case, the classical canonical partition function is

Z =

∫

e−βE d p dxh0

.

where E is the total energy of the oscillator and is given by E =p2

2m+

12

mω2x2.

∴ Z =1h0

∫ ∞

−∞e−

βp2

2m d p∫ ∞

−∞e−

12

mω2x2dx

.

Using the standard integral∫ ∞

−∞e−αx2

dx =

√πα

,

Z =1h0

[√

2πmβ

][√

2πmβω2

]

=2π

h0βω=

2πkTh0ω

.

This is the canonical partition function.The average energy of the oscillator is

〈E〉 = −∂ lnZ

∂β=

1β

= kT.

PROBLEM 2.6: Find the canonical partition function of a relativistic classical ideal gas withenergy-momentum relationship E = pc.

SOLUTIONFor a classical ideal gas, the canonical partition function is

Z =∫

e−βE d3N pd3Nxh3N

0= ξN ,

where ξ is the single particle partition function and is given by

ξ =∫

e−βpc d3 pd3xh3

0=

Vh3

0

∫ ∞

0d pe−βpc p2

∫ π

0dθsinθ

∫ 2π

0dφ =

4πVh3

0

∫ ∞

0d pe−βpc p2

=4πVh3

0

1(βc)3

∫ ∞

0d (βpc)e−βpc (βpc)2 =

4πVh3

0

(kTc

)3Γ(3)


Also ‘Z’ can be written asZ = ∑

n

NCne−βnw (iii)

Now the mean value of n is

n =∑n

nNCne−βnw

∑n

NCne−βnw =∑n

nNCne−βnw

Z= −

1w

∂ lnZ∂β

= −Nw

∂∂β

[

ln(

1+ e−βw)]

= Ne−βw

1+ e−βw

⇒ n+ne−βw = Ne−βw ⇒n

N −n= e

−wkT

When n N,n ≈ Ne

−wkT (iv)

PROBLEM 2.10: Show that the partition function of an extreme relativistic gas consisting of 3Nindistinguishable particles, with energy-momentum relationship ε = pc, c being the speed of lightand moving in one dimension is given by Z (N,L,T ) = 1

(3N)![2L

( kThc

)]3N , where L being the ‘length’of the space available.

SOLUTIONThe single particle canonical partition function is

ξ =1h

∫

e−βεdqd p =Lh

∫ ∞

0e−βpcd p = L

(kThc

)

(i)

Therefore, the canonical partition function of the system of N non-interacting indistinguishableparticles is

Z (N,L,T ) =ξ3N

(3N)!=

1(3N)!

[

L(

kThc

)]3N(ii)

PROBLEM 2.11: Consider a system in thermal equilibrium at temperature T− its two states withenergy difference 4.8×10−14 erg occur with relative probability e2 erg / deg. Calculate the temper-ature. Boltzmann’s constant k = 1.38×10−16 erg / deg.

SOLUTIONLet P1 and P2 be respectively the probabilities of the state of energies E1 and E2. Then

P1 = ce−E1kT and P2 = ce−

E2kT , where c is a constant of proportionality.

∴

P1

P2= e

(E2−E1)kT ⇒ T =

(E2 −E1)

k ln(

P1

P2

) =4.8×10−14

1.38×10−16 × lne2 = 173.913 K


PROBLEM 2.12: The first vibrational energy of a diatomic molecule is 600 cm−1 above the groundstate. Calculate the relative population of molecules in these two levels at T = 400 K.

SOLUTIONLet N0 and Eo be respectively the number of molecules and the energy of the ground state and

N1 and E1 be the corresponding values for the first excited state.

∴ N1 = N0e−(E1−E0)

kT (i)

Now

(E1 −E0) = hv =hλc

= hcv = 6.6×10−27 ×3×1010 ×600 = 1.2×10−13erg

∴

N1

N0= e−

(E1−E0)kT = e−

(1.2×10−13

1.38×10−16×400

)

∼= 0.1 ⇒ N1 = 0.1×N0

PROBLEM 2.13: Consider a system which can take only three different energy states E1 = 0,E2 = 1.38×10−14 ergs,E3 = 2.76×10−14 ergs. Find the probability that at temperature 100 K thesystem may be (a) in one of the microstates of energy E3, (b) in the ground state E1.

SOLUTIONLet P1, P2 and P3 be respectively the probabilities with which the microstates can occur in three

energy values E1, E2 and E3.Then P1 = ce−

E1kT = c

P2 = ce−E2kT = ce−

1.38×10−141.38×10−16×100 = ce−1

andP3 = ce−

E3kT = ce−

2.76×10−141.38×10−16×100 = ce−2

where c is a constant of proportionality.Again these three states can occur in 2, 5 and 4 different ways respectively. Then P1 = 2c,

P2 = 5ce−1 and P3 = 4ce−2. But3

∑i=1

Pi = 1 ⇒ (P1 +P2 +P3) = 1 ⇒(2c+5ce−1 +4ce−2) = 1

⇒ c =1

2+1.84+0.54⇒ c =

14.38

Since e = 2.72.(a) The probability for the system to be in one of the microstates of energy E3 is

P3 = 4ce−2 =4

4.38× (2.72)2 = 0.12.


Again

E = −∂ lnZ

∂β(vi)

Comparing (v) and (vi)

Fβ = − lnZ⇒ F = −kT lnZ

(vii)

Finally, using (vii) in (iii)

∑r

Pr lnPr = −EkT

+FkT

=1

kT(F −E) =

1kT

(−TS) = −Sk

⇒ S = −k∑r

Pr lnPr [Proved] (viii)

In case of microcanonical ensemble, we have a group of Ω states (which are equally likely tooccur) for each member. The value of Pr is then 1

Ω for each of these states and 0 for all others.Consequently,

S = −kΩ

∑r=1

1Ω

ln(

1Ω

)

= k lnΩ [Proved] (ix)

PROBLEM 2.17: The states of a system are (i) a group of g1 equally likely states with a commonenergy ε1, and (ii) a group of g2 equally likely states with a common energy ε2 = ε1.

(a) Show that the entropy of the system is given by S = −k[

P1 ln(

P1

g1

)

+P2 ln(

P2

g2

)]

(b) Also show that S = k[

lng1 + ln

1+

(g2

g1

)

e−x

+x

1+(g1/g2)ex

]

, where x =ε2 − ε1

kT,

assumed positive.(c) Check that at T → 0,S → k lng1.

SOLUTION(a) The canonical partition function is

Z = ∑r

gre−βEr (i)

and

Pr =gre−βEr

Z(ii)

∴

Pr

gr=

e−βEr

Z⇒ ln

(Pr

gr

)

= − lnZ −βEr

⇒ ∑r

Pr ln(

Pr

gr

)

= − lnZ −βE (iii)


The Helmholtz free energy isF = E −T S

⇒ dF = dE −T dS−SdT = dE − (dE +PdV )−SdT = −PdV −SdT

∴

S = −(

∂F∂T

)

V

P = −(

∂F∂V

)

T

(iv)

Hence,

E = F +T S = F −T(

∂F∂T

)

V= F +β

(∂F∂β

)

V=

[∂

∂β(Fβ)

]

V(v)

Again

E = −∂ lnZ

∂β(vi)

Comparing (v) and (vi)F = −kT lnZ (vii)

Using (vii) in (iii), we obtain

∑r

Pr ln(

Pr

gr

)

=F −E

kT= −

Sk⇒ S = −k∑

rPr ln

(Pr

gr

)

∴ S = −k[

P1 ln(

P1

g1

)

+P2 ln(

P2

g2

)]

(Proved) (viii)

(b) We know that

P1 =g1e−βε1

Z

P2 =g2e−βε2

Z

(ix)

Using (ix) in (viii)

S = −k [P1(−βε1 − lnZ)+P2(−βε2 − lnZ)]

= k [β(P1ε1 +P2ε2)+(P1 +P2) lnZ]

= k [β(P1ε1 +P2ε2)+ lnZ] (∵ P1 +P2 = 1)

= k

[

β

(

g1ε1e−βε1 +g2ε2e−βε2

Z

)

+ lnZ

]

= k

[

β

(


g1e−βε1 +g2e−βε2

)

+ ln(g1e−βε1 +g2e−βε2)

]

= k

[

β

(


g1e−βε1 +g2e−βε2

)

+ ln(g1e−βε1)+ ln

1+g2

g1e−β(ε2−ε1)

]


If the temperature T is so high that the thermal energy kT is large compared to the separation hνbetween energy levels, i.e., βhν = hν

kT 1,

〈E〉 ≈ hν

(

12

+1hνkT

)

= hν(

12

+kThν

)

≈ hν×kThν

= kT −→ classical result.

PROBLEM 2.19: A classical point particle is moving in a 3-dimensional harmonic oscillator po-tential well, V (r) = 1/2Kr2 = 1/2K(x2 + y2 + z2) at absolute temperature T . Obtain a formula forthe probability that the particle is in between r and r +dr from the centre of attraction. Also obtaina formula for the mean square distance of the particle from the centre of attraction and check yourresult by comparison with the equipartition principle.

