DEPARTMENT OF PHYSICS AND ASTROPHYSICS

UNIVERSITY OF DELHI

B.Sc.(H) PHYSICS (CBCS)

SEMESTER - VI

STATISTICAL MECHANICS

Model Problem Set

1
1 Classical Statistics
1.1 Macrostate and Microstate, Phase Space, Ensemble, Thermodynamic
Probability

Problem 1: Consider a particle undergoing simple harmonic motion such that the position of the
particle changes with time as x = x0 cos(ωt + φ), where the phase φ is completely unknown, and therefore
the position of the oscillator is not known. One therefore has to resort to determining the probability
that the position of the oscillator lies between x and x + dx.

(a) This probability must be proportional to the time the oscillator spends between x and x + dx. Find
the speed of the oscillator at position x as a function of x, ω and x0 . Using this expression, determine
the probability p(x)dx that the position of the oscillator is between x and x + dx.

(b) Let the energy of the oscillator lie between E and E + ∆E, where ∆E << E. Sketch the phase space
and the region accessible to the particle, calculating the volume of the accessible region. Next,
compute the ratio of the volume of the accessible phase space corresponding to the position of the
particle lying between x and x + dx and the total volume of the accessible phase space. What does
this result signify?

Solution 1:

(a) The energy of the oscillator is E = mω 2 x20 /2. If the position of the oscillator is x and its speed is v,
then
1 1 1
mv 2 + mω 2 x2 = mω 2 x20
2 2 2q
=⇒ v = ω x20 − x2

Since dx/dt = v, therefore the time it spends between x and x + dx is

dx
dt =
v
dx
= p
ω x20 − x2

The probability that the position is between x and x + dx is then of the form
N dx
P (x)dx = p
ω x20 − x2
R x0
where N is a normalisation constant, to be determined by the constraint −x0 dxP (x) = 1. Evaluating
the integral gives N = ω/π. Then
dx
P (x)dx = p 2
π x0 − x2

(b) Given the relation

p2 1
+ mω 2 x2 = E
2m 2
the phase space trajectory corresponding to energy E forms an ellipse in the p − x plane, with the
region of the phase space corresponding to energy lying between E and E + ∆E being the region
between two ellipses. To compute the volume of this region, we first compute the area enclosed by the

2
 elliptical region corresponding to the energy lying between zero and E. Given the above equation, this
volume (area, in this case) is easily computed to be
2πE
Γ=
ω
Then the volume of the region corresponding to energy lying between E and E + ∆E is
2π∆E
∆Γ =
ω
The volume of the region corresponding to position lying between x and x + dx is

δΓ = dxdp

where dp is computed using the relation between p, x and E. For a given x, p has two values, one
positive and one negative. The volume is then twice the volume computed assuming that p is positive,
given by p
p = 2mE − m2 ω 2 x2
Then
m∆E
dp = √
2mE − m2 ω 2 x2
Using the fact that E = mω 2 x20 /2, we get
2∆E dx
δΓ = p
ω x20 − x2
Then
δΓ dx
= p 2
∆Γ π x0 − x2
which is the same as the probability of the oscillator’s position lying between x and x + dx.

Problem 2: Consider an isolated system of four non-interacting spins labelled 1, 2, 3, and 4, each with
magnetic moment m, interacting with an external magnetic field B. Each spin can be parallel (‘up’) or
antiparallel (‘down’) to B, with the energy of a spin parallel to B equal to  = −mB and the energy of a
spin antiparallel to B equal to  = +mB. Let the total energy of the system be E = −2mB.
(a) How many microstates of the system correspond to this macrostate? Enumerate these microstates.

(b) What is the probability that the system is in a given microstate in equilibrium?

(c) What is the probability that a given spin points up? Use this probability to campute the mean
magnetic moment of a given spin in equilibrium.

(d) What is the probability that if spin 1 is ‘up’, spin 2 is also ‘up’ ?

