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A1: Thermodynamics, Kinetic Theory and Statistical
Mechanics

Toby Adkins

June 20, 2016

Contents
1 Basic Thermodynamics

1
...
2 The Zeroth Law
1
...
3
...
3
...
3
...
4 The Second Law
1
...
1 Carnot's Theorem
1
...
2 Back to Clausius and Kelvin
1
...
3 Clausius' Theorem and Entropy
1
...
4 Calculating Entropy Changes
1
...
5 The Joule Expansion
1
...
5
...
5
...
5
...
5
...
6 Thermodynamics of Other Materials
1
...
1 Elastic Rod
1
...
2 Liquid Film
1
...
3 Paramagnetism

2 Kinetic Theory

2
...
1
...
2 Distributions and Isotropy
2
...
1 Isotropic Distributions
2
...
2 The Maxwellian Distribution
2
...
3
...
3
...
4 Collisions
2
...
1 The Characteristic Velocity
2
...
2 Collisions in a Mixture
2
...
5
...
5
...
5
...
5
...
6 Local Equilibrium
1

3

4
6
7
7
9
11
15
15
16
17
18
20
21
21
22
23
24
26
26
27
28

30

31
31
33
33
34
36
36
39
41
41
42
43
43
44
46
47
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2
...
1 The Kinetic Equation
2
...
2 Conservation Laws
2
...
3 Self-Diusion 2
...
1 Basic Principles
3
...
1 States of a System
3
...
2 Gibbs Entropy
3
...
3 Entropy Maximisation
3
...
4 Stirling's Formula
3
...
2
...
2
...
2
...
3 The Grand Canonical Ensemble
3
...
1 Chemical Potential
3
...
2 The Grand Potential
3
...
3 Multi-species Systems
3
...
4
...
4
...
4
...
4
...
4
...
5 Quantum Gases
3
...
1 Occupation Number Statistics
3
...
2 Density of States
3
...
3 Standard Calculations
3
...
4 Degenerate Fermi Gas
3
...
5 Degenerate Bose Gas
3
...
6
...
6
...
1 Real Gases
4
...
1 Virial Expansion
4
...
2 Van-der-Waals Gas
4
...
3 Dieterici Gas
4
...
4 Critical Points
4
...
5 Expansions of Real Gases
4
...
2
...
2
...


Basic Thermodynamics

This chapter aims to cover the basic concepts of Thermodynamics, including:
• Some Denitions
• The Zeroth Law
• The First Law
• The Second Law
• Thermodynamic Potentials
• Thermodynamics of Other Materials

Students will nd that a lot of the material covered in this chapter has an almost purely
experimental basis; many relations in Thermodynamics come from the results of experiment, rather than derivation from some fundamental concept, such as the value of the
adiabatic constant for particular substances
...
It simply means more results that can be quoted without proof
...

¯

3

Toby Adkins

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1
...
Generally, we are only interested in changes to the system, rather than the
surroundings

• Observables - There are two main types of observables in Thermodynamics
...
Conversely, microscopic observables correspond to the
properties of individual entities within the system, such as the position and velocity
of all the particles within a gas
...
We can describe the state of the system in terms of
either macroscopic or microscopic observables, though evidently the former is simpler
to do
...
For example, if a gas is at a particular temperature and pressure, there is a very large
number of possibilities for the number of combinations of the energies and velocities
of the individual particles that may create this

• Thermodynamic Equilibrium - This occurs where the macroscopic observables are in-

dependent of time (not changing)
...
However, the micro-states of the system are always well-dened,
as they do not depend on a consideration of the average quantities of the system

• Function of State - These are functions of parameters q = q1 , q2 ,
...
These are by denition exact dierentials,
such that
∆f (q) = f (q f ) − f (q i )

They have a well dened value in every state of thermodynamic equilibrium, independent of how the system arrived at that equilibrium
...
The most well-known function of state
(in Thermodynamics at least) is the Ideal Gas Law
pV = nm RT

(1
...
3144598
JK−1 mol−1 is the universal gas constant
...
These two assumptions
are important in deriving many results in Kinetic Theory, as supposed to Thermodynamics, but in this chapter we will only be working with ideal gases

• Quasi-static - A change in the system is said to be quasi-static if it is done slowly

enough to preserve thermodynamic equilibrium
...
A good way of thinking about a quasi-static change
4

Toby Adkins

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is as making a very large number of innitesimal changes to the system, where each
barely modies the system, but their net eect is some overall change in the system
• Reversible - A change in the system is said to be reversible if it does not increase the

entropy of the Universe (entropy will be covered as a concept more formally later)
...
This is the case when the system is thermally isolated
...


5

Toby Adkins

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1
...
If A is to be in thermal equilibrium with C, then they both must satisfy some
function of state
fac (a1 , a2 , a3 ,
...
)

Similarly, if A is to be in thermal equilibrium with B, then they both must satisfy some
function of state
fbc (b1 , b2 , b3 ,
...
)

We can re-arrange these functions for c1 such that, for new functions gac and gbc ,
gac (a1 , a2 , a3 ,
...
) = gbc (b1 , b2 , b3 ,
...
)

(1
...
, b1 , b2 , b3
...
3)

without assuming any correspondence between the states of A and B
...
3) and substitute them into (1
...
2)
...
, and so we can dene two new functions such that
Θa (a1 , a2 , a3 ,
...
)

The function Θ is thus the property that is shared by all materials that we call temperature; a macroscopic observable that is a function of the state parameters
...
They function by placing two objects in thermal contact until they both reach thermal equilibrium
...
Thus, a simple alternative statement to the
Zeroth Law is that 'thermometers work'
...
3 The First Law
The First Law of Thermodynamics is simply a statement of energy conservation as

Energy is conserved, and both heat and work are forms of energy
Let U be the internal energy of the system; this can include the kinetic energy of the
particles, the rotational energy, the chemical potential energy, the electrical energy, and so
on
...
Then we can write:
(1
...
We can then dene the quantity
Q as the dierence between the adiabatic work, and the actual work done on the system
...

Consider an ideal gas inside a thermally isolated cylinder at some pressure p, constrained by a piston of area
A
...

Then, we can re-write (1
...
4)

Figure 1
...
3
...
As such,
we can write
dQ = C dT
¯
(1
...

Using (1
...
6)

Toby Adkins

A1

At constant volume, dV = 0
...
7)
V

Similarly, at constant pressure dp = 0
...
8)
∂T

∂V

V

T

∂T

p

∂T

p

Just looking at these two expressions, it is clear that generally Cp > CV ; this means that it
takes more energy to change the temperature at constant pressure than it does at constant
volume
...
Suppose that the internal energy is a
function of temperature only U = U (T ), such that
3
U = nm RT
2

We will derive this result in the chapter on Kinetic Theory
...
9)

This is a constant which appears a lot when considering heat changes of systems
...
Re-arraging the ideal gas law for T and substituting it
into (1
...


From an initial state (p1 , V1 ) a gas is cooled at constant pressure to (p1 , V2 )
...
Calculate the ratio of change in heat during the
rst process to the second
...
The rst process is at constant pressure, and so
Cp
pdV
R
V2
Cp
=
p1
dV
R
V1
Cp
=
p1 (V2 − V1 )
R

Q1 =

8

Toby Adkins

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Similarly, the second process is at constant volume, so
CV
V dp
R
p2
CV
=
dp
V2
R
p1
CV
=
V2 (p2 − p1 )
R

Q2 =

Thus, we arrive at the desired result
Cp p1 (V1 − V2 )
Q1
(V1 /V2 ) − 1
=
·
=γ
Q2
CV V2 (p2 − p1 )
(p2 /p1 ) − 1

1
...
2 Thermodynamic Changes
We can now examine isothermal and adiabatic processes for an ideal gas
...

Thus
0 = dQ + dW
¯
¯
dQ = − dW
¯
¯
∆Q = −

dW
¯
V2

=

pdV
V1
V2

nm RT
dV
V

=
V1

Integrating this, we arrive at the useful expression
∆Q = nm RT log

V2
V2

(1
...
There is no heat exchange with the surroundings, and
so dQ = 0
...
9) for an ideal gas, and taking the exponential of both sides, we
nd that
T V γ−1 = constant
(1
...
As a little
aside, we have derived this equation just by manipulation of mathematical identities; it
has not really told us anything about how the gas actually behaves
...
3
...

9

Toby Adkins

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From this, we can calculate the work done in an adiabatic process for an ideal gas
...
Then, recalling that
pV γ = constant from Equation (1
...
3
...
The most common set of axes to use for this are pressure and
volume, and so diagrams of this type are often referred to as p -V diagrams
...

Thermodynamic processes that form a closed loop when graphed on a p -V diagram are
known as cycles, as they can be repeated
...
2: The Carnot Cycle
This is known as the Carnot Cycle
...
As the Carnot Cycle is reversible, so is the Carnot Engine
...
3)
...


Figure 1
...
In this case, the work done is W out , while the heat supplied is Q1
...
12)

A geothermal power station is proposed that will extract power from a mass M of hot rock
of specic heat capacity C by cooling it from a initial temperature Ti to a nal temperature
Tf
...
Find an expression
for the total work extracted from the engine
...
We shall assume that the system behaves as a
¯
Carnot engine, such that
dQ2
¯
dQ1
¯
=
T1
T2

12

Toby Adkins

A1

Thus, the work obtained is given by
T2
T2
dQ1 = −M C dT1 1 −
¯
¯
T1
T1

dW = −M C dT1 −
¯
¯

Integrating, it follows that the nal expression for the work done is
∆W = −M C(Tf − Ti ) + M CT2 log

Tf
Ti

Carnot Engines in Reverse
One type of application of heat engines is where the engine is run in reverse, requiring the
input of work in order to move heat around
...

• Refrigerator - This moves heat from a cooler body to a hotter body
...
As such, we dene eciency in this case as
η=

Q
W

where Q is the heat taken from the lower body
...
13)

where T is the temperature of the cooler body, and Th is the temperature of the
hotter body
...
Suppose that we want to add Qh to the second body
...
14)

In both cases, the eciency may achieve values greater than unity (this has to be the
for the heat pump as Qh < W in order to conserve energy), which is why they are such
attractive methods for moving heat around
...
The heat pump consumes power W , and the
building loses heat to its surroundings at a rate α(T − T0 )
...

Using (1
...
e a stable temperature), the rate of heat loss to
the surroundings has to be equal to the output heat of the pump
...
4 The Second Law
Let us begin by considering two alternative statements of the Second Law of Thermodynamics
...


