Introduction to Thermodynamics and Statistical Physics phần 3 pdf

1.7. Problems Set 1
a) Sho w that α is given by
α =
a
2
·
1+tanh
µ
Fa
2τ
¶¸
. (1.135)
b) Show that in the limit of high temperature the spring constant is
given approximately by
k '
4τ
Na
2
. (1.136)
N
α
N
α
19. A long elastic molecule can be modelled as a linear chain of N links. The
state of each link is charac terized by two quantum numbers l and n.The
length of a link is either l = a or l = b. The vibrational state of a link
is modelled as a harmonic oscillator whose angular frequency is ω
a
for a
link of length a and ω
b

for a link of length b. Thus, the energy of a link
is
E
n,l
=
½
}ω
a
¡
n +
1
2
¢
for l = a
}ω
b
¡
n +
1
2
¢
for l = b
, (1.137)
n =0, 1, 2,
The chain is held under a tension F . Show that the mean length hLi of
the chain in the limit of high temperature T is given by
hLi = N
aω
b
+ bω

a
ω
b
+ ω
a
+ N
Fω
b
ω
a
(a −b)
2
(ω
b
+ ω
a
)
2
β + O
¡
β
2
¢
, (1.138)
where β =1/τ.
20. The elasticity of a rubber band can be described in terms of a one-
dimensional model of N polymer molecules linked together end-to-end.
The angle between successive links is equally likely to be 0
◦
or 180

◦
.The
length of eac h polymer is d and the total length is L. The system is in
thermal equilibrium at temperature τ. Show that the force f required to
main tain a length L is given by
f =
τ
d
tanh
−1
L
Nd
. (1.139)
21. Consider a sys tem which has two single particle states both of the same
energy. When both states are unoccupied, the energy of the system is
Eyal Buks Thermodynamics and Statist ical Physics 27
Chapter 1. The Principle of Largest Uncertainty
zero; when one state or the other is occupied by one particle, th e energy
is ε. We suppose that the energy of the system is much higher (infinitely
higher) when both states are occupied. Show that in thermal equilibrium
at temperature τ the av erage number of particles in the level is
hNi =
2
2+exp[β (ε − µ)]
, (1.140)
where µ is the chemical potential and β =1/τ.
22. Consider an array of N tw o-lev el particles. Each one can be in o ne of two
states, having energy E
1
and E

2
respectively. The n umbers of particles
instates1and2aren
1
and n
2
respectively, where N = n
1
+ n
2
(assume
n
1
À 1andn
2
À 1). Consider a n energy exchange with a reservoir at
temperature τ leading to population changes n
2
→ n
2
−1andn
1
→ n
1
+1.
a) Calculate the en tropy change of the two-level system, (∆σ)
2LS
.
b) Calculate the entropy change of the reservoir, (∆σ)
R

.
c) What can be said about th e relation bet ween (∆σ)
2LS
and (∆σ)
R
in
thermal equilibrium? Use your answer to express the ration n
2
/n
1
as
a function of E
1
, E
2
and τ.
23. Consider a lattice containing N sites of one ty pe, which is denoted as A ,
and the same number of sites of another type, which is denoted as B. The
lattice is occupied by N atoms. The number of atoms occupying sites of
type A is denoted as N
A
, w hereas the number of atoms occupying atoms
of type B is denoted as N
B
,whereN
A
+ N
B
= N.Letε be the energy
necessary to remove an atom from a lattice site of type A to a lattice

site of type B. The system is in thermal equilibrium at temperature τ .
Assume that N,N
A
,N
B
À 1.
a) Calculate the entropy σ.
b) Calculate the average number hN
B
i of atoms occupying sites of type
B.
24. Consider a microcanonical ensem ble of N quantum harmonic oscillators
in thermal equilibrium at temperature τ. The resonance frequency of all
oscillators is ω. The quantum energy levels of each quantum oscillator is
given by
ε
n
= }ω
µ
n +
1
2
¶
, (1.141)
where n =0, 1, 2, is integer. The total energy E of the system is given
by
E = }ω
µ
m +
N