Solution 2: Since the energy of the system is E = −2mB, the total magnetic moment of the system is
M = +2m, which corresponds to one spin ‘down’ and three spins ‘up’.
(a) The total number of microstates is then
4!
Ω =
3! × 1!
= 4

3
(b) The probability of any given microstate is
1
Pr =
Ω
1
=
4

(c) Given that a particular spin is ‘up’, there are three microstates corresponding to this (corresponding
to one of the remaining spins being ‘down’). Therefore, the probability is
3
Pu =
4
and the probability of the given spin being ‘down’ is
1
Pd =
4
The mean magnetic moment of a give spin is

m = m × Pu + (−m) × Pd
m
=
2

(d) Given that spin 1 is up, there are in all three microstates corresponding to this. Of these, two
correspond to spin 2 also being up. Therefore, the propbability that spin 2 is also up is
2
P1u|2u =
3

Problem 3: Consider a system of four non-interacting distinguishable particles, with each particle
localised to a lattice site. The energy of each particle is is restricted to values  = 0, 0 , 20 , 30 , .... The
system is divided into two subsystems A and B, subsystem A consisting of particles 1 and 2, and B
consisting of particles 3 and 4 respectively. A and B are initially thermally insulated from each other,
with energies EA = 50 and EB = 0 . What are the possible microstates of the composite system? Now,
suppose the two subsystems are allowed to thermally interact with each other, so that they can exchange
energy without the total energy of the system changing. After equilibrium is attained, enumerate the
possible microstates of the composite system. In equilibrium, what is the probability that subsystem A
has energy EA , for EA = 0, 0 , 20 , .., 60 ? For what value of EA is the probability maximum?

Solution 3: Let us denote the microstates of the composite system as (−, −|−, −) where the first two
slots are for system A and the second two slots for system B. Given that system A has energy 50 and
system B has energy 0 gives the following 12 microstates for the composite system:
(0, 5|0, 1), (0, 5|1, 0), (5, 0|0, 1), (5, 0|1, 0), (1, 4|0, 1), (1, 4|1, 0), (4, 1|0, 1), (4, 1|1, 0), (2, 3|0, 1), (2, 3|1, 0), (3, 2|0, 1)
and (3, 2|1, 0), where all numbers are in units of 0 . Once the system reaches equilibrium, the total
number of microstates is given by the total number of ways of distributing energy 60 among 4 particles.
(6 + 4 − 1)!
Ω =
6!(4 − 1)!
9!
=
6!3!
= 84

The probability that system A has energy EA is given by the product of the number of microstates of
system A corresponding to its energy being EA and the number of microstates of system B corresponding

4
to its energy being EB = E − EA , divided by the total number of microstates (here, E = 60 ). For
example, let EA = . Then, the number of microstates accessible to A corresponding to this is 2. The
energy available to system B is 5, which can be distributed among 2 particles in 6 different ways. Then,
the total number of microstates corresponding to system A having this energy is Ω(EA = 0 ) = 2 × 6 = 12.
The probability of this is then
Ω(EA = 0 )
P (EA = 0 ) =
Ω
12
=
84
1
=
7
Similarly, probabilities of other possible values of energy can be computed, and it can be checked that the
probability is maximum for EA = 30 , and is equal to 4/21.

1.2 Entropy and Thermodynamic Probability

Problem 1: Consider a system of N particles (which could be interacting with each other) with energy
E and occupying a volume V . The entropy of the system is known to be extensive. Suppose the energy
of the system is changed, such that the new energy is λE, where λ is a multiplicative factor. Can you say
that the new entropy will be λS, where S is the original entropy? If not, what other changes will be
needed such that this is true?

Solution 1: Since entropy is extensive, therefore, entropy will be λS under N → λN, V → λV and
E → λE simultaneously.

Problem 2: Consider a system of N >> 1 weakly interacting particles, each of which can be in quantum
states with energies 0, , 2, 3, .... Given the system has a certain energy, the temperature of the system
is given by
1 ∂S
=
T ∂E
∆S
'
∆E
where ∆S is the change in the entropy of the system due to the change in the energy of the system by
∆E.

