Course 443, Problem Set, Michaelmas Term, 2005
Given the partition function for a (non-relativistic) ideal gas in the classical canonical ensemble
1 V N
.
(1)
Z=
N ! λ3
Derive expressions for the internal energy, U , the specific heat at constant
volume, cV and the entropy, S.
Internal Energy
1
ln(Z)
β
#
"
1
V N eN
√
= − ln
β
λ3 N
2π
V
1
1 √
1
= − N ln( 3 ) − N + ln( 2π)
β
λ N
β
β
F =
Using U =
∂
∂(
1
T
)
F
T
from the notes:
∂
(βF )
∂β
√ ∂
V
=
−N ln 3
− N + ln 2π
∂β
λN
U =

2πmkT
∂ 
=
−N ln
∂β
h2
3 1
N
2 β
3
=
N kT
2
=
Specific Heat Capacity
Then cV =
Entropy
∂U
∂T
= 23 N k.
1
!3/2 

∂F
∂T "
#
V
1 √
∂
1
1
= −
− N ln 3
− N + ln 2π
∂T
β
λ N
β
β
S = −
= −

2πmkT
1
∂ 1
N ln(V ) − ln
∂T β
β
h2
!3N/2
−
√

1
1
1
N ln(1/N ) − N + ln( 2π)
β
β
β
√
3N kT
3N k
3
2πmk
+
N
kln(1/N
)
+
kN
−
kln
+
ln
= N kln(V ) + N kln(T ) +
2π
2
2T
2
h2
!
3
3
1 √
2πmkT
= N kln(V ) + N kln
+ N k − N k(ln(N ) − 1 + ln 2π)
2
2
h
2
N
2
!
Sketch the phase diagram (external magnetic field as a function of temperature) for the 1D Ising model.
Sketch the isotherms (lines of constant temperature) on a plot of of mean
magnetisation as a function of external field for 3 cases: T < Tc , T = Tc , T >
Tc .
Draw a sketch of the 2D Ising model and write down its Hamiltonian in
the presence of an external field and with nearest neighbour spin interactions.
external magnetic field as a function of temperature
h
spin up
spin down
Tc
T
Figure 1: Phase diagram: magnetic field strengh as a function of temperature
in the ID Ising model
magnetisation as a function of external field
Mean magnetic moment is
M =−
∂F
sinh(βB)
=q
.
∂B
sinh2 (βB) + e−4βg
3
1
g/kT = 0.1
g/kT = 1.0
g/kT = 2.0
magnetisation
0.5
0
-0.5
-1
-2
-1
0
magnetic field
1
2
Figure 2: Magnetisation of the 1D Ising model
Note that the values of βg in the plot correspond to isotherms. At finite
temperature T > 0, the magnetisation is always zero for zero magnetic field.
A non-zero magnetisation appears only in reponse to an applied magnetic
field.
At high temperatures the Curie law is recovered, χ ∼ 1/T while at low
temperatures there is an exponential divergence as zero temperature is approached. Recall, χ = ∂M
.
∂B
2D Ising
This was subsequently covered in notes (see section on Markov methods)
4
Consider the cluster expansion of the partition function in the canonical
ensemble
1
QN
(2)
Z=
N !λ3N
with
Z
Y
(3)
QN = d3 x (1 + fij ).
i<j
Write an expression for the contribution from Q4 . Write explicitly (in terms
of fij ) the contribution from S(1, 0, 1, 0). Calculate the symmetry factor for
this term and show that it agrees with that derived from an explicit writing
of the term.
Q4 =
Z
d 3 x1
Z
d 3 x2
Z
d 3 x3
Z
d3 x4 (1+f12 )(1+f13 )(1+f14 )(1+f23 )(1+f24 )(1+f34 )
(4)
The term S(1, 0, 1, 0) is given by
S(1, 0, 1, 0) =
Z
d 3 x1
Z
d 3 x2
Z
d 3 x3
Z
d3 x4 [f12 f13 + f12 f14 + f12 f23 + f12 f34 + f13 f14
+f13 f23 + f13 f34 + f14 f24 + f14 f34 + f23 f24 + f23 f34 + f24 f34
+f12 f23 f13 + f12 f14 f24 + f23 f34 f24 + f13 f34 f14 ]
The above can be written as the sum of four 3-clusters.
f12 f24 + f12 f14 + f14 f24 + f12 f14 f24
f12 f23 + f13 f12 + f13 f23 + f12 f23 f13
f23 f34 + f24 f34 + f24 f23 + f23 f34 f24
f13 f34 + f14 f13 + f14 f34 + f13 f34 f14
Each of these combinations gives the same contribution to S(1, 0, 1, 0)
since the labels represent integration variables only. So, 4 3-clusters contribute equally. Calculating the symmetry factor:
N!
ml
l (l!) ml !
4!
=
1
(1!) 1!(3!)1 1!
S = Q
5
4!
3!
= 4
=
as expected from the above.
6