1
...
1 Carnot's Theorem
Carnot's Theorem states that
No engine operating between two temperature reservoirs can be more efcient than the Carnot Engine
That is, the most ecient engine is the Carnot Engine
...
Run E
between two reservoirs at temperatures T1 and T2 such that it's output work powers C in
reverse, as shown below
...
4: The engine E and a Carnot engine C run in reverse
By the original assumption, we have that ηe > ηc
...
Thus, reductio ad absurdum, we have shown the result
...
Connect it to a Carnot Engine as shown
below
...
5: A Carnot Engine C and a reversible engine R run in reverse
From the First Law of Thermodynamics,
Q2 = Q1 − W
Q2 = Q1 − W

Now use our original assumption that
ηr < ηc
1−

Q
Q2
<1− 2
Q1
Q1
W
W
<
Q1
Q1
Q1 < Q1

But this contradicts Clausius' statement, and so the result follows
...
4
...
First, consider a
hypothetical engine K that violates Kelvin's statement (converting all heat into work) and
use it to drive a Carnot Engine in reverse
...
Thus, Kelvin's statement implies Clausius'
statement
...
6: A hypothetical 'Kelvin violator' powering a Carnot Engine in reverse

Figure 1
...
Consider an engine L that violates Clausius'
statement and connect to a Carnot engine as shown in Figure (1
...
Consider the net heat
ows in and out of the engine
...
We have thus shown that the statements are completely equivalent
...
4
...
15)

where the equality sign applies when the change is reversible
...
In the case where the equality applies, we know
that the integrand must be an exact dierential
...
Consider (1
...


b
a

dQ
¯
+
T

a
b

This means we can write
dS =

dQ
¯
T
dQ rev
¯
T
b
dQ
¯
a T
dQ
¯
T

≤0
≤0
b

≤
a

dQ rev
¯
≤
;
T

dQ rev
¯
dQ
¯
≥
T
T

17

dQ rev
¯
T

(1
...
This leads to a more
universal statement of The Second Law of Thermodynamics as

For any thermally isolated system, the entropy must either stay the same
or increase: dS ≥ 0
This is often stated as 'the entropy of the Universe must increase' as this is technically
true; if we consider the Universe as a thermally isolated system, it's entropy can do nothing
but increase (as there are no truly adiabatic processes)
...
This means we can re-write the First Law of Thermo¯
dynamics as
dU = T dS − pdV
(1
...
For an adiabatic
process, the total change in entropy is zero as dQ = 0 by denition
...


1
...
4 Calculating Entropy Changes
Consider a large reservoir at temperature Tr that is placed in thermal contact with a smaller
system at temperature Ts that has heat capacity C that is independent of temperature
...
5)
...
18)

Similarly for the system
dQ rev
¯
T
d(CT )
=
T
dT
=C
T

∆Ss =

using (1
...
Thus,
∆Ss = C log

Tr
Ts

(1
...


18

Toby Adkins

A1

Initially, a cube of heat capacity C is cooled from an initial temperature Ta to a nal temperature Tb using a heat-bath at temperature Tb
...

If in the interval Ta − Tb there are n baths, then each bath creates a temperature change
of (Ta − Tb )/n
...
The temperature at which we evaluate the bath is
given by
Ta − Tb
n

Ti = Ta − i

The total change in entropy for the baths is thus
Ta − Tb
∆Sb = C
n
n

=C

n

i=1

Ta − i

i=1

Ta −Tb
n

1

Ta
i=1 Ta −Tb
n

=C

1

n−i

1
Rn − i

We have let R = Ta /(Ta − Tb )
...

Rn−1

∆Sb = C
j=n

1
j

Rn − 1
n
1
= C log R −
n
≈ C log

Now, the change in entropy of the cube is always the same, regardless of the value of n
...

Ta
=2
Ta − Tb
Ta = 2Ta − 2Tb
Ta = 2Tb

This means that the ratio that we require for the entropy to vanish is Ta /Tb = 2, such as
Ta = 200K and Tb = 100K
...
Consider entropy as a function of T and V
...
31)
...
20)

Thus, the entropy of an idea gas increases with both T and V
...
4
...
Initially, container A is lled with one mole of ideal gas at pi and Ti
...
There is also no work done, which implies
¯
dW = 0
...

¯
dU = T dS + dW
¯
0 = T dS − pdV
RdV
pdV
dS =
=
T
V

This means we nd that the entropy of the expansion is
∆S = R log

Vf
Vi

The only way to get the gas back into chamber A is to compress it
...

Vi

∆W = −

Vi

pdV = −
Vf

Vf

RT
dV = RT log
V

Vf
Vi

This means that for The Joule Expansion, the entropy change is simply
∆S =

∆W
T

20

(1
...
5 Thermodynamic Relations
Before diving into the material for this section, it would be useful to recall two important
results from partial calculus
...
We can write
∂x
∂y

dx =

∂x
∂z

z

y

∂y
∂z

dy +

x

dx

and
∂y
∂x

dy =

dx +
z

dz

Substituting the second of these into the rst:
dx =

∂x
∂y

z

∂y
∂x

∂x
∂y

dx +
z

∂y
∂x

z

+
x

∂x
∂z

dz
y

For dz = 0, we nd the reciprocity relation
∂y
∂x

=
z

∂x
∂y

−1

(1
...
23)

y

We will use these extensively when manipulating partial derivatives in this section
...
5
...
The surroundings
will do work dW and transfer dQ to the system
...
24)

such that
dA ≤ 0

in equilibrium
...

21

Toby Adkins

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1
...
2 Thermodynamic Potentials
Thermodynamic potentials are scalar quantities used to represent the thermodynamic state
of a system
...


Internal Energy
Internal Energy U (S, V ) is dened as
(1
...
It follows that
∂U
T =
and
p=−
∂S V

∂U
∂V

S

• Internal energy corresponds to the heat absorbed by the system under isochoric
(constant V ) expansion: dU = CV dT
• If a system is thermally isolated at a xed volume, dA = dU
...
26)

H = U + pV
• It has natural variables S and p
...
For example, if a chemical reaction is exothermic, dH < 0
• If a system is thermally isolated with xed pressure, dA = dH
...
27)

F = U − TS
• It has natural variables V and T
...
If dF > 0, then

work is done on the system by the surroundings

• If a system has a xed temperature and volume, then dA = dF
...
28)

G = U − T S + pV
• It has natural variables T and p
...
This means that

the Gibbs function is conserved in any phase transition under these constraints

• If a system has a xed temperature and pressure, the dA = dG
...
5
...
This means that for some general function
df = Adx + Bdy −→

∂A
∂B
=
∂y
∂x

Considering the thermodynamic potentials:
• Internal Energy - dU = T dS − pdV
∂T
∂V

=−
S

∂p
∂S

(1
...
30)

• Helmholtz Free Energy - dF = −SdT − pdV
∂S
∂V

=
T

(1
...
32)
p

Toby Adkins

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1
...
4 Some Useful Thermodynamic Relations
We can use these thermodynamic potentials to derive some very useful thermodynamic
relations
...
Consider S(T, V )
∂S
∂T
∂S
∂T

dS =
=

∂S
∂V
∂p
∂T

dT +
V

dT +
V

dV
T

dV
V

from (1
...
Let dS = 0:
∂S
∂T

−

∂p
∂T

dT =
V

∂T
∂V

dV
V

∂p
∂T

=−
S

∂S
∂T

V

−1
V

Recall (1
...
Then:
∂T
∂V

=−
S

T
CV

∂p
∂T

(1
...
This can be used to derive (1
...
Consider U (T, V )
∂U
∂T

dU =

∂U
∂V

dT +
V

∂U
∂V

= CV dT +

dV
T

dT
T

Using the First Law of Thermodynamics, and substituting the result for dS from
above
dU = T dS − pdV
CV dT +

∂U
∂V

∂S
∂T

dT = T
T

∂p
∂T

dT +
V

dV − pdV
V

Let dT = 0
∂U
∂V

=T
T

∂p
∂T

−p
V

dU = CV dT + T

∂p
∂T

− p dT
V

Now, letting dU = 0, and re-arranging, we arrive at
∂T
∂V

=−
U

1
CV

T

∂p
∂T

−p

(1
...
Consider H(T, p)
∂H
∂T

dH =

p

∂H
∂p

= Cp dT +
∂S
∂T

dS =

∂H
∂p

dT +

dp
T

dp
T

∂V
∂T

dT −
p

dp
p

using (1
...
Substitute these into the explicit form of dH
∂H
∂p

Cp dT +
dp

∂H
∂p

∂V
∂T

+T
T

dp =
T

−V

∂S
∂T

∂V
∂T

dT −
p

∂S
∂T

p

p

∂V
∂T

= dT T

dp + V dp
p

p

− Cp

Let dT = 0
∂H
∂p

∂V
∂T

=V −T
T

p

dH = Cp dT + V − T

dp

Letting dH = 0, and re-arranging, we arrive at
∂T
∂p

1
T
Cp

=
H

∂V
∂T

(1
...

4
...
31) and a cyclical dierential identity:
∂S
∂V

=
T

∂p
∂T

=−
V

∂p
∂V

T

∂V
∂T

p

Substituting this in, and using the known denitions of βp and κT , we arrive at
Cp − CV =

25

2
V T βp
κT

(1
...
6 Thermodynamics of Other Materials
Up until this point, we have just considered Ideal Gases, writing the First Law of Thermodynamics in a particular form, and deriving many results from this
...
Instead of
writing dW = −pdV , we write
¯
dW = X dx
¯
(1
...
Some examples include
X

x

uid
pressure (−p)
volume (V )
elastic rod
tension (f )
length (L)
liquid lm surface tension (γ ) surface area (A)
dielectric
electric eld (E )
polarisation (p)
magnetic magnetic eld (B ) magnetisation (m)
In these more general cases, we dene the heat capacity at constant 'blah' by
C blah = T

∂S
∂T

(1
...
Do not forget the thermodynamic relations, particularly the
Helmholtz free energy in isothermal calculations involving entropy
...
6
...
39)

dU = T dS + f dL

Young's isothermal modulus is dened as the ratio of the stress σ to the strain ε
...

• Adiabatic Stretching - Suppose that we increase the length without a change in

entropy, what is the response of the system?

dQ = 0
¯
dU = dW
¯
CT dT = f dL

Thus, as CT > 0, f > 0, meaning that for an increase in length, temperature will
also increase
26

Toby Adkins

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• Contraction under constant f - If we warm the band under constant tension, it will

contract
...
This means it will contract
• Entropy changes under stretching - How does the entropy change under stretching?

Again using a reciprocity relation
∂f
∂T

=−
L

∂f
∂L

T

∂L
∂T

S

= −AET αf

From the Helmholtz function,
S=−

∂F
∂T

and f =
L

∂F
∂L

T

This means that
∆Q = T ∆S = αf AET T ∆L

Stretching the rod increases entropy if αf > 0 for the substance
...
This contains many small crystallites which have low entropy
...
However, for materials like rubber rubber αf < 0, and hence
an isothermal extension means that heat is emitted
...


1
...
2 Liquid Film
In this case, the equation of state for the internal energy is given by:
dU = T dS + γdA

From the Helmholtz free energy,
dF = −SdT + γdA
∂S
∂A

=−
T

∂γ
∂T

A

Consider the case where a cloud of droplets condenses isothermally to form a single droplet
...
6
...
We dene the magnetic
susceptibility as
C
M
=
(1
...
The equation of state for the internal energy
is
dU = T dS + B · dm
(1
...
This means that entropy decreases upon isothermal magnetisation; this is because the magnetic domains within the material become aligned, reducing
the number of micro-states that the system can occupy, and so reducing disorder
...


28

Toby Adkins

A1

The same paramagnetic material is initially at temperature T1
...