(a) If the system is in its ground state, what is its entropy?

(b) If the total energy of the system is , what is its entropy?

(c) What is the change in entropy of the system if the total energy of the system is increased from  to
2?

(d) Given the above definition of temperature, what is the temperature of the system if its total energy is
?

Solution 2:

5
(a) The ground state corresponds to a unique microstate, in which all the particles have energy 0. Its
entropy, therefore, is zero.
(b) If the total energy is , there are N possible microstates corresponding to this, in which one particle
has energy  and others have energy 0. The entropy is S = kB ln N .
(c) If the total energy is 2, this corresponds to either two particles with energy  (which corresponds to
N (N − 1)/2 possible microstates) or one particle with energy 2 (which corresponds to N possible
microstates). Therefore, the total number of microstates is Ω = N + N (N − 1)/2 = N (N + 1)/2.
Then, the entropy of the system is

S 0 = kB ln (N (N + 1)/2)
' kB ln N + kB ln(N/2) (∵ N >> 1)

Then, the change in entropy is ∆S = kB ln(N/2).

(d) The temperature of the system is given by
1 ∆S
'
T ∆E
kB
= ln(N/2)

Therefore,
 1
T =
kB ln(N/2)

Problem 3: A system of four weakly interacting distinct particles is such that each particle can be in
one of four states with energies , 2, 3 and 4 respectively. If the system has total energy 15, what is
the entropy of the system? For what possible values of total energy is the entropy of the system zero?

Solution 3: Since the total energy of the system is 15, this corresponds to three particles with energy
4 and one particle with energy 3. There are four microstates corresponding to this. Therefore, the
entropy of the system is S = kB ln 4. The entropy of the system is zero if either the total energy of the
system is 4 or 16, since these macrostates correspond to one microstate each.

Problem 4: Consider a lattice of N non-interacting distinguishable particles, with each particle localised
to a lattice site. The energy of each particle is restricted to values  = 0, 0 , 20 , 30 , .... The system is in
equilibrium.
(a) If the energy of the system is E, what is the number of microstates of the system?
(b) Find an expression for the entropy of the system as a function of energy and simplify it using
Sterling’s approximation ln n ' n ln n − n for n >> 1.
(c) Using the relation
1 ∂S
=
T ∂E
determine a relation between the energy of the system and its temperature.
Hint: The problem of determining the number of microstates can be reduced to counting the number of
ways of arranging a certain number of sticks and a certain number of dots along a line.

Solution 4:

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(a) A microstate of the system can be represented as the set (n1 , n2 , .., nN ) which represents a state in
which the first particle has energy n1 0 , second has energy n1 0 ,... Let M = E/. Then,

n1 + n2 + ..nN = M

The total number of microstates is the number of ways of choosing integers n1 , n2 , .., nN such that
their sum is M . This is the same as the number of ways of partitioning M into N parts. This can be
visualised as the number of ways M dots and N − 1 sticks can be arranged in a line, the sticks
creating partitions. Therefore, the number of microstates is
(M + N − 1)!
Ω(E) =
M !(N − 1)!
(E/ + N − 1)!
=
(E/)!(N − 1)!
(E/ + N )!
'
(E/)!N !
assuming E, N >> 1.

(b) The entropy of the system is

S = kB ln Ω
    
E E E E
' kB + N ln + N − ln − N ln N
   
where Sterling’s approximation has been used.

(c) The temperature of the system is given by

1 ∂S
=
T ∂E  
kB N
= ln 1 +
 E
which is inverted to give
N
E=
e/kB T −1

1.3 Maxwell Boltzmann Distribution

Problem 1: Consider atomic hydrogen in thermal equilibrium at temperature T . Estimate the ratio of
the number of atoms with energy E = −3.4 eV to the number of atoms with energy E = −13.6 eV for
T = 1000◦ K.

