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