Using another cyclic relation:
∂T
∂B

=−
S

∂T
∂S

B

∂S
∂B

T

Thus, we can write, using the Helmholtz function
∂T
∂B

CBV
T
−
cB V
µo T 2
S
CB
=
µo cB T
dT
T CB
=
2
dB
µ a 1+ B C
−

=

o

µo a

Re-arranging and integrating:
T2

dT
=
T

T1

log

T2
T1

=

0
B

CB
dB
µ0 a + B 2 C

1
log µo a + B 2 C
2

B2C
1
= − log 1 +
2
µo a

Thus, we arrive at the expression
T1

T2 =

1+

29

B2C
µo a

0
B

2
...
It has the further advantage of always being dened, regardless of the
state of the system
...
Note that
we will be making extensive use of probability distributions in this chapter, so it is recommended that readers are familiar with some of their basic properties and implementation
...
1 Statistical Description of a Gas
In this treatment of Kinetic Theory, we will make the following assumptions about the
Ideal Gas that we are considering:
• The number of particles is very large, but their separation is large in comparison to

their molecular size
...
Evidently, we will relax some of these assumptions
later, but they will hold unless otherwise stated
...
One way to do this would be to know the initial position and velocity of
every molecule of gas in the container, and then solve the resultant equations of motion
...
There is too much information; as there are about 1023 molecules per m3 , we would
require about 1012 Tb of data to store this, which is more data than the entire web
generated last year
2
...
However, even
if we did manage to do this, a small error would change the evolution by an order
of unity in a very small time-scale, as particles experience about 109 collisions every
second
...


2
...
1 The Thermodynamic Limit
We solve this problem by assuming that particle motion is random, meaning that v is a
random variable, and there is a suciently large number of particles to describe the system
by average quantities
...

Let us consider an example
...
Evidently, the exact internal energy of the gas is given by the
sum of the kinetic energies if the individuals molecules
...

σ 2 = (Ue − U )2
2
= Ue − U 2

=
i

1
mv 2
2 i


1
= m2 
4

j

1
mv 2
2 j

2

1
2
m vi
2

−
i

2
2
vi

4
vi +
i

2
vj −

2
vi




i

i=j

1
= m2 N v 4 + N (N − 1) v 2
4
1
2
= N m2 v 4 − v 2
4

2

− N 2 v2

2

Now consider the ratio of the standard deviation to the actual energy:
1
σ
=√
Ue
N

v4
v2 2

1
−1∝ √
N

This means that for very large N , U ≈ Ue
...


32

Toby Adkins

A1

2
...
To do this, we need to introduce the concept of a velocity distribution
function f (v)d3 v
...
It represents
the faction of molecules that have velocities in the cube [v, v + d3 v] in phase space
...
2
...
We can thus make a transformation to polar coordinates in phase
space:
f (v)d3 v = f (v)v 2 sin θ dvdθdφ

This means that while there is azimuthal symmetry, there is in fact a dependence on θ
...


Figure 2
...
Then
2π

f (v) =

π

dθ f (v)v 2 sin θ = 4πv 2 f (v)

dφ
0

0

We thus nd the important relationship between the velocity and speed disbributions of
f (v) = 4πv 2 f (v)

33

(2
...
As the system
is isotropic, we can say that
n
n
n
vx = vy = vz = 0

For example, it is easy to show by direct integration that
|vi | =

1
v
2

and

2
vi =

1 2
v
3

Suppose that we have some general moment
v i1 v i2 v i3
...
If n is odd, then we can immediately say the expression
is zero; this is because f (v) is an even function in v , and so the integral will evaluate to
zero
...
For example, suppose that we want to calculate
vi vj vk vl

There are six dierent possible tensor combinations
...
By exchanging arbitrary combinations of the indices, it becomes clear that
α = β = γ
...
2
...
We require that the distribution function is normalised
over all phase space
...
2)

Using (2
...
3)

This is known as the Maxwell-Boltzmann distribution
...
2: The Maxwell-Boltzmann distribution, showing v max , v and

v2

Moments of the Maxwellian
Using our knowledge of probability distributions from statistics, we know that we can nd
the expectation value of some v n by
∞

vn =

dv v n f (v)
0

However, evaluating these integrals can become very long and tedious for higher powers
of n as integration by parts has to be used multiple times
...
4)

Evidently, these formulae do not have to be remembered, but they are just here for reference
to check one's evaluation of moments
...

2
v = √ vth
π

3
v 2 = vth 2
2

35

4
v 3 = √ vth 3
π

Toby Adkins

A1

2
...
2) and (2
...
In order to be able to nd this, we need to consider the ux of
particles on the walls of the container
...
Particles will only hit the wall
if they are less than vz away in some unit time
...
This means that the number of particles in the phase-space cube [v, v + d3 v]
that hit the wall per-unit-area, per-unit-time is given by
dΦ(v) = nvz f (v)d3 v
= n(v cos θ)f (v)v 2 sin θ dvdθdφ

Our expression for particle ux thus is
dΦ(v) = n cos θ sin θ v 3 f (v) dvdθdφ

(2
...
This means

that it preferentially selects molecules with higher velocities, as these are more likely
to be going quickly enough to hit the wall
...
6)

This means that the most probable speed is
∂
:
∂v

0 = 3v 2 −
vp =

2 4
2 v
vth

3
vth
2

This is higher than the most probable speed for the rest of the gas vth
...
This is a manifestation

of the idea of the angular size of the area element as seen from the perspective of a
given particle; if it is travelling almost perpendicular to the wall, the angular size of
the are element will be very small, and so is less likely to hit it
...
7)
The most probable angle is clearly π/4, which is in a sense intuitively obvious

2
...
1 Pressure
Pressure is a form of momentum ux; it is a measure of the average amount of momentum
imparted to the walls of the container by the gas
...
Particles perpendicularly incident on
the wall at a speed vz will rebound with the same speed in the opposite direction, meaning
that the impulse imparted to the wall is
∆p = 2mvz

36

Toby Adkins

A1

where m is the mass of an individual particle
...
5), the number of
molecules doing this per-unit-area, per-unit-time is dΦ(v)
...
1)
...
8)

This is in accordance with our expectations; a higher pressure may be a result of more gas
molecules, or the fact that the molecules are moving more quickly
...
Note that N = nm NA where NA is Avogadro's constant
...
It is also clear from the expression for vth that
1
kB T = mvth 2
2

This means that we can now dene temperature as the energy of a particle moving at the
most probable speed, giving us a more concrete idea than we originally had when dening
temperature in Section (1
...
We also see that the energy is given by
3
U = N kB T = C V T
2

(2
...


Adiabatic Expansion
We are now equipped to consider the adiabatic expansion of a gas as in Section (1
...
2)
...
The piston is very slowly pulled out at some
37

Toby Adkins

A1

velocity u vth
...

In the rest frame of the piston, the particle is incident at velocity
vi = vz − u

We assume that the velocity of the piston is unchanged as m piston
velocity of the particle in this frame is thus

m particle
...
Let d2 ΦA (vz ) be the ux of molecules hitting the
surface area element dA of the piston per unit time
...
This can be written as
dU
2 U
=−
u dA
dt
3 V

However, u dA is simply the rate at which the volume increases
...

38

Toby Adkins

A1

2
...
2 Eusion
Suppose now that we open a small hole of dimension d
λmf p (the typical distance
travelled by a given particle, as we will see later) in the side of the container
...
10)
2πmkB T

Once again, the distribution of eusing particles has the properties outlined in Section
(2
...
We have derived this expression assuming that all incident angles are possible
...
Then, instead of integrating θ over the interval [0, π/2], we would integrate up
to the corresponding solid angle
...
For example, we can nd the vapour
pressure of a gas inside a container by measuring the rate of change of mass inside that
container
...
Then the rate of
change of mass is
dM
=mΦA
dt
m
2kB πT

=pA
p=

2kB πT 1 dM
m
A dt

A closed vessel is partially lled with liquid mercury; there is a hole of area A = 107 m2
above the liquid level
...
4 × 105 kg
...
(The relative molecular mass of mercury is 200
...

∆M
≈ 9
...
3 × 10−25 kg

→ p ≈ 0
...

39

Toby Adkins

A1

Conditions for Equilibrium
Suppose that we have two containers joined by a hole of radius a
...

• a

λmf p - In this case, we require that the pressures balance, namely

(2
...
Evidently, we require that the eusive ux is

the same in both directions, namely
n1

T1 = n2

T2

(2
...


Some results of Eusion
Suppose that we have a container with a small hole of area A
...

• Number Density - The rate of particles escaping from the container is evidently
dN
1
= −ΦA = − A v n
dt
4

Dividing through by the volume in the container, we nd that the dierential equation
for the number density is given by
dn
1A
=−
v n
dt
4V

(2
...

• Energy - We need to calculate the energy ux J
1
dJ = dΦ(v) · mv 2
2
1
3
J = nm v
8

This means that we can write the rate of change of the internal energy as
dU
= −JA
dt

(2
...
9), we nd that U = U (n, T )
...


40

Toby Adkins

A1

2
...
Let the two species have radii r1 and r2 respectively, and the same number density n
...
The particles that it will collide with are contained within the cylinder of
volume π(r1 + r2 )2 vt for some characteristic speed of the system v
...
15)

This is known as the collision cross section
...
The collision time t = τc occurs when the number of
particles in this cylinder is one
...
16)

This is known as the mean free path of the particles
...

Typically, τc 10−9 , λmf p 10−6 and r 10−10
...
4
...
As
we have assumed that all particles (apart from the one that we were considering) were
stationary, it makes the most sense for v to be the relative velocity between the particles
...
Then:
2
vr = |v 1 − v 2 |2
2
2
= v1 + v2 − 2 v 1 · v 2

= 2 v2
√
v≈ 2 v

We have made the important assumption that the velocities of the particles prior to the
collision are independent; that is, the particles have collided enough times with other particles prior to a second collision that we can neglect any eect of the previous collision
...

It turns out that we do not need to be this specic when specifying this characteristic
velocity v
...

41

Toby Adkins

A1

2
...
2 Collisions in a Mixture
Consider a more general case where we have two species of particle in the same container,
one with eective collision radius r1 and number density n1 , and the other with r2 and n2
...
Once again, we will assume that the
particle velocities are independent, namely that
f (v 1 , v 2 ) = f (v 1 )f (v 2 )

Assume that both species have a Maxwellian distribution
...
Then
2
2
v1
v2
1
2
2
2 + u2 = 2k T V (m1 + m2 ) + vr
u1
B
2
1
2
V 2 (m1 + m2 ) + vr
=
2kB T
1
2
=
V 2 (Mt ) + vr (µ)
2kB T
2
vr
V2
+
=
(2kB T )/Mt (2kB T )/µ
V2
v2
= 2 + r
u2
ut
µ

m1 v1 +m2 v2
m1 +m2

such that

m1 m2 + m2 m2
2
1
(m1 + m2 )2
m1 m2
m1 + m2

Thus, the nal expression that we want to evaluate is
vr =
=
=
=

1
(πut uµ )3

−

d3 V d3 v r vr e

1
2
2
d3 V √
e−V /ut
3
( πut )
1
2
2
d3 V x √
e−Vx /ut
3
( πut )
vr
2
2
d3 v r √
e−vr /uµ
( πuµ )3

v2
V2
+ r
u2
u2
µ
t

vr
2
2
e−vr /uµ
d3 v r √
3
( πuµ )
3

vr
2
2
d3 v r √
e−vr /uµ
3
( πuµ )

2
= √ uµ
π

This means that we obtain a nal expression for the collision frequency νc = 1/τc as
2
2
νc = (r1 + r2 )

42

8πkB T
µ

(2
...
5 Transport
This section focusses on examining how a gas transports quantities, which could be momentum, energy or other particles themselves, from one place to another
...
In all
derivations, we assume that the pressure is not too high (λ d, where d is the molecular
diameter) such that we can neglect collisions involving more than two particles, and the
pressure is not too low (λ L, where L is the length scale of the container) such that the
particles mainly collide with one another and not the walls of the container
...
5
...
By the conservation of energy, we know that J has to have the
opposite sign to T
...


