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Phy 504 hw 4 solution
March 27, 2013
1 The Mayer Linked Cluster Expansion
1, The Grand Canonical Ensemble is
Z(µ, V, T ) =1X
N=0
z
N
Z(N,V, T ) (1)
Therefore the mean particle number is
hNi =P1
N=0 Nz
N
ZP1N=0 z
N
Z
=@ lnZ@ ln z
(2)
The z = e
�µ is fugacity. The pressure P is
P (µ, V, T ) =kT
V
lnZ (3)
So
hNiV
=1
v
=z
V
@
@z
lnZ (4)
Therefore
1
v
=1
V
@
@ ln z
✓PV
kT
◆= kT
@
@µ
✓PV
kT
◆(5)
or in P (v, T, µ)
P (v, T, µ) = kT
Z@(�µ)
v
(6)
2, For the classical, the grand partition function is
Z(µ, V, T ) =1X
N=0
1
N !
⇣z
�
3
⌘N
Q
N
(7)
where
Q
N
=
ZNY
i=1
d
3r exp[�
X
i<j
�U(|rij
|)]
� =
s2⇡~2mk
B
T
(8)
1
For pairwise interaction U(rij
), we define f
ij
= exp(��U(|rij
|))� 1 and write
Q
N
=
ZNY
i=1
d
3r
i
Y
i<j
(1 + f
ij
)
=
ZNY
i=1
d
3r
i
(1 +X
i<j
f
ij
+X
i<j,k<l
f
ij
f
kl
+ · · · ) (9)
For a given term in the series, let n
l
be the number of l-clusters. Note that that n
l
satisfyN =
Pl
ln
l
. Summing over all such sets {nl
}, we can write
Q
N
=X
{nl}
=N !Q
l
n
l
!(l!)nl(10)
where
W ({nl
}) = N !Ql
n
l
!(l!)nl(11)
is the number of times terms with {nl
} appear in Q
N
. It follows that the grand partitionfunction is
Z(µ, V, T ) =1X
N=0
X
{nl}
1
N !
✓z
�
3T
◆N
N !Ql
n
l
!(l!)nl
Y
l
b
nll
=X
{nl}
✓z
�
3T
◆Pl lnl Y
l
b
nll
n
l
!(l!)nl
=Y
l
1X
nl=0
1
n
l
!
✓z
l
b
l
�
3lT
l!
◆
= exp
X
l
z
l
b
l
�
3lt
l!
!(12)
Hence from
P (µ, V, T ) =kT
V
lnZ
=kT
V
X
l
b
l
�
3lT
l!z
l (13)
we see that the pressure written as a series expansion in powers of z has contribution onlyfrom the linked diagrams.
3, To the third order we have
P =kT
�
3T
1X
l=1
b
l
z
l =kT
�
3T
�b1z + b2z
2 + b3z3 + · · ·
�(14)
2
b1 = 1
b2 =1
2�3T
V
Zd
3x1d
3x2f12
=1
2�3T
V
Zd
3x1d
3x2
⇣e
��U(|~x1�~x2|) � 1⌘
b3 =1
3�6T
V
Zd
3x1d
3x2d
3x3(f12f13 + f12f23 + f12f23 + f12f13f23)
=1
6�6T
V
Zd
3x1d
3x2d
3x3
⇣e
��(U12+U13+U23) � e
��U12 � e
��U13 � e
��U23 + 2⌘
(15)
4,
PV = kT lnZ (16)
Therefore insert P here
Z = e
Vz3
Pl blz
l
= e
V�3 (b1z+b2z
2+b3z3+··· ) (17)
The exponential functions are analytic, so are the products of exponentials eaeb = e
a+b. So z isanalytic in z. Furthermore, the exponential function e
↵x has no zeros for x > 0, provided that↵ > 0
5, The derivative of the pressure with respect to z is
@
@z
P (µ, V, T ) =kT
V
@ lnZ@z
=kT
V
1
Z@
@z
=kT
V
hNiz
> 0 (18)
which means that the pressure is monotonically increasing with respect to z. SImilary
@
@z
✓1
v
◆=
1
V
@
@z
✓z
Z@Z@z
◆
=1
V
@
@z
1
Z
1X
N=1
Nz
N
Z
N
!