SOLUTIONThe probability that the particle will be in phase space volume element d3rd3 p

=e−

βp22m −β 1

2 Kr2d3 pd3r∫

e−βp22m −β 1

2 Kr2d3 pd3r=

e−βp22m −β 1

2 Kr2d3 pr2 sinθ dθ dφ dr∫

e−βp22m −β 1

2 Kr2d3 pr2 sinθ dθ dφ drTherefore, the probability that the particle will lie in the range r and r +dr is

P(r)dr =

∫

θ

∫

φ

∫

pe−

βKr22 r2 sinθ dθ dφ dr e−

βp22m d3 p

∫

θ

∫

φ

∫

p

∫

re−

βKr22 r2 sinθ dθ dφ dr e−

βp22m d3 p

=e−

βKr22 r2dr

∞∫

0e−

βKr22 r2dr

=e−

βKr22 r2dr

14√

π(

2Kβ

) 32

∴ P(r)dr =4√

π

(Kβ2

) 32

e−12 βKr2

r2dr (i)

The mean square distance is

〈r2〉 =

∫r2P(r)dr

∫P(r)dr

=

∞∫

0r4e−

βKr22 dr

∞∫

0r2e−

βKr22 dr

=

38√

π(

2βK

) 52

14√

π(

2βK

) 32

=32

(2

βK

)

=3kTK

(ii)

Therefore, the average potential energy = 12 K〈r2〉 = 1

2 K 3kTK = 3

2 kT , which is consistent with theequipartition law.

PROBLEM 2.20: Suppose a system of N non-interacting particles at temperature T in an externalmagnetic field −→H pointing along the z-direction. Assume that each atom has spin 1/2 and anintrinsic magnetic moment −→µ .


1. If the magnetic moment of each atom can point either parallel or anti-parallel to the externalfield −→H , what is the mean magnetic moment 〈µH〉 (in the direction of −→H , i.e., z-direction) ofsuch atom ?

2. Calculate the partition function of one atom, the mean magnetic moment along the fielddirection and hence the susceptibility of the system, first by treating the system classicallyand then quantum mechanically.

SOLUTIONEach atom can be in two possible states: the state (+) where its spin points up (i.e., parallel to

−→H ) and the state (–) where its spin points down (i.e. anti-parallel to −→H ). In the (+) state, −→µ isparallel to −→H and so µH = µ. The corresponding magnetic energy of the atom is E+ = −µH. Theprobability of finding the atom in this state is

P+ = Ce−βE+ = CeβµH (i)where C is a constant of proportionality and β = 1

kT . Similarly, in the (–) state, −→µ is anti-parallel to−→H and so µH =−µ. The corresponding magnetic energy of the atom is E− = +µH. The probabilityof finding the atom in this state is

P− = Ce−βE− = Ce−βµH (ii)∴ The mean value of µH is

〈µH〉 =P+ (µ)+P− (−µ)

P+ +P−= µ

eβµH − e−βµH

eβµH + e−βµH = µtanh(βµH) = µtanh(

µHkT

)

.

When the system is treated classicallyThe energy of a magnetic dipole in the presence of the external magnetic field is

E = −−→µ ·−→H = −µH cosθ (iii)

The canonical partition function of the system isZ = ξN (iv)

where ξ is the single particle canonical partition function and is given by

ξ = ∑θ

exp(βµH cosθ) =∫ 2π

φ=0

∫ π

θ=0eβµH cosθ sinθ dθ dφ = 4π

sinh(βµH)

βµH(v)

∴ Z =

[

4πsinh(βµH)

βµH

]N(vi)

The average magnetic moment along the field direction is given by

µH = 〈µcosθ〉 =

∑θ

µcosθexp(βµH cosθ)

∑θ

exp(βµH cosθ)=

1β

∂∂H

lnξ = µ[

coth(βµH)−1

βµH

]

= µL(βµH) (vii)


where L(x) is the so-called Langevin function:

L(x) = cothx−1x

(viii)

Therefore, the magnetization of the system is given by

M = NµH = NµL(x) (ix)

For magnetic fields so strong (or temperatures so low) that the parameter x 1, L(x) ∼= 1 and soM ∼= Nµ−→ saturation. On the other hand, for high temperatures (or the magnetic fields so weak)that the parameter x 1,L(x) ∼=

x3

and so

M ∼=Nµ2

3kTH (x)

Therefore, the high temperature susceptibility of the system is given by

χ(T ) = LimitH→0

(∂M∂H

)

=Nµ2

3kT=

CT

(xi)

This is Curie law of paramagnetism, the constant C being the Curie constant of the system.

When the system is treated quantum mechanicallyIn this case, the major modification arises from the fact that the magnetic dipole moment −→µ and itscomponent µH along the field direction cannot have arbitrary values.

The magnetic moment −→µ is related to the total angular momentum −→J through the relation

−→µ = gµB−→J (xii)

where µB = e

2mc , known as Bohr magneton and g is the so-called g factor of the atom, e = charge ofan electron, m = mass of an electron and c = velocity of light in vacuum.

Using (xii) in (iii), we have

E = −gµB−→J ·

−→H = −gµBHJz (xiii)

since −→H is along the z-direction. In quantum mechanical description, the values which Jz canassume are discrete and are given by Jz = m, where m = −J,−(J−1), . . . ,(J−1),J.∴ The possible energies of the atom are

Em = −gµBHm (xiv)

Then the partition function of one atom is

Z = ∑m

e−βEm =J

∑m=−J

eβgµBHm (xv)


Let x = βgµBH =gµBH

kT−→ a dimensionless parameter. Therefore,

Z =J

∑m=−J

exm

=[

e−xJ + e−x(J−1) + · · ·+ exJ]

=e−xJ − ex(J+1)

1− ex =e−x(J+ 1

2)− ex(J+ 12)

e− x2 − e x

2=

sinh(J + 1

2)

x

sinhx2

∴ Z =sinh

(J + 1

2)

βg µBHsinh 1

2 βg µBH(xvi)

Now the probability of finding the atom in a state labelled m is given byPm = Ce−βEm (xvii)

where C is a constant of proportionality, determined from the normalization condition, ∑m

Pm = 1.

Therefore, the mean value of the magnetic moment along the field direction (i.e. z-direction) is

〈µH〉 =∑m

µHPm

∑m

Pm=

J∑

m=−JµHeβgµBHm

J∑

m=−JeβgµBHm

(xviii)

Since the field −→H is along the z-direction, µH = µz = gµBJz = gµBm, from (iv).

∴ 〈µH〉 =

J∑

m=−JgµBmeβgµBHm

J∑

m=−JeβgµBHm

=1β

J∑

m=−J(βgµBm)eβgµBmH

J∑

m=−JeβgµBmH

=1β

1Z

∂Z∂H

=1β

∂ lnZ∂H

=1β

∂ lnZ∂x

∂x∂H

= gµB∂ lnZ

∂x= gµB

∂∂x

[

lnsinh(

J +12

)

x− lnsinhx2

]

= gµB

[(J + 1

2)

cosh(J + 1

2)

xsinh

(J + 1

2)

x−

12 cosh x

2sinh x

2

]

⇒ 〈µH〉 = gµBJBJ (x)

where BJ (x) =1J

[(

J +12

)

coth(

J +12

)

x−12

cothx2

]

, known as Brillouin function. Therefore,

the magnetization of the system is given byM = N 〈µH〉 = NgµBJBJ (x) (xix)

For strong fields and low temperatures (x 1), the function BJ (x) ∼= 1 for all J and soM ∼= NgµBJ −→ magnetic saturation.


On the other hand, for high temperatures and weak fields (x 1),BJ (x) ∼=13

(

1+1J

)

x, so that

M ∼= N(gµBJ)2

3kT

(

1+1J

)

H = Ng2µ2

BJ (J +1)

3kTH (xx)

Therefore, the susceptibility is given by

χ(T ) =Ng2µ2

BJ (J +1)

3kT∝

1T

, which shows that the Curie law is again valid.

PROBLEM 2.21: Show that entropy increases when two ideal gases at the same temperature andpressure diffuse into each other. Discuss Gibbs paradox in this connection.

SOLUTION

According to Boltzmann’s relation, the entropy S of a system is related to the thermodynamicprobability Ω and is given by

S = k lnΩ (i)

From Maxwell-Boltzmann statistical count

Ω = N!∏i

gnii

ni!⇒ lnΩ = lnN!+∑

i(ni lngi − lnni!)

= lnN!+∑i

(ni lngi −ni lnni +ni)

= lnN!+∑i

ni (lngi − lnni +1)

Ω = lnN!+∑i

ni

(

lngi

ni+1

)

(ii)

Again for most probable state, Ω or lnΩ is maximum and then we have

ni = e−αgie−EikT (iii)

[see equation (3.1.7)].