2
¶
, (1.142)
where
Eyal Buks Thermodynamics and Statist ical Physics 28
1.8. Solutions Set 1
m =
N
X
l=1
n
l
, (1.143)
and n
l
is state number of oscillator l.
a) Calculate the number of states g (N,m) of the system with total
energy }ω (m + N/2).
b) Use this re sult to calculate the en tropy σ of the system with total
energy }ω (m + N/2). Appro ximate the r esult by assuming t hat N À
1andm À 1.
c) Use this result to calculate (in the same limit of N À 1andm À 1)
the average energy of the system U as a function of the temperature
τ.
25. The energy of a donor level in a semiconductor is −ε when occupied
by an electron (and the energy is zero otherwise). A donor level can be
either occupied by a sp in u p electron or a spin do wn electron, however, it
cannot be simultaneously occupied by two electrons. Express the average
occupation of a donor state hN
d

i as a function of ε and the chemical
potential µ.
1.8 Solutions Set 1
1. Final answers:
a)
N!
(
N
2
)
!
(
N
2
)
!
¡
1
2
¢
N
b) 0
2. Final answers:
a)
¡
5
6
¢
N
b)

¡
5
6
¢
N−1
1
6
c) In general
∞
X
N=0
Nx
N−1
=
d
dx
∞
X
N=0
x
N
=
d
dx
1
1 − x
=
1
(1 − x)
2

,
thus
¯
N =
1
6
∞
X
N=0
N
µ
5
6
¶
N−1
=
1
6
1
¡
1 −
5
6
¢
2
=6.
3. Let W (m) be the probability for for taking n
1
steps to the right and
n

2
= N − n
1
steps to the left, where m = n
1
− n
2
,andN = n
1
+ n
2
.
Using
Eyal Buks Thermodynamics and Statist ical Physics 29
Chapter 1. The Principle of Largest Uncertainty
n
1
=
N + m
2
,
n
2
=
N −m
2
,
one finds
W (m)=
N!

¡
N+m
2
¢
!
¡
N−m
2
¢
!
p
N+m
2
q
N−m
2
.
It is convenient to employ the moment generating function, defined as
φ (t)=

e
tm
®
.
In general, the following holds
φ (t)=
∞
X
k=0
t

k
k!

m
k
®
,
thus from the kth derivative of φ (t) one can calculate the kth moment
of m

m
k
®
= φ
(k)
(0) .
Using W (m)onefinds
φ (t)=
N
X
m=−N
W (m) e
tm
=
N
X
m=−N
N!
¡
N+m

2
¢
!
¡
N−m
2
¢
!
p
N+m
2
q
N−m
2
e
tm
,
or using the summation variable
n
1
=
N + m
2
,
one has
φ (t)=
N
X
n
1

=0
N!
n
1
!(N − n
1
)!
p
n
1
q
N−n
1
e
t(2n
1
−N)
= e
−tN
N
X
n
1
=0
N!
n
1
!(N − n
1
)!

¡
pe
2t
¢
n
1
q
N−n
1
= e
−tN
¡
pe
2t
+ q
¢
N
.
Using p = q =1/2
Eyal Buks Thermodynamics and Statist ical Physics 30
1.8. Solutions Set 1
φ (t)=
µ
e
t
+ e
−t
2
¶
N

=(cosht)
N
.
Thus using the expansion
(cosh t)
N
=1+
1
2!
Nt
2
+
1
4!
[N +3N (N −1)] t
4
+ O
¡
t
5
¢
,
one finds
hmi =0,

m
2
®
= N,


m
3
®
=0,

m
4
®
= N (3N − 2) .
4. Using the binomial distribution
W (n)=
N!
n!(N − n)!
p
n
(1 − p)
N−n
=
N (N − 1)(N −1) × (N −n +1)
n!
p
n
(1 − p)
N−n
∼
=
(Np)
n
n!
exp (−Np) .

