# Final solutions ```Final solutions
Problem 1
(a) We can write
Si Sj = −m2 + m(Si + Sj ) + δSi δSj
The last term is due to fluctuations. We then have the mean field Hamiltonian
X
1
Si
HM F = N zJm2 − (zJm + B)
2
i
(b) The partition function is
N

Z = e−βF = e
− 21 βN zJm2
X

eβ(zJm+B)S 
S=−1,0,+1
from which we get
F =
1
zJm + B
N zJm2 − N kB T ln 1 + 2 cosh
.
2
kB T
In terms of dimensionless quantities we get
m+h
1 2
f = m − θ ln 1 + 2 cosh
.
2
θ
Problem 2
The first condition gives
∂P
∂v
=−
T
RT
3a
+ 4 =0
2
(v − b)
v
which is equivalent to
RT = 3a
(v − b)2
.
v4
The second condition gives
2 2RT
12a
∂ P
=
− 5 =0
∂v 2 T
(v − b)3
v
1
or, equivalently
(v − b)3
.
v5
Equating the two equations for RT we get vc = 2b. Substituting this result in
anyone of the equations for RT we obtain
RT = 6a
Tc =
3a
.
16b2 R
Substituting vc and Tc in the equation of state we get
a
.
16b3
Pc =
Finally
RTc
3
=
Pc v c
2
which is a very low number.
Problem 3
As explained in class, we can assume that the chemical potential is zero in the
region of Bose-Einstein condensation. The number of particles in the excited
state is thus
Z ∞
N̄ g()d,
Nexc =
0
where
N̄ =
and
g()d =
1
e/kB T
−1
,
V 4πp2 dp
V 4π 2/s 1 1/s−1
= 3
d
3
h
h
A
s A
so that
Z
Nexc = const &times; V
0
∞
3/s−1
d,
e/kB T − 1
If we set x = /kB T , we get
Nexc = const0 &times; V (kB T )3/s .
Therefore, since TB is determined by the condition Nexc = N we find that
TB ∝
N
V
s/3
T
TB
.
The fraction
Ngs
=1−
N
2
3/s
.
Finally, we find that
Z
∞
N̄ g()d ∝ T 3/s+1 ,
U=
0
from which we find
CV ∝ T 3/s ,
and
Z
S=
0
T
CV dT
∝ T 3/s .
T
3
```