fisika_statistik

Lecture notes on

Physics 18324312

STATISTICAL PHYSICS

Tjipto Prastowo and Mita Anggaryani

Department of PhysicsFaculty of Mathematics and Natural Sciences

State University of Surabaya

Physics Department 11 February 2008

TO THE STUDENT WE LOVE

Here it is, lecture notes on Fisika Statistik, a series of lectures taught once a week intahun ke tiga, Jurusan Fisika, Fakultas Matematika dan Ilmu Pengetahuan Alam (FMIPA),Universitas Negeri Surabaya (Unesa). Please note that we do not intend to replace anystandard book of statistical physics as listed in the Bibliography. Rather, we would like tohelp students understand materials covered in this course through simple, but systematicguidance.

To use math and other basic theory needed in this course, you need not just knowledge,but skill. This can be obtained only through continual practice. You may obtain a superficialknowledge by listening to lectures in the class, but you cannot reach the skill expected bythat way. It is common to come across student conversation outside the class, for example,something like “I understand it but I can’t do the problem !” This student feels uncomfortableto do some problem although it looks so easy when the lecturers do it in the class.

Such a statement shows lack of practice and hence lack of skill. Our dearest students,please always study with pencil and paper at hand. You will find that the more able youare to choose effective methods of solving problems the easier it will be for you to masternew materials. This costs you nothing but practice, practice and again practice. Please doremember that the best way to learn to solve problems is to solve them.

We welcome good comments for further improvements because penyusunan materi ajarini merupakan salah satu upaya untuk meningkatkan kualitas layanan publik. Semoga halini bermanfaat bagi segenap sivitas akademik di lingkungan Jurusan Fisika FMIPA, Unesa.

Best wishes,Kampus Ketintang, 12 Mei 2008Mita Anggaryani, T jipto Prastowo

ii

iii

General Guidance

PHYSICS 18324312: STATISTICAL PHYSICS

Pre-requisites: Fundamental Physics (I and II) and Mathematical Physics (I and II)

Lecturers: Tjipto Prastowo, Mita Anggaryani

References: An Introduction to Statistical Physics for Students (A. J. Pointon)

Time and Place: Monday, 07.00 - 09.40am, D1

Marking Scheme: NA = 20%P + 20%UTS + 30%T + 30%UAS

NA=Final Mark, P=Presence, UTS=Mid-Exam, T=Homework, UAS=Final Exam

Notes:

1. Students are not allowed to join the class for being late (a maximum of 15 minutesfrom the starting time), except for reasonable arguments.

2. Each lecturer contributes an equal proportion of mark to the final mark.

3. P is possibly reduced to a minimum.

4. UTS = 50%Q1 + 50%Q2 Q = Quiz

5. T = 50%T1 + 50%T2 T = Assignment

6. UAS normally contains 4, but possibly has 5 problems.

7. Assignments will be distributed to class members and all students are required to handthe completed assignments in within a limited time. Various penalties will be given forany delay, i.e, 25% discounted mark for a one-day delay and 50% for a two-day delay.There will be no mark given for more than two-day delays.

8. No additional assignments or examinations after Final Exam, except for specifiedreasons with very limited permission or medical examination required.

9. Students are allowed to have their own notes (2 pages only), written in a folio paper,in both Mid-Exam and Final Exam.

10. Other issues will be determined during the course, including details of problems andexam time-tables. Students are strongly encouraged to be active and well-prepared. Ifpossible, tutorial is available for a further, detailed description of each topics.

Course Contents:

1. Chapter One: Introduction (Lectures 1 and 2)

• overview, system of many particles, distribution function, description of assemblyand phase space, the scope of statistical physics

2. Chapter Two: Maxwell-Boltzmann Statistics (Lectures 3, 4 and 5)

• fundamental concepts, classical systems, applications of MB statistics, Boltzmannpartition function

3. Chapter Three: Bose-Einstein Statistics (Lectures 6 and 7)

• fundamental concepts, boson systems, applications of BE statistics

4. Chapter Four: Fermi-Dirac Statistics (Lectures 8 and 9)

• fundamental concepts, fermion systems, applications of FD statistics

5. Quiz 1: Chapters Two, Three and Four (Lecture 10)

6. Chapter 5: Thermodynamics of Gases (Lectures 11 and 12)

• concept of entropy, Gibb’s paradox, semi-classical perfect gas

7. Chapter 6: Applications of Statistical Thermodynamics (Lectures 13 and 14)

• system of two-energy level, harmonic oscillator, diatomic molecule

8. Quiz 2: Chapters 5 and 6 (Lecture 15)

9. Final Exam Preparation: Discussion on Quiz 1 and 2 (Lecture 16)

iv

Contents

1 Introduction 11.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Distribution function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Description of assembly and phasa space . . . . . . . . . . . . . . . . . . . . 31.4 The scope of statistical physics . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Maxwell-Boltzmann Statistics 52.1 The speed distribution function . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 The momentum and energy distribution functions . . . . . . . . . . . . . . . 92.3 Applications of Maxwell-Boltzmann statistics . . . . . . . . . . . . . . . . . 10

2.3.1 The mean, rms and most probable velocities . . . . . . . . . . . . . . 102.3.2 Equipartition principle of energy . . . . . . . . . . . . . . . . . . . . 112.3.3 Specific heats . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.4 The Boltzmann partition function . . . . . . . . . . . . . . . . . . . . . . . . 142.5 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Bose-Einstein Statistics 173.1 Weight of configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.2 Population of particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.3 The Bose-Einstein gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.4 Black-body radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.5 The specific heat of solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.6 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4 Fermi-Dirac Statistics 294.1 Weight of configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294.2 Population of fermions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304.3 The Fermi-Dirac gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304.4 The electron gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324.5 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

5 Termodinamika Gas 355.1 Konsep entropi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355.2 Gas klasik . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.3 Paradoks Gibbs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395.4 Gas semi klasik . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405.5 Latihan soal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

v

vi CONTENTS

6 Aplikasi Distribusi Statistik 456.1 Sistem dua tingkat energi . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456.2 Osilator harmonik . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476.3 Gas diatomik . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496.4 Latihan soal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

Bibliography 55

Chapter 1

Introduction

1.1 Overview

This course is devoted to discuss systems containing very many particles (of order 1023),

which include gases, liquids and solids. The inherent problem of these systems appears to

be complex owing to the complexity of interactions of such a large number of particles. The

behaviour of these particles as a whole often reveals different qualitative features from that

of each of the individual particles. Since it is not possible to study the detailed behaviour

of each particle, a statistical approach then becomes plausible. The question is that what

kind of statistics is appropriate for describing the dynamics of the system? Can common

statistics handle this problem? Or, do we need to seek for an alternative method? If so,

what should we do then?

At this stage, it is useful to distinguish microscopic systems (i.e., small-scale systems of

order 10−9 m or less), whose description is of fundamental interest, from macroscopic systems

(i.e., large-scale systems of order 10−6 m or greater). In the latter systems, macroscopic

parameters, such as temperature, volume and pressure, are usually used to characterise the

systems. For an isolated, macroscopic system, these parameters may not vary with time

and the associated condition is said to be in equilibrium. Whereas for an open, macroscopic

system where interactions between particles constituting the system and the surroundings are

allowed, macroscopic parameters depend upon time. The time variations of these parameters

lead to mean values over a period of large times corresponding to a final equilibrium state.

The discipline in which fundamental laws of physics describing the relationship between

macroscopic parameters are examined is the subject of thermodynamics. A fundamental

approach, statistical physics provides insights into macroscopic systems from a microscopic

point of view. In this approach, a wide range of physical principles from basic laws to

advanced methods are used to examine systems under consideration.

1

2 1. Introduction

1.2 Distribution function

In this section, we discuss discrete and continuous distribution functions that represent

a physical system. For any statistical data associated with this system, three important

quantities involved are: the average or mean value, the root-mean-square (rms) value and

the most probable value. If the data are given in a discrete form of xi, then these values are

written as xave, xrms and xmax, respectively, and are in general different from each other. In

practice, such values are obtained by introducing a discrete distribution function fi defined

as fi = ni/N , where ni is the occurrence of the individual data and N is the total number

of the data. The followings are useful definitions associated with this discrete system.

∑i

fi =∑i

niN

=1

N

∑i

ni = 1, (1.1)

xave =∑i

xi fi =1

N

∑i

xi ni, (1.2)

(x2)ave =∑i

x2i fi =

1

N

∑i

x2i ni xrms =

√(x2)ave, (1.3)

xmax is obtained when fi is maximum. (1.4)

Let us examine further the above definitions using an explicit example: 10 students taking

Final Exam on Statistical Physics. The Final Marks are 20, 20, 35, 60, 60, 60, 60, 80, 80

and 90. From these marks we have discrete data in the form of xi = 20, 20, 35, 60, 60, 60,

60, 80, 80, 90, and N = 10 for which

∑i

fi =1

N

∑i

ni =1

10

(n1 + n2 + n3 + n4 + n5

)=

1

10

(2 + 1 + 4 + 2 + 1

)= 1,

xave =1

N

∑i

xi ni =1

10

(20× 2 + 35× 1 + 60× 4 + 80× 2 + 90× 1

)= 56.5,

xrms =1

N

∑i

x2i ni =

1

10

(202 × 2 + 352 × 1 + 602 × 4 + 802 × 2 + 902 × 1

)= 61.1,

xmax = 60 for f1 = n1/N = 0.2, f2 = n2/N = 0.1, f3 = n3/N = 0.4, f4 = n4/N = 0.2 and

f5 = n5/N = 0.1.

As statistical physics deals with systems of very many particles, the above formulations

no longer hold. In these systems, a continuous distribution function f(x) is used to describe

the dynamics of the systems. Hence, (1.1) through (1.4) are replaced by∫f(x) dx = 1, (1.5)

1.3. Description of assembly and phasa space 3

xave =

∫x f(x) dx, (1.6)

(x2)ave =

∫x2 f(x) dx xrms =

√(x2)ave, (1.7)

xmax occurs when the first derivative of f(x) is equal to zero. (1.8)

In addition, a useful quantity, the standard deviation σ, is introduced as a measure of the

spread of data about the average value (i.e., the scatter of the measurements). The standard

deviation of the distribution is defined as

σ =√

(x2)ave − x2ave, (1.9)

where σ is, by definition, positive since xrms ≥ xave. For a normal or Gaussian distribution,

two-thirds of the values constituting the data are expected to fall within the range xave± σ.

There will always be random and systematic errors in the measurements. Mean values thus

include some error, which could also be indicated by the standard error of the mean σm,

defined as

σm =σ√N

(1.10)

It is therefore possible to write the results of the measurements as xave ± σm.

1.3 Description of assembly and phasa space

Here an assembly is defined as a group of individual components forming a physical body.

For example, in the case of a gas in a closed container such components are gas atoms or

gas molecules while the assembly is the gas itself. The state of an assembly is determined

by position and momentum of each of individual components in a six-dimensional, Γ space

(known as phasa space). In a Cartesian coordinate system, the phasa space consists of

Euclidean space (x, y, z) and momentum space (px, py, pz). Thus, any quantity can be

written in terms of position and momentum coordinates. For example, element volume dΓ

of a single-particle system in phasa space is written as

dΓ = dx dy dz × dpx dpy dpz, (1.11)

where dxdydz is element volume in Euclidean space and dpxdpydpz is element volume in

momentum space. The (non-relativistic) kinetic energy ε of this system is written as

ε =1

2m

(p2x + p2

y + p2z

), (1.12)

4 1. Introduction

where m is the mass of the system, and px, py and pz are the components of linear momentum

of the system in x, y and z directions, respectively.

The simple method above could be extended for a system consisting of many particles.

The corresponding element volume dΓN and total kinetic energy E for N particles are

respectively written as

dΓN = dx1 dy1 dz1 dpx1 dpy1 dpz1 × · · · · · · dxN dyN dzN dpxN dpyN dpzN

=N∏i=1

dxi dyi dzi dpxi dpyi dpzi(1.13)

and

E = ε1 + ε2 + ε3 + · · · · · · εN

=N∑i=1

1

2m

(p2xi + p2

yi + p2zi

),

(1.14)

where i in both (1.13) and (1.14) refers to individual particles.

1.4 The scope of statistical physics

While modern statistical physics in general covers a broad spectrum of materials, the main

theme considered in this course involves classical and quantum statistics, with their simple

applications being introduced. There are fundamental differences in the basic assumptions

made regarding the distinct behaviour of classical and quantum assemblies. For a classical

assembly, each of individual components is considered to be completely distinguishable. For

a quantum assembly, however, the particles constituting the assembly is considered to be

indistinguishable. Another difference between the classical and quantum statistics is that

energy and momentum of classical systems spread over a continuum spectrum while in the

quantum systems particles are distributed over discrete energy levels. The quantum systems

studied here are divided into two groups of particles: bosons and fermions. The differences

between these systems will be detailed in the next chapters.

The main reference to which we refer is Pointon (1978) as this book is much simpler for

undergraduate students without losing basic ideas behind the scene. The students are also

encouraged to have a read the topics considered in other sources from a fundamental level

(Tipler, 1999) to an advance level (Reif, 1985; Huang, 1987), as well as from related materials

in a standard book of modern or quantum physics (Beiser, 1988; Gasiorowicz, 1996).

Chapter 2

Maxwell-Boltzmann Statistics

It is desired to find a continuous distribution function that characterises classical systems.