Solution 1: It is useful to determine the temperature equivalent to energy equal to 1ev. This
temperature is
1 ev
Tev =
kB
= 1.16 × 104 ◦ K

There are two microstates corresponding to the atom in its ground state (with energy −13.6 eV)
corresponding to the two spin states of the electron. Similarly, there are eight microstates corresponding

7
to the first excited states (with energy −3.4 eV). Then, the relative probability of the atom being in one
of the first excited states is
Pexc 8 × e3.4ev/kB T
=
Pgr 2 × e13.6ev/kB T
= 4e−10.2ev/kB T
= 4e−10.2 Tev /T

= 4e−118.32
which is vanishingly small.

Problem 2: A system of N weakly interacting particles, each of mass m, is in thermal equilibrium at

temperature T . The system is contained in a cubical box of side L, whose top and bottom surfaces are
parallel to the Earth’s surface, where the acceleration due to gravity is g. A coordinate system is set up
with the origin at the centre of the base of the box and the positive z axis along the vertical direction,
such that the ranges of coordinates accessible to any particle are −L/2 ≤ x ≤ L/2, −L/2 ≤ y ≤ L/2,
0 ≤ z ≤ L.
(a) What is the probability that a given particle has velocity in the range (vx , vy , vz ) and
(vx + dvx , vy + dvy , vz + dvz )?
(b) What is the probability that a given particle has x coordinate between x and x + dx?
(c) What is the probability that a given particle has y coordinate between y and y + dy?
(d) What is the probability that a given particle has z coordinate between z and z + dz?
(e) From the above probability distributions, calculate the mean kinetic and potential energies of a
particle.

Solution 2: The probability of a microstate of a particle corresponding to its position lying between
(x, y, z) and (x + dx, y + dy, z + dz) and velocity between (vx , vy , vz ) and (vx + dvx , vy + dvy , vz + dvz ) is
given by the Maxwell Boltzmann distribution
2 2 2
P (x, y, z, vx , vy , vz )dxdydzdvx dvy dvz = N e−β[m(vx +vy +vz )/2+mgz]
R
where N is a normalisation constant such that P dxdydzdvx dvy dvz = 1. The integral over velocity
components is Gaussian and gives (2π/mβ)3/2 . The integral over position coordinates gives
Z L/2 Z L
I = dxdy dze−βmgz
−L/2 0
L2  
= 1 − e−βmgL
βmg
Then, the normalisation constant is determined to be
mβ 3/2
 
βmg
N=
2π L2 (1 − e−βmgL )
(a)
Z
P (vx , vy , vz )dvx dvy dvz = dvx dvy dvz dxdydzP (x, y, z, vx , vy , vz )
 3/2
mβ 2 2 2
= e−βm(vx +vy +vz )/2 dvx dvy dvz
2π
which is just the Maxwell velocity distribution.

8
(b)
Z
P (x)dx = dx dydzdvx dvy dvz P (x, y, z, vx , vy , vz )
dx
=
L

(c)
Z
P (y)dy = dy dxdzdvx dvy dvz P (x, y, z, vx , vy , vz )
dy
=
L

(d)
Z
P (z)dz = dz dxdydvx dvy dvz P (x, y, z, vx , vy , vz )
βmg
= e−βmgz dz
(1 − e−βmgL )

(e) The mean kinetic energy is

1
K = 3 × mvx2
2 Z
1
= 3 × m dvx P (vx )vx2
2
mβ 1/2
  Z
1 2
= 3× m dvx vx2 e−βmvx /2
2 2π
3
=
2β
The mean potential energy is

U = mgz
Z L
= mg dz zP (z)
0
Z L
βmg
= mg dzz e−βmgz
(1 − e−βmgL ) 0
 
1 βmgL
= 1 − βmgL
β (e − 1)