5.
a)
∞
X
n=0
W (n)=e
−λ
∞
X
n=0
λ
n
n!
=1
b) AsinEx.1-6,itisconvenienttousethemomentgeneratingfunction
φ (t)=

e
tn
®
=
∞
X
n=0
e
tn
W (n)=e
−λ
∞
X

n=0
λ
n
e
tn
n!
= e
−λ
∞
X
n=0
(λe
t
)
n
n!
=exp
£
λ
¡
e
t
− 1
¢¤
.
Using the expansion
exp
£
λ
¡

e
t
− 1
¢¤
=1+λt +
1
2
λ (1 + λ) t
2
+ O
¡
t
3
¢
,
one finds
hni = λ.
c) Using the same expansion one finds

n
2
®
= λ (1 + λ) ,
thus
D
(∆n)
2
E
=


n
2
®
− hni
2
= λ (1 + λ) −λ
2
= λ.
Eyal Buks Thermodynamics and Statist ical Physics 31
Chapter 1. The Principle of Largest Uncertainty
6.

R
2
®
=
*Ã
N
X
n=1
r
n
!
2
+
=
N
X
n=1


r
2
n
®
|
{z}
=l
2
+
X
n6=m
hr
n
· r
m
i
|
{z }
=0
= Nl
2
7.
20
X
n=11
20!
n!(20− n)!
0.2
n
× 0.8

20−n
=5.6 × 10
−4
. (1.144)
8. Using
N
+
+ N
−
= N, (1.145a)
N
+
− N
−
=
M
m
, (1.145b)
one has
N
+
=
N
2
µ
1+
M
mN
¶
, (1.146a)

N
−
=
N
2
µ
1 −
M
mN
¶
, (1.146b)
or
N
+
=
N
2
(1 + x) , (1.147a)
N
−
=
N
2
(1 − x) , (1.147b)
where
x =
M
mN
.
The number of states having total magnetization M is giv en by

Ω (M)=
N!
N
+
!N
−
!
=
N!
£
N
2
(1 + x)
¤
!
£
N
2
(1 − x)
¤
!
. (1.148)
Since all states have equal proba bility one has
f (M)=
Ω (M)
2
N
. (1.149)
Taking the natural logarithm of Stirling’s formula one finds
log N!=N log N − N + O

µ
1
N
¶
, (1.150)
Eyal Buks Thermodynamics and Statist ical Physics 32
1.8. Solutions Set 1
th us in the limit N À 1 one has
log f = −log 2
N
+ N log N − N
−
·
N
2
(1 + x)
¸
log
·
N
2
(1 + x)
¸
+
·
N
2
(1 + x)
¸
−

·
N
2
(1 − x)
¸
log
·
N
2
(1 − x)
¸
+
·
N
2
(1 − x)
¸
= −N log 2 + N log N
−
·
N
2
(1 + x)
¸
log
·
N
2
(1 + x)
¸

−
·
N
2
(1 − x)
¸
log
·
N
2
(1 − x)
¸
=
µ
−
N
2
¶½
−2log
N
2
+(1+x)log
·
N
2
(1 + x)
¸
+(1− x)log
·
N

2
(1 − x)
¸¾
=
µ
−
N
2
¶·
−2log
N
2
+(1+x)
µ
log
N
2
+log(1+x)
¶
+(1− x)
µ
log
N
2
+log(1− x)
¶¸
=
µ
−
N

2
¶µ
log
¡
1 − x
2
¢
+ x log
1+x
1 − x
¶
.
(1.151)
The function log f (x) has a sharp peak near x = 0, thus we can approx-
imate it by assuming x ¿ 1. To lowest order
log
¡
1 − x
2
¢
+ x log
1+x
1 − x
= x
2
+ O
¡
x
3
¢