These systems are allowed to have arbitrary values of energy and momentum, for which they

obey Maxwell-Boltzmann (MB) statistics. We will derive this statistical distribution using

a simple way, independent of the details of molecular interactions. We first construct the

classical velocity distribution function for each direction, with which we derive the MB speed

distribution function in §2.1. In order to obtain a full description of a classical assembly,

we then examine the momentum and energy distribution functions in §2.2. This is followed

by the applications of the MB statistical distribution in §2.3. The fundamental concept of

partition function is also introduced in §2.4.

2.1 The speed distribution function

Here we are particularly interested in finding the speed distribution function of a classical

system containing very many particles. Such a system can be an ideal gas in a closed volume.

Once this function is obtained, other distribution functions (i.e., energy and momentum

distribution functions) can be directly derived.

We begin with describing the dynamics of molecules of such a gas using a Cartesian

coordinate system for simplicity. The density ρN of the gas molecules per unit volume in

velocity space is determined by the product of the total number N of the molecules and

velocity distribution function for all motional directions,

ρN = N f(vx) f(vy) f(vz), (2.1)

where f(vx), f(vy) and f(vz) are velocity distribution functions, and vx, vy and vz are the

speeds of the molecules in x, y and z directions, respectively.

5

6 2. Maxwell-Boltzmann Statistics

The total derivative dρN of (2.1) with respect to vx, vy and vz is

dρN =∂ρN∂vx

dvx +∂ρN∂vy

dvy +∂ρN∂vz

dvz, (2.2)

which can be, due to (2.1), rewritten as

dρN = N(f ′(vx) f(vy) f(vz) dvx + f(vx) f

′(vy) f(vz) dvy + f(vx) f(vy) f′(vz) dvz

). (2.3)

It is expected to have separated parameters for each of the terms on the RHS of (2.3). Thus

we divide (2.3) by ρN , resulting in

dρNρN

=f ′(vx)

f(vx)dvx +

f ′(vy)

f(vy)dvy +

f ′(vz)

f(vz)dvz.

As the density is constant, i.e., dρN = 0, the above equation becomes

0 =f ′(vx)

f(vx)dvx +

f ′(vy)

f(vy)dvy +

f ′(vz)

f(vz)dvz. (2.4)

We assume that the gas molecules are separated by distances that are sufficiently large

compared with the molecule diameters and that there is no external forces acting on the

molecules except when they collide. In the absence of such forces, there is no preferred

direction for the velocity of each molecule and the speed is, of course, constant. We can then

write

v.v = v2 = v2x + v2

y + v2z = constant,

which can be derived with respect to vx, vy and vz to obtain

vx dvx + vy dvy + vz dvz = 0, (2.5)

as a restrictive condition for (2.4). Thus, we have so far a set of two-differential equations,

(2.4) and (2.5). The solution of these equations can be obtained by introducing a constant

parameter λ, to be determined later, by which we multiply (2.5) to obtain

λ vx dvx + λ vy dvy + λ vz dvz = 0. (2.6)

We then add (2.6) to (2.4) to obtain(f ′(vx)

f(vx)+ λ vx

)dvx +

(f ′(vy)

f(vy)+ λ vy

)dvy +

(f ′(vz)

f(vz)+ λ vz

)dvz = 0,

where dvx, dvy and dvz are all not zero for non-trivial solutions. Instead, a system of

2.1. The speed distribution function 7

homogeneous differential equations below are required,

f ′(vx)

f(vx)+ λ vx = 0,

f ′(vy)

f(vy)+ λ vy = 0,

f ′(vz)

f(vz)+ λ vz = 0,

for which f(vx), f(vy) and f(vz) will be determined. For the x-component, the above

expression means

df(vx)

dvx= −λ vx f(vx) or

df(vx)

f(vx)= −λ vx dvx,

which can be easily solved. Integrating both sides and applying simple calculus yield

`nf(vx) = −1

2λ v2

x + `nC = `nCe−12λv2x .

We thus have

f(vx) = Ce−12λv2x , (2.7)

where C is an integration constant, and will be determined below. The next step is thus to

determine the exact value of C by normalising (2.7) such that∫ ∞−∞

f(vx) dvx = 1 or C

∫ ∞−∞

e−12λv2x dvx = 1. (2.8)

As e−12λv2x is an even function, (2.8) can then be written as

2× C∫ ∞

0

e−12λv2x dvx = 1

(see Appendix 6 of Pointon, 1978 for the details of the properties of Γ function integrals).

We here refer to ∫ ∞0

xn e−ax2

dx =1

2a(n+1)/2Γ(n+ 1

2

)(2.9)

and use Γ(1/2) =√π to calculate C as

C

(2

λ

)1/2√π = 1 or C =

√λ

2π.

Hence, (2.7) becomes

f(vx) =

√λ

2πe−

12λv2x . (2.10)


Using (2.10), we can calculate the mean square velocity as

(v2x)ave =

∫ ∞−∞

v2x f(vx) dvx =

∫ ∞−∞

v2x

√λ

2πe−

12λv2x dvx =

1

λ, (2.11)

where we have again used (2.9) and a simple relation, Γ(n+ 1) = nΓ(n).

According to the kinetic theory of a perfectly ideal gas, for every degree of freedom in

each direction translational kinetic energy is equal to the thermal energy of kT/2 such that

1

2m (v2

x)ave =1

2kT,

1

2m (v2

y)ave =1

2kT,

1

2m (v2

z)ave =1

2kT, (2.12)

where m is the total mass of molecules, k is the Boltzmann’s constant (1.38 × 10−23 J K−1)

and T is the equilibrium temperature of the molecules. We thus have

(v2x)ave =

kT

m(2.13)

for the x-component of the velocity. Combining both (2.11) and (2.13) yields λ = m/kT

and similar results are obtained for the y and z directions. We can therefore rewrite the

x-component of the velocity distribution function in (2.10) as

f(vx) =

√m

2πkTe−

12mv2x/kT . (2.14)

In the same manner, the velocity distribution function for the other degrees of freedom can

be obtained,

f(vy) =

√m

2πkTe−

12mv2y/kT and f(vz) =

√m

2πkTe−

12mv2z/kT . (2.15)

Having derived the velocity distribution function for each direction, we are now ready to

derive the corresponding speed distribution function by defining the molecule density ρN ,

previously given in (2.1), as

ρN =dNvx vy vz

dvx dvy dvz, (2.16)

where dNvx vy vz is the number of gas molecules having velocity components in the range vx

to vx + dvx in the x-direction, vy to vy + dvy in the y-direction, vz to vz + dvz in the

z-direction, and dvxdvydvz is element volume in velocity space in a Cartesian coordinate

system. Substituting (2.16) into (2.1) gives

dNvx vy vz

dvx dvy dvz= N f(vx) f(vy) f(vz) (2.17)

2.2. The momentum and energy distribution functions 9

or

dNvx vy vz = N

(m

2πkT

)3/2

e−12m(v2x+v2y+v2z)/kT dvx dvy dvz. (2.18)

Equation (2.18) can be written in a spherical coordinate system as follows

dNvr vθ vφ = N

(m

2πkT

)3/2

e−12mv2/kT v2 sin θ dθ dφ dv. (2.19)

Integrating (2.19) over the whole space and using (2.17) written in terms of v yield

dNv = N f(v) dv = N

(m

2πkT

)3/2

e−12mv2/kT v2 dv

∫ π

0

sin θ dθ

∫ 2π

0

dφ,

or

dNv = N f(v) dv = 4π N

(m

2πkT

)3/2

e−12mv2/kT v2 dv, (2.20)

where dNv is the number of gas molecules having speed in the range v to v + dv. It is clear

from (2.20) that

f(v) = 4π

(m

2πkT

)3/2

e−12mv2/kT v2, (2.21)

which is known as the MB speed distribution function. This function is more convenient to

use than the velocity component distribution functions, defined in (2.14) and (2.15), in that

it is taken independent of direction.

2.2 The momentum and energy distribution functions

As previously mentioned, the speed distribution function defined in (2.21) can be used to

derive other forms of distribution function. For convenience, we will first derive momentum

distribution function, then energy distribution function. But first, we rewrite (2.21) as

f(v) dv = 4π

(m

2πkT

)3/2

e−12mv2/kT v2 dv, (2.22)

where f(v) dv is the MB probability speed distribution function. Integral of this function

over the whole space is, by definition, unity since such integral describes the total probability

to find a molecule with speed in the range v to v + dv in the velocity space. The Euclidean

volume V of each molecule is absent in the present discussion. It follows that the properties

of the molecules of ideal gases are independent of the geometry of the molecules.

Using a simple relation from classical physics p = mv and its associated differential form

dp = m dv, we can write an expression for the MB probability momentum distribution


function f(p) dp as

f(p) dp = 4π

(1

2πmkT

)3/2

e−p2/2mkT p2 dp, (2.23)

which characterises classical assemblies having momentum in the range p to p + dp. As is

the case of the probability speed distribution function given in (2.22), the integral of (2.23)

over the whole space equals to one. It is understood that

f(p) = 4π

(1

2πmkT

)3/2

e−p2/2mkT p2, (2.24)

where f(p) is known as the MB momentum distribution function.

Again, using a simple formula from classical physics but this time relating momentum to

kinetic energy p =√

2mε and its associated differential form dp =√m/2ε dε, we can write

an expression for the MB probability energy distribution function f(ε) dε as

f(ε) dε = 2π

(1

πkT

)3/2

e−ε/kT ε1/2 dε, (2.25)

where f(ε) dε describes classical assemblies with energy in the range ε to ε + dε. Again, as is

the case of the probability speed and momentum distribution functions given in (2.22) and

(2.23), respectively, the integral of (2.25) over the whole space equals to one. It then follows

that

f(ε) = 2π

(1

πkT

)3/2

e−ε/kT ε1/2, (2.26)

where f(ε) is known as the MB energy distribution function.

2.3 Applications of Maxwell-Boltzmann statistics

2.3.1 The mean, rms and most probable velocities

As mentioned in §1.2, there are three dynamic quantities that can be drawn from a given

distribution function. These quantities are the mean, rms and most probable values. For

example, we will use the MB speed distribution function given in (2.21) to derive the mean,

rms and most probable velocities of an ideal gas in a closed container having total volume V

and N molecules. It is expected that maximum information can be obtained by calculating

such values.

Here we calculate the mean velocity vave, which is by definition written as

vave =

∫ ∞0

v f(v) dv = 4π

(m

2πkT

)3/2 ∫ ∞0

e−12mv2/kT v3 dv. (2.27)

2.3. Applications of Maxwell-Boltzmann statistics 11

Using (2.9) and a simple relation, Γ(n+1) = n!, the above equation yields

vave =

√8kT

πm, (2.28)

where m and T are the mass and temperature of such a gas, respectively. The rms velocity

can be derived in a similar manner to that used for the mean velocity. Thus, we have

(v2)ave =

∫ ∞0

v2 f(v) dv = 4π

(m

2πkT

)3/2 ∫ ∞0

e−12mv2/kT v4 dv. (2.29)

Again, applying (2.9) and Γ(n+ 1) = nΓ(n) to (2.29) yields

(v2)ave =3kT

m, vrms =

√(v2)ave =

√3kT

m. (2.30)

The above result for the rms velocity is in line with the sum of the equations in (2.12). To

make this clear, we here rewrite it as

1

2m (v2)ave =

1

2m (v2

x)ave +1

2m (v2

y)ave +1

2m (v2

z)ave =3

2kT. (2.31)

The most probable velocity occurs when the the speed distribution function f(v) defined

in (2.21) is maximum, i.e., the first derivative of this function is zero. Thus, we have

df(v)

dv= 4π

(m

2πkT

)3/2(2v − mv3

kT

)e−

12mv2/kT = 0, (2.32)

for which we obtain

vmax =

√2kT

m. (2.33)

The results for the mean, rms and most probable velocities confirm that they are in general

different. The rms velocity is the one, which is directly related to the mean energy of a

system having expression for energy in a special form that will be discussed below.

2.3.2 Equipartition principle of energy

In this section, we discuss the equipartition principle for systems having energy in the

form of quadratic terms. Such systems can be an ideal gas with the same probability of

translational movements in all directions, in which molecular interactions are negligible, or

a harmonic oscillator where potential interactions between individual components exist. For

convenience, we first examine the application of the principle to an ideal gas, and later

extend the result to a harmonic oscillator.


The mean energy of a perfectly ideal gas having energy between ε and ε + dε can be,

with the help of (2.26), calculated as

εave =

∫ ∞0

ε f(ε) dε = 2π

(1

πkT

)3/2 ∫ ∞0

e−ε/kT ε3/2 dε. (2.34)

Using a functional approach of Γ function below,∫ ∞0

xn e−ax dx =1

an+1Γ(n+ 1), (2.35)

we obtain

εave = 2π

(1

πkT

)3/2(kT)5/2 3

4

√π =

3

2kT. (2.36)

When the gas is in equilibrium, the total mean energy will be equally distributed among

the translational kinetic energies associated with momentum in the x, y and z directions.

This is well described by (2.12) and (2.31). Thus, we have

εave =1

2m (v2)ave = 3× 1

2m (v2

x)ave =3

2kT, (2.37)

and hence1

2m (v2

x)ave =1

2kT. (2.38)

The result shows that each component of momentum that appears as a squared term in the

expression for the energy will contribute 12kT per molecule.