=1
zV
�hN2i � hNi2
�
� 0 (19)
So 1/v is also monotonically increasing function of z.6, We know
P
kT
=1
�
3T
1X
l=1
b
l
z
l
1
v
=1
�
3T
1X
l=1
lb
l
z
l (20)
As V ! 1 , bl
! b̄
l
3
Therefore
PV
kT
=1X
m=0
a
m
(1X
n=1
nb̄
n
z
n)m =
P1l=1 b̄lz
l
P1l=1 lb̄lz
l
(21)
Therefore
1X
m=0
a
m
(1X
n=1
nb̄
n
z
n)m+1 =1X
l=1
b̄
l
z
l
= a0(b̄1z + 2b̄2z2 + · · · ) + a1(b̄1z + 2b̄2z
2 + · · · )2 + a2(b̄1z + 2b̄2z2 + · · · )3 + · · ·
(22)
Therefore
z
1 : b̄1 = a0b̄1
z
2 : b̄2 = 2a0b̄2 + a1b̄21
z
3 : b̄3 = 3b̄3 + 4a1b̄1b̄2 + a2b̄31 (23)
Therefore
a0 = 1, a1 = �b̄2, a2 = 4b̄22 � 2b̄3 (24)
7,
a1 = �b̄1 = � 1
2�3
Zd
3r12(e
��u12 � 1)
= � 1
2�3(
Z 1
a
4⇡r2dr(e�U0r60r
�6
� 1)�Z
a
04⇡r2dr)
=2⇡
3�3T
✓a
3 � U0r60
kTa
3
◆(25)
2 Statistical Mechanics of a Lattice Gas
1 The grand partition function
Z =1X
N=0
z
N
Z
N
(26)
where Z
N
is the canonical partition function. Consider the M ⇥M lattice by summing over theposition (n,m) for each particle 1 · · ·N of the gas and by summing over the pairs to account forall the configurations, where our Boltzmanm term is the usual
U
ij
=
8><
>:
1, i=j
�U0, i and j are nearest neighbours
0, else
(27)
The summation can be normalized by N ! and by a
3N ,
e
��U(rij) ⇡ 1 + f
ij
(28)
4
Therefore
Q
N
=X
r1
X
r2
X
r3
e
��
Ppairs U(rij)
=X
r1
X
r2
· · ·X
rN
(1 + (f12 + f13 + · · · ) + (f12f13 + · · · ) + · · · ) (29)
Each term can be represented by a graph with N points in the following way
= (X
1,2
f12)(X
2,3,4
f23f34) · · · (30)
whereP0
(1,2,··· ) is the sum over all configurations which link particles 1, 2, · · · with the restrictionthat no two particles are on the same site.
Each graph can be written as a product of connected graphs, i.e. in the form
NY
l=1
(l!M2b
l
)al (31)
where a
l
is the number of l-connected subgraphs and b
l
is defined as
b
l
=1
M
2l!{sum of all possible l-connected graphs} (32)
With these definitions, the configurational sum in the grand partition function can be writtenas
0X
al
X
PN
NY
l=1
(l!M2b
l
)al (33)
whereP
PNis the sum over all permutations of the N particle labels which lead to the same
graph.Putting this back in the the grand partition function
Z =1X
N=0
z
N
X
{al}
NX
l=1
(M2b
l
)al
a
l
!
=1X
a1=0
1X
a2=0
· · ·1Y
l=1
1
a
l
!(M2
b
l
z
l)al
=1Y
l=1
1X
al=0
1
a
l
!(M2
b
l
z
l)al
= exp
1X
l=1
M
2b
l
z
l
!(34)
2)
F = �kT lnZ = �kT
Xz
l
b
l
l!
= �kT
✓b1z +
b2
2z
2 +b3
6z
3 +b4
24z
4 + · · ·◆
(35)
5
Figure 1: The calculation of b3 and b4
b1 = 1
b2 =1
2M2
X
(1,2)
f12 = 2(e�U0 � 1)
b3 =1
6M2(3X
(1,2,3)
f12f23 +X
(1,2,3)
f12f23f13)
= 2(e�U0 � 1)2
b4 =3M2
24M218(e��U0 � 1) =
9
4(e��U0 � 1) (36)
3) As before it is true that lnZ = PL
2
kT
and n = @(lnZ)@ ln z
P =kT
a
20
1X
l=1
b
l
z
l (37)
The specific volume
1
v
=1
a
20
1X
l=1
lb
l
z
l (38)
Therefore
PL
2
kT
=1X
l=1
a
l
(T )
✓a
20
L
2
◆l�1
(39)
This is similar to the interacting gas example, so
a1 = b1 = 1
a2 = �b2 = �2(e��U0 � 1)
a3 = 4b22 � 2b3 = 16(e��U0 � 1)2 � 4(e��U0 � 1)
a4 = �20b32 + 18b2b3 � 3b4
= �160(e��U0 � 1)3 + 72(e��U0 � 1)2 � 27
4(e��U0 � 1) (40)
So ↵ = Na
20
This was determined by taking the lattice spacing a0 and dividing it by the specific volume.It does not depend on the de Broglie wavelength because this system is not governed by quantummechanics and no kinetic energy considered with the translational DOF.
6