Problem 3: A two-dimensional solid at temperature T contains N negatively charged impurity ions per
unit area, the negative ions replacing some ordinary atoms of the solid. The solid as a whole is
electrically neutral, since each negative ion with charge −e has in its vicinity one positive ion with charge
+e. The positive ion, much smaller, is free to move between each of the four equidistance sites A, B, C
and D surrounding the stationary negative ion, as shown. The spacing between the these sites is a and
the energy of interaction of the positive ion with the stationary negative ion is −0 for each lattice site

9
(a) What are the relative probabilities of the positive ion being found at the four lattice sites?
(b) The solid is placed in a region of a uniform electric field of magnitude E, as illustrated above. Taking
the origin at the location of the negative ion, determine the interaction energy of the system with the
external electric field at the four lattice sites (the interaction energy is Eint = −~ ~ where p~ is the
p·E
dipole moment of the system).
(c) What now are the relative probabilities of the positive ion being found at the four lattice sites?
(d) The mean polarisation of the solid is the mean dipole moment per unit area along the direction of the
electric field. Calculate the polarisation of the solid as a function of temperature and the external
electric field E.
(e) Calculate the expression for the polarisation at ‘high’ temperatures. What temperatures are ‘high’ ?

Solution 3: The system has four microstates, corresponding to the four possible locations of the positive
ion.
(a) In absence of an external electric field, each microstate has the same energy. Therefore, they are all
equiprobable.
(b) The magnitude
√ of the dipole moment of the system corresponding to each position of the positive ion
is p0 = ea/ 2. Given that the interaction energy of a dipole is Eint = −~ ~ the energies of the four
p · E,
microstates are respectively A = D = +eaE/2 and B = C = −eaE/2.
(c) The probability of the ion being found at sites A and D are equal. Similarly, the probability of the ion
being found at sites B and C are equal. The reltive probabilities of the ion being found at sites A and
B is
PA e−βA
= −β
PB e B
= e−βeaE

(d) If the polarisation of the dipole at a lattice site is p~, the mean polarisation is the average value of
~ The value of this quantity at the four lattice sites is
pk = p~ · Ê, where Ê is a unit vector along E.
pkA = pkD = −ea/2 and pkB = pkC = +ea/2. Given that PA + PB + PC + PD = 1 and given the ratio
PA /PB , the probabilities are computed to be
 
1 1
PA =
2 1 + eβeaE
 βeaE 
1 e
PB =
2 1 + eβeaE
Then,
pk = 2PA pkA + 2PB pkB
ea eβeaE − 1
 
=
2 eβeaE + 1
 
ea βeaE
= tanh
2 2

10
 Since there are N impurities per unit area, therefore, the mean polarisation is

P = N pk
 
N ea βeaE
= tanh
2 2

‘High’ temperatures correspond to the condition βeaE/2 << 1. Under such conditions,
 
N ea βeaE
P '
2 2
2
(ea) E
= N
4kB T

Problem 4: A sensitive spring balance consists of a quartz spring with spring constant k. This balance
is used to measure the mass of very tiny, light objects by suspending them from the balance and
observing the extension in the spring. Consider a tiny object of mass m suspended from the spring. The
object is in an environment which is at temperature T , and gets ‘kicked’ around by it, reaching
equilibrium with the environment.
1. What is the potential energy of the system if the spring is extended by x?
2. What is the probability that the spring is extended by x relative to its equilibrium length?
3. Calculate the mean extension x and the mean squared extension (x − x)2 .
4. Comparing the square root of the mean squared extension with the mean extension, extimate the
minimum mass that can be reliably measured.