When an equilibrium system is in rotational and vibrational motions as well, the total

energy of the system is the sum of all possible forms of the energy associated with types

of motions. Potential interaction between the individual particles may lead to vibrational

motion, where the corresponding potential energy is in the form of a quadratic term. For

example, the energy of a one-dimensional harmonic oscillator can be written as

ε =p2x

2m+

1

2µx2, (2.39)

where µ denotes the restoring force per unit displacement. Now we introduce a symbol 〈〉,which is actually the same as ( )ave used before. Thus, we have

〈 ε 〉x =

⟨p2x

2m

⟩+

⟨1

2µx2

⟩=

1

2kT +

1

2kT = kT.

(2.40)

Notice that the LHS of (2.40) describes the mean energy corresponding to the x-component

2.3. Applications of Maxwell-Boltzmann statistics 13

of momentum of 1-D harmonic oscillator. For 3-D harmonic oscillator, we have

〈 ε 〉x y z =

⟨p2x

2m

⟩+

⟨p2y

2m

⟩+

⟨p2z

2m

⟩+

⟨1

2µx2

⟩+

⟨1

2µy2

⟩+

⟨1

2µz2

⟩=

1

2kT +

1

2kT +

1

2kT +

1

2kT +

1

2kT +

1

2kT = 3kT.

(2.41)

2.3.3 Specific heats

In this section, we derive a useful quantity, namely specific heat, by defining the total energy

of a classical system of N particles as E = N〈 ε 〉. The specific heat discussed here is that of

perfectly ideal gases. For the ideal gas with total N molecules and the mean energy 〈 ε 〉, as

given in (2.36), we have

E = N〈 ε 〉 =3

2NkT =

3

2nNokT =

3

2nRT, (2.42)

where n denotes the number of moles, No = 6.02 × 1023 mole−1 is the Avogrado’s number

and R = 8.314 J mole−1 K−1 is the universal gas constant. We generalise this result for a

given system with total f degrees of freedom at temperature T to calculate the total energy

for the system as follows,

E = f × 1

2NkT = f × 1

2nRT. (2.43)

It is useful to introduce the molar specific heat at constant volume V , defined as the heat

capacity per unit mole. Unlike the heat capacity, the molar specific heat depends only on

the nature of the system under consideration. The specific heat at constant volume cV is

thus written as

cV =1

n

(∂E

∂T

)V

= f × 1

2Nok = f × 1

2R. (2.44)

For monoatomic molecules, cV = 32R, as directly measured from classical thermodynamics.

For diatomic molecules, there are two additional degrees of freedom regarding rotational

motions. Therefore, the diatomic gases have a total of 5 degrees of freedom, for which their

specific heat at constant volume cV = 52R. Another quantity, the molar specific heat at

constant pressure P is defined as

cP =1

n

(∂E

∂T

)P

= cV +R, (2.45)

by which cP is always greater than cV .

The above relation is of practical importance as calculations by statistical physics are

usually performed for a fixed volume, while experimental measurements are carried out

under conditions of constant pressure. Therefore, cV is a theoretically calculated value and


cP is an experimentally measured parameter. The determination of these quantities provides

information about internal energy, and hence molecular structure. When heat is added to

a gas at constant volume, there will be no work done and so the amount of heat added is

converted to increase internal energy of the gas. When the heat is added at constant pressure,

work is done. Hence, less amount of heat is used to increase the internal energy, resulting in

a smaller increase in temperature change, compared with that at constant volume.

2.4 The Boltzmann partition function

In this section, we introduce a special function that links statistical expression for microscopic

systems to the corresponding thermodynamic functions. The detailed properties of this

function, together with the discussion of the Helmholtz free energy, will be later discussed in

Chapter 5. But for now, it is sufficient to state that the Boltzmann partition function Z

describes how energy is distributed among systems in the classical assembly. This function

is defined as

Z =

∫ ∞0

g(ε) e−ε/kT dε, (2.46)

where g(ε) dε is the number of states having energy between ε and ε + dε, and e−ε/kT is

known as the Boltzmann factor. Based on physical argument, g(ε) dε can be expressed as

the product of the density of states B and element volume dΓ in phasa space, or simply

g(ε) dε = B × dΓ. Thus, (2.46) can be written as

Z =

∫ ∞0

B e−ε/kT dΓ =

∫ ∞−∞

B e−(p2x+p2y+p2z)/2mkT dx dy dz dpx dpy dpz

= BV

∫e−p

2/2mkT p2 sin θ dθ dφ dp = 4πBV

∫ ∞0

e−p2/2mkT p2 dp

= BV(2πmkT

)3/2.

(2.47)

The above partition function can be used to calculate pressure P , a measured quantity

of thermodynamic functions. With the help of this function, the total pressure of a classical

system containing N particles is written as

P = NkT

(∂ `nZ

∂V

)T

= NkTd `n V

dV

=NkT

V.

(2.48)

The last expression for P is well-known as the equation of state of a perfectly ideal gas,

PV = NkT = nRT , as frequently found in standard books of fundamental physics.

2.5. Exercise 15

2.5 Exercise

1. Diketahui fungsi distribusi untuk sistem kontinu sebagai berikut:

f(x) =4x

λ2e−

2λx for x ≥ 0

(a) Tunjukkan bahwa total peluang adalah 1.

(b) Hitunglah nilai x rata-rata, rms dan yang paling mungkin dari sistem tersebut.

2. Hitunglah kecepatan rms untuk atom gas Helium pada suhu 300 K.

3. Hitunglah pada suhu berapa molekul gas Hidrogen memiliki kecepatan yang sama.

4. Diketahui fungsi distribusi untuk sistem kontinu sebagai berikut:

f(v) = Av2e−12mv2/kT , A adalah konstanta

(a) Tentukan A untuk atom yang tiap mol-nya memiliki massa sebesar 36 gram pada

suhu 0◦Celcius.

(b) Tentukan laju rata-rata partikel.

(c) Tentukan waktu bebas rata-rata, bila diketahui lintasan bebas rata-rata partikel

adalah 60 A.

5. Satu mol gas Neon bermassa 20,15 gram dan berjari-jari 1,25 A. Pada suhu 27◦Celcius,

tekanannya 1,013 × 104 Nm−2.

(a) Tentukan jarak rata-rata antara 2 atom Neon.

(b) Tentukan jumlah tumbukan rata-rata per detik yang dialami oleh satu atom Neon.

6. Diketahui satu mol gas beratom tunggal berjari-jari atom 1,05 A bermassa 4 gram

pada volume 17,5 liter, tekanan 1,01× 105 Nm−2 dan suhu 27◦Celcius.

(a) Tentukan jarak rata-rata antar atom.

(b) Tentukan jarak bebas rata-rata.

(c) Tentukan waktu bebas rata-rata

(d) Jika terdapat lubang seluas 10−2 mm2, tentukan jumlah atom persatuan waktu

yang menerobos keluar dari lubang tersebut.

7. Gas ideal terperangkap dalam ruang tertutup berupa silinder homogen dengan luas

penampang A dan tinggi L (searah terhadap sumbu z). Tentukan energi rata-rata

molekul gas ideal tersebut. Bagaimana bila energi potensialnya diperhitungkan?


8. A flux of 1012 neutrons/m2 emerges each second from a port of nuclear reactor. If these

neutrons have a Maxwell-Boltzmann energy distribution corresponding to temperature

T = 300 K, calculate the density of neutrons in the beam (taken from Beiser, 1988,

Ch.15, problem 7).

9. By direct integration of∫∞

0ε n(ε) dε with n(ε) taken from equation 2.59, show that the

total energy of the assembly is 32RT where R = Nk (taken from Pointon, 1978, Ch.2,

problem 2).

10. Calculate the mean values of vx and v2x (taken from Pointon, 1978, Ch.3, problem 2).

11. Show that the integration of equation 3.10 over all polar angles leads to distribution

in equation 3.8 (taken from Pointon, 1978, Ch.3, problem 3).

12. Derive the variation of pressure with height in a column of a gas at temperature T

(a) using the fact that the change in pressure over a height dh is -ρg dh, where ρ is the

density of the gas

(b) using the Boltzmann factor to give the concentration gradient of the molecules

(taken from Pointon, 1978, Ch.3, problem 1).

13. By what factor must the absolute temperature of a gas be increased to double the rms

speed of its molecules? (taken from Tipler, 1999, Ch.18, p.563, problem 35).

14. A mole of He molecules is in one container and a mole of CH4 molecules is in a second

container. Both are in standard conditions. Which molecules have the greater mean

free path? (taken from Tipler, 1999, Ch.18, p.563, problem 37).

15. The molecules of a monoatomic ideal gas are escaping by effusion through a small hole

in a wall of an enclosure maintained at absolute temperature T.

(a) By physical reasoning (without actual calculation) do you expect the mean kinetic

energy 〈 εo 〉 of a molecule in the effusing beam to be equal to, or greater than, or less

than the mean kinetic energy 〈 εi 〉 of a molecule within the enclosure?

(b) Calculate 〈 εo 〉 for a molecule in the effusing beam

(taken from Reif, 1985, Ch.7, p.287, problem 7.30).

Chapter 3

Bose-Einstein Statistics

In this chapter, we discuss Bose-Einstein (BE) statistics. This statistics is used to describe

the statistical behaviour of bosons - quantum systems having integer or zero spin. As the

spin classification of particles determines the nature of the energy distribution over the

available energy states, the bosons do not obey the Pauli exclusion principle. Thus, there

is no restriction for how many bosons are allowed to occupy the same quantum state. This

chapter is outlined as follows. The concept of weight of configuration, or the number of ways

of distributing particles over the available states, is first introduced in §4.1. The number

of particles occupying a certain state is derived in §3.2, followed by the discussion of the

condition under which a gas could be considered as a classical gas, or as a Bose-Einstein

gas in §3.3. As real examples, the applications of the BE statistics in boson systems, such

as electromagnetic radiation in space and elastic waves in solid, are discussed. The classical

Planck’s radiation formula is derived in §3.4 while the Debye theory for specific heats of solid

is examined in §3.5.

3.1 Weight of configuration

There are two principal differences between classical and boson systems. In the MB statistics,

the basic assumption made is that particles are distinguishable. Thus, the interchange of

two classical particles leads to a new arrangement. Whilst, in the BE statistics the assembly

consists of indistinguishable particles, hence the interchange of two identical particles gives

no new configuration. Another difference between these systems arises from the nature of

the available energy states. Rather than a continuous form of energy, bosons are distributed

over discrete levels of energy.

Before further discussing bosons in detail, it is necessary to calculate the number of

ways of arranging particles within the available energy states. Let us consider two classical

particles, labelled as a and b, occupying two energy states. Based on the MB statistics, there

17

18 3. Bose-Einstein Statistics

will be four possible arrangements for these particles,

| a | b |︸︷︷︸arrangement 1

| b | a |︸︷︷︸arrangement 2

| ab | |︸︷︷︸arrangement 3

| | ab |︸︷︷︸arrangement 4

However, the first two arrangements are, according to the BE statistics, completely identical.

Thus, they are considered to be united, resulting in only three arrangements for the system.

| a | a |︸︷︷︸arrangement 1

| aa | |︸︷︷︸arrangement 2

| | aa |︸︷︷︸arrangement 3

How about if such particles are distributed over three energy states. You will find easily

that 9 configurations are possible for classical arrangements while only 6 are found for boson

arrangements. The details of these are left for students for exercise. This difference in the

number of arrangements between classical and boson systems is associated with different

statistical formulations for calculation of possible configurations within each assembly. For

classical statistics, the number of possible configurations W (also known as the weight of

configuration) is given by

W = N !∏i

gNiiNi !

, (3.1)

where gi and Ni are the number of states and the number of particles in the ith sheet,

respectively, and N is the total number of particles. For boson statistics, this quantity is

written as

W =∏i

(Ni + gi − 1) !

Ni ! (gi − 1) !. (3.2)

We can then calculate the number of ways of arranging particles using the above formulations.

3.2 Population of particles

Having discussed the weights of configurations for both the MB and BE statistics, it is now

necessary to find an expression for the number of particles distributed over a particular sheet.

We begin with the classical case by writing (3.1) and taking a natural logarithm of it,

`nW = `nN ! + `n∏i

gNiiNi !

= `nN ! +∑i

`ngNiiNi !

= `nN ! +∑i

(Ni `n gi − `nNi !

)(3.3)

3.2. Population of particles 19

It is useful to apply Stirling’s approximation: `nN ! ≈ N `nN − N to the above equation.

Equation (3.3) becomes

`nW = N `nN −N +∑i

(Ni `n gi −Ni `nNi +Ni

)= N `nN +

∑i

(Ni `n gi −Ni `nNi

),

where we have used∑Ni = N .

The most probable configuration corresponds to the condition under which `nW reaches

maximum. This is simply written as d `nW = 0, or

0 = d(N `nN) +∑i

d(Ni `n gi −Ni `nNi

)= `nN dN +N

dN

N+∑i

(`n gi × dNi − `nNi × dNi −Ni ×

dNi

Ni

)=∑i

(`n gi − `nNi

)dNi,

(3.4)

where∑

dNi = d∑Ni = dN = 0 because N is constant (see also below). It is important

to mention here that the above condition goes together with the imposed limitations on the

values of total particle N and total energy E, respectively, for which the followings hold,∑i

Ni = N = constant∑i

dNi = d∑i

Ni = dN = 0∑i

Niεi = E = constant∑i

d(Niεi) = d∑i

Niεi = dE = 0(3.5)

Thus we can write

d `nW + α dN + β dE = 0, (3.6)

where α and β are Lagrange multipliers that are to be later determined. Now we have all

ingredients needed to develop Lagrangian method using (3.4), (3.5) and (3.6). Equation (3.6)

becomes ∑i

(`n gi − `nNi + α + βεi

)dNi = 0, (3.7)

which requires

`n gi − `nNi + α + βεi = 0.