= −κ T +
...
18)

J = −κ T

where κ is known as the coecient of thermal conductivity
...
As the particles travel an average of λmf p ≡ λ between
collisions, the distance travelled parallel to the z -axis is δz = λ cos θ (see Figure (2
...

These particles arriving at z from z − δz will bring some extra energy δE
...
Then
δE = cn T (z − δz) − cn T (z)
= cn T (z) −
= −cn

∂T
δz
∂z

− cn T (z)

∂T
λ cos θ
∂z

The total energy ux in the z -direction is
Jz =

dΦ(v) δE

= −cn λ

∂T
∂z

dΦ(v) cos θ

1
∂T
= − ncn v λ
3
∂z

Comparing this with the z component of (2
...
19)

where cv = 3/2nkB = ρcm is the heat capacity per unit volume
...

43

Toby Adkins

A1

The thermal conductivity of argon (atomic weight 40) at S
...
P
...
6 × 102 Wm1 K1
...
Solid argon has a close packed cubic structure, in which, if the
atoms are regarded as hard spheres, 0
...
The density
of solid argon is 1
...
Find the eective radius in this case, and compare the two results
Using the fact that CV = 3/2N kB , we can rearrange (2
...
93 × 10−10 m
4πnλ
λ=

Now for solid argon,
N V atom = 0
...
74
3 2
m
4 2
πr2 = 0
...
95 × 10−10 m

These answers are dierent because the hard sphere approximation used to derive λ becomes less accurate
...


2
...
2 Viscosity
Let us consider a one-dimensional shear ow
u = ux (t, z) x

That is, a ow in the x direction whose magnitude depends on the value of z
...
Then particles that travel
to z from z − δz will bring some extra momentum with them, as in Figure (2
...

We know that Πzx = 0 for ux = 0, and (by the conservation of energy) that Πzx must
have the opposite sign to ux
...

∂z

44

∂ux
+
...
3: Transport due to a temperature and velocity gradient
This means that to rst order
Πzx = −η

∂ux
∂z

(2
...

Let us calculate the extra momentum brought by each particle
...
20) yields
1
η = nmλ v
3

(2
...
How can this be? At lower
pressures, less particles will be moving through this plane of constant z , but they will have
a longer λ
...

Evidently, this will start to break down when λ is on the order of magnitude of the size of
the container
...
For nite dierences in velocity
and separation, we can write that
F
∆u
= −η
A
∆z

(2
...


Two plane disks, each of radius 5 cm, are mounted coaxially with their adjacent surfaces 1
mm apart
...
T
...
(viscosity 2
...
One of them rotates with an angular velocity of 10 rad s1
...

We want to consider the force exerted on the disk in the θ direction along the edge of the
disk
...
As uθ is the velocity at r, uθ = ωr
...
Integrating:
τ = −η

ω πa4
z 4

We thus nd that τ ≈ 2 × 10−6
...


2
...
3 Self-Diusion
A gas can move itself around, in a sense
...
As these particles move, they set up a 'number gradient' n∗
...
We can thus write
Φ∗ = Φ∗ ( n∗ )
≈ Φ∗ (0) + Φ∗ (0) n∗ +
...


This means that to rst order

Φ∗ = −D∗ n∗

where D∗ is the coecient of self-diusion
...
23)

Toby Adkins

A1

Now, consider a particle travelling with some component of velocity in the positive z
direction that crosses a plane of constant z
...
23) yields
1
D∗ = λ v
3

(2
...
Observing that ρ = nm, we obtain the useful relationship that D∗ ρ = η
...
Then the rate of change of
the number density n∗ within the volume will be opposite and equal to the number of the
labelled particles that leave the volume via this surface
...
24)
...
25)

This is known as the self-diusion equation
...
5
...
We now want to nd how
temperature ows within the gas, which is directly related to the rate at which energy
is transported throughout the gas
...
Then the rate at which energy changes
in the volume is going to be opposite and equal to the amount of energy that leaves the
volume through this surface
...
As we are integrating over an arbitrary volume, we can write
∂ε
+
∂t

·J =0

This is the energy conservation equation for the gas
...
Assume that there is no loss of the gas
particles, meaning that n is time-independent
...
18):
∂(cv T )
+
∂t

· (−κ T ) = H
cv

∂T
=κ
∂t

2

T +H

Re-arranging, we arrive at the heat diusion equation
∂T
=α
∂t

2

T+

H
cv

(2
...


Newton's Law of Cooling
Newton's law of cooling states that the temperature of a cooling body falls exponentially
towards the temperature of its surroundings with a rate which is proportional to the area
of contact between the body and the environment
...
Mathematically, we can express this as
J = h∆T

(2
...


Some Examples
Let us take a look at some examples of problems concerning the heat equation
...
A cylindrical wire of thermal conductivity κ, radius a and resistivity ρ uniformly carries a current I
...
Find the temperature T (r)
...
We
know that in the steady state, the heat loss to the surroundings must be opposite
and equal to the heat generated
...

JA = V H
2πa (h(T (a) − To )) = πa2 H
a
h(T (a) − To ) = H
2
a
T (a) = To +
H
2h

48

Toby Adkins

A1

This is our boundary condition at the surface
...
This means that the
heat equation becomes
H
κ
dT
H
r
=−
dr
κ
dT
Hr2
r
=−
+ c1
dr
2κ
Hr2
T (r) = −
+ c1 log(r) + c2
4κ
2

1 d
r dr

T =−

We require the solution to be nite at r = 0, meaning that c1 = 0
...
Consider a thick, uniform layer of material with coecient of thermal diusivity α
...
Let c1 =
T (0, ω)
...
28)

w

skin depth, and is a measure of the attenuation of the propagating wave
...
28), we nd that
Tn e−x/δn cos nt −

T (x, t) = To +
n

•

x
δn

The interor of the material is subject to a space-dependant sinusoidal temperature variation given by
T (x, 0) = T0 +

Tn cos(nx)
n

Suppose that the solution is of the form
T (k, t)eikx

T (x, t) =
k

Taking the FT of both sides of the heat equation:
∞

∂
∂t

dk e−ikx

−∞
∞

∂T
=
∂t

∞

dk e−ikx

−∞

dk e−ikx T = α(ik)2 T

−∞

∂T
= −αk 2 T
∂t
2
T (k, t) = T (x, 0)e−αk t

50

∂2T
∂x2

Toby Adkins

A1

Thus, the nal solution is of the form
2

T (x, 0)e−αk t eikx

T (x, t) =

(2
...


51

Toby Adkins

A1

2
...

Up until this point, we have assumed that the distribution function of the particles depended only on their velocity v
...
With this in mind, allow the distribution to depend on both position (r) and time (t):
F (t, r, v)
...

Let our macroscopic scales of the system be and t
...
Suppose
that we now break the system into small uid 'elements' of size δ and observe them for a
time δt
...
This means that we arrive at
the local distribution function
n(t, r) −|v−u|2 /v2
th
FM (t, r, v) = √
e
( πvth )3

(2
...
Evidently, global equilibrium occurs where n, u and T are constant
...
This means that in fact these local
distributions are not quite Maxwellian
...
6
...
Particles with these
properties will either be those that move into this location, or those that are scattered into
this velocity
...
31)

F = C[F ]

where v is the velocity of the particles, and C[F ] is known as the collision operator
...


The simplest model that satises this is:
C[F ] = −

F − FM
1
= − δF
τc
τc

(2
...


2
...
2 Conservation Laws
Now that we have an equation for the evolution of the local distributions, we can use it to
derive some local conservation equations for the gas
...

• Number Density
∂n
∂
=
∂t
∂t

d3 v F
∂F
+ C[F ]
∂z

=

d3 v

−vz

=−

∂
∂z

d3 v vz F +

no mean velocity
in z direction

d3 v C[F ]
collisions do not
change particle no
...

∂ε
∂
=
∂t
∂t

1
1
mv 2 F − mnu2
x
2
2
1
∂F
∂ux
= d3 v mv 2 −vz
+ C[F ] − mnux
2
∂z
∂t
∂Πzx
∂
1
d3 v mv 2 vz F + ux
=−
∂t
2
∂z
d3 v

Let v = ux x + w
...
26) by letting ux = 0

2
...
3 Self-Diusion 2
...
Let us compare the relative magnitude of the terms in (2
...
We know
that δF FM , and that variations in δF occur over very large time-scales as supposed to
those of FM
...
This means that we can neglect

(2
...
31)
...

d3 v vF ∗ =
=

∗
d3 v v(FM + δF ∗ )

d3 v vδF ∗

∗
as the mean velocity for FM is zero
...
33),

d3 v vδF ∗ =

d3 v v · (−τc v FM )

= −τc

∗
d3 v v 2 FM
∗
d3 v v 2 FM

= −τc
= −τc v 2

n∗

This means the equation becomes
∂n∗
− τc v 2
∂t

· ( n∗ ) = 0
∂n∗
= D∗
∂t

2 ∗

n

where D∗ = τc v 2 is the expression for the self-diusion coecient
...
5
...

We can take the rst moment of the Kinetic Equation in order to nd the momentum
conservation equation of the labelled particles
...

F friction = −

55

1 ∗ ∗ ∗
m n u
τc

3
...
Evidently, Quantum Mechanics will always give the correct answer,
but it will become increasingly dicult - and almost impossible - to compute the required
problem
...
This author certainly enjoyed this course immensely,
and hopes that this is reected in these notes
...
1 Basic Principles
Evidently, as we saw in Chapter (1), Thermodynamics allows us to make a large set of
predictions about a system; it can give us the equation of state, the entropy, the energy,
the heat capacity, and so on
...
For ideal gases, we
can get this from Kinetic Theory, but we need to be able to extend this kind of formalism
to other systems
...


3
...
1 States of a System
Suppose that a system consists of Ω microstates (that is, a possible states of all the particles
of the system), where for each microstate α we have an associated probability pα , and
associated energy Eα
...
This means that we can
dene certain quantities of the system in terms of their mean quantities, such as the mean
energy :
U=
pα Eα
(3
...
Note that we evidently require the probabilities are properly normalised,
namely that α pα = 1
...
Consider
an system undergoing some adiabatic process
...
From (1
...
,pΩ

pα
α

∂Eα
dV
∂V

However, during an adiabatic process, we know that dU = −pdV
...
2)

This kind of calculation can also be repeated for non-pV systems
...