, (1.152)
thus
f (M)=A exp
µ
−
M
2
2m
2
N
¶
, (1.153)
where A is a normalizatio n constant, which is determined by requiring
that
1=
∞
Z
−∞
f (M)dM. (1.154)
Using the iden tity
∞
Z
−∞
exp
¡
−ay
2
¢
dy =
r

π
a
, (1.155)
one finds
1
A
=
∞
Z
−∞
exp
µ
−
M
2
2m
2
N
¶
dM = m
√
2πN , (1.156)
Eyal Buks Thermodynamics and Statist ical Physics 33
Chapter 1. The Principle of Largest Uncertainty
thus
f (M)=
1
m
√
2πN

exp
µ
−
M
2
2m
2
N
¶
, (1.157)
The expectation value is giving by
hMi =
∞
Z
−∞
Mf (M)dM =0, (1.158)
and the v ariance is giv en b y
D
(M −hMi)
2
E
=

M
2
®
=
∞
Z
−∞

M
2
f (M)dM = m
2
N. (1.159)
9. The probability to have n steps to the righ t is given by
W (n)=
N!
n!(N − n)!
p
n
q
N−n
. (1.160)
a)
hni =
N
X
n=0
N!n
n!(N − n)!
p
n
q
N−n
(1.161)
= p
∂
∂p
N

X
n=0
N!
n!(N − n)!
p
n
q
N−n
= p
∂
∂p
(p + q)
N
= pN (p + q)
N−1
= pN .
Since
X = an − a (N −n)=a (2n − N) (1.162)
we find
hXi = aN (2p − 1) = aN (p − q) . (1.163)
b)

n
2
®
=
N
X
n=0
N!n

2
n!(N − n)!
p
n
q
N−n
=
N
X
n=0
N!n (n −1)
n!(N −n)!
p
n
q
N−n
+
N
X
n=0
N!n
n!(N −n)!
p
n
q
N−n
= p
2
∂
2

∂p
2
N
X
n=0
N!
n!(N −n)!
p
n
q
N−n
+ hni
= p
2
∂
2
∂p
2
(p + q)
N
+ hni = p
2
N (N − 1) + pN .
Eyal Buks Thermodynamics and Statist ical Physics 34
1.8. Solutions Set 1
Thus
D
(n −hni)
2
E

= p
2
N (N − 1) + pN −p
2
N
2
= Npq , (1.164)
and
D
(X − hXi)
2
E
=4a
2
Npq . (1.165)
10. The total energy is given by
E =
kx
2
2
+
m ˙x
2
2
=
ka
2
2
, (1.166)
where a is the amplitude of oscillations. The time period T is given by

T =2
Z
a
−a
dx
˙x
=2
r
m
k
Z
a
−a
dx
√
a
2
− x
2
=2π
r
m
k
, (1.167)
thus
f(x)=
2
T | ˙x|
=
1

π
√
a
2
− x
2
. (1.168)
11. The six experiments are independen t, thus
σ =6×
µ
−
2
3
ln
2
3
−
1
3
ln
1
3
¶
=3.8191 .
12. The random variable X obtains t he value n with proba b ility p
n
= q
n
,
where n =1, 2, 3, ,andq =1/2.

a) The entropy is given by
σ = −
∞
X
n=1
p
n
log
2
p
n
= −
∞
X
n=1
q
n
log
2
q
n
=
∞
X
n=1
nq
n
.
This can be rewritten as
σ = q

∂
∂q
∞
X
n=1
q
n
= q
∂
∂q
µ
1
1 − q
− 1
¶
=
q
(1 − q)
2
=2.
b) A series of questions is of the form: ”Was a head obtained in the 1st
time ?”, ”Was a h ea d obtain ed in the 2nd time ?”, etc. The expected
number of questioned required to find X is
1
2
× 1+
1
4
× 2+
1