After several simple steps, we obtain the particle population Ni in the ith sheet,

Ni =gi

e−(α+βεi). (3.8)


From the discussion of the Boltzmann partition function in §2.4 we have Z in a continuous

form,

Z =

∫ ∞0

g(ε) e−ε/kT dε.

Rewriting this Boltzmann partition function into a discrete form gives

Z =∑i

gi e−εi/kT . (3.9)

Dividing both sides of (3.9) by N results in

Z

N=

∑gi e−εi/kT∑Ni

. (3.10)

In order to derive the number of particles occupying a particular sheet, we here define

N

Z≡ eα (3.11)

and substitute it into (3.10) to obtain

Ni =gi

e−(α−εi/kT ). (3.12)

It is clear from (3.8) and (3.12) that

β = − 1

kT. (3.13)

The same result can also be obtained from thermodynamics. It is important to keep it

in mind that either (3.8) or (3.12) is frequently referred to as classical distribution (also

known as classical population number), which describes population of particles within

each sheet in the classical assembly.

In a similar manner to the above steps, we can derive the number of bosons, or boson

population number, occupying available quantum states by first taking a natural logarithm

of (3.2),

`nW = `n∏i

(Ni + gi − 1) !

Ni ! (gi − 1) !

=∑i

(Ni + gi − 1

)`n(Ni + gi − 1

)−Ni `nNi − (gi − 1) `n (gi − 1),

(3.14)

where we have applied Stirling approximation to the above derivation.

The most probable configuration for boson distribution is obtained when (3.14) reaches

3.3. The Bose-Einstein gas 21

maximum, or d `nW = 0. Again, using (3.5) and (3.6) we can write

0 = d `nW + α dN + β dE

=∑i

(`n (Ni + gi − 1)− `nNi + α + βεi

)dNi.

(3.15)

The solution of (3.15) requires

0 = `n (Ni + gi − 1)− `nNi + α + βεi

= `n (Ni + gi)− `nNi + α + βεi,(3.16)

where we have again used Stirling approximation, i.e., Ni � 1 to simplify calculation. After

simple algebra, we obtain the population number of bosons in the form of

Ni =gi

e−(α−εi/kT ) − 1, (3.17)

where we have used β = −1/kT . Equation (3.17) is frequently referred to as quantum

distribution for bosons (also known as boson population number), which describes

population of bosons within each sheet in the boson system. It can then be shown easily

that (3.17) approaches to (3.12) when gi � Ni. It follows that the number of states is much

larger than the population of particles. Note that α has not been yet determined, and that

it is the purpose of the next section to derive the actual form of α.

3.3 The Bose-Einstein gas

In this section, we examine conditions under which a gas may obey either the MB or BE

statistics. In doing so, we need to assume that each available energy state in the quantum

assembly occupies a definite volume in phasa space. From the discussion of the Boltzmann

partition function in §2.4, it is defined that g(ε) dε = B dΓ, where g(ε) dε denotes the number

of available energy states having energy in the range ε to ε + dε and B is a unique constant

associated with the definite volume, which will be determined below.

For a single allowed energy state (i.e., g = 1) occupying element volume dΓ in phasa

space, we can write

1 = B dΓ

= B dx dy dz dpx dpy dpz

= B dx dpx dy dpy dz dpz.

Assume that the changes in position and momentum in each direction are finite, the above


expression can then be written as

1 = B ∆x∆px︸︷︷︸h

∆y∆py︸︷︷︸h

∆z∆pz︸︷︷︸h

= B h3,

for which B = 1/h3. Here we have taken ∆x∆px = h from the Heisenberg’s uncertainty

principle. Similar expressions for the other two directions also hold. Thus, equation (2.47)

can be written as

Z =V

h3

(2πmkT

)3/2. (3.18)

Hence, equation (3.11) becomes

N

Z=

Nh3

V(2πmkT

)3/2= eα = A, (3.19)

where A is constant. The necessary condition under which the gas can be considered as a

classical gas is closely related to the value of A. In most cases, this value is sufficiently small

(i.e., A−1 � 1) such that A−1 eε/kT − 1 ≈ A−1 eε/kT . If this condition is satisfied then boson

population (3.17) approaches to the classical distribution (3.12).

Consider here, for example, Helium gas with molar mass of 4 gram per mole−1 at standard

atmospheric pressure of 1 atm. The value of A for the He gas is approximately 3 × 10−6

at T = 300 K, drops to 0.15 at T = 4 K. Thus, the He gas at room temperature obeys

classical distribution, and even at low temperature of T = 4 K, the He gas will behave, to

some degree, as a classical gas.

3.4 Black-body radiation

Here we consider electromagnetic radiation within an enclosure of volume V with enclosure

walls being maintained at temperature T . The radiation can be considered as a collection of

photons, and is assumed to obey the BE statistics. When the photons set in motion within

the container, some of them are continuously absorbed and reemitted by the walls. In this

context, the number of photons inside the enclosure is not conserved. Because (3.6) must

hold, it follows that α must be zero, leading to A = 1. Thus, (3.17) can be rewritten in the

form of

N(ε) dε =g(ε) dε

eε/kT − 1, (3.20)

where we have invoked e−α = A−1 = 1. N(ε) dε and g(ε) dε denote the number of photons

and the number of available states with energy in the range ε to ε + dε, respectively.

3.4. Black-body radiation 23

Note that the energy of a photon depends on the frequency ν or equivalently on the

wavelength λ of the radiation, i.e., ε = hν = hc/λ, and that it is radiated in space over the

whole values of either frequency or wavelength. Thus, equation (3.20) can be written in two

forms,

N(ν) dν =g(ν) dν

ehν/kT − 1

N(λ) dλ =g(λ) dλ

ehc/λkT − 1,

(3.21)

where N(ν) dν and g(ν) dν denote the number of photons and the number of available states

with frequency in the range ν to ν + dν, respectively. Similar definitions hold for N(λ) dλ

and g(λ) dλ in the second equation of (3.21).

The energy distribution of the photons can be obtained by rewriting the number of

available energy states g as

g = 2dΓ

h3= 2

dx dy dz dpx dpy dpzh3

= 2V 4πp2 dp

h3, (3.22)

where the multiplier “ 2 ” in front of element volume dΓ arises from the fact that there are

two independent directions of spin orientation, that is parallel and anti-parallel to the photon

linear momentum.

Another physical argument based on which the multiplier 2 is derived is that according

to the particle-wave duality, a photon with a definite spin state corresponds to a propagating

electromagnetic wave having two independent directions of polarisation, with each direction

being perpendicular to the wave propagation. For simplicity, it is assumed that photons are

linearly polarised.

The number of available states having wavelength in the range λ to λ + dλ can be derived

with the help of the De Broglie’s hypothesis, p = h/λ. Substituting this relation into (3.22)

yields

g(λ) dλ = V8π

λ4dλ. (3.23)

Here we define the density of states G having wavelength between λ and λ + dλ as the

number of available states g(λ) dλ per unit volume V such that (3.23) becomes

G(λ) dλ =8π

λ4dλ. (3.24)

The photons are distributed over the available states, in which the associated energy density

E in the range of wavelength between λ and λ + dλ is defined as the product of the photon

energy ε and the density of photons N (i.e., the number of photons per unit volume, N/V )


as follows,

E(λ) dλ = ε×N

=hc

λ× G(λ) dλ

ehc/λkT − 1

=8πhc

λ5

dλ

ehc/λkT − 1.

(3.25)

The above expression is known as the Planck’s radiation formula for the spectral distribution

of radiant energy within a thermal equilibrium enclosure at temperature T .

Two possible, interesting approximations can be directly derived from the last expression

of (3.25). At sufficiently long wavelengths, or when hc/λ� kT , we have

ehc/λkT − 1 = 1 +hc

λkT+ · · · · · · −1

ehc/λkT ≈ 1 +hc

λkT

for which (3.25) becomes

E(λ) dλ =8πkT

λ4dλ, (3.26)

which is known as the classical Rayleigh-Jeans’ formula for the spectral distribution of

radiant energy within an enclosure. On the other extreme, at sufficiently short wavelengths,

or when hc/λ� kT , we have

ehc/λkT − 1 ≈ ehc/λkT

for which (3.25) becomes

E(λ) dλ =8πhc

λ5

dλ

ehc/λkT, (3.27)

which is known as the classical Wien’s formula for the distribution of radiant energy within

an enclosure.

Thus, it has been shown that the Planck’s formula (3.25) for black-body radiation is valid

over the whole spectrum of wavelengths, or accordingly frequencies. The formula covers the

empirically observed findings of the spectral distribution of electromagnetic radiation that

are previously unexplained by classical electrodynamic and thermodynamic theories.

Let us now assume that there is a small hole in the enclosure wall, giving a way for the

photons to radiate out. The number of photons emitted per unit time per unit area of the

hole (i.e., the flux of photons crossing the hole) is written as

Nrad(λ) dλ =1

4c × N (λ) dλ, (3.28)

because the photon is always in move with the velocity of light c in free space.

3.4. Black-body radiation 25

By analogy, the energy radiated out per unit time per unit area of the hole is then given

by

Erad(λ) dλ =1

4c × E(λ) dλ. (3.29)

Substituting the Planck’s formula (3.25) into (3.29) results in

Erad(λ) dλ =2πhc2

λ5

dλ

ehc/λkT − 1. (3.30)

Integrating both sides of (3.30) gives∫ ∞0

Erad(λ) dλ =

∫ ∞0

2πhc2

λ5

dλ

ehc/λkT − 1,

and arranging the integral leads to

Erad =2πk4T 4

h3c2

∫ ∞0

(hc/λkT )3

ehc/λkT − 1d

(hc

λkT

),

where Erad denotes the total energy radiated out of the hole per unit time per unit area.

Now let us introduce x = hc/λkT as a variable of integration such that

Erad =2πk4T 4

h3c2

∫ ∞0

x3

ex − 1dx. (3.31)

The integral can be evaluated with the help of some steps (see Gasiorowicz, 1996, p.5),∫ ∞0

x3

ex − 1dx =

∫ ∞0

x3 dx e−x∞∑n=0

e−nx =∞∑n=0

1

(n+ 1)4

∫ ∞0

y3 e−y dy = 6∞∑1

1

n4=π4

15.

Equation (3.31) then becomes

Erad =

(2π5k4

15h3c2

)T 4 = σ T 4, (3.32)

where σ is called the Stefan’s constant (5.7 × 10−8 Watt m−2 K−4). The above equation is

known as the Stefan-Boltzmann law for the energy density of electromagnetic radiation at

temperature T . It is found empirically that an additionally constant e, the emissivity of a

substance should be added to the RHS of (3.32). The magnitude of this quantity depends on

the characteristics (or surface) of the object, which is valued between 0 and 1. Objects that

are grouped into perfect heat absorbers (or heat radiators) have unit emissivity. Conversely,

imperfect heat absorbers (or heat radiators) correspond to the value of emissivity between

0 and 1. Black holes in the universe are considered as perfect black-body.


3.5 The specific heat of solid

As previously discussed in §3.4, electromagnetic radiation is quantised as photons. In this

section, the energy of elastic waves propagated in solids is quantised as phonons, which obey

the BE statistics. The energy of a phonon also depends upon frequency. For convenience,

we use the phonon frequency ν as a dynamic variable. Thus, the phonon population having

frequency between ν and ν + dν is written as

N(ν) dν =g(ν) dν

ehν/kT − 1. (3.33)

Here we use Debye approximation, g(ν) dν = Cν2 dν for ν ≤ νm, where C is a constant

and νm is the maximum frequency corresponding to three modes of vibration, that is one

longitudinal polarisation and two transverse relative to the direction of propagation of the

waves. It should be noted that the former polarisation is associated with a compressional

wave and the latter corresponds to shear waves. Hence, for a solid having N atoms, there

exists a total of 3N allowed states. It follows that

3N =

∫ νm

0

g(ν) dν = C

∫ νm

0

ν2 dν,

for which C = 9N/ν3m. Thus, equation (3.33) becomes

N(ν) dν =9N

ν3m

ν2 dν

ehν/kT − 1for ν ≤ νm. (3.34)

The total energy of phonons in solid is determined by the energy of a phonon ε = hν and

the number of phonons N(ν) dν with frequency in the range ν to ν + dν,

E =

∫ νm

0

εN(ν) dν =9Nh

ν3m

∫ νm

0

ν3 dν

ehν/kT − 1. (3.35)

In the same manner as the definition in (2.44), the specific heat of solid can be derived with

the help of (3.35),

cV =1

n

(∂E

∂T

)V

=9Noh

2

ν3m

1

kT 2

∫ νm

0

ν4 ehν/kT dν(ehν/kT − 1

)2 . (3.36)

We introduce a new variable, x = hν/kT and a new parameter, namely Debye temperature

Θ = hνm/k to replace (3.36) by

cV = 9Nok

(T

Θ

)3 ∫ Θ/T

0

x4 ex dx(ex − 1

)2 . (3.37)

3.6. Exercise 27

There will be two regions of temperature for which (3.37) is evaluated. For a region of

high temperatures, it is understood that Θ/T � 1 and ex ≈ 1 + x ≈ 1. In this condition,

equation (3.37) becomes

cV = 9Nok

(T

Θ

)3 ∫ Θ/T

0

x2 dx = 3R, (3.38)

where R replaces Nok. The above result is the same as that derived from the classical

equipartition of energy.