∴

ni

gi= e−

(

α+EikT

)

⇒ lngi

ni= α+

Ei

kT(iv)

Using (iv) in (ii)

lnΩ = lnN!+∑i

ni

(Ei

kT+α+1

)

= lnN!+1

kT ∑i

niEi +α∑i

ni +∑i

ni


SB =32

Nk +Nk ln

[

V(

2πmkTh2

0

) 32]

.

Hence, the total entropy is simply

SA +SB = 3Nk +2Nk ln

[

V(

2πmkTh2

0

) 32]

(ix)

We now remove the partition very slowly so that the gases diffuse into each other. Therefore, thetotal number of molecules 2N of the two gases occupy a volume 2V . Since there is no change inpressure and temperature, the entropy of the system after the partition is removed is

S = 3Nk +2Nk ln

[

(2V )

(2πmkT

h20

) 32]

(x)

Now, according to the additive property of entropy

S = SA +SB (xi)

But from (ix) and (x), we see that

S− [SA +SB] = 2Nk ln2 > 0. (xii)

which is not equal to zero as required by (xi).This paradox was first discussed by Gibbs and is commonly referred to as the ‘Gibbs paradox’.The root of the difficulty embodied in the Gibbs paradox is that we treated the gas molecules as

distinguishable, as though interchanging the positions of two molecules would lead to a physicallydistinct state of the gas. This is not so. Indeed, if we treated the gas by quantum mechanics,the molecules would have to be regarded as indistinguishable (see Chapter 3). In this case, N!permutations of the molecules among themselves do not lead to physically distinct situations, sothat we should subtract the term k lnN! from (vi), so that

S =52

Nk +Nk ln

[

VN

(2πmkT

h2

) 32]

(xiii)

This has been verified experimentally at high temperatures and is known as Sackur-Tetrodeequation.

Equation (xii) can be written as

S = Nk ln

[

VN

(2πmkT

h2

) 32

e52

]

(xiv)

Using this expression, it can be easily verified that SA + SB − S = 0 and hence the paradox is re-moved.

PROBLEM 2.22: Imagine that a system R1 has probability P1r of being found in a state of r and

a system R2 has probability P2s of being found in a state s. Then one has S1 = −k ∑

rP1

r lnP1r and


S2 = −k ∑r

P2s lnP2

s . Each state of the composite system R(= R1 +R2) can be labelled by a pair ofnumbers r, s. Let the probability of R being found in this state be denoted by Prs. Then the entropyis defined by S = −k ∑

r∑s

Prs lnPrs. If R1 and R2 are weakly interacting so that they are statistically

independent, then Prs = P1r P2

s . Show that under these circumstances, the entropy is simply additive,i.e., S = S1 +S2.

SOLUTIONWe have

S = −k∑r

∑s

Prs lnPrs = −k∑r

∑s

P1r P2

s ln(P1

r P2s) (

∵ Prs = P1r P2

s)

=

(

∑s

P1s

)(

−k∑r

P1r lnP1

r

)

+

(

∑r

P1r

)(

−k∑s

P2s lnP2

s

)

= −k∑r

P1r lnP1

r +

(

−k∑s

P2s lnP2

s

) (

∵ ∑r

P1r = 1 = ∑

sP2

s

)

∴ S = S1 +S2 (Proved)

PROBLEM 2.23: Suppose that a system R1 has probability P1r of being found in a state r and a

system R2 has probability P2s of being found in a state s. The entropy of R1 is S1 =−k ∑

rP1

r lnP1r and

that of R2 is S2 =−k ∑s

P2s lnP2

s . Each state of the composite system R(= R1 +R2) is then labelled by

the pair numbers r, s. Let the probability of R being found in this state be Prs; its entropy is definedby s = −k ∑

r∑s

Prs lnPrs. Assume that R1 and R2 are not weakly interacting, so that Prs = P1r P2

s . Of

course P1r = ∑

rPrs and P2

s = ∑r

Prs. Moreover, all the probabilities are properly normalized so that

∑r

P1r = 1, ∑

rP2

s = 1, ∑r

∑s

Prs = 1. Show that

(a) S− (s1 +S2) = k ∑r,s

Prs ln(

P1r P2

sPrs

)

, and (b) S ≤ S1 +S2.

SOLUTION(a) We have

P1r = ∑Prs (i)

p2s = ∑Prs (ii)

S1 = −k∑r

P1r lnP1

r (iii)

S2 = −k∑s

P2r lnP2

s (iv)


Using (i) in (iii)S1 = −k∑

r∑

sPrs lnP1

r (v)

Again using (ii) in (iv)S2 = −k∑

r∑

sPrs lnP2

s (vi)

Adding (v) and (vi)S1 +S2 = −k∑

r∑

sln(P1

r P2s ) (vii)

Again according to the questionS = −k∑

r∑

sPrs lnPrs (viii)

[(viii) - (vii)] gives

S− (S1 +S2) = k∑1

+∑s

Prs lnP1

r P2s

Prs(Proved) (ix)

(b) Using the inequality, lnx ≤ x−1

lnP1

r P2s

Prs≤

(P1

r P2s

Prs−1

)

⇒ Prs lnP1

r P2s

Prs≤ Prs

(P1

r Ps2

Prs−1

)

= (P1r P2

s −Prs) ≤ 0

⇒ k∑r

∑s

Prs lnP1

r P2s

Prs≤ 0

⇒ [S− (S1 +S2)] ≤ 0

∴ S ≤ (S1 +S2) (x)

PROBLEM 2.24: Consider a system of N very weakly interacting particles at a temperature Tsufficiently high so that classical statistical mechanics is applicable. Each particle has mass m andis free to perform one-dimensional oscillation about its equilibrium position. Calculate the heatcapacity of this system, if the restoring force is proportional to x3.

SOLUTIONThe restoring force is given by

F = −Kx3 (i)where K is the constant of proportionality.

Hence, the potential energy of a particle is

V =14

Kx3 (ii)

Therefore, the total energy of a particle is

ε =p2

2m+

14

kx3 (iii)


Y

Z

0

L

LL

X

Now

L∫

0

e−mgzkT dz =

kTmg

(

1− e−mgLkT

)

and

L∫

0

ze−mgzkT dz =

(kTmg

)2 (

1− e−mgLkT

)

−kT Lmg

e−mgLkT

∴ 〈mgz〉 = mg×

(kTmg

)2 (

1− e−mgLkT

)

−kT Lmg

e−mgLkT

kTmg

(

1− e−mgLkT

) = kT −mgLe−

mgLkT

1− e−mgLkT

⇒ 〈mgz〉 = kT +mgL(

1− emgLkT

)−1(Ans.)

(ii)

The average kinetic energy of a molecule is given by

⟨p2

2m

⟩

=

∫ p2

2me−β

(p22m +mgz

)

d3 pd3r∫

e−β(

p22m +mgz

)

d3 pd3r=

∫ p2

2me−

βp22m d3 p

∫

e−βp22m d3 p

=

−∂

∂β

[(2πm

β

) 32]

(2πm

β

) 32

=3

2β=

32

kT (iii)

PROBLEM 2.26: Calculate the rotational partition function for heteronuclear and homonuclearmolecules separately.


SOLUTIONThe rotation of a diatomic molecule is specified by the angles (θ,φ) and the corresponding

momenta (Pθ,Pφ) and its kinetic energy assumes the form

εrot =P2

θ2I

+P2

φ

2I sin2 θ.

For heteronuclear diatomic molecule:

ξrotheteronuclear =

1h2

∫

e−3/kT ld pθd pφdθdφ

=1h2

2π∫

φ=0

dφ∞∫

pφ=0

e−p20

2IkT d pθ

π∫

θ=0

dθ

∞∫

pφ=0

e−p2φ

2IkT sin2 θ d pφ

=1h2 ×2π×

12√

2πIkT ×

π∫

θ=0

dθ12√

2πIkT sinθ

∴ ξrotheteronuclear =

2π2IkTh2 (i)

For homonuclear diatomic molecule:For homonuclear diatomic molecule, φ ranges from 0 to π (see Section 4.12.2).

∴ ξrothomonuclear =

π2IkTh2

PROBLEM 2.27: Study the thermodynamics of a system of N non-interacting diatomic molecules,each having an electric dipole moment µ, placed in an external electric field E. Assume that (i) thesystem is classical and (ii) | µE | kT , where T is the absolute temperature of the system.