8
× 3+ =2,
which is exactly the entropy σ.
13.
Eyal Buks Thermodynamics and Statist ical Physics 35
Chapter 1. The Principle of Largest Uncertainty
a) Consider an infinitesimal change in the va riable z = z (x, y)
δz =
µ
∂z
∂x
¶
y
δx +
µ
∂z
∂y
¶
x
δy . (1.169)
For a process for which z is a constant δz =0,thus
0=
µ
∂z
∂x
¶
y
(δx)
z
+

µ
∂z
∂y
¶
x
(δy)
z
. (1.170)
Dividing by (δx)
z
yields
µ
∂z
∂x
¶
y
= −
µ
∂z
∂y
¶
x
(δy)
z
(δx)
z
= −
µ
∂z
∂y

¶
x
µ
∂y
∂x
¶
z
= −
³
∂y
∂x
´
z
³
∂y
∂z
´
x
.
(1.171)
b) Consider a process for which the variable w is k ept constant. An
infinitesimal change in the va riable z = z (x, y) is expressed as
(δz)
w
=
µ
∂z
∂x
¶
y

(δx)
w
+
µ
∂z
∂y
¶
x
(δy)
w
. (1.172)
Dividing by (δx)
w
yields
(δz)
w
(δx)
w
=
µ
∂z
∂x
¶
y
+
µ
∂z
∂y
¶
x

(δy)
w
(δx)
w
. (1.173)
or
µ
∂z
∂x
¶
w
=
µ
∂z
∂x
¶
y
+
µ
∂z
∂y
¶
x
µ
∂y
∂x
¶
w
. (1.174)
14. We have found in class that

hUi = −
µ
∂ log Z
gc
∂β
¶
η
, (1.175)
hNi = −
µ
∂ log Z
gc
∂η
¶
β
. (1.176)
Eyal Buks Thermodynamics and Statist ical Physics 36
1.8. Solutions Set 1
a) Using relation (1.125) one has
hUi = −
µ
∂ log Z
gc
∂β
¶
η
= −
µ
∂ log Z
gc

∂β
¶
µ
−
µ
∂ log Z
gc
∂µ
¶
β
µ
∂µ
∂β
¶
η
= −
µ
∂ log Z
gc
∂β
¶
µ
−
η
β
2
µ
∂ log Z
gc
∂µ

¶
β
= −
µ
∂ log Z
gc
∂β
¶
µ
+ τµ
µ
∂ log Z
gc
∂µ
¶
β
.
(1.177)
b) Using Eq. (1.176) one has
hNi = −
µ
∂ log Z
gc
∂η
¶
β
= −
µ
∂µ
∂η

¶
β
µ
∂ log Z
gc
∂µ
¶
β
= τ
µ
∂ log Z
gc
∂µ
¶
β
,
(1.178)
or in terms of the fugacity λ, which is defined by
λ =exp(βµ)=e
−η
, (1.179)
one has
hNi = τ
µ
∂ log Z
gc
∂µ
¶
β
= τ

∂λ
∂µ
∂ log Z
gc
∂λ
= λ
∂ log Z
gc
∂λ
.
(1.180)
15. The canonical partition function is given by
Z
c
= Z
N
1
, (1.181)
where
Z
1
=exp
µ
βε
2
¶
+exp
µ
−
βε

2
¶
=2cosh
µ
βε
2
¶
. (1.182)
Eyal Buks Thermodynamics and Statist ical Physics 37
Chapter 1. The Principle of Largest Uncertainty
Thus
hUi = −
∂ log Z
c
∂β
= −N
∂ log Z
1
∂β
= −
Nε
2
tanh
βε
2
, (1.183)
and
τ =
ε
2tanh

−1
³
−
2hUi
Nε
´
. (1.184)
The negative temperature is originated by our assumption that the en-
ergy of a single magnet has an upper bound. In reality this is nev er the
case.
16. The canonical partition function is given by
Z
c
= Z
N
1
, (1.185)
where
Z
1
=exp
µ
−
β}ω
2
¶
∞
X
n=0
exp (−β}ωn) (1.186)