At low temperatures, however, Θ/T � 1 hold and the upper limit approaches to infinity.

Thus, equation (3.37) becomes

cV = 9R

(T

Θ

)3 ∫ ∞0

x4 ex dx(ex − 1

)2 . (3.39)

The integral can be evaluated with the help of some steps below,∫ ∞0

x4 ex dx(ex − 1

)2 = 24∞∑1

1

n4=

4π4

15.

Substituting the above expression into (3.39) results in

cV =12

5π4R

(T

Θ

)3

. (3.40)

Note that the above results for high and low temperatures derived from (3.37) are in good

agreement with experimental measurements of the specific heat of solids. The equation is

therefore usually referred to as the Debye theory for the specific heat of solid, and is applicable

to most metals.

3.6 Exercise

1. Jelaskan dengan dua cara pendekatan dalam hal apa Distribusi Bose-Einstein (3.17)

menuju ke Distribusi Maxwell-Boltzmann (3.12).

2. Persamaan radiasi elektromagnetik Planck untuk foton dalam ruang tertutup pada

temperatur T dinyatakan oleh

E(λ) dλ =8πhc

λ5

dλ

ehc/λkT − 1

(a) Carilah bentuk persamaan radiasi Planck dalam variabel frekuensi ν.


(b) Jika λm menyatakan besarnya panjang gelombang radiasi pada kerapatan energi

maksimum, tunjukkan dari persamaan radiasi Planck bahwa λmT = 2, 9× 10−3 mK.

(c) Kerapatan energi maksimum cahaya matahari terjadi pada panjang gelombang

radiasi 4,84 × 10−7 m. Dengan menganggap matahari sebagai benda hitam, berapakah

temperatur permukaan matahari ?

3. Using equation (4.20) of Pointon (1978), or alternatively the equation below

N (λ) dλ =8π

λ4

dλ

ehc/λkT − 1

show that the total number of photons per unit volume of an enclosure at temperature

T is given by

16π

(kT

hc

)3 ∞∑n=1

1

n3

where the n’s are positive integers (taken from Pointon, 1978, Ch.4, problem 1).

Chapter 4

Fermi-Dirac Statistics

In this chapter, we examine a group of quantum particles having half-integer spin which

obey the Pauli exclusion principle. Such particles are called fermions and the quantum

statistics used to describe the fermions is commonly known as Fermi-Dirac (FD) statistics.

This chapter is outlined as follows. The number of ways of arranging fermions or the weight

of configuration is shortly introduced in §4.1, followed by the discussion of the fermion

population in §4.2. The discussion of these sections is mostly similar to that for the BE

statistics, and thus many details are left for students. The concepts of the Fermi energy

and Fermi function, as well as their properties under ideal conditions, are examined in §4.3.

Finally, the application of the Fermi distribution to real cases is discussed. Because of the

high concentration of conduction electrons in a metal, an electron gas cannot be treated as

a classical gas. It will be later shown in §4.4 that the FD statistics is appropriate for the gas

in a metal.

4.1 Weight of configuration

Let us consider an isolated, quantum assembly containing N non-interacting fermions with

total energy E. These fermions are distributed over energy sheets, which include all available

states gi. As with bosons, fermions are indistinguishable particles, hence any rearrangement

of these particles within the available states will make no configuration. Unlike bosons, the

number of fermions occupying a single state is limited by the exclusion principle. It follows

that there exists only one particle or no particle per unit state. Thus, the number of possible

configuration for fermions can be written as

W =∏i

gi !

Ni ! (gi −Ni) !, (4.1)

where Ni is the number of particles in the sheet i, which will be derived in the next section.

29

30 4. Fermi-Dirac Statistics

4.2 Population of fermions

Recall again that in the Lagrange method (3.6), we have

d `nW + α dN + β dE = 0,

where α and β are the Lagrange multipliers. In a manner similar to the steps discussed in

Chapter 3, we derive the fermion population by first taking the natural logarithm of (4.1),

and then applying (3.5) to the above expression. After several simple algebra, we obtain

Ni =gi

e−(α+βεi) + 1. (4.2)

The above expression is called the Fermi-Dirac distribution or fermion population

number, which may be written in the form of

N(ε) dε =g(ε) dε

e−(α+βε) + 1= f(ε) g(ε) dε, (4.3)

where f(ε) is

f(ε) =1

e−(α+βε) + 1. (4.4)

Here β = −1/kT as before, but this time α is defined as εF/kT (different from that for

bosons), where εF is known as the Fermi energy. With this new definition for α, we can

rewrite (4.4) as

f(ε) =1

e(ε−εF )/kT + 1, (4.5)

which is generally called as the Fermi function. This function provides the probability of

finding fermions with energy ε in the given sheets.

4.3 The Fermi-Dirac gas

It is widely known that the absolute zero of temperature has never been achieved, and so

the condition under which the Fermi-Dirac gas behaves at T = 0 K is considered to be an

ideal case. Therefore, we here examine the behaviour of the Fermi function and derive the

Fermi energy at this temperature for comparison. We begin with analysing f(ε) at T = 0 K

as follows.

f(ε) =1

e−∞ + 1= 1 for ε < εF (0)

and

f(ε) =1

e∞ + 1= 0 for ε > εF (0)

4.3. The Fermi-Dirac gas 31

The above values for f(ε) at T = 0 K imply that all states with energy smaller than εF (0)

are occupied while those having energy greater than εF (0) are empty. These values are used

to evaluate the Fermi energy at T = 0 K.

In doing so, we first write explicitly the number of available states for fermions and to

relate it to the volume occupied by these states in phasa space. Recall again that the relation

between the number of states g(ε) dε and element volume in phasa space dΓ is given by

g(ε) dε = B dΓ

= 2dx dy dz dpx dpy dpz

h3

= 2V 4πp2 dp

h3,

where the multiplier “ 2 ” arises from the fact that for fermions there are two independent

directions of spin orientation. Based on classical linear momentum-kinetic energy relation

p2 = 2mε, the above expression can be rewritten as

g(ε) dε =4π V

h3(2m)3/2 ε1/2 dε. (4.6)

Now integrating both sides of (4.3) and inserting (4.6) yields∫ ∞0

N(ε) dε =

∫ ∞0

f(ε) g(ε) dε,

where the LHS of the above integral denotes the total number of fermions and the RHS can

be separated into two parts as follows.

N =

∫ εF (0)

0

f(ε) g(ε) dε+

∫ ∞εF (0)

f(ε) g(ε) dε

Substituting the values of Fermi function f(ε) at T = 0 K and equation (4.6) to the above

expression results in

N =4π V

h3(2m)3/2

∫ εF (0)

0

ε1/2 dε

= 4π V

(2m

h2

)3/22

3εF (0)3/2,

from which we obtain the Fermi energy at T = 0 K below,

εF (0) =

(3N

8πV

)2/3h2

2m(4.7)


Fermi-Dirac gas εF (0) in eV TF in K

Helium-3 atoms 0.94 × 10−3 10electron gas in Lithium 4.7 5.4 × 103

electron gas in Potassium 2.1 2.4 × 103

Table 4.1: The values of the Fermi energies at T = 0 K and Fermi temperatures forsome Fermi-Dirac gases.

We may equate the Fermi energy at T = 0 K to thermal energy kT to derive a Fermi

temperature TF as follows,

TF =

(3N

8πV

)2/3h2

2mk. (4.8)

We here provide examples of the values of εF (0) and TF for some different Fermi-Dirac gases

in Table 4.1. It is clear from Table 4.1 that molecular gases, such as He-3 atoms which

consist of fermions, have a low Fermi temperature compared with room temperature.

However, it has been already shown in §3.3 that the He-4 gas behaves as a classical gas

at T = 300 K. It means that the energy distribution of the gas molecules at ordinary (room)

temperature will approximate to the classical, MB statistics. The condition under which

molecular gases obey classical distribution does not hold for an electron gas in a metal. It

is therefore necessary to discuss the electron gas in the next section.

4.4 The electron gas

The fact that the Fermi temperature for the electron gas in a metal is high (see Table 4.1)

leads to a fundamental idea: an increase in temperature from absolute zero to a value around

room temperature may only affect the electrons with the energy near the Fermi energy. This

can be shown below,

for ε = εF − kT f(ε) =1

e−1 + 1= 0.73

for ε = εF f(ε) =1

e0 + 1= 0.50

for ε = εF + kT f(ε) =1

e1 + 1= 0.27

The mean values of dynamics properties of the electrons can then be evaluated easily.

For example, the mean energy of the electron gas at T = 0 K is given by

⟨ε (0)

⟩=

∫∞0εN(ε) dε∫∞

0N(ε) dε

, (4.9)

4.5. Exercise 33

where N(ε) dε is given in (4.3). After several simple steps, we obtain 35εF (0) for the mean

energy at T = 0 K. In a similar manner to the above steps, the mean velocity of the electron

at T = 0 K can also be evaluated using

⟨v (0)

⟩=

∫∞0v N(ε) dε∫∞

0N(ε) dε

, (4.10)

and the result is 34vF , where vF is the velocity of the electron at the Fermi energy.

Note that equations (4.7) and (4.9) hold for T = 0 K only. In fact, the Fermi energy and

the mean energy of the gas electrons are a function of temperature. Thus, it is necessary to

find the general form of the Fermi energy εF and the mean energy of the electrons 〈 ε 〉 at

temperature T . The details of calculations are left for students (see Appendix 7 of Pointon,

1978 for help), but the main results are here presented. The Fermi energy is

εF = εF (0)

(1− π2

12

(T

TF

)2)

(4.11)

and the mean energy of the electrons is

〈 ε 〉 = εF (0)

(3

5+π2

4

(T

TF

)2). (4.12)

The second terms in the bracket of both (4.11) and (4.12) are considered as the correction

terms that are in fact relatively small compared with the first terms in each equation.

4.5 Exercise

1. Jelaskan dengan dua cara pendekatan dalam hal apa Distribusi Fermi-Dirac (4.2)

menuju ke Distribusi Maxwell-Boltzmann (3.12).

2. Jelaskan perbedaan antara Statistik Bose-Einstein and Statistik Fermi-Dirac.

3. Gas isotop Helium-3 mengikuti Statistik Fermi-Dirac dengan fungsi distribusi sebagai

berikut

Ni =gi

e(ε−εF )/kT + 1

(a) Turunkan energi Fermi pada temperatur 0 K.

(b) Tunjukkan bahwa energi rata-rata elektron pada temperatur 0 K adalah 35εF (0).

(c) Hitunglah energi rata-rata elektron pada temperatur 0 K.


4. Perak mempunyai rapat massa 10,5 × 103 kgm−3 dan pada temperatur 0 K, 1 mol

atomnya mengandung 107,88 gram. Elektron logam perak tunduk pada Statistik

Fermi-Dirac.

(a) Hitunglah kerapatan elektron logam perak pada temperatur 0 K.

(b) Hitunglah energi Fermi pada temperatur 0 K.

(c) Hitunglah energi rata-rata elektron pada temperatur 0 K.

(d) Hitunglah energi Fermi pada temperatur 300 K.

5. Show that the mean velocity of an electron in an electron gas at the absolute zero of

temperature is 34vF , where vF is the velocity of an electron at the Fermi energy (taken

from Pointon, 1978, Ch.5, problem 1).

6. Show that, for a gas in which the molecules behaves as fermions, the value of the Fermi

energy is approximately

εF ≈ kT `n

(Nh3

V (2πmkT )3/2

)for reasonably high temperatures (taken from Pointon, 1978, Ch.5, problem 3).

Chapter 5

Termodinamika Gas

Distribusi statistik klasik (MB) maupun statistik kuantum (BE dan FE) melibatkan dua

besaran makroskopik, yaitu energi total E dan temperatur mutlak T . Pembahasan kedua

macam distribusi statistik tersebut (klasik dan kuantum) mengasumsikan bahwa tidak ada

interaksi antara sistem dan lingkungan, sedemikian sehingga jumlah total partikel N dan

energi total E dapat dianggap tetap, jadi berlaku dN = 0 dan dE = 0. Namun seperti

telah dibahas pada Bab 3 untuk radiasi elektomagnetik yang menembus lubang pada ruang

tertutup, jumlah partikel (foton) berubah setiap saat, jadi dN = 0 tidak lagi berlaku.

Pembahasan lebih jauh memberikan kemungkinan interaksi antara sistem dan lingkungan,

sedemikian sehingga bisa saja berlaku dN = 0 tetapi dE 6= 0, atau bisa juga berlaku dN 6= 0

tetapi dE = 0, atau keduanya dN 6= 0 dan dE 6= 0.

Rincian topik bahasan Bab 5 ini adalah sebagai berikut. Konsep entropi untuk sistem

tertutup diperkenalkan pada pasal 5.1, diikuti dengan pembahasan penerapan entropi pada

gas klasik pada pasal 5.2. Salah satu contoh gejala fisis yang tidak dapat dijelaskan dengan

konsep statistik gas klasik adalah peristiwa yang dikenal sebagai Paradoks Gibbs, yang akan

dibahas pada pasal 5.3. Konsep statistik gas semi klasik akan dibahas pada pasal 5.4 sebagai

alternatif solusi untuk mengatasi masalah Paradoks Gibbs.