3
...
2 Gibbs Entropy
The next logical question to ask is as to how to calculate these {pα } given that we know
the set of energies {Eα }
...
What about the {pα }? Suppose that we
57

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know nothing about our system at all
...
The only fair way to assign probabilities to the microstates of the
system is to say that they are equally likely; we have no information to tell use otherwise
...

Suppose that our system consists of N particles, where N
Ω
...
In this way, we arrive at having Nα particles in the
microstate α, and associated probability
pα =

Nα
N

for

Nα = N
α

Now what is the most likely outcome of our assignment {Nα }? The number of ways E of
carrying out our un-ordered assignment is given by
W =

N!
N1 !
...
This means that the most likely
outcome for {Nα } will be the one that maximises W , or rather log W
...
5)
...
NΩ !)
N log N − N −

(Nα log Nα − Nα )
α

=−

Nα
N

Nα log
α

Recalling the denition of pα above, and dividing through by N , we arrive at the expression
SG = −

pα log pα

(3
...
We will see it's full signicance shortly,
but for now, we understand it as a function that is related to the possible assignments of
particles to microstates within a given system
...
1
...
This corresponds directly to SG being maximised, subject
to a set of constraints Fi = 0 that correspond to our knowledge about the system
...
, p Ω ,

λ 1 ,
...
This is because the number of constraints cannot be greater than the
number of microstates, as m = Ω corresponds to total knowledge of the system
...

Let us suppose that we are completely ignorant about our system; we know that it exists,
but we really don't know anything about it
...
4)

=0

α

Maximising unconditionally:
dSG − dλ

pα − 1

−λ

α

(log pα + 1 + λ) dpα − dλ

−

dpα = 0
α

pα − 1

α

=0

α

Setting each of these terms individually to zero:
log pα + 1 + λ = 0
pα = e−(1+λ)
Ω

e−(1+λ)

=1

α independent of α

→ pα =

1
Ω

Thus, we conrm our assumption in the previous section that the microstates of an isolated
system in equilibrium are all equally likely
...
1
...
It states that
N log N − N

log N !

(3
...
In fact, it turns out that N 50 is enough to reduce
the error in using this formula to 2%, so in the Thermodynamic limit, this is essentially
an exact expression
...
Let us expand the integrand around the maximum
...
We can thus Taylor
expand the integrand as
∞

dx ef (x)

N! =
0

N log N −N

∞

=e
e

N log N −N

1

2 +
...
It follows that
log N !

N log N − N +

1
log(2πN )
2

If N is suciently large, we can ignore the last term, as it scales as log N rather than as
N
...


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Toby Adkins

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3
...
This means that we now add the
constraint that
pα Eα − U

β

(3
...
Maximising
SG unconditionally with respect to both constraints:
(log pα + 1 + λ + βEα )dpα − dλ

−

pα − 1

α

−β

pα Eα − U

=0

α

α

Setting each of the terms identically to zero:
log pα = −1 − λ − βEα −→ pα = e−(1+λ+βEα )
1
e−(1+λ+βEα ) = 1 −→ e−(1+λ) =
e−βEα
α
α

Putting these two expressions together, we nd that the probability of a particular micostate α is given by
pα =

e−βEα
Z(β)

for Z(β) =

(3
...
This means that if we know all Eα 's, we can calculate
the probability of each microstate from this expression
...
Substitute our new-found expression for pα into the normal expression for SG
...
2), this reduces to
¯
SG = β (dU + pdV ) = β dQrev

...
Then, by the denition of dQrev , we nd that
¯
β=

1
kB T

(3
...


3
...
1 Some Important Relationships
Our formalism is thus clear: maximise entropy subject to constraints, nd the distribution, nd the partition function, relate to thermodynamic quantities through entropy, and
dierentiate to nd everything
...
Now that we
have an expression for β , we can now derive some useful thermodynamic quantities from
our partition function Z(β)
...
9)

Heat Capacity
At this stage, is would be useful to nd a relationship between derivatives with respect to
T and those with respect to β , as we work mainly with the former in Thermodynamics,
and the latter in Statistical Mechanics
...

CV =

∂U
∂T

=−

β
T

∂U
∂β

V

Substituting our denition of U from above, we arrive at
CV =

β ∂ 2 log(Z)
T
∂β 2

62

(3
...

∂ 2 log(Z)
∂
=
2
∂β
∂β
=

1
Z

−

1
Z

Eα e−βEα
α

2
Eα e−βEα −
α
2
pα Eα −

=

1
Z2

Eα e−βEα
α

γ

pγ Eγ

pα Eα

α

Eγ e−βEγ

γ

α

Recalling our denition of the expectation value of some quantity, we see that the above
expression is clearly the variance of the energy (∆E)2
...
We examine similar conditions in Section (3
...
2)
...
11)

This expression is always greater than or equal to zero, as Z > 0 for probabilities to be
well-dened (positive)
...
This means that often the hardest part
about Statistical Mechanics is nding the partition function; the rest is merely taking
derivatives!

3
...
2 Stability and Equilibria
We are now going to derive the conditions that are required for a system to be in equilibrium
...
This means that Si = Si (Ui , Vi ), where Ui is the internal
energy of the system (the dierence between the total and kinetic energy)
...


Equilibria
Maximising the expression for entropy and requiring that all of the individual constraints
are zero allows us to obtain relationships between the quantities in each of our subsystems
...

This means that in equilibrium, we cannot have an gradients of temperature, pressure, or
velocity inside the total system, as we could have anticipated
...
This is because
we do not in fact know whether the extremum found when maximising S was a maximum,
minimum, or saddle point
...
Then, this means that all Si 's would be maximised by decreasing
their argument as much as possible; that is, by increasing their kinetic energy, ying
o to innity
...

Note that this restriction does not apply for systems who's motion is constrained,
and prevented from ying apart (such as a chain of molecules)

3
...
3 Some Common Examples
Let us now examine two common systems that can be treated with the Canonical Ensemble
...

Before we do so, a quick note about systems that are composed of many individual units,
such as the particles in a gas, or atoms within a lattice
...
12)

This follows from the denition of probabilities; the probability pα of having N particles
in a state α is given by pα N (where pα is the probability for a single unit), meaning that
the partition function must be modied in the above way
...


The Two-Level System
Suppose that we have a system that has two energy levels, +∆ and −∆
...
Suppose that we have N spin- 2 particles within
our system
...
13)

Toby Adkins

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The quantity we are most often interested in with the spin- 1 paramagnet is its heat capacity
2
CV
...
9):
1
U = −N µB B tanh(βµB B) = − N kB ΘB tanh
2

ΘB
2T

where we have dened ΘB = 2µB B , which is the excitation temperature for the paramagkB
netic eects
...

• High Temperature Limit (T

ΘB ):
eΘB /T

• High Temperature Limit (T

1 −→ cB ∝

1
T2

ΘB ):

eΘB /T

1 −→ cB ∝

1 −Θ/T
e
T2

As CB tends to zero in both limits, this means that there must be some maximum temperature for which the magnetic heat capacity of the system is maximised
...
6K
...
This means that the magnetisation can be easily
obtained as
M =−

1
V

∂F
∂B

=
T

N µB
tanh
V

µB B
kB T

This once again behaves sensibly; we re-obtain Curie's Law in the high temperature limit,
and nd that the magnetisation is constant in the low temperature limit
...
14)

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This expression is actually used within the Einstein model of the heat capacity of solids
...
Note that there is some constant energy U0
in the Hamiltonian of the system, but we are going to ignore this as it does not eect the
properties of the solid (apart from the mean energy)
...
For N particles in a 3-D lattice, it follows that
N
3N
Z(β) = (Z 1D )3 = (Z1 )3 = Z1

Recalling (3
...


67

ω

eβ
(eβ

ω

ω

− 1)2

1 + β ω
...
3 The Grand Canonical Ensemble
We are now going to make N (the number of particles) a variable parameter within our
formalism, such that we can treat systems where numbers of particles are exchanged, or
where there are various inhomogeneities
...
15)

=0

α

¯
where N is the mean particle number for the system
...
16)

α

This is known as the grand canonical ensemble, with Z(β, µ) being the grand partition
function
...


3
...
1 Chemical Potential
In a similar fashion to Section (3
...
Substitute (3
...

¯
pα [−β(Eα − µNα ) − log(Z)] = βU − βµN + log(Z)

SG = −
α

Taking the total derivative:
¯ $
¯ β N dµ hhhhhh dZ
¯ h
dSG = βdU − βµdN − $$$ + (U − µN )dβ +
Z
dZ
$ − hhhhh )dβ]
=
pα [−β(dEα − $α dµ) (Eα − µNα h
N$
hh
Z
α

where we have restricted dNα = 0
...
2), we arrive at the equation
¯
T dS = dU + pdV − µN

(3
...
This means that
we can write µ as
µ = −T

∂S
¯
∂N

U,V

We can thus interpret −µ as the heat from adding an extra particle to the system at constant U and V
...
It is known as the chemical potential of a system
...

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Let us now consider the Gibbs Free Energy
...
28) and (3
...

¯
This means that if we scale N by some constant λ, G must also scale in the same way
...
If we set λ = 1,
we nd that
G
µ= ¯
N

(3
...
This means that it is an intensive quantity, and so that µ = µ(p, T )
¯
(as it cannot depend on system size, and thus N )
...
Then:
∂S1
∂S1 ¯
∂S2 ¯
∂S2
dU1 + ¯ dN1 +
dU2 + ¯ dN2
∂U1
∂U2
∂ N1
∂ N2
∂S1
∂S1
∂S2
∂S2
¯
=
−
+
¯ − ¯ dN1
∂U1 ∂U2
∂ N1 ∂ N2
1
1
µ1 µ2
¯
=
−
dU1 + − +
dN1
T1 T2
T1
T2

dS =

Setting dS = 0, we nd that when a system is in equilibrium T1 = T2 (as before) and
µ1 = µ2
...
Suppose now that the
system is in thermal equilibrium, but not chemical equilibrium
...
This means that matter moves from large to
small µ
...


3
...
2 The Grand Potential
For open systems, we dene a new thermodynamic quantity known as The Grand Potential,
dened as
¯
Φ = F − µN = −kB T log(Z) = −pV
(3
...
18), and simplifying the resultant expression
...
Taking the total derivative:
¯
dΦ = −SdT − pdV − N dµ

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This results in the relations:
∂Φ
∂T

¯
N =−

µ,T

∂Φ
∂µ

p=−

µ,V

∂Φ
∂V

S=−

T,V

(3
...
21)
(3
...


3
...
3 Multi-species Systems
Now that we have set up our apparatus for dealing with multiple component systems, let
us put this into practise
...
For each microstate of
the system α, let us associate an energy Eα , and particle numbers
N1α , N2α ,
...
, µm

We are again going to consider the Gibbs Free Energy of the system, and scale by an
arbitrary factor λ
...
, λNm ) = λ G(p, T, N1 ,
...
All such reactions can be written
as
νs as = 0

(3
...
Conventionally, the products are dened to have negative µs
...

In the rst chapter, we learnt that equilibrium was reached for a system held at a given p
and T when the Gibbs Free Energy was minimised
...
23), implies that as the chemical reaction progresses
¯
¯
¯
dN1 : dN2 : · · · : dNm = ν1 : ν2 : · · · : νm

That is, the particle numbers and the coecients νs will be in the same ratio
...
24)

νs µs = 0
s

This is one statement of what is known as The Law of Mass Action, which determines the
behaviour of chemical reactions
...