Solution 4:
(a) If the position of the mass relative to the relaxed position of the spring is x, then its potential energy
is
1 2
U (x) = kx + mgx
2
1 m2 g 2
= k(x + x0 )2 −
2 2k
where x0 = mg/k.
(b) The probability of the position of the mass lying between x and x + dx is

P (x)dx = N e−βU (x) dx

R∞
where N is determined by the normalisation condition −∞ dxP (x) = 1. The integral is a Gaussian,
resulting in
r
βk −βk(x+x0 )2 /2
P (x)dx = e dx
2π

(c) The mean extension is clearly x = −x0 . The mean square extension is
Z ∞
2
(x − x) = dx(x + x0 )2 P (x)
−∞
kB T
=
k

11
 q
(d) The fluctuation in position is ∆x = (x − x)2 . For the measurement to be reliable, ∆x << x0 . This
results in the condition
1p
m >> kB T k
g

1.4 Partition Function, Heat Capacity, Entropy

Problem 1: Consider a single particle system with six states. There is one state with energy 0, two
states with energy  and two states with energy 2. The system is in equilibrium with a heat bath at
temperature T .

(a) Calculate the partition function for the system.

(b) Calculate the mean energy and heat capacity of the system as functions of temperature.

(c) What is the relative probability of the system having energy 2 and ?

Problem 2: The partition function of a system is given by

lnZ = aT 4 V

where T is the absolute temperature, V is the volume of the system and a is a constant. Evaluate the
mean energy, pressure and entropy of the system.

Problem 3: Consider a simplified model of graphite, in which each carbon atom acts as a harmonic
oscillator, oscillating with frequency ω within the layer and frequency ω 0 perpendicular to it. The
oscillations in the three directions are independent, such that the expression for energy of a carbon atom
is
1 2 m
E= (px + p2y + p2z ) + (ω 2 x2 + ω 2 y 2 + ω 02 z 2 )
2m 2
where coordinates x, y are in the plane of the layer and z is perpendicular to it. The sample is at
temperature T , such that ~ω >> T and ~ω 0 << T (the restoring forces in the plane of the layer are much
stronger than those perpendicular to it).

(a) Given the temperature conditions, one kind of the oscillations (in the plane or perpendicular to it)
can be treated classically, and the other quantum mechanically with only the ground and first excited
states appreciably populated. Identify the corresponding oscillations.

(b) Taking into account the above considerations, calculate the partition function and show that it
factorises into three factors, two of which are identical.

(c) Find an expression for the molar specific heat of the system as a function of temperature, using
approximations appropriate to the temperature conditions stated above.

Solution 3:

(a) Given the temperature conditions, the oscillations perpendicular to the plane can be treated
classically, and those in the plane need to be treated quantum mechanically.

(b) Since the expression for energy is additive in the contributions along the three independent directions,
the partition function for each atom will factorise as

Z = Zx Zy Zz

12
 where further, Zx = Zy . Given that the motion perpendicular to the plane can be treated classically, it
follows that
Z
2 02 /
Zz = dzdpz e−β(pz /2m+mω z 2)
2π
=
ω0β
For the motion in the plane, given that ~ω >> T , only the ground and the first excited state
contributions are relevant. Then
1
X
Zx = e−β(n+1/2)~ω
n=0
−β~ω/2
= e + e−3β~ω/2 

= e−β~ω/2 1 + e−β~ω

Finally, the partition funciton for a single atom is

Z = Zx2 Zz
 2  2π 
−β~ω −β~ω
= e 1+e
ω0β

(c) The logarithm of the partition function is

βω 0
   
−β~ω
ln Z = −β~ω + 2 ln 1 + e − ln
2π

The mean energy per atom is

∂ ln Z
 = −
∂β
2~ω
= ~ω + kB T +
1 + eβ~ω
The heat capacity per atom is
∂
c =
∂T " #
 2
~ω 1
= kB 1 + 2
2
kB T 1 + e~ω/kB T

Therefore, the molar specific heat of the system will be

" #
~ω 2
 
1
C =R 1+2
2
kB T 1 + e~ω/kB T

Problem 4: N diatomic molecules are stuck on a surface. Each molecule can either lie flat on the
surface (in which case it can orient itself either along the x or the y direction) or it can stand up
perpendicular to the surface (along the z direction). Assume that the flat configurations have zero energy
and the configuration perpendicular to the surface has energy  > 0. The system is in thermal
equilibrium at temperature T > 0.