5.1 Konsep entropi

Satu dari dua pengali Lagrange, yaitu β telah diturunkan (lihat kembali persamaan 3.13),

dimana diperoleh β = −1/kT . Besaran ini digunakan untuk merumuskan ulang bentuk

persamaan diferensial Lagrange (3.6) yang diterapkan untuk sistem tertutup, dimana jumlah

partikel N dan volume V dapat dianggap tetap, jadi berlaku dN = 0 dan dV = 0. Hukum

pertama Termodinamika dalam bentuk diferensial menyatakan bahwa

dQ = dE + P dV, (5.1)

35

36 5. Termodinamika Gas

dimana dQ adalah perubahan panas yang terjadi pada sistem, dE adalah perubahan energi

total sistem (energi total E juga dikenal sebagai energi dalam U dalam ilmu termodinamika)

dan P dV adalah kerja luar akibat adanya perubahan volume dV . Mengingat tidak ada

perubahan volume (dV = 0) pada sistem tertutup, maka berlaku dQ = dE. Jadi persamaan

diferensial Lagange (3.6) berikut,

d `nW + α dN + β dE = 0,

berubah menjadi

d `nW + β dQ = 0. (5.2)

Pada tahap ini, konsep entropi diperkenalkan sebagai derajat keteraturan sistem. Seperti

halnya temperatur mutlak T dan energi total E, maka entropi sistem S juga merupakan

fungsi keadaan yang nilainya secara umum berbeda untuk keadaan yang berbeda. Dalam

hal ini, perubahan entropi dS merupakan besaran penting yang menggambarkan dinamika

sistem, dimana nilainya hanya ditentukan oleh keadaan awal dan keadaan akhir (dikenal

sebagai diferensial eksak). Perubahan entropi dS dari satu keadaan ke keadaan yang lain

dirumuskan sebagai

dS =dQ

Tatau ∆S = Sakhir − Sawal =

∫dQ

T(5.3)

Pembahasan detil tentang perubahan entropi dapat dipelajari dalam kuliah termodinamika,

dimana dibahas perubahan entropi berbagai macam proses, baik proses reversibel maupun

non-reversibel.

Dengan bantuan definisi β dan persamaan diferensial (5.3) di atas, maka persamaan (5.2)

dapat diubah menjadi

d `nW − 1

kdS = 0,

dengan mana kita bisa memperoleh

S = k `nW. (5.4)

Perumusan entropi di atas sering dikenal sebagai relasi Boltzmann, yang menghubungkan

peluang termodinamik W (besaran mikro) dan fungsi termodinamik entropi S (besaran

makro). Pada kuliah ini, pembahasan ditekankan pada kegunaan (5.4) untuk menjelaskan

beberapa gejala makroskopis dari kacamata mikroskopis. Akan ditunjukkan bahwa (5.4)

memberikan hasil yang sama dengan pengamatan makroskopis untuk sistem gas klasik.

Bagaimanapun keberhasilan persamaan ini bergantung pada bentuk perumusan peluang

termodinamik W yang dibangun berdasarkan asumsi-asumsi dasar tertentu.

5.2. Gas klasik 37

5.2 Gas klasik

Pada bagian ini, akan ditunjukkan bahwa persamaan keadaan gas ideal dapat diturunkan

dengan bantuan perumusan peluang termodinamika klasik (3.1),

W = N !∏i

gNiiNi!

,

yang disubstitusikan ke dalam persamaan (5.4). Kita akan peroleh bentuk seperti di bawah

ini,

S = k `n

(N !∏i

gNiiNi!

)= k

(`nN ! + `n

∏i

gNiiNi!

)

= k

(N`nN −N +

∑i

`ngNiiNi!

)= k

(N`nN −N +

∑i

`n gNii −∑i

`nNi!

)= k

(N`nN −N +

∑i

Ni `n gi −∑i

Ni `nNi +∑i

Ni

)= k

(N`nN +

∑i

Ni `n gi −∑i

Ni `nNi

).

Kemudian kita substitusikan persamaan (3.10) berikut

Ni =N

Zgi e−εi/kT

untuk memunculkan secara eksplisit fungsi partisi Z, energi total (dalam) E dan temperatur

mutlak T , sedemikian sehingga perumusan entropi di atas menjadi

S = k

(N`nN +

∑i

Ni `n gi −∑i

Ni `nN

Zgi e−εi/kT

)= k

(N`nN +

∑i

Ni `n gi −∑i

Ni `nN +∑i

Ni `nZ −∑i

Ni `n gi e−εi/kT

)= k

(∑i

Ni `n gi +N `nZ −∑i

Ni `n gi −∑i

Ni `n e−εi/kT

)= k

(N `nZ +

∑i

Ni εi/kT

)= k

(N `nZ + E/kT

).

Adalah sangat jelas dari bentuk terakhir di atas bahwa

S = Nk `nZ +E

T. (5.5)


Sekarang kita siap untuk menurunkan persamaan keadaan gas ideal. Langkah pertama

adalah mengalikan kedua ruas persamaan (5.5) dengan T , kemudian mengatur suku-suku

persamaan tersebut sedemikian sehingga diperoleh bentuk berikut ini,

E − TS = −NkT `nZ.

Dalam ilmu termodinamika, ruas kiri dari bentuk perumusan di atas dikenal sebagai fungsi

bebas Helmholtz F. Jadi kita sekarang punya bentuk berikut,

F = −NkT `nZ. (5.6)

Dengan mensubstitusikan fungsi partisi Boltzmann (2.47), Z = BV(2πmkT

)3/2, ke dalam

persamaan (5.6) di atas, kita akan peroleh bentuk eksplisit dari energi bebas Helmholtz,

yaitu

F = −NkT `nBV(2πmkT

)3/2. (5.7)

Persamaan keadaan gas ideal dapat diperoleh dengan memanfaatkan salah satu definisi

yang diturunkan dari hukum pertama Termodinamika, yaitu

P = −(∂F

∂V

)T

= NkT

(∂ `nZ

∂V

)T

= NkTd `n V

dV

=NkT

V.

Bentuk ungkapan di atas tidak lain merupakan persamaan keadaan gas ideal, PV = NkT .

Selain persamaan keadaan gas ideal tersebut di atas, kita bisa juga menurunkan ungkapan

energi total (dalam) gas ideal E dengan memanfaatkan definisi berikut,

E = −T 2

(∂F/T

∂T

)V

= NkT 2

(∂ `nZ

∂T

)V

=3

2

NkT 2

T

=3

2NkT.

Sekali lagi kita bisa membuktikan keampuhan perumusan entropi sistem S pada (5.5)

yang menjelma menjadi energi bebas Helmholtz F pada (5.7), sebelum digunakan untuk

menurunkan beberapa kuantitas gas ideal. Namun demikian, perumusan entropi pada (5.5)

tidak dapat digunakan untuk menjelaskan gejala pencampuran gas dalam ruang tertutup

bersekat. Gejala fisis ini akan kita bahas pada pasal berikutnya.

5.3. Paradoks Gibbs 39

5.3 Paradoks Gibbs

Untuk menguji lebih lanjut apakah perumusan entropi sistem pada (5.5) berlaku umum,

maka pada bagian ini dibahas situasi fisis dimana dua gas klasik dengan jumlah partikel

masing-masing N1 dan N2, massa total masing-masing m1 dan m2 mula-mula ditempatkan

pada dua buah ruang terpisah bersekat yang masing-masing volumenya adalah V1 dan V2.

Misalkan bahwa temperatur T kedua ruang dijaga tetap. Adalah menarik untuk dipelajari

apa yang akan terjadi bila sekat pemisah kedua ruang dibuka, dengan demikian memberikan

kemungkinan interaksi antara partikel-partikel dari kedua gas tersebut. Dalam konteks ini,

pembicaraan ditekankan pada nilai entropi sistem dua gas tersebut sebelum dan sesudah

sekat dibuka.

Marilah kita mulai pembahasan dengan menghitung nilai entropi total sebelum sekat

dibuka. Dengan mensubstitusikan fungsi partisi Boltzmann pada (2.47), Z = BV(2πmkT

)3/2,

ke dalam persamaan (5.5), kita akan peroleh bentuk eksplisit entropi sistem, yaitu

S = Nk `nB +Nk `nV +Nk `n(2πmkT

)3/2+E

T. (5.8)

Berdasarkan persamaan (5.8) di atas, kita bisa tuliskan entropi masing-masing gas sebelum

sekat dibuka, yaitu

S1 = N1k `nB +N1k `n V1 +N1k `n(2πm1kT

)3/2+E1

T

S2 = N2k `nB +N2k `n V2 +N2k `n(2πm2kT

)3/2+E2

T.

(5.9)

Dengan demikian entropi total sistem dua gas tersebut sebelum sekat dibuka adalah

S = S1 + S2 = (N1 +N2) k `nB + (N1 +N2) k `n V +N1 k `n(2πm1kT

)3/2

+N2 k `n(2πm2kT

)3/2+E1 + E2

T, (5.10)

dimana telah diasumsikan bahwa V1 = V2 = V .

Sekarang kita akan menghitung nilai entropi sistem pada saat sekat telah dibuka. Perlu

diketahui bahwa sesudah sekat dibuka, volume total menjadi V1 + V2 = 2V , jumlah partikel

masing-masing gas tetap N ′1 = N1 dan N ′2 = N2, massa total masing-masing gas tetap

m′1 = m1 dan m′2 = m2, energi dalam masing-masing gas tetap E ′1 = E1 dan E ′2 = E2.

Dengan demikian entropi total sistem dua gas sesudah sekat dibuka adalah

S ′ = (N1 +N2) k `nB + (N1 +N2) k `n 2V +N1 k `n(2πm1kT

)3/2

+N2 k `n(2πm2kT

)3/2+E1 + E2

T. (5.11)


Dari kedua persamaan (5.10) dan (5.11) tersebut di atas jelas bahwa perubahan entropi

total sistem adalah

∆S = S ′ − S = (N1 +N2) k (`n 2V − `n V ). (5.12)

Misalkan bahwa kedua gas yang kita bicarakan adalah berjenis sama (dengan kerapatan

jenis ρ1 = ρ2 = ρ), maka perubahan entropi total sistem adalah ∆S = 2Nk `n 2. Hasil

ini sangat menarik mengingat tidak sesuai dengan akal sehat yang menyatakan bahwa de-

rajat keteraturan suatu sistem semestinya sama berapapun volume ruang yang ditempati

oleh sistem tersebut. Dengan kata lain, jika kedua jenis gas yang mengisi masing-masing

ruang adalah sama maka semestinya tidak boleh ada perubahan entropi pada saat sekat

dibuka, atau secara matematis diharapkan ∆S = 0. Tetapi bila hal ini diterima, maka

timbul pemikiran bahwa seolah-olah entropi sistem bukan saja bergantung pada keadaan

awal dan akhir, melainkan juga bergantung pada ‘riwayat’ sistem tersebut. Dalam hal ini,

apakah sistem fisis yang ditinjau terdiri atas gas-gas berjenis sama ataukah tidak. Hal inilah

yang merupakan paradoks, karena bertentangan dengan prinsip dasar termodinamika yang

menyatakan bahwa entropi adalah fungsi keadaan saja.

5.4 Gas semi klasik

Gibbs adalah orang pertama yang mengamati paradoksial entropi sistem dua gas dalam

ruang tertutup bersekat, dan mengajukan usulan perumusan ulang peluang termodinamika

untuk mengatasi masalah paradoks tersebut. Untuk lebih jelasnya, kita ingat kembali bahwa

perumusan entropi (5.5)

S = Nk `nZ +E

T

diperoleh dari perumusan peluang termodinamika klasik (3.1) sebagai berikut

W = N !∏i

gNiiNi!

.

Nampaknya perumusan peluang termodinamika klasik tersebut di atas yang menjadi

biang keladi masalah. Perumusan tersebut berdasarkan asumsi dalam Distribusi Statistik

Klasik yang menyatakan bahwa gas klasik terdiri dari partikel-partikel yang dapat dibedakan.

Beliau mengusulkan jumlah cara untuk mengatur sistem ke dalam tingkat-tingkat energi yang

tersedia, atau yang dikenal sebagai peluang termodinamik W seharusnya N ! kali lebih kecil

dari yang diberikan oleh persamaan (3.1). Jadi seharusnya

W =∏i

gNiiNi!

. (5.13)

5.4. Gas semi klasik 41

Hal ini hanya akan bisa dipenuhi bila gas terdiri atas partikel-partikel yang tak terbedakan.

Gas semacam ini yang memiliki peluang termodinamik yang mirip dengan persamaan (3.1)

kemudian dikenal dengan istilah gas semi klasik dan persamaan (5.13) disebut dengan

peluang termodinamik semi klasik.

Sekarang saatnya untuk menghitung entropi sistem dengan menggunakan perumusan

peluang termodinamik semi klasik (5.13). Perlu diketahui bahwa relasi Boltzmann (5.4) yang

menghubungkan besaran makro entropi sistem S dan besaran mikro peluang termodinamik

W adalah bentuk dasar perumusan entropi sistem, dan oleh karena itu manjur digunakan

untuk semua situasi baik jenis gas yang ditinjau adalah gas klasik maupun semi klasik.

Substitusi persamaan (5.13) ke dalam persamaan (5.4) menghasilkan

S = k `n∏i

gNiiNi!

= Nk +Nk `nB +Nk `nV

N+Nk `n

(2πmkT

)3/2+E

T

= Nk −Nk `nN +Nk `nB +Nk `nV +Nk `n(2πmkT

)3/2+E

T.

(5.14)

Jelas sekali terlihat bahwa dua suku pertama pada ruas kanan persamaan (5.14) merupakan

“pembeda” yang membedakan persamaan entropi semi klasik (5.14) dari persamaan entropi

klasik (5.8).