SOLUTIONThe energy of a diatomic molecule in the presence of the electric field E is given by

ε =P2

2m+

P2θ

2I+

P2φ

2I sin2 θ

−µE cosθ (i)

where I is the moment of inertia of the molecule.Now the single particle canonical partition function is given by

ξ =1h3

∫ ∫ ∫ ∫ ∫ ∫

e−βεdr dθ dφ d p d pθ d pφ (ii)

⇒ ξ =1h3

V 13∫

r=0

dr

×

2π∫

φ=0

dφ

×

∞∫

P=0

d pe−βP22m

×

∫

Pθ=0

∞d pθe−βPθ2

2m


×

∞∫

θ=0

dθeβµE cosθ

∞∫

Pφ=0

d pφe−βp2

φ2I sin2 θ

=1h3V

13 ×2π×

(

12

√

2mπβ

)

×

(

12

√

2Iπβ

)

×

∞∫

θ=0

dθeβµE cosθ·12

√

2Iπβ

sinθ

=

√2mIπ 5

2

2h3 ·1

β 32·

π∫

0

dθsinθeβµE cosθ

=

√2mIπ 5

2

2h3 ·1

β 32·

1βµE

·(

eβµE − eβµE)

∴ ξ =

√2mIπ 5

2

µEh31

β 52

sinh(βµE) (iii)

Therefore, the canonical partition function of the system is

Z = ξN =

[√2mIπ 5

2

µEH3

]N [

sinhβµEβ 5

2

]N

(iv)

⇒ lnZ = N ln

(√2mIπ 5

2

µEh3

)

+N[

lnsinhβµE −52

lnβ]

(v)

The mean energy of the system is given by

ε = −∂ lnZ

∂β=

52

Nβ−NµE coth(βµE) =

52

NkT −NµE sinh(

µEkT

)

(vi)

The specific heat at constant volume is given by

CV =

(∂ε∂T

)

V=

52

Nk−Nk(

µEkT

)2cosec h2

(µEkT

)

(vii)

For T → 0, i.e., x(

=µEkT

)

→ ∞

Cv ∼=52

Nk(

∵

Ltx → ∞

cosec hx = 0)

(viii)

Now for µE/kT 1, i.e., for high temperature

cosec h2(

µEkT

)

∼=1

(µEkT

) −13

+115

(µEkT

)2(ix)


and

∑3 =∞

∑n=0

n2e−u(n+ 12) = e−

u2

∞

∑n=0

n2e−ν

= e−u2[e−u +22e−2u +32e−3u + · · ·

]

= e−u2 · e−u [

1+22e−u +32e−2u + · · ·]

= e−3u2 ·

(1+ e−u)

(1− e−u)3 (iv)

Substituting (ii), (iii) and (iv) in (i)

ξ =e− u

2

1− e−u

[

1+ xu

e−u (1+ e−u)

(1− e−u)2 +e−u

(1− e−u)+

14

]

=e u

2

1− e−u

[

1+ xu

2e−u

(1− e−u)+

14

] (v)

Therefore, the canonical partition function for the system is given byZ = ξN

⇒ lnZ = N lnξ = N

[

−u2− ln

(1− e−u)+ ln

1− xu

(

2e−u

(1− e−u)2 +14

)]

(vi)

Then the internal energy of the system is

E = −∂ lnZ

∂β= kT 2 ∂ lnZ

∂T= −hv

∂ lnZ∂u

= −Nhv

−

12−

e−u

1− e−u +

x

2(

e−u

1−e−u

)2− ue−u

(1−e−u)2 −2u(e−u)

2

(1−e−u)3+ 14

1+ xu

2e−u

(1−e−u)2+ 14

= E0 +Ecorrection (vii)where

E0 =12

Nhv+Nhv

eu −1=

12

Nhv+Nhv

e hvkT −1

(viii)

The corresponding specific heat at constant volume is

(CV )0 =

(∂E0

∂T

)

V= Nk

(hvkT

)2 e hvkT

[

ehvkT 1

]2 (ix)

Ecorrection = −xNhv

2(

e−u

1−e−u

)2− ue−u

(1−e−u)2 −2u(e−u)

2

(1−e−u)3+ 14

1− xu

2e−u

(1−e−u)2 + 14

(x)


Upto the first order in x, (x) reduces to

Ecorrection = −xNhv

[

2e−u

(1− e−u)2 −2ue−u

(1− e−u)2 −4ue(e−u)

2

(1− e−u)3 + 14

]

(xi)

The corresponding specific heat is

(Cv)correction = xNku2 2e−u

(1− e−u)2 +12u(e−u)

2

(1− e−u)4 −4(e−u)

2

(1− e−u)3− 4e−u

(1−e−u)3

(xii)

This is the correction to the specific heat upto the first order in x.Comment: For u

(= hv

kT) 1, i.e., for high temperature,

(CV )correction = xNk×4u

= xNk×4kThv

= x(

4Nk2

hv

)

T ∝ T,

which shows that the correction due to the anharmonicity increases with temperature, T .

PROBLEM 2.29: Calculate the canonical partition function and hence the specific heat at constantvolume for a classical system of N non-interacting diatomic molecules enclosed in a box of volumeV at temperature T .

SOLUTIONThe energy of a diatomic molecule can be written as

E (p1,p2,r1,r2) =1

2m(

p21 + p2

2)+

12

K |r1 −r2 |2 (i)

where p1,p2,r1 and r2 are the momenta and position coordinates of the two atoms in a diatomicmolecule.

Let R =r1 + r2

2andρ =r1 −r2.

Then the Jacobian is

J =

∣∣∣∣∣∣

∂r1∂R

∂r2∂R

∂r1∂ρ

∂r2∂ρ

∣∣∣∣∣∣

=

∣∣∣∣

1 112 − 1

2

∣∣∣∣= −1 ⇒| J |= 1 (ii)

∴ d3r1d3r2 = d3Rd3ρ (iii)Now the single particle canonical partition function is

ξ =∫

e−βHd3r1d3r2d3 p1d3 p2 =∫

e−βHd3r1d3r2d3Rd3ρ

=

[∫

e−βp21d3 p1

][∫

e−βp22d3 p2

][∫

d3R][∫

e−βkp22d3ρ

]

=

(2πm

β

) 32

×

(2πm

β

) 32

×V ×

(2πβK

) 32

= V

(

8π3m2

K

32)

×1

β 92


Therefore, the canonical partition function for the system is

Z = ξN = V N(

8π3m2

K

) 3N2

×1

β 9N2

(iv)

⇒ lnZ = ln

[

V N(

8π3m2

K

) 3N2]

−9N2

lnβ (v)

The mean energy of the system is

E = −∂ lnZ

∂β=

9N2β

=92

NkT (vi)

Hence, the specific heat at constant volume is

CV =

(∂E∂T

)

V=

92

Nk (vii)

PROBLEM 2.30: In Problem 2.31, calculate the mean square molecular diameter, 〈|r1 −r2|2〉.

SOLUTIONThe mean square molecular diameter is

〈|r1 −r2|2〉 =

∫|r1 −r2|

2e−βEd3r1d3r2d3 p1d3 p2∫

e−βEd3r1d3r2d3 p1d3 p2

=

∫ρ2e−β[ 1

2m(p21+p2

2)+ 12 Kρ2]d3r1d3r2d3 p1d3 p2

e−β[ 12m(p2

1+p22)+ 1

2 Kρ2]d3r1d3r2d3 p1d3 p2

=

∫ρ2e− 1

2 βKρ2d3ρ∫

e− 12 βKρ2d3ρ

=

∞∫

0ρ4e− 1

2 βKρ2dρ

∞∫

0ρ2e− 1

2 βKρ2dρ=

2kTK

∞∫

0x 3

2 e−xdx

∞∫

0x 3

2 e−xdx=

2kTK

Γ(5

2)

Γ( 3

2) =

3kTK

〈|r1 −r2|2〉 ∞ T.

PROBLEM 2.31: Consider a classical system of N non-interacting diatomic molecules enclosedin a box of volume V and temperature T . The energy for a single molecule is E (p1,p2,r1,r2) =

12m

(p2

1 + p22)+ ε|r12 − r0|, where p1,p2,r1 andr2 are the momenta and position coordinates of the

two atoms in a molecule. ε and r0 are given positive constants and r12 = |r1 −r2|. Find the specificheat of the system at a constant volume as a function of temperature.


SOLUTIONThe energy of a molecule is

E (p1,p2,r1,r2) =1

2m(

p21 + p2

2)+ ε|r12 − r0| (i)

Let R =r1 +r2

2andρ =r1 −r2.

Then the Jacobian is

J =

∣∣∣∣∣∣∣∣∣

∂r1

∂R∂r2

∂R

∂r1

∂ρ∂r2

∂ρ

∣∣∣∣∣∣∣∣∣

=

∣∣∣∣

1 112 − 1

2

∣∣∣∣= −1 ⇒| J |= 1 (ii)

∴ d3r1d3r2 = d3Rd3ρ (iii)

The single particle canonical partition function is

ξ =∫

e−βHd3r1d3r2d3 p1d3 p2 =∫

e−βHd3r1d3r2d3Rd3ρ

=

[∫

e−βp2

12m d3 p1

]

×

[∫

e−βp2

22m d3 p2

]

×

[∫

d3R]

×

[∫

ρ−βε|ρ−r0|d3ρ]

(iv)

=

[

4π(

2mβ

) 32 1

2τ(

32

)]

×

[

4π(

2mβ

) 32 1

2τ(

32

)]

×V ×

4π

(βε)3

∞∫

0

y2e−|y−x|dy

where y = βερ and x = βεr0.Now ∫

y2e−ydy = y2ey −

∫

2yey +2∫

eydy = y2ey −2yey +2ey (v)

Similarly,∫

y2e−ydy = −y2e−y −2ye−y2e−y (vi)

∴

∞∫

x

y2e−ydy = x2e−x +2xe−y2e−x (vii)

Again∞∫

0

y2e−|y−x|dy =

x∫

0

y2e−(y−x)dy+

∞∫

0

y2e−(y−x)dy = e−x∞∫

x

y2e−ydy+ ex∞∫

x

y2e−ydy

= ex [x2ex −2xex +2ex −2

]+ ex [

x2e−x +2xe−x +2e−x] (viii)

∴

∞∫

0

y2e−|y−x|dy = 2x2 +4−2e−x (ix)


Now, when the cylinder is rotated, it becomes energetically favourable for the molecules to movetowards the edge of the cylinder, so we expect the density to increase with radius r.