13
(a) Calculate the partition function of the system.
(b) Calculate the mean energy of the system. What is the largest possible value for this energy (attained
by changing the temperature)?
(c) Calculate the heat capacity of the system as a function of temperature.
(d) What is the probability of a given molecule ‘standing up’ ?

Solution 4: For each molecule, there are three microstates corresponding to it oriented along the x, y or
z directions. The energies of these microstates are x = y = 0, z = .
(a) The partition function for a single molecule is
Z1 = 1 + 1 + e−β
= 2 + e−β
Then, since the molecules are independent, the partition fucntion of the system is
Z = Z1N
 N
= 2 + e−β

(b) The mean energy of the system is

∂ ln Z
E = −
∂β
N
=
1 + 2eβ
As T → ∞, β → 0. In this limit, E → N /3.
(c) The heat capacity of the system is
∂E
C =
∂T
2
eβ


= 2N kB
kB T 1 + 2eβ

(d) The probability of a given molecule standing ‘up’ is

1 −β
Pz = e
Z1
1
=
1 + 2eβ

1.5 Negative Temperatures

Problem 1: Consider an isolated system of N >> 1 weakly interacting, distinct particles in equilibrium.
Each particle can be in one of three states, with energies 0,  and 2 respectively. Given the system has a
certain energy, the temperature of the system is given by
1 ∂S
=
T ∂E
∆S
'
∆E

14
where ∆S is the change in the entropy of the system due to the change in the energy of the system by
∆E.

(a) Let the entire system be in its ground state. What is its entropy? If the anergy ∆E =  is added to
the system, what is its entropy? Given the definition of temperature above, what can you say about
the temperature of the system if it is in the ground state?

(b) Let the total energy of the system be 2N  − . What is the entropy of the system? What is the
entropy of the system if energy ∆E =  is added to it? If the system has energy 2N  − , what can
you say about the temperature of the system?

Solution 1:
(a) If the system is in its ground state, there is only one microstate corresponding to this (all particles
with energy 0). Therefore, the entropy of the system is zero. If energy ∆E =  is added to it, the
energy of the system is . This corresponds to N possible microstates, in which one particle has energy
 and all others have energy 0. Then, the entropy of the system is S = kB ln N . The temperature of
the system in the ground state is positive, since an increase in energy increases entropy.

(b) If the total energy of the system is 2N  − , this corresponds to N microstates in which one particle
has energy  and all others have energy 2. The entropy of the system is then S = kB ln N . If energy
 is added to the system, the number of microstates reduces to one, corresponding to all the particles
having energy 2. The entropy therefore reduces to zero. Since addition of energy decreases entropy,
the temperature of the system is negative.

Problem 2: Consider an isolated system of N >> 1 weakly interacting, distinct particles in equilibrium.
Each particle can be in one of M states with energies 0 , 20 , ..., M 0 . Can this system exhibit negative
temperatures? If so, give a value of energy corresponding to which the temperature of the system is (a)
positive (b) negative. (c) If M −→ ∞, will the system exhibit negative temperatures? Give a physical
argument.

Solution 2: The system does exhibit negative temperatures, since the energy of the system is bounded
from above, the maximum possible energy being N M 0 . (a) The system in the ground state has a positive
temperature, since adding energy 0 to the system increases its entropy by ∆S = kB ln N . (b) The system,
when its energy is M N 0 − 0 , has a negative temperature, since increasing its energy by 0 decreses its
entropy by ∆S = −kB ln N . (c) As M → ∞, increasing the energy of the system will always increase its
entropy. Therefore, the system will always have a positive temperature.

Problem 3: Consider two systems A and B, system A consisting of NA >> 1 weakly interacting
particles, each of which can be in one of an infinite number of possible quantum states with energies
0, , 2, 3, .... System B on the other hand consisting of NB >> 1 weakly interacting particles, each of
which can be in one of two quantum states with energies 0, . Initially, these systems are insulated from
each other, with system A having total energy NA  and B having energy 3NB /4.
(a) What can you say about the sign of the temperatures of these two systems?