Berdasarkan persamaan (5.14) tersebut di atas, maka entropi masing-masing gas sebelum

sekat dibuka adalah

S1 = N1k −N1k `nN1 +N1k `nB +N1k `n V1 +N1k `n(2πm1kT

)3/2+E1

T

S2 = N2k −N2k `nN2 +N2k `nB +N2k `n V2 +N2k `n(2πm2kT

)3/2+E2

T

(5.15)

Dengan demikian entropi total sistem gas semi klasik sebelum sekat dibuka adalah

S = 2Nk − 2Nk `nN + 2Nk `nB + 2Nk `nV + 2Nk `n(2πmkT

)3/2+E1 + E2

T, (5.16)

dimana sekali lagi telah diasumsikan V1 = V2 = V dan N1 = N2 = N serta m1 = m2 = m

(untuk jenis gas yang sama).

Analogi dengan gas klasik, sekarang kita menghitung entropi total sistem gas semi klasik

sesudah sekat dibuka sebagai berikut,

S ′ = 2Nk − 2Nk `nN + 2Nk `nB + 2Nk `nV + 2Nk `n(2πmkT

)3/2+E1 + E2

T. (5.17)

Jelas sekali terlihat dari kedua persamaan (5.16) dan (5.17) tersebut di atas bahwa tidak


ada perubahan entropi sistem sebelum dan sesudah sekat dibuka. Secara matematis, hal ini

dituliskan sebagai ∆S = 0. Tidak adanya perubahan entropi total sistem gas semi klasik

pada saat sekat dibuka dikarenakan “specific volume” V/N sistem tersebut bernilai sama

sebelum dan sesudah sekat dibuka.

Menarik untuk dibicarakan di sini bahwa perumusan entropi sistem gas semi klasik pada

(5.14) dapat dituliskan dalam bentuk

S = Nk

(`nV(2πmkT

)3/2

Nh3+

5

2

). (5.18)

Perumusan di atas dikenal sebagai persamaan Sackur-Tetrode, yang diperoleh dengan

jalan mensubstitusikan B = 1/h3 dan E = 32NkT ke dalam persamaan (5.14).

Juga menarik untuk dikaji bahwa tidak ada cara untuk memahami secara klasik mengapa

kita harus membagi perumusan peluang termodinamik klasik (3.1) dengan N ! agar diperoleh

perhitungan entropi total sistem dua gas dengan benar. Bagaimanapun secara kuantum,

hal tersebut dapat dimengerti mengingat atom-atom gas tersusun oleh partikel-partikel tak-

terbedakan yang dibentuk oleh fungsi gelombang yang simetri dan anti-simetri terhadap

pertukaran dua partikel. Detil matematis pembahasan masalah ini adalah di luar cakupan

kuliah Fisika Statistik.

5.5 Latihan soal

1. Gas yang terdiri atas N buah partikel dengan massa per partikelnya adalah m memiliki

temperatur T dan menempati ruang dengan volume V . Partikel-partikel gas tersebut

mengikuti hukum Distribusi Semi Kasik.

(a) Carilah bentuk fungsi energi bebas Helmholtz.

(b) Turunkan bentuk persamaan keadaan gas tersebut.

(c) Turunkan ungkapan entropi gas tersebut.

(d) Turunkan ungkapan kapasitas kalor gas tersebut.

2. (a) Jelaskan perbedaan antara Statistik Klasik (Maxwell-Boltzmann), Statistik Semi

Klasik dan Statistik Kuantum (Bose-Einstein dan Fermi-Dirac).

(b) Jelaskan secara fisis apa yang dimaksud dengan Paradoks Gibbs, dan bagaimana

cara mengatasi masalah tersebut.

3. Show that, when two different gases A and B of volumes VA and VB and total molecules

of NA and NB, respectively, are mixed together at constant temperature T to form a

5.5. Latihan soal 43

total volume of VA + VB, there is an increase in the total entropy given by the mixing

term as follows

k((NA +NB) `n (VA + VB)−NA `n VA −NB `n VB

).

Hence deduce that, when VA = VB and NA = NB = N , the mixing term is 2Nk `n 2


4. Derive the equation of state and the total energy of a semi classical perfect gas from

the expression for the entropy for such a gas. Explain why the results obtained are the

same as those for a classical perfect gas (taken from Pointon, 1978, Ch.7, problem 3).

5. Show that the total partition function derived from equation (7.45) of Pointon (1978)

on the assumption that the energy of the N sub-systems is

E =p2x1

2m+p2y1

2m+p2z1

2m+ · · · · · · +

p2xN

2m+p2yN

2m+p2zN

2m

agrees with that obtained by substituting (7.28) into (7.31) of Pointon (1978), i.e.,

Z =V N

N !h3N

(2πmkT

)3N/2


Chapter 6

Aplikasi Distribusi Statistik

Bab ini merupakan bab terakhir dari rangkaian pembahasan pada kuliah Fisika Statistik,

yang meliputi aplikasi fisika statistik pada beberapa sistem fisis sederhana, dimana jumlah

partikel dan energi totalnya dapat dianggap tetap (dN = 0 dan dE = 0). Sistem fisis dengan

partikel-partikel yang saling berinteraksi tidak ikut dibahas pada bab ini. Namun demikian,

hasil penurunan matematis untuk sistem fisis yang ditinjau akan berlaku secara umum.

Adapun rincian topik bahasan Bab 6 adalah sebagai berikut. Model fisis dengan dua

tingkat energi akan diperkenalkan terlebih dahulu pada pasal 6.1, kemudian diikuti dengan

pengenalan konsep osilator harmonik menurut distribusi statistik kuantum pada pasal 6.2.

Pembahasan molekul diatom dengan melibatkan gabungan gerak translasi, rotasi dan vibrasi

akan diberikan pada pasal 6.3.

6.1 Sistem dua tingkat energi

Bagian ini membahas sistem N partikel yang terdistribusi ke dalam dua tingkat energi ε1

dan ε2 yang tak-terdegenerasi (g1 = g2 = 1), dimana ε2 - ε1 = ε. Misalkan bahwa tingkat

energi ε1 ditetapkan berisi N1 partikel dan tingkat energi ε2 ditetapkan berisi N2 partikel,

maka dapat dituliskan N1 +N2 = N .

Berikut akan ditunjukkan bahwa distribusi partikel pada masing-masing tingkat energi

merupakan fungsi dari eε/kT . Langkah-langkah untuk menurunkan fungsi distribusi partikel

tersebut adalah sebagai berikut. Mula-mula kita mengasumsikan bahwa partikel-partikel

yang menghuni kedua tingkat energi adalah partikel-partikel yang terbedakan, agar N1 dan

N2 bersifat tetap. Jadi, distribusi statistik yang digunakan untuk memecahkan masalah

ini adalah Distribusi Maxwell-Boltzmann, dimana jumlah cara untuk mengatur partikel ke

dalam tingkat-tingkat energi yang tersedia dinyatakan menurut persamaan (3.1).

Karakteristik termodinamik sistem tersebut di atas digambarkan oleh fungsi entropi

45

46 6. Aplikasi Distribusi Statistik

melalui relasi Boltzmann (5.4) sebagai berikut,

S = k `nW = k `nN !

N1 !N2 !. (6.1)

Dengan memanfaatkan pendekatan Stirling, persamaan (6.1) berubah menjadi

S = k(N `nN −N1 `nN1 −N2 `nN2

). (6.2)

Seperti telah dibahas pada pasal 5.2, fungsi energi bebas Helmholtz dituliskan sebagai

F = E − TS. Dengan demikian energi bebas Helmholtz sistem tersebut adalah

F = E − kT(N `nN −N1 `nN1 −N2 `nN2

)= N1 ε1 +N2 ε2 − kT

(N `nN −N1 `nN1 −N2 `nN2

).

(6.3)

Agar sistem dalam keadaan setimbang, maka nilai energi bebas Helmhotz haruslah minimum.

Dengan kata lain, turunan parsial dari energi bebas Helmholtz (6.3) terhadap N1 atau N2

di atas haruslah berharga nol. Jadi,

∂F

∂N1

= 0 atau∂F

∂N2

= 0. (6.4)

Substitusi persamaan (6.3) ke dalam salah satu persamaan (6.4) menghasilkan

0 = ε2 − ε1 − kT(`nN1 − `nN2

)= ε2 − ε1 − kT

(`nN1

N2

).

Dengan mengatur suku-suku pada persamaan di atas, maka akan diperoleh

`nN1

N2

=ε2 − ε1kT

`nN1

N2

= `n e(ε2−ε1)/kT

N1

N2

= e(ε2−ε1)/kT = eε/kT .

Bentuk perumusan di atas adalah seperti yang diharapkan pada temperatur yang cukup

tinggi, dimana partikel cenderung mengisi tingkat energi yang lebih tinggi (N1 � N2).

Substitusi N1 = N − N2 ke dalam perumusan tersebut di atas dan mengatur suku-suku

sehingga akan diperoleh populasi partikel pada tingkat energi ε2 sebagai berikut

N2 =N

1 + eε/kT. (6.5)

6.2. Osilator harmonik 47

Energi total sistem dapat diturunkan dengan mudah dengan bantuan (6.5), yaitu

E = N1 ε1 +N2 ε2

= (N −N2) ε1 +N2 ε2

= Nε1 + (ε2 − ε1)N2

= Nε1 +Nε

1 + eε/kT.

(6.6)

Berbekal persamaan (6.6) dapat ditunjukkan dengan mudah bahwa kapasitas panas zat yang

terdiri atas sistem dua tingkat energi pada volume tetap adalah

CV =

(∂E

∂T

)V

=Nε2

kT 2

eε/kT

(1 + eε/kT )2. (6.7)

6.2 Osilator harmonik

Banyak sekali masalah fisika yang dapat dipelajari dengan pemodelan sistem fisis yang

memanfaatkan osilator harmonik untuk menggambarkan sifat dinamik dari sistem yang

bersangkutan. Perlu diketahui di sini bahwa pendekatan osilator harmonik dapat dilakukan

dengan dua cara, yaitu mekanika klasik dan mekanika kuantum.

Ingat kembali pembahasan ekipartisi energi pada sub-pasal 2.3.2, dimana ditemukan

energi rata-rata osilator harmonik satu dimensi adalah kT (persamaan 2.40). Hal ini berarti

energi kinetik osilator dapat berharga berapa saja (terdistribusi secara kontinu) bergantung

pada besarnya temperatur. Sebaliknya, pendekatan mekanika kuantum memberikan batasan

harga untuk energi osilator yaitu

εn =

(n+

1

2

)hν, (6.8)

dimana n = 0, 1, 2, ......, dan ν adalah frekuensi osilator. Jadi, menurut mekanika kuantum

energi osilator terdistribusi secara diskrit dalam tingkat-tingkat energi yang diperbolehkan

(merupakan kelipatan setengah dari hν). Tingkat energi terendah diperoleh untuk n = 0,

yaitu ε = 12hν. Tingkat energi terendah ini disebut dengan tingkat energi keadaan dasar.

Tingkat-tingkat energi lain yang lebih tinggi berkaitan dengan harga n > 0 disebut dengan

tingkat energi keadaan eksitasi.

Dalam paragraf berikut akan dicoba diturunkan energi rata-rata osilator harmonik satu

dimensi menurut mekanika kuantum dengan memanfaatkan persamaan (6.8) di atas. Hasil

yang diperoleh akan dibandingkan dengan hasil menurut mekanika klasik. Langkah pertama


adalah menuliskan kembali persamaan (3.9),

Z =∑i

gi e−εi/kT . (6.9)

Mengingat tingkat-tingkat energi osilator harmonik adalah tak-terdegenerasi (g = 1) dan

dengan bantuan persamaan (6.8), maka persamaan (6.9) di atas dapat dituliskan ulang

dalam bentuk

Z =∞∑n=0

e−(n+1/2)hν/kT = e−12hν/kT

∞∑n=0

e−nhν/kT . (6.10)

Substitusi deret berikut∑xn = (1 − x)−1 dimana x = e−hν/kT ke dalam persamaan (6.10)

di atas akan menghasilkan

Z =e−

12hν/kT

1− e−hν/kT. (6.11)

Energi rata-rata osilator harmonik dapat dengan mudah dihitung melalui perumusan yang

telah diperkenalkan sebelumnya pada bagian akhir pembahasan pasal 5.2,

〈 ε 〉 = kT 2

(∂ `nZ

∂T

)V

= kT 2 ∂

∂T

(`n

e−12hν/kT

1− e−hν/kT

)= kT 2 ∂

∂T

(`n e−

12hν/kT − `n

(1− e−hν/kT

) )= kT 2

(1

2

hν

kT 2+

hν

kT 2 (ehν/kT − 1)

)= hν

(1

2+

hν

ehν/kT − 1

).

(6.12)

Pada temperatur cukup tinggi berlaku hν � kT , sedemikian sehingga

ehν/kT ≈ 1 +hν

kT+

1

2

(hν

kT

)2

sampai pada suku order kedua. Substitusi deret tersebut di atas ke dalam persamaan (6.12)

akan menghasilkan

〈 ε 〉 = hν

(1

2+kT

hν

(1− hν

kT

))= kT. (6.13)

Hasil tersebut di atas menunjukkan bahwa perumusan energi rata-rata osilator harmonik

menurut mekanika kuantum pada temperatur tinggi akan sama dengan perumusan menurut

mekanika klasik.