In a frame of reference rotating with a gas, a molecule of mass m at a distance r from the axisexperiences a centrifugal force

Fc = mω2r (ii)away from the axis.∴ The centrifugal potential energy is

Vc = −12

mω2r2 (iii)

From the Boltzmann distribution law, we expect the probability of a molecule to be proportional toexp

(− Vc

kT)

and so we can write

σ(r) = Aexp(

mω2r2kT

)

(iv)

where A is a constant of proportionality.Now the total mass of the gas within the cylinder is given by

M = 2πLa∫

0

σ(r)rdr = 2πLAa∫

0

r exp(

mw2r2

2kT

)

dr (v)

The integral can be easily evaluated by noting thatddr

[

exp(

mω2r2

2kT

)]

=mω2r2

2kTexp

(mω2

r2

)

(vi)

∴ M =2πLAkT

mω2

[

exp(

mω2a2

2kT

)

−1]

(vii)

Equating (i) and (vii), we have the expression for A:

A =σ0mω2a2

2kT1

[

exp(

mω2a2

2kT

)

−1] (viii)

Using (viii), we get the expression for σ(r) as

σ(r) =σ0mω2a2

2kT

exp(

mω2a2

2kT

)

[

exp

(

mω[

exp(

mω2a22kT

)

−1]2

a2

2kT

)

−1

] (ix)

Let R =ra

and Ω =mω2a2

2kT.

∴

σ(R)

σ0= Ω

exp(ΩR2)

[exp(Ω)−1](x)


Writing the function as a series expansion, we have

σ(R)

σ0= Ω

(1+ΩR2 + 1

2! Ω2R4 + . . .

)

Ω+ 12! (Ω2 + . . .)

=

(1+Ω+ 1

2! Ω2 + . . .

)

(1+ +Ω

2 + . . .) (xi)

For small ω (i.e., for small Ω), we can write (xi) approximately as

σ(R)

σ0∼=

(1+ΩR2)

(1+ Ω

2) (xii)

which shows that σ(R) = σ0, i.e., the density remains unchanged when R2 = 12 ⇒

r2

a2 = 12 ⇒ r = a√

2 .Thus for small ω, the density remains unchanged at a distance a√

2 from the axis of the cylinder.

PROBLEM 2.33: Consider an ideal gas at absolute temperature T in a uniform gravitational fielddescribed by acceleration g. By writing the condition of hydrostatic equilibrium for a slice of thegas located between heights z and z+dz, derive an expression for n(z), the number of molecules percm3 at height z. Also show that this result is identical with that derived from statistical mechanics.

SOLUTIONLet us consider a slice between heights z and z+dz. Let the pressure at z be p and that at z+dz

be p+d p.

z + dz, p + dp

z, p

Then for hydrostatic equilibrium,

pA− (p+d p)A = mngAdz (i)

where, A = cross-sectional area of the slicen = number densitym = particle massg = acceleration due to gravity.

∴ d p = −mngdz (ii)

Again treating the gas as an ideal gas, we can write

p = nkT ⇒ d p = kT dn (iii)

Using (iii) in (ii), we havednn

= −mgkT

dz


On integrationlnn = −

mgkT

z+ constant (iv)

Let at z = 0,n = n(0).∴ n = n(0)e−

mgzkT (v)

Using ideal gas law: p = nkT, p(0) = n(0)kT

p = p(0)e−mgzkT (vi)

Now the probability P(r,p)d3rd3 p that the molecule has position lying in the range betweenr andr +dr and momentum in the range between p and p+dp is given by

P(r,p)d3rd3 p ∝d3rd3 p

h3 e−β(

p22m +mgz

)

(vii)

Therefore, the probability P(z)dz that a molecule is located at a height between z and z+dz is

P(z)dz ∝1h3 ×

∫

(x,y)

dx dy

×

∞∫

P=0

π∫

θ=0

2π∫

φ=0

e−β P22m P2 sinθ d p dθ dφ

× e−βmgz dz (viii)

But∫

(x,y) dx dt = A, cross-sectional area of the slice.

∴ P(z)dz = constant×Ah3

(2πm

β

)3/2e−

mgkT z (ix)

Let at z = 0,P(z)|z=0 = P(0).

∴ P(0) = constant×Ah3

(2πm

β

)

(x)

⇒ P(z) = P(0)e−mgkT z (xi)

Thus, the probability of finding a molecule at height z decreases exponentially with height.Now

n(z) dz ∝ p(z) dz = constant×P(0)e−mgzkT dz (xii)

At z = 0,n = n(0) ⇒ constant×P(0) = n(0)

∴ n(z) = n(0)e−mgzkT (xiii)

which is same as (v).

PROBLEM 2.34: Consider a system consists of N weakly interacting particles, each of which canbe in either of two states with respective energies ε and ε2 (ε2 > ε1). Calculate explicitly the meanenergy E and heat capacity CV of this system.


SOLUTIONConsider a pair of one +ve and one –ve particle.

a

a

+ +

a

2

+ +

+ +

+ +

The adjoining figure shows that there are 4 possible states of this pair. Since the potential energyof a charged particle of charge in an external electric field is V (x) = −eεx, the average separationbetween a +ve ion and a –ve ion is given by

〈x〉 =2 · a

2 eβ a2 εe −2 · a

2 e−β a2 εe

2eβ a2 εe +2e−β a

2 εe =a2

tanhβeεa

2(i)

If there are N number of pairs per unit volume, the electric polarization or average dipole momentper unit volume is given by

〈µ〉 = Ne〈x〉 = Nea2

tanhβeεa

2(ii)

PROBLEM 2.36: Calculate the canonical partition function and hence the specific heat of a classicalideal two-dimensional gas.

SOLUTIONThe energy of a single particle is

ε =p2

x + p2y

2m(i)

Then the single particle canonical partition function is

ξ =1h2

∫

e−β

2m(p2x+p2

y)dx dy d px d py =Ah2

[∫ ∞

−∞e−

βp2x2m d px

]

×

[∫ ∞

−∞e−

βp2y2m d py

]

=Ah2

(2πm

β

)

(ii)

where A is the area.


Hence, the canonical partition function of the system is given by

Z =AN

h3N

(2mπ

β

)N(iii)

The mean energy of the system is

E = −

(∂ lnZ

∂β

)

V= NkT

∴ The specific heat at constant volume is

CV =

(∂E∂T

)

V= Nk (iv)

PROBLEM 2.37: Suppose that a wire of radius r0 is placed along the axis of a metal cylinder ofradius R and length L. The wire is maintained at a positive potential V with respect to the cylinderand the whole system is at some high absolute temperature T . As a result, electrons emitted fromthe hot metals form a dilute gas filling the cylindrical container and in equilibrium with it. Thedensity of these electrons is so low that their mutual electrostatic interaction can be neglected.

Usine Gauss’ theorem, obtain an expression for the electrostatic potential at a radial distancer from the wire (r0 < r < R). [Assume the cylinder to be infinitely long so that end effects areneglected.] Also find the density of the electron gas filling the space between the wire and cylinder,as a function of the radial distance in equilibrium.

SOLUTION

r0

R

L

V = 0 V


SOLUTIONIn the rotating frame of reference, the molecules feel a centrifugal force mω2r radially outward.

This force can be obtained from a potential

Fr = mω2rr = −σVσr

r (i)

where r is the unit vector in the radial direction.Therefore, the potential energy of a molecule in the rotating frame is

V = −12

mω2r2 (ii)

so that the total energy of a molecule in the rotating frame is given by

ε =p2

2m+

(

−12

mω2r2)

(iii)

The probability of finding the molecule in the energy range ε to ε+dε is then given by

P(ε) =1Z

ep2

2mkT emω2r2

2kT (iv)

Then the probability of finding the molecule between r and r + dr is obtained by integrating overall the other coordinates and is given by

P(r)dr = P(0)emω2r2

2kT dr (v)where P(0) is a constant which correctly normalizes the distribution.

Equation (v) is the radial distribution of molecules. As expected, the molecules crowd towardsthe larger radii. It should be noted that we have taken the rotating speed to be very high so that theeffect of the Coriolis force is ignored.