Given that the chemical potential of a non-relativistic electron/positron gas can be written
as
µ = −kB T log (2s + 1)

nQ
n

where nQ =

me kB T
2π 2

3/2

nd the density of positrons created from spontaneous pair-production dened by the equation
photon(s)

e− + e+

Remember to take into account the relative concentration of the positron and electron pairs
with respect to the density of normal electrons in matter (without pair production) n0
...

Now, we will see the expressions for µ and nQ in later sections, so do not be too worried
about their signicance here
...
4) and (3
...
3)
...
In this limit, the energy of the system is
so high that the number of the electron-positron pairs dominates in comparison to the
number of electrons in normal matter, and so we can neglect n0 as it is very small
...
The equilibrium
concentrations can thus be calculated using the results of Section (3
...
3), though this is
left as an exercise for the reader once they have covered Quantum Gases
...
We now need to take into account the chemical
1
potential that results from the rest mass
...
This means that the equilibrium concentration of the positrons is given by
n+

4
n0

me kB T
2π 2

3

2 )/(k

e−(2me c

BT )

This author tried calculating this value (estimating n0 a−3 , where a0 is the Bohr radius),
0
but it came out to be zero
...


72

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3
...
If we treat the ideal gas
classically, it's energy levels are given by
εk =

2 k2

2m

Theoretically, we could just plug this expression into (3
...

However, the problem that we encounter is that sums of this type are very dicult to
calculate; it is much easier to calculate an equivalent integral
...
Then, we are restricted to the wavenumbers that
satisfy
2π
2π
2π
ix ,
iy ,
iz
Lx
Ly
Lz

k=

for integers ix , iy , iz in order to full appropriate (periodic) boundary conditions at the
extremities of the box
...
Then, we can approximate the
desired sum by an integral:
e−βεk =

Z1 =
k

k

Lx Ly Lz 2π 2π 2π −βεk
e
(2π)3 Lx Ly Lz
V
(2π)3

V
(2π)3

d3 k e−β

2 k 2 /2m

∆kx ∆ky ∆kz

Assuming that the system is isotropic, we can transform to polar coordinates:
V
(2π)3

d3 k e−β

2 k 2 /2m

=

V
(2π)3

∞

dk 4πk 2 e−β

2 k 2 /2m

0

∞

=

dk g(k) e−β

2 k 2 /2m

0

The quantity g(k) = (V k2 )/(2π 2 ) is known as the density of states, which tells us the number of microstates α per k or (with appropriate substitution) per ε
...

Evaluating this integral, we obtain the single particle partition function
Z1 =

V
λ3
th

for λth =

2π
mkB T

(3
...

This means that this expression must hold in the relativistic case, except with a dierent
denition of λth (it is left as an exercise to the reader to nd out exactly the denition is
modied)
...
4
...
12) to calculate the partition function for the entire system
...
This is a result of the fact that we have determinate momenta for the particles, but
indeterminate position; this essentially means that every particle is everywhere, and so we
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Toby Adkins

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cannot distinguish between them
...
This means that we can write the N
particle partition function as
Z=

N
Z1
N!

(3
...
But how much greater? The
number of single particle states can be roughly approximated by the single particle partition
function, as only the dominant terms remain in the sum
...
It is very easy to show
th
quantitatively that at room temperatures that this condition is indeed satised
...
4
...
First, the Helmholtz Free Energy:
F = −kB T log(Z) = −N kB T 1 − log

n
nQ

where we have used Stirling's Formula
...
27)

This is known as the Sackur-Tetrode equation
...
We can also use this equation to nd the
adiabatic equations (see (3
...
Next, the heat capacity:
3
3
U = F + T S = N kB T −→ CV = N kB
2
2

Thankfully, this is the same expression that we have derived via both Thermodynamics
and Kinetic Theory
...
The dependence of F on the volume comes from the
dependence Z ∝ V N , and so the equation of state only depends on the factor of volume in
the partition function
...


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3
...
3 Internal Degrees of Freedom
Thus far, we have only taken into account the translational degree of freedom for the ideal
gas, as we have implicitly assumed that it is monatomic gas
...
If we
model a diatomic molecule as two hard spheres connected by a weak spring that are free to
rotate around their centre of mass, then evidently we also have vibrational and rotational
degrees of freedom
...
28)

Z = Ztrans
...
Zrot
...
We now simply have to nd expressions
for the remaining two terms
...
Using (3
...
=
1 − e−Θvib
...


(3
...
= ω/kB , which is the temperature
at which the vibrational modes get excited
...


Rotational Freedom
For the rotational degrees of freedom, we treat the molecule as a rigid rotor
...
The degeneracy of each energy level is (2 + 1)
...
=

∞

(2 + 1)e

−βε

(2 + 1)e−

=

=0

( +1)Θrot
...
= 2 /(2IkB )
...
Performing the integration, by noting that d 2 + 1 = d( ( + 1)), we nd that
Zrot
...


(3
...
In the lower temperature limit, it is clear that the contribution to the heat capacity is approximately zero; the
partition function is dominated by the large Boltzmann factor, and so eectively tends to
zero
...

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Total Heat Capacity
As we calculate the Helmholtz Free Energy by taking the log of the partition function,
it is simply the sum of the separate Helmholtz Free energies for each degree of freedom
...
Evidently, which modes are excited depends on
temperature, and so the resultant heat capacity takes the form:

Figure 3
...
4
...

2
Assume that the energy of the system is given by the sum of quadratic modes
q

αi x2
i

E=
i=1

Assuming the probabilities to be Gibbsian, the mean energy can be calculated as follows:

...
dxn

...
dxn exp −β

q
2
i=1 αi xi

q
2
j=1 αj xj

∞
2
2
−∞ dxi αi xi exp −βαi xi
∞
2
−∞ dxi exp −βαi xi

αi x2
i

=
i=1

1
= qkB T
2

Note that this is only valid at high temperatures
...

One particular application of the equipartition theorem is in the prediction of the heat
capacity of gases
...
31)

Evidently, the value of f will depend on whether the translational (3), rotational (2) or
vibrational (2) modes are excited, as we saw in the previous section
...
4
...
} such that k nk = N
...
32)

βµ

Z = eZ1 e

where Z1 is the single-particle canonical partition function
...
However, now we can nd the chemical potential of an ideal gas
...

Φ = −kB T log(Z) = −kB T Z1 eβµ

Then, the mean particle number is given by (3
...
Z1internal
...
33)

As a sanity check, let us examine whether µ is large and negative (as it should be in the
classical limit)
...
This means that it is indeed large and negative
...
Starting from the grand canonical distribution,
show that the probability to nd exactly N particles in this volume is a Poisson distribution
...
Letting α = αN such that Nα = N :
pN =

eβµN
Z

e−βEαN
N

sum of energies in N

77

=

N
eβµN
eβµN Z1
ZN =
Z
Z N!

Toby Adkins

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It follows directly from (3
...
Then:
¯ N
¯
N
N
¯
Z1 −N log Z1
N N e−N
¯
=
e e
=
pN
N!
N!
¯
which is indeed a Poisson distribution with mean N
...


ZN
¯ βN
= 1 e−N e
N!

1
β

log

¯
N
Z1

Equilibria of Inhomogeneous Systems
Thus far, we have considered the entire system of the ideal gas to be homogeneous
...
We are going to make use of the fact that in equilibrium, the only parameter that is allowed to vary in space is n; we showed that the other
parameters must be xed in Section (3
...
2)
...
This means that we can write the partition function as
φ
Z1 = Z1 (0) Z1 = Z1 (0) e−βεφ

where Z1 (0) is the partition function of the system in the absence of the potential
...
We can thus arbitrarily set µ = 0, as we are only
looking at changes in the system, meaning that:
log

nλ3
th

=−

Z internal
1

εφ
kB T

So that in general the density distribution is given by
(3
...
Find the density distribution of the gas in equilibrium
...
In this frame, the
particles will be subject to a "centrifugal potential" given by
1
1
φ(r) = − (Ωr)2 −→ εφ = − m(Ωr)2
2
2

Using (3
...
Now, we have to nd our normalisation constant n(0)
...
Thus:
2

!

N=

dV n(r) = 2πh n(0)

2

dr reAr = 2πh n(0)

78

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Re-arranging, we can write that
2

n(r) = n
¯

AR2 eAr
eAR2 − 1

for n = N/(πR2 h)
...
To make things easier, consider the value of n at r = R
...
This means
that we can Taylor expand the exponential factor in A:
n(R)

n
¯

AR2
=n
¯
1 + AR2 − 1

This makes sense, as at very high temperatures, the thermal energy will be so high
that the gas does not care about the eects of the rotation
• Low temperature limit (T → 0) - In this limit, β becomes very large, and so the

exponential factor will dominate
...
This we would also expect as the
molecules would be dominated by the eect of the rotation, and would cluster in a
small cylindrical shell around the rim of the cylinder
...
5 Quantum Gases
Before jumping into the very interesting world of quantum gases, we rst have to look at
a particularly important property of the sorts of particles that we will be considering
...
Let us now 'swap' the two particles
...
|1, 2 = |2, 1 for bosons that have integer spin
2
...
This means that
for fermions |1, 1 = − |1, 1 = 0; that is, no two fermions can occupy the same quantum
state
...

This actually has an important eect on the behaviour of quantum gases
...
As such, the occupation numbers (the
amount in said state) of particular energy states are bounded for fermions (ni = 0, 1) by
the Pauli Exclusion Principle, but in principle unbounded for bosons (ni = 0, 1,
...

This crucial dierence between Fermi and Bose gases leads to quite dierent behaviours
under certain conditions, as we will soon see
...
5
...
Let εi be the single particle energy levels
...
35)

log 1 ± e−β(εi −µ)

log Z = ±
i

where the "+" sign corresponds to fermions, and the "−" sign corresponds to bosons
...

Let us now go back to considering our microstates {α}
...
Let us calculate the mean occupation numbers for each of these states
...
, nj ,
...
35), we arrive at the very important results for the mean occupation
numbers of
1
ni = β(ε −µ)
¯
(3
...
If you are going to memorize any result in the Statistical
Mechanics of quantum gases, it would be this one, as it allows you go calculate everything
else
...
5
...

As we saw in Section (3
...

=
i

(2s + 1)

V
(2π)3

= (2s + 1)

spin components k

from mesh size

d3 k =

(2s + 1)V
2π 2

∞

dk k 2
0

moving to polar coordinates

This means that the density of states is
g(k) =

(2s + 1)V 2
k
2π 2

This is essentially the same as we found before, except we are now taking into account the
spin of the particles, which is an intrinsically quantum mechanical eect
...

• Non-relativistic limit (kB T

mc2 ) - In this case, the energy relation is clearly
√
2 k2
p2
2m 1
√ dε
ε=
=
→ dk =
2m
2m
2 ε

Recalling that g(k) dk = g(ε) dε, we nd that the density of states is clearly
g(ε) =

(2s + 1)V m3/2 √
√
ε
2π 2 3

(3
...
In the ultra-relativistic case, ε
particle is ε = m
kc
...
38)

Numerical constants aside, we can see that the energy component of these states diers by
a factor of ε3/2 , which means that we will observe dierent sorts of behaviour for Quantum
Gases that are in the non-relativistic and ultra-relativistic limits
...