6.3. Gas diatomik 49

6.3 Gas diatomik

Model fisis lain yang dapat digunakan untuk menggambarkan dinamika sistem adalah model

molekul diatomik. Dalam model ini, energi total sistem terdiri dari energi yang berasal dari

gerak translasi, rotasi, vibrasi, orbital dan spin. Gerak orbital yang dimaksud di sini adalah

gerakan elektron disekeliling inti atom, sedangkan gerak spin adalah gerak berputar inti

atom terhadap porosnya. Gerak translasi dapat didekati dengan mekanika klasik, namun

keempat gerak yang lain secara umum mensyaratkan adanya kuantisasi (diskritisasi) energi.

Fungsi partisi total sistem molekul diatomik dapat dituliskan sebagai

Z = Zt × Zr × Zv × Ze × Zn, (6.14)

dimana indeks t, r, v, e dan n masing-masing untuk gerak translasi, rotasi, vibrasi, elektron

dan nuklir. Kita akan turunkan satu per satu kelima bentuk fungsi partisi tersebut dimulai

dari yang paling mudah.

Fungsi partisi untuk gerak translasi adalah yang paling mudah mengingat kita sudah

menurunkan sebelumnya pada persamaan (2.47),

Zt = BV(2πmkT

)3/2=V

~3

(2πmkT

)3/2. (6.15)

Berikutnya yang juga mudah adalah fungsi partisi untuk gerak vibrasi karena gerak vibrasi

molekul diatomik sebagai satu kesatuan dapat dipandang sebagai osilator harmonik satu

dimensi. Jadi fungsi partisi gerak vibrasi dapat didekati oleh persamaan (6.11),

Zv =e−

12hν/kT

1− e−hν/kT, (6.16)

dimana ν menyatakan frekuensi karakteristik vibrasi molekul diatomik.

Kemudian fungsi partisi dari kontribusi gerak orbital elektron dapat ditentukan langsung

dari persamaan (6.9) berikut,

Ze =∑i

gi e−εi/kT ≈ go + g1 e

−ε1/kT , (6.17)

dimana telah diambil dua suku pertama dari deret fungsi partisi yang bersangkutan sebagai

pendekatan yang cukup baik. Dalam hal ini go menyatakan derajat degenerasi keadaan

dasar molekul diatomik, g1 menyatakan derajat degenerasi keadaan tereksitasi pertama,

dan ε1 menyatakan tingkat energi eksitasi pertama. Pada umumnya tingkat-tingkat energi

tereksitasi berharga jauh lebih besar dari energi termal kT , kecuali untuk temperatur tinggi,

dimana energi termal tersebut menjadi dominan.


Perhitungan fungsi partisi gerak rotasi agak sedikit sulit. Kita bayangkan tiap molekul

gas terdiri atas 2 atom bermassa sama m terpisah pada jarak 2a. Momen inersia sistem

ini terhadap sumbu yang melewati titik tengah garis hubung kedua atom tersebut adalah

J = 2ma2, dimana a adalah jarak dari salah satu atom menuju ke titik pusat sistem. Bila

sistem ini tidak berada dalam pengaruh medan potensial apapun, maka energi totalnya hanya

merupakan energi kinetik rotasi saja. Jadi, kita dapat menuliskan persamaan Hamilton

dalam bentuk operator untuk menggambarkan dinamika rotasi sistem diatomik tersebut

sebagai berikut,

H ψ = (E + V )ψ = (E + 0)ψ = E ψ, (6.18)

dimana H adalah operator Hamiltonian yang menyatakan energi total sistem, E menyatakan

operator energi kinetik, V menyatakan operator energi potensial, dan ψ menyatakan fungsi

gelombang sistem yang bersangkutan. Analogi dengan gerak translasi, energi kinetik gerak

rotasi dapat digambarkan oleh

E =L2

2J, (6.19)

dimana L adalah operator momentum sudut orbital, yang bekerja terhadap fungsi gelombang

menurut

L2 ψ = ~2 `(`+ 1)ψ,

dimana ` adalah bilangan kuantum momentum sudut orbital (`= 0, 1, 2, ......). Substitusikan

bentuk di atas ke dalam persamaan (6.19) akan diperoleh

E ψ =~2 `(`+ 1)

2Jψ. (6.20)

Koreksi kecil dilakukan untuk perumusan momentum sudut. Dengan mempertimbangkan

momentum sudut spin, maka bilangan kuantum momentum sudut orbital ` diganti dengan

bilangan kuantum momentum sudut total j. Dengan demikian persamaan 6.20) berubah

menjadi

E ψ =~2 j(j + 1)

2Jψ. (6.21)

Secara implisit, persamaan (6.21) di atas menyatakan bahwa tingkat-tingkat energi rotasi

yang diijinkan adalah

εj =~2 j(j + 1)

2J, (6.22)

dimana j = ` ± 12. Perlu diketahui bahwa untuk setiap harga j akan terdapat (2j + 1)

kemungkinan harga mj (bilangan kuantum magnetik total), yaitu mj = -j,...... 0, ......j.

Kemungkinan sebanyak itu dapat dipandang sebagai akibat degenerasi kuantum. Jadi dalam

hal ini derajat degenerasinya adalah gj = (2j+1) sedemikian sehingga persamaan (6.9) dapat

6.3. Gas diatomik 51

dituliskan ulang menjadi

Zr =∞∑j=0

gj e−εj/kT =

∞∑j=0

(2j + 1) e−εj/kT

=∞∑j=0

(2j + 1) e−j(j+1)K/kT ,

(6.23)

dimana K = h2/8π2J .

Fungsi partisi dari kontribusi gerak orbital elektron dikembangkan dari persamaan (6.9)

berikut,

Ze =∑i

gi e−εi/kT ≈ go + g1 e

−ε1/kT , (6.24)

dimana telah diambil dua suku pertama dari deret fungsi partisi yang bersangkutan sebagai

pendekatan yang cukup baik. Dalam hal ini go menyatakan derajat degenerasi keadaan

dasar molekul diatomik, g1 menyatakan derajat degenerasi keadaan tereksitasi pertama,

dan ε1 menyatakan tingkat energi eksitasi pertama. Pada umumnya tingkat-tingkat energi

tereksitasi berharga jauh lebih besar dari energi termal kT , kecuali untuk temperatur tinggi,

dimana energi termal tersebut menjadi dominan.

Yang terakhir adalah fungsi partisi gerak spin Zn dari inti atom. Fungsi partisi ini bukan

merupakan fungsi temperatur, melainkan merupakan konstanta biasa. Tidak begitu penting

membicarakan berapa nilai sebenarnya dari fungsi partisi ini.

Setelah memperoleh kelima bentuk fungsi partisi (lihat kembali persamaan 6.14), maka

fungsi partisi sistem molekul diatomik secara lengkap dapat dituliskan sebagai

Z =V

~3

(2πmkT

)3/2× e−12hν/kT

1− e−hν/kT×∞∑j=0

(2j+1) e−j(j+1)K/kT ×(go+g1 e

−ε1/kT)×Zn. (6.25)

Perlu dipahami di sini bahwa kita meninjau gas semi klasik yang terdiri dari N buah molekul

diatomik yang identik satu sama lain. Entropi sistem gas semacam ini dinyatakan oleh

persamaan (5.14), yang dapat dituliskan ulang dalam bentuk

S = Nk +Nk `nB +Nk `nV

N+Nk `n

(2πmkT

)3/2+E

T

= Nk +Nk `nZ

N+E

T.

(6.26)

Dengan bantuan persamaan (6.26) dan definisi fungsi energi bebas Helmholtz F = E − TS,

maka kita dapat menurunkan fungsi Helmholtz untuk gas semi klasik dalam bentuk

F = −kT `n ZN

N !. (6.27)


Cobalah bandingkan persamaan (6.27) dengan bentuk fungsi Helmholtz untuk gas klasik

pada persamaan (5.6),

F = −NkT `nZ = −kT `nZN .

Jelas terlihat sekali lagi bahwa terdapat perbedaan faktor N ! antara besaran makro energi

bebas Helmholtz F menurut distribusi klasik dan semi klasik. Sebelumnya adalah perbedaan

(dengan faktor yang sama N !) antara besaran mikro peluang termodinamik W menurut

distribusi klasik dan semi klasik (lihat kembali pembahasan pasal 5.4).

Dengan membandingkan perumusan energi bebas Helmholtz untuk sistem klasik dan

semiklasik, maka dapat didefinisikan besaran fungsi partisi total Z sistem semi klasik sebagai

berikut

Z =ZN

N !. (6.28)

Substitusikan persamaan (6.28) ke dalam persamaan (6.27), akan diperoleh

F = −kT `nZ. (6.29)

Dengan bantuan persamaan (6.29) kita bisa menghitung besarnya energi total gas diatomik

semi klasik sebagai berikut,

E = −T 2

(∂F/T

∂T

)V

= kT 2

(∂ `nZ∂T

)V

= NkT 2

(∂ `nZ

∂T

)V

= NkT 2

(∂ `nZt∂T

+∂ `nZr∂T

+∂ `nZv∂T

+∂ `nZe∂T

+∂ `nZn∂T

).

Suku terakhir ruas kanan jelas berharga nol mengingat Zn adalah konstanta. Keempat suku

pertama ruas kanan memberikan hasil berikut ini,

E =3

2NkT +NkT 2 ∂

∂T`n

∞∑j=0

(2j + 1) e−j(j+1)K/kT

+Nhν

(1

2+

hν

ehν/kT − 1

)+N g1 ε1 e

−ε1/kT

go + g1 e−ε1/kT. (6.30)

Pada tahap ini akan sangat bermanfaat bila didefinisikan suatu besaran karakteristik,

dengan mana kita bisa membandingkan kontribusi masing-masing gerak dalam persamaan

(6.30) di atas. Besaran karakteristik yang dimaksud adalah temperatur karakteristik rotasi,

vibrasi dan orbital elektronik berturut-turut dinyatakan sebagai berikut,

θr =Kk, θv =

hν

k, θe =

ε1k. (6.31)

Pada umumnya nilai ketiga besaran karakteristik tersebut di atas memenuhi hubungan

6.4. Latihan soal 53

berikut θr � θv � θe. Pada temperatur rendah (T � θr), tiga suku pada ruas kanan

persamaan (6.30) menjadi dominan, yaitu suku pertama 32NkT , suku kontribusi dari energi

keadaan dasar gerak vibrasi 12Nhν, dan suku kontribusi dari energi keadaan dasar gerak

orbital elektron Ngo. Dari ketiga suku ini, hanya suku pertama saja yang merupakan fungsi

temperatur. Suku ini merupakan kontribusi dari gerak translasi. Jadi kapasitas panas CV

gas diatomik pada temperatur rendah dinyatakan sebagai

CV =

(∂E

∂T

)V

=3

2Nk =

3

2nNok =

3

2nR. (6.32)

Dengan kata lain, kalor jenis cV gas diatomik pada temperatur rendah dituliskan sebagai

cV =CVn

=3

2R. (6.33)

Hasil tersebut di atas sama dengan hasil yang diperoleh melalui perhitungan distribusi klasik.

Bagaimanapun, perhitungan semi klasik memungkinkan kita untuk menentukan kapasitas

panas ataupun kalor jenis gas diatomik pada temperatur sedang dan tinggi. Untuk sekedar

diketahui, pada temperatur sedang kalor jenis gas diatomik adalah 52R, sedangkan pada

temperatur tinggi 72R. Detil penurunan tidak diberikan dalam kuliah ini, namun demikian

mahasiswa diharapkan untuk melakukannya buat olahraga ringan.

6.4 Latihan soal

1. Calculate the maximum specific heat for an assembly of systems having two energy

levels separated by an energy hν if the upper level is doubly degenerate (taken from

Pointon, 1978, Ch.8, problem 6).

2. Two different groups of n fermions are distributed among m states of the same energy

where m > n. Show that the entropy is given by

2k(m`nm− n `nn− (m− n) `n (m− n)

)(taken from Pointon, 1978, Ch.8, problem 2).

3. Show that at high temperatures the free energy of an assembly of quantum mechanical,

1D oscillators approximates to kT `n (hν/kT ) per oscillator, corresponding to a mean

thermal energy of kT (taken from Pointon, 1978, Ch.8, problem 3).

4. Evaluate the partition function at temperature T for a classical 1D harmonic oscillator


having an energy of

ε =p2x

2m+

1

2µx2

and hence find the mean energy of such an oscillator at this temperature (taken from

Pointon, 1978, Ch.8, problem 4).

5. Show that the equation of state of a diatomic gas as derived from (8.27) of Pointon

(1978) is the same as that for a monoatomic gas (taken from Pointon, 1978, Ch.8,

problem 5).

6. Write out the partition function for a 3D harmonic oscillator assuming that it obeys

classical and quantum mechanics. Show that both give a mean energy of 3kT at high

temperatures (taken from Pointon, 1978, Ch.8, problem 7).

Bibliography

Beiser, A. 1988 Perspective of Modern Physics . London, UK: McGraw-Hill.

Gasiorowicz, S. 1996 Quantum Physics . New York, US: John Wiley and Sons.

Huang, K. 1987 Statistical Mechanics . New York, US: John Wiley and Sons.

Pointon, A. J. 1978 An Introduction to Statistical Physics . London, UK: Longmann.

Reif, F. 1985 Fundamentals of Statistical and Thermal Physics . Singapore: McGraw-Hill.

Tipler, P. A. 1999 Physics for scientists and engineers . New York, US: W. H. Freemann.

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