Now from (v), the concentration at r1 is

σ(r1) ∝ emω2r2

12kT (vi)

Similarly,

σ(r2) ∝ emω2r2

22kT

∴

σ(r1)

σ(r2)= e

12

mω2(r21−r2

2)

kT (vii)

⇒ m =2kT

ω2(r2

1 − r22) ln

[σ(r1)

σ(r2)

]

PROBLEM 2.40: In order to measure a certain physical quantity, it is necessary to know theequilibrium θ0 of a torsional pendulum. θ0 is determined by the minimum in the potential energyof the pendulum V (θ) = 1

2 K(θ−θ0)2. The pendulum, when rotating, has a kinetic energy T = 1

2 Iθ,where θ = dθ/dt. The precision with which θ0 can be found is limited because the pendulum is inthermal equilibrium with its environment at a temperature T .


(b) The degeneracy of the nth energy level is n2. The partition function for a single atom, ne-glecting the unbound states, is

Z = ∑states

exp(

−εi

kT

)

=∞

∑n=1

n2 exp( α

n2

)

(iii)

where α = A/

kT . Since α > 0, it follows that exp(α/

n2)

> 1 for all n. Using this, we can setthe lower bound of Z, but Z diverges since the lower bound diverges.

∴ Z >∞

∑n=1

n2, which diverges.

2.3 GRAND CANONICAL ENSEMBLEIt is a collection of large number of essentially independent systems, having same temperature T ,volume V and chemical potential µ. Thus, the individual systems of a grand canonical ensembleare separated by rigid, permeable and conducting walls.

µ,V,T

µ,V,T

µ,V,T

µ,V,T

µ,V,T

µ,V,T

µ,V,T

µ,V,T

µ,V,T

µ,V,T

µ,V,T µ,V,T

Fig. 2.7. Grand canonical ensemble.

As the separating walls are conducting and permeable, the exchange of heat energy as well asthat of particles between the systems take place in such a way that all the systems arrive at commontemperature T and the chemical potential µ.

2.3.1 Probability distribution and grand canonical partition functionLet us consider a system R of fixed volume V in contact with a large reservoir R′ with which thesystem can exchange not only energy, but particles also. We wish to find the probability Pr offinding the system R in any one particular microstate r of energy Er.

Neither the energy Er of R nor the number of particles Nr in R are fixed, but the total energyE0and total number of particles N0of the combined system R0 = R+R′are fixed, i.e.,

Er +E ′ = E0 = constantNr +N ′ = N0 = constant

(2.3.1)

where E ′ and N′ denote the energy and number of particles in the reservoir R′ respectively.


Let Ω′ (E ′N′) denote the number of states accessible to R′, when it contains N′ particles and hasan energy in the range near E ′. Then if R is in a particular state r, the number of states accessibleto the combined system R0 is 1×Ω′ (E ′,N′) = Ω′

(E0 −Er,N0 −Nr

). Hence, the probability of

finding the system R in this state is

Pr =1×Ω′

(E0 −Er,N0 −Nr

)

Ω0total

(2.3.2)

where Ω0total is the total number of states accessible to the combined system R0. It is independent of

r and is just a constant. Writing 1Ω0

total= c′ = constant,

Pr = c′Ω′(E0 −Er,N0 −Nr

)(2.3.3)


lnPr = lnc′ + lnΩ′(E0 −Er,N0 −Nr

)(2.3.4)

Since R is very small compared to R′, Er E0 and Nr N0, expanding lnΩ′ (E ′,N′)= lnΩ′

(E0 −Er,N0 −Nr

)about E ′ = E0 and N′ = N0, we have

lnΩ′(E0 −Er,N0 −Nr

)= lnΩ′

(E0,N0)−

[∂ lnΩ′

∂E ′

]

E0Er −

[∂ lnΩ′

∂N′

]

N0Nr . . .

Since Er E0 and Nr N0, we can neglect the higher order terms and so we have


)= lnΩ′

(E0,N0)−

[∂ lnΩ′

∂E ′

]

E0Er −

[∂ lnΩ′

∂N′

]

N0Nr (2.3.5)

The derivatives are evaluated for E ′ = E0 and N′ = N0 and so they are constants, characterizing thereservoir R′. We write

[∂ lnΩ′

∂E ′

]

E0= β and

[∂ lnΩ′

∂N′

]

= α.Then equation (2.3.5) becomes


)= lnΩ′

(E0,N0)−βEr −αNr (2.3.6)

⇒ Ω′(E0 −Er,N0 −Nr

)= Ω′

(E0,N0)e−βEr−αNr

Using equation (2.3.6), equation (2.3.3) becomes

Pr = c′Ω′(E0,N0)e−βEr−αNr

⇒ Pr = ce−βEr−αNr (2.3.7)where c = c′Ω′

(E0,N0), constant of proportionality, independent of r.

Equation (2.3.7) is called the grand canonical distribution. An ensemble of systems, dis-tributed according to this probability distribution, is called a grand canonical ensemble.

The parameter β is the temperature parameter of the reservoir. β = 1/kT , where T is theabsolute temperature of the reservoir and µ = −kT α, the chemical potential of the reservoir.

∴ Pr = ceµβNr−βEr (2.3.8)


The normalization condition for probabilities demands ∑r

Pr = 1. Applying this to equation (2.3.8),

we have c =1

∑r

eµβNr−βErand so

Pr =eµβNr−βEr

∑r

eµβNr−βEr=

eµβNr−βEr

Z (µ,V,T )(2.3.9)

where Z (µ,V,T ) = ∑r

eµβNr−βEr −→ grand canonical partition function.

2.3.2 Density and energy fluctuations in grand canonical ensembleIn this case, fluctuations in both E and N are present.

Let N be the average number of particles of the system R. Then the mean square deviation in itis given by

(∆N)2 =(N −N

)2= N2 −N2

Now

N =∑r

NrPr

∑r

Pr= ∑

rNrPr =

∑r

NreµβNr−βEr

∑r

eµβNr−βEr.

The grand canonical partition function is

Z (µ,V,T ) = ∑r

eµβNr−βEr ⇒1β

∂Z∂µ

= ∑r

NreµβNr−βEr .

∴ N =1β

1Z

∂Z∂µ

=1β

∂ lnZ∂µ

Again

N2 =∑r

N2r Pr

∑r

Pr= ∑

rN2

r Pr =∑r

N2r eµβNr−βEr

∑r

eµβNr−βEr=

∑r

N2r eµβNr−βEr

Z (µ,V,T )=

1Z

1β2

∂2Z∂µ2

Hence, (∆N)2 =1β2

1Z

∂2Z∂µ2 −

(1β

1Z

∂Z∂µ

)2=

1β

∂∂µ

(1β

1Z

∂Z∂µ

)

= kT∂N∂µ

.

∴ The root mean square fluctuation in N is

∆∗N =[

(∆N)2]1/2

=

[

kT∂N∂µ

]1/2

T,V.

The relative fluctuation in N is then

∆∗NN

=1N

[

kT∂N∂µ

]1/2

T,V(2.3.10)


and

E2=

∑r

E2r eµβNr−βEr

∑r

eµβNr−βEr=

∑r

E2r eµβNr−βEr

Z (µ,V,T )

=1Z

(∂2Z∂β2

)

µ,V=

∂∂β

(1Z

∂Z∂β

)

+1

Z2

(∂Z∂β

)2

=∂

∂β

(∂ lnZ

∂β

)

+

(∂ lnZ

∂β

)2= −

∂E∂β

+(E

)2

∴ (∆E)2 = E2 −E2= −

(∂E∂β

)

µ,V= kT 2

(∂E∂T

)

µ,V

andN =

1β

(∂ lnZ

∂µ

)

V,T= −

(∂ lnZ

∂(−µβ)

)

V,T(2.3.14)

Eliminating µ from (2.3.13) and (2.3.14), we see that E = E(N,V,T

), so that

(∂E∂T

)

µ,V=

(∂E∂T

)

N,V+

(∂E∂N

)

T,V×

(∂N∂T

)

µ,V= CV +

(∂E∂N

)

T,V×

(∂N∂T

)

µ,V

where CV =(

∂E∂T

)

N,V, the specific heat of the system at constant volume.

∴ (∆E)2 = kT 2

[

CV +

(∂E∂N

)

T,V

(∂N∂T

)

µ,V

]

Again from equation (2.3.13)(

∂E∂(−µβ)

)

V,T=

(∂N∂β

)

µ,Vand hence

(∂N∂T

)

µ,V=

1T

(∂E∂µ

)

T,V.

∴ (∆E)2 = kT 2

[

CV +1T×

(∂E∂N

)

T,V

(∂E∂µ

)

T,V

]

= kT 2CV + kT(

∂E∂N

)

T,V

(∂E∂µ

)

T,V

= kT 2CV + kT(

∂E∂N

)2

T,V

(∂N∂µ

)

T,V

But (∆N)2 = kT(

∂N∂µ

)

T,V.