=
exp
³
−
β}ω
2
´
1 − exp (−β}ω)
=
1
2sinh
β}ω
2
.
a)
hUi = −
∂ log Z
c
∂β
= −N
∂ log Z
1
∂β
=
N}ω
2
coth
β}ω
2
(1.187)

b)
D
(∆U)
2
E
=
∂
2
log Z
c
∂β
2
= N
∂
2
log Z
1
∂β
2
=
N
¡
}ω
2
¢
2
sinh
2
β}ω
2

(1.188)
17. The canonical partition function is given by
Z
c
=[exp(βε)+1+exp(−βε)]
N
=[1+2cosh(βε)]
N
, (1.189)
where β =1/τ.
a) Thus the average energy is
hUi = −
∂ log Z
c
∂β
= −
2Nεsinh (βε)
1+2coshβε
, (1.190)
b) and the va riance is
D
(U −hUi)
2
E
=
∂
2
log Z
c
∂β

2
= −
∂ hUi
∂β
=2Nε
2
cosh (βε)+2
[1 + 2 cosh (βε)]
2
.
(1.191)
Eyal Buks Thermodynamics and Statist ical Physics 38
1.8. Solutions Set 1
18. Each section can be in one of two possible sates with corresponding en-
ergies 0 a nd Fa.
a) By denition, isthemeanlengthofeachsegment,whichisgiven
by
=
a exp (Fa)
1+exp(Fa)
=
a
2
ã
1+tanh
à
Fa
2
ảá
, (1.192)

where =1/.
b) At high temperature Fa 1 the length of the chain L = N is
given b y
L =
Na
2
ã
1+tanh
à
Fa
2
ảá
'
Na
2
à
1+
Fa
2
ả
, (1.193)
or
F = k
à
L
Na
2
ả
, (1.194)
where the spring constant k is given by

k =
4
Na
2
. (1.195)
19. The average length of a single link is given by
hli =
a exp (Fa)

X
n=0
exp
Ê
}
a
Ă
n +
1
2
ÂÔ
+ b exp (Fb)

X
n=0
exp
Ê
}
b
Ă
n +

1
2
ÂÔ
exp (Fa)

X
n=0
exp
Ê
}
a
Ă
n +
1
2
ÂÔ
+exp(Fb)

X
n=0
exp
Ê
}
b
Ă
n +
1
2
ÂÔ
(1.196)

=
a exp(F a)
2sinh
}
a
2
+
b exp(F b)
2sinh
}
b
2
exp(Fa)
2sinh
}
a
2
+
exp(Fb)
2sinh
}
b
2
.
To rst order in
hli =
a
b
+ b
a

b
+
a
+
F
b

a
(a b)
2
(
b
+
a
)
2
+ O
Ă

2
Â
. (1.197)
The average total length is hLi = nhli .
20. The length L is given by
L = N hli ,
where hli is the average contribution of a single molecule to the total
length, which can be either +d or d. The probability of each possibility
Eyal Buks Thermodynamics and Statist ical Physics 39
Chapter 1. The Principle of Largest Uncertainty

is determined by the Boltzmann factor. The energy change due to flipping
of one link from 0
◦
to 180
◦
is 2fd,therefore
hli = d
e
βfd
− e
−βfd
e
βfd
+ e
−βfd
= d tanh (βfd) ,
where β =1/τ.Thus
L = Ndtanh ( βfd) ,
or
f =
τ
d
tanh
−1
L
Nd
.
21. The grand partition function Z
gc
is given by

Z
gc
=1+2exp[β (µ − ε)] , (1.198)
thus
hNi =
1
β
∂ log Z
gc
∂µ
=
2
2+exp[β (ε −µ)]
. (1.199)
22.
a)
(∆σ)
2LS
=log
N!
(n
2
− 1)! (n
1
+1)!
− log
N!
n
2
!n

1
!
=log
n
2
n
1
+1
' log
n
2
n
1
.
(1.200)
b)
(∆σ)
R
=
E
2
− E
1
τ
(1.201)
c) For a small change near thermal equilibrium one expects (∆σ)
2LS
+
(∆σ)
R