(b) The systems are now made to interact with each other, till they reach equilibrium. What is the sign
of the temperature of each system after equilibrium is attained?

Solution 3:

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(a) Since the energy of system A is not bounded from above, it will always have a positive temperature.
System B, on the other hand, can exhibit both positive and negative temperatures. It will have a
negative temperature for energies greater than NB /2, since for energies greater than this, increasing
the energy of the system will decrease its entropy. Given that its energy is 3NB /4, it has a negative
temperature.

(b) After the systems interact and attain equilibrium, since system A always has a positive temperature,
system B will be driven to a positive temperature as well, as a result of losing a part of its energy to
system A to maximise the overall entropy.

1.6 Equipartition Principle

Problem 1: Consider a classical system of N >> 1 independent oscillators, each of which has energy
given by
p2 1
= + kx2
2m 2
where x and p are the position and momentum of the particle. If the system is in equilibrium at
temperature T , what is the molar specific heat of the system? If the expression for energy has a small
correction which is not quadratic in x, what will be the qualitative change in the behaviour of the specific
heat of the system?

Solution 1: From the Equipartition Principle, the molar specific heat of the system will be cv = R. If
there is a non-quadratic correction, the Equipartition Principle is not applicable, and the molar specific
heat will in general depend on the temperature.

Problem 2: Consider a classical system of N >> 1 weakly interacting particles, each of which has
energy given by
p2 1
= + kx2 + λx4
2m 2
where x and p are the position and momentum of the particle. The system is in equilibrium at
temperature T .

(a) If λ = 0, what is the moler specific heat of the system?

(b) If λ is not zero, but the quartic term is very ‘small’ compared to the quadratic term, what will be the
variation in the specific heat of the system with temperature? Hint: Use eu ' 1 + u for ‘small’ u.

Solution 2:

(a) If λ = 0, the energy is quadratic in position and momentum. Therefore, the Principle of Equipartition
is valid, and the molar specific heat of the system will be cv = R.

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(b) If λ 6= 0, the partition function of the system will be
Z
2 2 4
Z = dxdp e−β(p /2m+kx /2+λx )
Z
2 2
dxdp e−β(p /2m+kx /2) 1 − βλx4


' since the correction to energy is ‘small’
r r Z ∞
2π m 2mπ 2
= − βλ × dx x4 e−βkx /2
β k β −∞
√  
2mπ 3 π 2 5/2
r r
2π m
= − βλ ×
β k β 4 βk
r  
2π m 3λ
= 1−
β k βk 2
Therefore
 
3λ
ln Z = ln Z0 + ln 1 −
βk 2
3λ
' ln Z0 − (∵ correction is ‘small’)
βk 2
where Z0 is the partition function without the correction. The mean energy per particle is
∂ ln Z
 = −
∂β
3λ(kB T )2
= 0 −
k2
where 0 is the mean energy paer particle if λ = 0. The heat capacity per particle is
∂
c =
∂T
6kB λ(kB T )
= c0 −
k2
Therefore, the molar specific heat of the system is
cv = NA × c
6Rλ(kB T )
= R−
 k2 
6λ(kB T )
= R 1−
k2

Problem 3: Consider a weakly interacting system of particles, such that the expression for energy of any
one particle consists of n terms, quadratic in position and momentum components. If classical physics is
an adequate description of this system, what is the molar specific heat of the system? If the temperature
of the system is progressively lowered, will the experimentally measured molar specific heat be in
agreement with this result? Explain.

Solution 3: If Classical Physics is an adequate description, then the Principle of Equipartition can be
applied, which gives the moolar specific heat to be cv = nR/2. As the temperature is lowered, eventually,
quantum effects become significant, and the molar specific heat will be found to be temperature dependent,
in disagreement with the classical result.

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