3
...
3 Standard Calculations
This section will simply cover some of the standard calculations that students need to
be familiar with when it comes to quantum gases
...
37), but results will be included for both cases
where convenient
...
We will have to do this implicitly; that
is, we will have to determine µ as the analytic solution to some equation for N (µ, T )
...
39)

A similar one can be obtained for the relativistic gas, except that nQ and f (β, µ) will have
dierent denitions (watch out for this when doing calculations!)
...

82

Toby Adkins

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Mean Energy
Again, we are going to have to convert our sum into an integral
...
However, when it is written
in this form, we can see that we require the quantity in brackets to reduce to 3/2 in the
classical limit, that we will examine in Section (3
...
3)
...

∞

Φ = −kB T log Z =

kB T

dε g(ε) log 1 ± e−β(

0

=

N kB T

∞

nQ 2
√
n
π

√

dx
0

2
=3

−µ)

x

log 1 ± e−β(

−µ)

d
(x3/2 )
dx

Integrating by parts:
2
Φ = − N kB T
3

∞

nQ 2
√
n
π

dx
0

x3/2
ex−βµ ± 1

2
=− U
3

from above
...
19) to nd the equation of state as
p=ξ

U
V

2
3
1
3

where ξ =

for non-relativistic
for relativistic

(3
...
We are starting to see that the dierences in the density of states
between non-relativistic and ultra-relativistic gases leads to a dierence in coecients in
the common expressions
...
5
...
Now consider an adiabatic process; we know that in
such a process that both S and N must remain constant
...
This means that
n
= constant
nQ

for the appropriate denition of nQ
...
40), we arrive at the result of
pV 1+ξ = constant

83

(3
...
We now ask the question what is the classical limit
for quantum gases? In general, the classical limit will occur where the gas is hot (T → ∞)
and dilute (n → 0)
...
Then:
2
f (β, µ) = √
π

√

∞

dx
0

x
x−βµ ± 1
e

2
√
π

∞

dx

√

x e−x eβµ = eβµ

0

Using the denition of f (β, µ), this means that the condition for the classical limit of a
quantum gas is
n
1
(3
...
This is in fact the same condition that we found
in Section (3
...
1)
...


Degeneration
In this context, degeneration occurs when the number of quantum states available to a
single particle becomes comparable to the number of particles in the system
...
This means that we would expect degeneration
appear when n is on the order of nQ , namely that
n
nQ

1

By writing nQ explicitly and n in terms of the pressure p, we can show that air at STP is
safely non-degenerate (n/nQ 10−6 ), where as electrons in metals are very much degenerate (n/nQ 104 ) under everyday conditions
...
The following two sections will examine how Fermi and Bose gases
behave when they become degenerate
...
5
...
Then, considering carefully the behaviour
of the exponential factor, the mean occupation numbers will behave according to
ni =
¯

1
eβ(ε−µ)

+1

→

1 for ε < µ(T = 0)
0 for ε > µ(T = 0)

This is shown in Figure (3
...
The consequence of this is that when Fermi gases become
degenerate, the electrons will begin to 'stack up' and occupy all the available single-particle
states from the lowest-energy one to maximum energy equal to the value of the chemical
potential at T = 0
εF = µ(T = 0)

This quantity εF is known as the Fermi energy, and tells us what the maximum energy
per particle is at T = 0
...

84

Toby Adkins

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Figure 3
...

εF

N=
0

εF

2
dε g(ε) = nQ V β 3/2 √
π

dε

√

0

2
ε = nQ V β 3/2 √
π

2 3/2
ε
3 F

as step function

Rearranging, we nd that the Fermi energy is given by
6π 2 n
2s + 1

εF =

2/3

2

2m

(3
...
Looking at Figure (3
...
This can be interpreted as roughly the
temperature at which the gas becomes degenerate
...
This does make sense; at absolute zero,
it would require a vanishingly small amount of energy to increase the temperature of the
system
...
2)
...

6

(3
...
This a result
that is worth remembering for Fermi gases, as it allows us to calculate quantities to greater
precision
...
It is a useful exercise to check that the results match those predicted
by rst order theory
...
Let us begin
by calculating the chemical potential from N :
N=

N
3/2

(2/3)εF

Observing that µ

2 3/2 π 2 1
µ +
√ (kB T )2 +
...
, we can re-arrange this expression to obtain the equation
µ = εF 1 −

π2
12

kB T
εF

2

(3
...


Evidently, this expression is well-behaved; it reduces to µ = εF at T = 0, which it has to
by the denition of εF
...
We see that it becomes very
large and negative at high temperatures, as anticipated by (3
...


Figure 3
...

U=

N
3/2

(2/3)εF

2 5/2 π 2 3 √
3
5π 2
µ +
µ (kB T )2 +
...


Toby Adkins

A1

where we have again used that µ
of heat capacity that

εF +
...
45)
...
46)

Thus, the correction to our zero-temperature result is initially linear in T for T
as can be seen in the gure below
...
4: The heat capacity of a degenerate Fermi gas
We could also have anticipated this result qualitatively
...
For small corrections, the number of excited
particles ∆Nexc
...


g(εF )∆ε

Looking at Figure (3
...
∆ε

kB T
...

All that remains is to calculate the equation of state
...
40) and our
expression for U from above
...


(3
...


Derive the Chandrasekhar mass limit for white dwarfs
...

87

Toby Adkins

A1

Figure 3
...
The answer lies in degeneracy
...
43), and
recalling that TF Tdeg , we can see that the degeneration temperature scales inversely
with mass
...
But
why does this matter? Looking at the gure above, it is clear that the pressure of a degenerate gas is much more than that of a non-degenerate gas, and at the density in stars,
we can essentially treat the electron gas as being at T = 0 as TF is very large
...

Now that this is understood, we can now begin with the derivation
...
First, let us
nd the potential energy of the system by building the star out of spherical shells
...
It is easy to show that the Fermi energy
for an ultra-relativistic gas is given by
εF = c

6π 2 n
2s + 1

1/3

−→ g(ε) =

N
ε2
(1/3)ε3
F

where we have re-written the density of states(3
...
Then:
Ue =

N
(1/3)ε3
F

εF
0

3
c
dε ε3 = N εF = 2 (3π 2 )4/3
4
4π

4
π
3

−1/3

N 4/3 R−1 =

A
R

where we have used the fact that V = (4/3)πR3
...
Re-arranging this equality, we nd the
expression for the mass as
M=

√
3 π
m2
p

5
16

3/2

c
G

3/2

1
...
We can do similar
calculations for non-relativistic white-dwarfs, and we obtain a stable radius R that scales
with M −1/3
...
It is not immediately obvious, or intuitive, that this should be the case!

3
...
5 Degenerate Bose Gas
We are now going to consider the behaviour of a degenerate Bose gas
...
This means that µ < εmin = ε0 = 0, where we have
arbitrarily chosen the lowest energy state to be that of zero energy
...
In this limit:
n0 =
¯

1
e−βµ

−1

= N −→ µ(T → 0) = −kB T log 1 +

1
N

−

kB T
N

0

This means that the chemical potential for a Bose gas is vanishingly small for these low
temperatures, as seen below
...
6: The chemical potential of a degenerate Bose gas
We know that the density of states g(ε) ∝ εp for some power p (not necessarily an integer)
...
This is
not surprising, seeing as the use of continuous approximation for a sum implicitly assumed
that there were a small number of particles in each state
...

We thus must come up with another way of describing the system
...
39)
...


Figure 3
...
This means that there
is some critical temperature Tc above which our original theory is valid; that is, some
temperature that satises n/nQ = f (0)
...
48)

For all T < Tc , we thus set µ = 0
...
= N
V nQ

T
Tc

3/2

Thus, the number of particles that remain in the groundstate is given by
n0 = N 1 −
¯

T
Tc

3/2

(3
...

As Tc is very small, this eect can only really be observed at very low temperatures, and
complete condensation can only occur at absolute zero
...

Astute readers may now be asking the question, what about the occupation number of
the rst excited state? Will that not be quite large as well, and change the behaviour of
90

Toby Adkins

A1

Figure 3
...
The smallest possible wave-number allowed in our system of volume V is
given by
2π
2π
= 1/3
L
V

kmin =

Using our classical expression for energy:
ε1 =

At low temperatures, 1

βε1

2 k2

=

2π 2

2 n2/3

1
N 2/3

2m
m
βµ
...
36),
n1
¯

1
−1

kB Tc
N 2/3

1
∝ N 2/3
βε1

eβε1

This means that n1 /¯ 0 1/N 1/3
...
As a result, we do not need to give special consideration to
particles in the rst excited state as a being a condensate, as they are only a small fraction
in comparison to those in the groundstate
...

This example will demonstrate the signicance of having a dierent density of states for
the system; it can completely change the way it behaves
...

Putting it another way, there is no unique critical temperature Tc dened by this equation
...
9: The pressure of a degenerate
Bose gas

Figure 3
...
Assuming T < Tc ,
we can set µ = 0
...
72N kB Tc

T
Tc

5/2

(3
...
Using (3
...
51nkB Tc

T
Tc

5/2

(3
...
In fact, careful calculation reveals that this equation is asymptotic to the
classical limit from below, as shown in the following gure above
Lastly, the heat capacity also follows very quickly from the denition of (3
...
93N kB

T
Tc

5/2

(3
...
Evidently, all of these quantities have
been calculated assuming the non-relativistic case, but all that changes are the numerical
factors, rather than the dependences on Tc
...

The easiest way to tackle this question is via the Grand Potential, and take derivatives to
nd χm
...
Writing the former in terms of the latter, we nd that
1
1
1
Φ = Φ0 (µ − 2µB s1 B) + Φ0 (µ − 2µB s2 B) + Φ0 (µ − 2µB s3 B)
3
3
3
1
1
1
= Φ0 (µ) + Φ(µ − 2µB B) + Φ0 (µ + 2µB B)
3
3
3

Evidently, we obtain the trivial equality that Φ = Φ0 when we set B = 0, as expected
...
of B

The magnetisation of the system is dened as
M=

∂Φ
∂B

1
V

=
T,V

¯
∂N
∂µ

8µ2 B ∂ 2 Φ0
8µ2 B
B
=− B
3 V ∂µ2
3 V

T,V

¯
Calculating an expression for N for a degenerate Bose gas for all T , and setting s = 1, we
arrive at the expression
∂M
∂B

χm =

In the classical limit, eβµ
∞
0

T,V

nλ3
th
√

√

∞

8µ2 2 ∂
= 3B √
λth π ∂µ

dx

x
ex−βµ − 1

−x

βµ

0

1, and so

x
dx x−βµ
e
−1

∞
βµ

e

dx

√

√
xe

=e

0

π
2

Substituting this into the expression for χm , we nd that the magnetic susceptibility reduces to Curie's Law in the classical limit, as it should
...
This means that the
system must have innite magnetic susceptibility at T = Tc
...
49)
...
However, as
soon as B is non-zero, this makes a particular alignment energetically favourable, and so
all the particles in the groundstate will align as such
...