∴ (∆E)2 = kT 2CV +

(∂E∂N

)2

T,V(∆N)2 = (∆E)2

canonical +

(∂E∂N

)2

T,V(∆N)2 (2.3.15)


This is the desired result, which says that the mean square fluctuation in the energy E of a systemin the grand canonical ensemble is equal to the value it would have in the canonical ensemble plusa contribution arising due to the fluctuation in the particle number. Under ordinary circumstances,the relative fluctuation in the energy of the system is negligible. However, in the region of phasetransition, the fluctuation in the energy is large because of the second term.

2.3.3 Thermodynamic quantities in terms of the grand canonical partition function1. Internal energyThis is given by

E = −

(∂ lnZ

∂β

)

µ,V= kT 2

(∂ lnZ∂T

)

µ,V(2.3.16)

2. Specific heatThe specific heat at constant volume is given by

CV =

(∂E∂T

)

N,V(2.3.17)

where E is given by (2.3.16).

3. Thermodynamic pressureThe grand canonical partition function is Z = Z (µ,V,T ). For the sake of convenience, let us writeZ as a function of α(= −µβ), V and β(= 1/kT ), i.e., Z = Z (α,V,β).

∴ d lnZ (µ,V,T ) = d lnZ (α,V,β) =

(∂ lnZ∂α

)

dα+

(∂ lnZ∂V

)

dV +

(∂ lnZ

∂β

)

dβ (2.3.18)

Now

N =1β

(∂ lnZ

∂µ

)

V,T= −

(∂ lnZ∂α

)

V,T(2.3.19)

Using (2.3.16) and (2.3.19) in (2.3.18)

d lnZ = −Ndα+

(∂ lnZ∂V

)

dV −Edβ

⇒ d[lnZ +Nα+Eβ

]= αdN +βdE +

(∂ lnZ∂V

)

dV

⇒ d[

k

lnZ −NµkT

+EkT

]

=1T

[

dE +

(∂ lnZ∂V

)

dV −µdN]

To interpret the second term on the right-hand side of this equation, we compare the expressionenclosed within the parenthesis with the statement of the first law of thermodynamics, viz.,

δQ = dE +δW −µdN (2.3.20)


We immediately identify (∂ lnZ/∂V )dV as δW , the work done by the system.

∴

δQT

= d[

k

lnZ −NµkT

+EkT

]

.

The quantity[

k

lnZ −NµkT

+EkT

]

is identified with the entropy S of the system. Accordingly,

S =

[

k

lnZ −NµkT

+EkT

]

⇒ lnZ =Sk

+NµkT

−EkT

=T S +Nµ−E

kTBut Nµ = G, the Gibbs free energy of the system [see Problem (2.46)]. Again G is equal to(E −T S +PV

). So finally

P =kTV

lnZ (µ,V,T ) (2.3.21)Thus, the grand canonical partition function directly gives the pressure as a function of µ,V

and T .

4. Helmholtz free energyThe Helmholtz free energy is given by

F ≡ Nµ−PV = NkT lnz− kT lnZ (µ,V,T )

where z = eµβ −→ fugacity.

F = −kT lnZ (µ,V,T )

zN(2.3.22)

which can be compared with the canonical ensemble formula F = −kT lnZ (N,V,T ).

5. EntropyThe entropy of the system is given by

S ≡E −F

T= kT

(∂ lnZ (µ,V,T )

∂T

)

µ,V+ k ln

Z (µ,V,T )

zN

∴ S = kT[

∂ lnZ (µ,V,T )

∂T

]

µ,V+ k lnZ (µ,V,T )−Nkµβ (2.3.23)

2.3.4 Equation of stateEliminating µ between (2.3.19) and (2.3.21), one obtains the equation of state, i.e., the (P,V,T )-relationship.

PROBLEM 2.42: Show that

(a) 〈N〉 = z∂∂z

lnZ(z,V,T ), (b) 〈N2〉−〈N〉2 = z∂∂z

z∂∂z

lnZ(z,V,T )

where z = eµβ, known as fugacity.


where g(T,P) is independent of Nj.

∴ µj =

(∂G∂Nj

)

T,P= g(T,P) =

GNj

(Proved) (xi)

i.e., the chemical potential per molecule is just equal to the Gibbs free energy per molecule.When several species are present, then G = G(T, p,N1, . . .Nm) and in general

µj =

(∂G∂Nj

)

T,P,N1...Ni−1...Ni+1...Nm

=GNj

.

PROBLEM 2.44: Show that the entropy of a grand canonical ensemble can be written as

S = −k∑r

Pr lnPr.

SOLUTION

The grand canonical partition function is

Z(µ,V,T ) = ∑r

eµβNr−βEr (i)

and the probability of finding the system in the rth microstate is

Pr =eµβNr−βEr

∑r

eµβNr−βEr=

eµβNr−βEr

Z(µ,V,T )(ii)

∴ lnPr = − lnZ +µβNr −βEr

⇒ ∑r

Pr lnPr =∑r

Pr lnPr

∑r

Pr= − lnZ

∑r

Pr

∑r

Pr+µβ

∑r

NrPr

∑r

Pr−β

∑r

ErPr

∑r

Pr

⇒ ∑r

Pr lnPr = µβN −βE − lnZ (iii)

Let us now introduce a new variable α = µβ.

∴ Z(α,V,β) = ∑r

eαNr−βEr (iv)

Now

d lnZ =∂ lnZ∂α

dα+∂ lnZ∂V

dV +∂ lnZ

∂βdβ = Ndα+ pβdV +βdE

⇒ d(lnZ −Nα+Eβ

)= −αdN + pβdV +βdE

⇒ d(lnZ −Nα+Eβ

)= β

[

dE + pdV −αβ

dN]

= β[dE + pdV −µdN

]


But T dS = dE + pdV −µdN.

∴ d(lnZ −Nα+Eβ) =dSk

whenceSk

= (lnZ −Nµβ+Eβ)

⇒ lnZ =Sk

+Nµβ−Eβ(v)

Using (v) in (iii), we finally get

S = −k∑r

Pr lnPr (Proved) (vi)

PROBLEM 2.45: Using S = −k ∑r

Pr lnPr, show that for a grand canonical ensemble, T dS = dE +

pdV −µdN.

SOLUTIONWe know that

S = −k∑r

Pr lnPr (i)

Now

Pr =eµβNr−βEr

Z (µ,V,T )⇒ lnPr = µβNr −βEr − lnZ (ii)

Using (ii) in (i)

S = −k[

µβ∑r

PrNr −β∑r

PrEr − lnZ ∑r

Pr

]

= −k[µβN −βE − lnZ

](

∵ ∑r

Pr = 1,∑r

PrEr = E,∑r

PrNr = N)

= k lnZ + kβE − kµβN

⇒ dS = k∂ lnZ

∂µdµ+ k

∂ lnZ∂V

dV + k∂ lnZ∂T

dT + kβdE + kEdβ− kβµdN − kβNdµ− kµNdβ

= kβNdµ+ kβpdV + kµNdβ− kEdβ+ kβdE + kEdβ− kβµdN − kβNdµ− kµNdβ(

∵

∂ lnZ∂T

dT =∂ lnZ

∂βdβ,

∂ lnZ∂µ

= βN,∂ lnZ∂V

= βp,∂ lnZ

∂β= µN −E

)

= kβdE + kβpdV − kβµdN

⇒ T dS = dE + pdV −µdN(

∵ β =1

kT

)



lnZ (µ,V,T ) =α

∑i=1

lnZ(

µ,Vα

,T)

= α · lnZ(

µ,Vα

,T)

⇒ lnZ(

µ,Vα

,T)

=1α· lnZ (µV,T )

This shows that lnZ (µ,V,T ) is linear in V .

PROBLEM 2.47: Show that Z (µ,V,T ) = eN for an ideal gas, if the gas molecules are treated asindistinguishable.

SOLUTIONThe grand canonical partition function is Z (µ,V,T ) = ∑

reµβNr−βEr . We can write this partition

function in another form:

Z (µ,V,T ) = ∑Nr

eµβN−βENr = ∑Nr

eµβNe−βENr =∞

∑N=0

eµβNZ (N,V,T )

where Z (N,V,T ) = ∑r

e−βENr −→ canonical partition function.

For an ideal gas, the canonical partition function is Z (N,V,T ) = ξN , where ξ is the single

particle canonical partition function and is given by ξ = Vh3

(2πm

β

)3/2.

∴ Z (µ,V,T ) =∞

∑N=0

eµβNξN (i)

If the gas molecules are treated as indistinguishable, then N! permutations of the molecules amongthemselves do not lead to physically distinct situations and hence equation (i) must be divided byN!.