=0,thus
n
2
n
1
=exp
µ
−
E
2
− E
1
τ
¶
. (1.202)
23. Th e number o f ways to select N
B
occupied sites of type B out of N sites
is N!/n!(N − n)!. Similarly the number of ways to select N
B
empty sites
of type A out of N sites is N! /n!(N − n)!.
Eyal Buks Thermodynamics and Statist ical Physics 40
1.8. Solutions Set 1
a) Thus
σ =log
µ
N!
N
B

!(N − N
B
)!
¶
2
' 2[N log N −N
B
log N
B
− (N −N
B
)log(N − N
B
)] .
(1.203)
b) The energy of the system is giv en by U = N
B
ε. Thus, the Helmholtz
free energy is gi ven by
F = U −τσ = U −2τ
·
N log N −
U
ε
log
U
ε
−
µ
N −

U
ε
¶
log
µ
N −
U
ε
¶¸
.
(1.204)
At thermal equilibrium (∂F/∂U)
τ
=0,thus
0=
µ
∂F
∂U
¶
τ
=1+
2τ
ε
·
log
U
ε
− log
µ
N −

U
ε
¶¸
, (1.205)
or
N −N
B
N
B
=exp
³
ε
2τ
´
, (1.206)
therefore
hN
B
i =
N
1+exp
¡
ε
2τ
¢
.
Alternatively, one can calculate the chemical potential from the re-
quirement
1=
N

A
N
+
N
B
N
, (1.207)
where
N
A
N
=
exp (βµ)
1+exp(βµ)
, (1.208a)
N
B
N
=
exp (βµ −βε)
1+exp(βµ−βε)
, (1.208b)
which is s atisfied when
µ =
ε
2
, (1.209)
thus
hN
B

i =
N
1+exp
¡
ε
2τ
¢
. (1.210)
Eyal Buks Thermodynamics and Statist ical Physics 41
Chapter 1. The Principle of Largest Uncertainty
24. In general,
g (N,m)={# o s ways to distribute m identical balls in N boxes}
Moreover
{#oswaystodistributem iden tical balls in N boxes}
= {#oswaystoarrangem identical balls and N − 1 identical pa rtitions in a line }
…
…
…
…
a) Therefore
g (N,m)=
(N −1+m)!
(N −1)!m!
. (1.211)
b) The entropy is given by
σ =log
(N −1+m)!
(N −1)!m!
' (N + m)log(N + m) −N log N −m log m,
(1.212)

or in terms of the total energy E = }ω (m + N/2)
σ =
·
N +
µ
E
}ω
−
N
2
¶¸
log
·
N +
µ
E
}ω
−
N
2
¶¸
− N log N −
µ
E
}ω
−
N
2
¶
log

µ
E
}ω
−
N
2
¶
.
(1.213)
c) The temperature τ is given by
1
τ
=
∂σ
∂E
=
1 − ln
2}ω
2E+N}ω
}ω
+
ln
2}ω
2E−N}ω
− 1
}ω
=
1
}ω
ln

µ
2E + N}ω
2E −N}ω
¶
.
(1.214)
Eyal Buks Thermodynamics and Statist ical Physics 42
1.8. Solutions Set 1
In the thermodynamical limit (N À 1, m À 1) the energy E and its
average U are indistinguishable, thus
exp
µ
}ω
τ
¶
=
2U + N}ω
2U −N}ω
, (1.215)
or
U =
N}ω
2
coth
}ω
2τ
. (1.216)
25. The grand canonical partition function is given by
ζ =1+2λ exp(βε) , (1.217)
where λ =exp(βµ)isthefugacity,thus

hN
d
i = λ
∂ log ζ
∂λ
=
2λe
βε
1+2λe
βε
=
1
1+
1
2
e
−β(ε+µ)
. (1.218)
Eyal Buks Thermodynamics and Statist ical Physics 43