93

Toby Adkins

A1

3
...
It follows that for a gas of N particles that the
energy density is given by
u = u(T ) = n ω

From Kinetic Theory, we know that the particle ux is Φ = nc/4, meaning that the energy
ux is given by P = uc/4
...
8), we can write the pressure of the
photon as as p = u/3
...


3
...
1 Thermodynamically
Starting from (3
...
Recalling
(1
...
This means that incident power per unit area is given
by
1
P = uc = σT 4
4

(3
...

However, to nd the actual value of the constant, we have to treat the photon gas using
the theory of Section (3
...


3
...
2 As a Quantum Gas
Photons are bosons, where their spin can take the values s = ±1
...
The density of
states for such a system is given by
g(k) dk =

2
polarisation states

V
V k2
4πk 2 dk = 2 dk
(2π)3
π

94

Toby Adkins

A1

Using the dispersion relation ω = ck, this can be written as
g(ω) dω =

V
ω 2 dω
π 2 c3

Assuming that there are a large enough number of photons to approximate them as having
a continuous spectrum, we can use (3
...
54)

When integrated, this will give the energy density of the photon gas
...
By comparison with (3
...
67 × 10−8 W m−2 K−4
60 3 c2

(3
...
56)

In the classical limit, β becomes very small, so we can Taylor expand, yielding the RayleighJeans law :
uλ (λ) =

8πkB T
λ4

This evidently diverges for low λ, giving innite energy density, which is known as the
ultraviolet catastrophe
...
56)
...

This gives rise to Wien's Law which states that
λmax T = constant

(3
...
This can be used to nd the temperature of stars, for
example
...


Further Thermodynamics

This chapter aims to extend the readers knowledge of Thermodynamics, covering
• Real Gases
• Phase Transitions

This quite short chapter builds on the ideas outlined in the rst three chapters to look
at some more accurate descriptions of gas behaviour, as well as look at some further
concepts about phase transitions between solids, liquids and gases
...
1 Real Gases
For most of the Thermodynamics that we have been doing with gases, we have merely
dealt with the Ideal Gas, as this is the simplest way to model gases
...
Before we do so, let us derive
a useful result for the energy of a gas
...
17), assuming a closed system (dN = 0):
∂U
∂V

=T
T

∂S
∂V

−p=T
T

∂p
∂T

−p
V

where we have made us of (1
...
Putting these together, this gives
dU = CV dT + T

∂p
∂T

− p dV

(4
...
It
is also a relation that one is commonly asked to derive, so it is well worth having in the
memory bank
...
We will use nm = N/NA to refer to the number
of moles of a substance
...
This means that we can write
the simple relation
N kB = n m R
(4
...


4
...
1 Virial Expansion
The equations describing real gases include some corrections for the fact that intermolecular
forces, that we have neglected with the Ideal Gas, exist
...

nm RT
V
V

(4
...

The Boyle temperature is dened as the temperature that satises B(Tb ) = 0; that is, the
temperature that means that the gas behaves like an ideal gas to rst order in nm
...


97

Toby Adkins

A1

4
...
2 Van-der-Waals Gas
This is the one of the most common real gas approximations
...

The Van-der-Waals equation of state is as follows:
p+

a n2
m
V2

(4
...
This
has been written in molar form as this allows easier comparison with (4
...
There are some key features to note:
• The number of nearest neighbours to a particular particle is proportional to nm /V

, and so attractive intermolecular interactions lower the total potential energy by
an amount proportional the number of atoms multiplied by the number of nearest
neighbours
...
This can be thought of as an eective
m
pressure causing the energy change, giving rise to the extra term shown

• The term bnm comes from considering the particles to have a nite size (instead of

point particles), meaning that we must exclude this volume in the equation of state

It is evident that this equation reduces to the Ideal Gas equation in the low density limit
nm /V
1
...


This means that the Boyle temperature for a Van der Waals gas is Tb = a/(bR)
...
1
...
5)

Once again, the bnm comes from the requirement of excluding the nite volume of the
particles from the calculation, and the exponential term regulates the strength of the
inter-particle interaction
...
Once again, this can also be
written as a virial expansion in the low density limit:
pV
e−(anm )/(RT V )
=
nm RT
1 − (bnm )/V

1+ b−

a
RT

nm
a2
ab
+ b2 +
−
2T 2
V
2R
RT

nm
V

2

+
...


4
...
4 Critical Points
In Thermodynamics, phases are regions of a system throughout which all physical properties
of a material are essentially uniform ; this is often used to refer to dierent states of a
particular substance that exist within the one system, such as a 'liquid phase' and 'gas
98

Toby Adkins

A1

Figure 4
...
With that in mind, consider the isotherms (lines of constant temperature) for the
Van der Waals gas
...

We remark how for a particular range of values of pressure (for a given isotherm), that
there are two possible values of volume that we shall call V1 and V2
...
We know from
Sections (3
...
2) and (3
...
1) that can only have equilibrium when temperature, pressure
and chemical potential are equal in both phases
...
6)

That is, the shaded areas under the curves must be equal
...
The region of the curve marked in red
in the diagram represents a region of instability; here, the isothermal compressibility
1
κT = − V ∂V
∂p T is negative, meaning that the more pressure you apply, the more the
gas is allowed to expand
...

Now, there is a particular critical point for which there is no point of inexion
...
This causes the equilibrium curve in the p-T plane to terminate at this critical point,
meaning that we can actually avoid 'sharp' phase transitions by going 'around' this point
...
As a
critical point is a point of inexion by denition, we can nd it by applying
∂p
∂V

= 0 and
T

∂2p
∂V 2

=0

(4
...
We
can then use these 'critical coordinates' to dene the compressibility factor
Z=

pc Vc
nkB Tc

This leads to what is known as the Law of Corresponding States ; that many dierent gases
all have the same compressibility factor
...
Writing the equation of state in terms of these reduced coordinates, according the the law of corresponding states, should give the same equation
regardless of the gas, with only the critical units changing from gas to gas
...

We can re-arrange (4
...

Then, it is just a few simple steps of algebra to show that
Tc =

a
4bR

and pc =

a
4e2 b2

˜
Writing T = T Tc , and similarly for p and V , it is again a step of relatively simple algebra
to show that
1
˜
p(2V − 1) = T exp 2 1 −
˜ ˜
˜V
T˜

4
...
5 Expansions of Real Gases
We can use the results derived in the previous sections to now look at some of the deviations
from Ideal Gas behaviour that can be observed in practise; this often involves the expansion
of gases
...


100

Toby Adkins

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The Joule Expansion
We have already met this expansion in Section (1
...
5)
...
The change in
internal energy is therefore zero
...
31), this becomes
µJ = −

1
CV

T

∂p
∂T

(4
...
As we have seen, µJ = 0 for an Ideal Gas, meaning that we
observe no change in temperature
...
This
m
we would expect, as for real gases the potential energy is raised by forcing the molecules
apart against intermolecular forces upon expansion, which lowers the kinetic energy, and
thus temperature
...
The gradient between the two chambers, as well as
the rate of ow, are kept constant using pistons
...


Figure 4
...
This means that we can write the energy change
as
U2 − U1 =

p1 V1
work done by p1

−

p2 V2
work against p2

101

−→

U1 + p1 V1 = U2 + p2 V2
H1

H2

Toby Adkins

A1

It is clear that the the enthalpy H is conserved in the process
...
We thus dene (you guessed it), the
Joule-Kelvin coecient
µJK =

∂T
∂p

∂T
∂H

=−
H

∂H
∂p

p

T

Recalling that dH = T dS + V dp, we have that
∂H
∂T

=T
p

∂S
∂T

∂H
∂p

= Cp and
p

=T
T

∂S
∂p

+V
T

Putting this together, we can write the Joule-Kelvin coecient as
µJK =

1
T
Cp

∂V
∂T

−V

(4
...

This is evidently again zero for the ideal gas
...
22), this can be written in the more convenient form
T

∂p
∂T

+V
V

∂p
∂V

=0

(4
...


102

Toby Adkins

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4
...
A very typical example
is that of the boiling of water; the phase transition is quite rapid, as it is only when the
boiling point is reached that the liquid phase becomes thermodynamically unstable, and
the gas phase thermodynamically stable
...
2
...
This is known as the latent heat (of evaporation,
melting, etc
...
11)
where T is used to refer to the temperature at which the phase transition occurs; the
change in entropy occurs instantaneously at this temperature
...

Ehrenfest's method of classifying phase transitions is that the order of a phase transition
is the order of the derivative of G or µ that is discontinuous
...
Second-order
phase transitions have no latent heat but may have discontinuous heat capacities or compressibility (2nd derivatives of G)
...
In Van der Waals gas, we can have either continuous
or discontinuous phase changes depending on whether we go 'directly across' the phase
boundary or around the critical point, as outlined in Section (4
...
4)
...
2
...
18) that G = µN
...
From
Section (3
...
1), we know that µ1 = µ2 for the phases to coexist in equilibrium, and so we
know that along the p-T boundary for the two phases
dµ1 = dµ2
dp
dp
−s1 + v1
= −s2 + v2
dT
dT
dp
s1 − s2
=
dT
v1 − v2

Substituting in our expression for latent heat, we obtain the Clausius-Clapeyron equation
dp
L
=
dT
T (V2 − V1 )

(4
...

103

Toby Adkins

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Solid-Liquid Boundary
The Clausius-Clapeyron equation can be simply re-arranged to give
LdT
T ∆V

dp =

where ∆V = V2 − V1
...
13)

The constants p0 and T0 are chosen such that p = p0 and T = T0 on the boundary
...


When lead is melted at atmospheric pressure the melting point is 327
...
01 × 103 to 10
...
5 kJ kg−1
...

We can use the above equation to write that
∆p =

L
log
∆V

T2
T1

−→ T2 = T1 e

∆p∆V
L

We are already given that ∆p = 99 atm
...
Then
the change in the specic volume (volume per unit mass) is given by
∆V = v liquid − v solid =

1
ρ liquid

−

1
ρ solid

3
...
75◦ C , which is a very small change in temperature
...
The former of these assumptions is not
particularly assumptive, as most incompressible uids undergo a large expansion when
moving to their gaseous phase
...
This equation can be
integrated to give
p = p0 exp −

L
nm RT

(4
...
This equation can be used to
solve that annoyingly typical example about boiling a cup of tea on the top of a mountain;
it is evident that the British simply cannot get away from their tea
...
At the phase boundary, we can write that ∆S = L/T
...
Using the assumption about the relative volumes of the liquid and the
gas, and substituting the Clausius-Clapeyron equation, we nd that
L
L
dL
= + ∆Cp −
dT
T
Vg

∂Vg
∂T

p

Assuming that the gas produced is ideal, then this equation simplies to
dL
= ∆Cp −→ L = L0 + ∆Cp T
dT

Substituting this result back into the Clausius-Clapeyron equation, and solving, yields:
p = p0 exp −
p-T

∆Cp log(T )
L0
+
nm RT
nm R

(4
...
Suppose that
a saturated gas expands adiabatically
...
This means that
when moving left (lower pressure and temperature associated with expansion) along the adiabatic curve, we need to be able to cross the
transition curve
...
3: The adiabatic compression of
a gas
∂p
∂S

<
S

dp
dT

105



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