∴ Z (µ,V,T ) =

∞∑

N=0eµβNξN

N!=

∞∑

N=0

(eµβξ

)N

N!= exp

(

eµβξ)

(ii)

Again, we know that

N =1β

∂ lnZ (µ,V,T )

∂µ=

1β

(

∂(eµβξ

)

∂µ

)

= eµβξ. (iii)

Using (iii) in (ii)Z(µ,V,T ) = eN (Proved)

PROBLEM 2.48: Consider a system of N molecules in a container of volume V . Assuming thatthere is no correlation whatsoever between the locations of the various molecules, calculate theprobability that a region of volume v contains exactly n molecules.


SOLUTIONThe probability that the region of volume v exactly contains n molecules is given by

Pn =∑r

eµβn−βEnr

Z (µ,v,T )=

eµβn ∑r

e−βEnr

Z (µ,v,T )=

eµβnZ (n,v,T )

Z (µ,v,T )(i)

where Z(n,v,T ) = canonical partition function = ξn/n!, ξ being the single particle canonical parti-tion function. n! appears because of the essential indistinguishability of the particles. Again, fromproblem 2.50, n = eµβξ and Z (µ,v,T ) = en.∴ Pn = (eµβξ)n / n!

en = (n)n

n! e−n. This is known as Poisson’s distribution.

PROBLEM 2.49: Consider a system of N molecules forming a substance which consists of twophases. The whole system is isolated so that its total energy E and its total volume V are both fixed.Find the necessary conditions for equilibrium between the two phases.

SOLUTIONLet the two phases be 1 and 2. The whole system is isolated, so that its total energy E, its total

volume V and its total number of particles N are fixed. Let Ni be the number of molecules of thesubstance in phase i and denote the energy of this phase by Ei and its volume by Vi.

Then we must have the conservation conditions:E1 +E2 = E = constant

⇒ dE1 = −dE2 (i)

V1 +V2 = V = constant

⇒ dV1 = −dV2 (ii)

N1 +N2 = N = constant

⇒ dN1 = −dN2 (iii)The equilibrium condition corresponds to the most probable situation, i.e., when the thermodynamicprobability Ω is maximum. But S = k lnΩ. Hence, the equilibrium condition is that the entropy ismaximum, i.e.,

S = S (E1,V1,N1; E2,V2,N2) = maximum (iv)Again S = S1 (E1,V1,N1)+S2 (E2,V2,N2).

Thus, the maximum condition (iv) yieldsdS = dS1 +dS2 = 0 (v)

Now S1 = S1 (E1,V1,N1).

∴ dS1 =

(∂S1

∂E1

)

V1,N1

dE1 +

(∂S1

∂V1

)

E1,N1

dV1 +

(∂S1

∂N1

)

E1,V1

dN1 (vi)

In a simple case, when the number N1 is fixed, the fundamental thermodynamic relation asserts that

dS1 =1T1

(dE1 + p1dV1) (vii)


where T1 and p1 are respectively the temperature and mean pressure of phase 1. Under this circum-stance, dN1 = 0. A comparison of the coefficients of dE and dV in (vi) and (vii) yields

(∂S1

∂E1

)

V,N1

=1T1

(viii)(

∂S1

∂V1

)

E1,N=

p1

T1(ix)

Let µ1 = −T1

(∂S1∂N1

)

E1,V1. The quantity µ1 is the chemical potential per molecule in phase 1.

Then (vi) can be written in the form

dS1 =1T1

dE1 +p1

T1dV1 −

µ1

T1dN1 (x)

Similarly,

dS2 =1T2

dE2 +p2

T2dV2 −

µ2

T2dN2 = −

1T2

dE1 −p2

T2dV1 +

µ2

T2dN1 (xi)

Using (x) and (xi) in (v)(

1T1

−1T2

)

dE1 +

(p1

T1−

p2

T2

)

dV1 −

(µ1

T1−

µ2

T2

)

dN1 = 0 (xii)

Since (xii) is to be valid for arbitrary variations dE1,dV1,dN1, it follows that the coefficients of allthese differentials must vanish separately. Thus, one obtains

(1T1

−1T2

)

= 0(

p1

T1−

p2

T2

)

= 0(

µ1

T1−

µ2

T2

)

= 0

whenceT1 = T2 (xiii)p1 = p2 (xiv)µ1 = µ2 (xv)

These are the necessary conditions for equilibrium between two phases. (xiii) is the condition forthermal equilibrium, (xiv) for mechanical equilibrium and (xv) for diffusive equilibrium.


where v is 2f-dimensional velocity of the representative points and n is a unit vector, outwardnormal to the point of location of the surface element ds of the surface s, enclosing the volume τand dτ = dq1 . . .dq f d p1 . . .d p f .

Applying Gauss’ divergence theorem∮

sρv · nds =

∫

τ∇ · (ρv)dτ (2.5.3)

Using (2.5.3) in (2.5.2)∫

τ

[∂ρ∂t +∇ · (ρv)

]

dτ = 0. Since τ is arbitrary,

∂ρ∂t

+∇ · (ρv) = 0 (2.5.4)

This is the well-known equation of continuity and is a statement of the conservation of density inphase space. Here ∇ is a 2f-dimensional del operator.

The second term of (2.5.4) can be written as

∇ · (ρv) =f

∑i=1

[∂

∂qi(ρqi)+

∂∂pi

(ρpi)

]

= ρf

∑i=1

[∂qi

∂qi+

∂pi

∂pi

]

+f

∑i=1

[∂ρ∂qi

qi +∂ρ∂pi

pi

] (2.5.5)

In view of the Hamilton’s equation (2.5.1), the first term isf

∑i=1

[∂2H

∂qi∂pi−

∂2H∂pi∂qi

]

= 0

∴ Equation (2.5.4) takes the final form

∂ρ∂t

+f

∑i=1

[∂ρ∂qi

qi +∂ρ∂pi

pi

]

= 0 (2.5.6)

Again using Hamilton’s equation (2.5.1), we can put (2.5.6) in another form:

∂ρ∂t

+f

∑i=1

[∂ρ∂qi

∂H∂pi

−∂ρ∂pi

∂H∂qi

]

= 0

⇒∂ρ∂t

+[ρ,H] = 0 (2.5.7)

where [ρ,H] =f

∑i=1

[∂ρ∂qi

∂H∂pi

−∂ρ∂pi

∂H∂qi

]

, the Poisson bracket.

Since ρ is a function of q’s, p’s and t, the left-hand side is the total time derivative,dρdt

.

∴

∂ρ∂t

=∂ρ∂t

+[ρ,H] = 0 (2.5.8)


This is known as Liouville’s equation.Thus, ρ(q1, . . .q f , p1, . . . p f , t) remains constant along the dynamic trajectories, i.e., the lines

whose parametric equations qi = qi (t), pi = pi (t) are derived from the equations of motion (2.5.1),in Γ-space. This is known as Liouville’s theorem.

2.5.1 Statistical equilibriumTo get an equilibrium distribution, we need

∂ρ∂t

= 0 (2.5.9)

which immediately implies that∂ρ∂qi

qi +∂ρ∂pi

pi = 0 (2.5.10)

Using the Hamilton’s equation (2.5.1)∂ρ∂qi

∂H∂pi

−∂ρ∂pi

∂H∂qi

= 0 (2.5.11)

which is easily satisfied, if we take

ρ(qi, pi) = ρ(H(qi, pi)) (2.5.12)

because then∂ρ∂qi

∂H∂pi

= 0 and∂ρ∂pi

∂H∂qi

= 0

Thus, there is a restriction (2.5.12) on the equilibrium distribution. In other words, the equilibriumdistribution corresponds to the situation where ρ is constant on surfaces of constant energy H inphase space. This is indeed the basic assumption of statistical mechanics.

QUESTIONS1. Give the concept of ensemble.2. What do you mean by microcanonical ensemble? For what type of system is it suitable?3. Show that Ω(E) ∼ E f , i.e., Ω(E) is a rapidly increasing function of the energy E of the

system.4. Discuss the specific heat behaviour of a two-level system. What is Schottky anomaly?5. What do you mean by canonical ensemble? For what type of system is it suitable?6. For a canonical system, derive the Boltzmann’s canonical probability distribution law,

pr = e−βEr

∑r

e−βErwhere β = 1

kT .

7. What is partition function? What is its importance in statistical mechanics?8. Show that the energy fluctuation in a canonical system is proportional to 1√

N9. How can you justify the equivalence between a microcanonical ensemble and a canonical

ensemble?


3. Consider a system of N particles at a temperature T . Each particle has mass m and free to ex-ecute one-dimensional simple harmonic motion about its mean position. If the restoring forceis proportional to x, calculate the specific heat of the system. Assume that the temperature Tis high enough so that classical statistics is applicable.

4. Find the canonical distribution function for a body rotating with angular velocity.5. Suppose that two atoms interact with the mutual potential energy of the form

U (x) = U0

[(ax

)12−2

(ax

)6]

, where x is the separation between the two particles. The

particles are in contact with a heat reservoir at an absolute temperature T such that kT U0,but high enough so that classical statistics is applicable. Calculate the mean separation x(T )

of the particles and use it to compute the expansion coefficient α =1x

∂x∂T

.

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