MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
8.044 Statistical Physics I
Spring Term 2013
Final Exam, Solutions
Problem 1 (20 points) Binary Alloy
a) Find the number of different ways of choosing the n α-sites to be vacated and occupied
by β atoms.
N!
#α =
n!(N − n)!
b) Find the number of different ways of choosing the n β-sites from which to take the β
atoms.
N!
#β =
n!(N − n)!
c) Find the entropy of the system as a function of n.
S(n) = kB ln Ω = kB ln(#α × #β ) = 2kB ln
N!
n!(N − n)!
d) Find U (T, N ).
S(n) = 2kB [N ln N − n ln n − (N − n) ln(N − n) − N + n + (N − n)]
1
=
T
∂S
∂U
=
N
∂S
∂n
∂n
2kB
=
[−1 − ln n + 1 + ln(N − n)]
∂U N
N
| {z }
1/
= − ln
2kB T
n
N −n
→
n
= e−/2kB T
N −n
Solving for n gives
n=
N
1 + e/2kB T
→
1
U (T, N ) =
N
1 + e/2kB T
Problem 2 (20 points) DNA Model
a) This is a classical system with N non-degenerate states with energies En = n.
Z1 =
N
X
e−n/kB T
n=0
b) When kT one need consider only the lowest two energy states; this becomes an
energy gap dominated situation.
N
X
hni =
n
n=0
e−n/kB T
Z
0 e−0 + 1 e−/kB T
e−/kB T
=
e−0 + e−/kB T
1 + e−/kB T
≈
≈ e−/kB T
c) When kB T one can approximate sums over n by integrals. For example
Z=
N
X
e
−n/kB T
=
n=0
kB T
X
N
−n/kB T
e
kB T
n=0
kB T
≈
∞
Z
|0
e−x dx =
{z }
1
d) In a similar manner
hni =
N
X
n
n=0
e−n/kB T
Z
N
N
X
n −n/kB T
kB T X n −n/kB T
=
e
=
e
k
T
k
T
kB T
B
B
n=0
n=0
Z
kB T ∞ x
≈
xe dx
0
| {z }
1
≈
kB T
Alternatively use Z1 = 1/(β).
<U >
1
hni =
=
−1 ∂Z
Z ∂β
2
1
=
−1
Z
−Z
β
=
kB T
kB T
Problem 3 (20 points) Spin Waves
a)
~ = Lx Ly Lz = V
D(k)
2π 2π 2π
(2π)3
b)
~
#(ω) = (volume of sphere in k-space) × D(k)
=
=
D(ω) =
D(ε)
4 3
V
πk (ω)
use k = (ω/a)1/2
3
3
(2π)
V ω 3/2
6π 2
a
d#(ω)
V
=
a−3/2 ω 1/2
2
dω
(2π)
c)
0
ε
∞
Z
h(ω)i D(ω) dω
U =
0
Z ∞
~ω
V
−3/2
ω 1/2 dω + Z.P. contribution
=
a
~ω/k
2
B T − 1)
(e
(2π)
0
3/2
Z ∞ 3/2
x
V
1
5/2
=
(k
T
)
dx + Z.P. contribution
B
(2π)2 ~a
ex − 1
0
|
{z
}
≡ I
CV (T, V ) =
∂U
∂T
V
5
= 2 kB V
8π
kB T
~a
3/2
I
d) There is no energy gap behavior (CV ∝ T n e−∆/kB T ) because of the integration over a
continuous distribution of gaps (∆ = ~ω), some of which are less than kB T for any physical
T.
3
Problem 4 (20 points) Graphene
a)
~ = Lx Ly = A
D(k)
2π 2π
(2π)2
b)
~
#() = 2 × (area of disk in k-space) × D(k)
= 2 × πk 2 ()
A
=
2π
1
~v
A
use k =
2
(2π)
~v
2
d#()
A
Dc () =
=
d
π
2
1
~v
2
c)
D(ε)
0
ε
d)
µ(T = 0) rests at the last filled state at T = 0 which is at the top of the valence band, so
µ(T = 0) = 0 .
D() is symmetric about = 0. If µ stays at = 0 the symmetry of hn(, T )i assures that
as T increases the number of electrons lost from the valence band is exactly equal to the
number of electrons appearing in the conduction band. Thus µ(T ) = 0 for all T covered by
this model.
4
e)
∞
Z
hn(, T )i D() d
U =
−∞
Z
∞
hn(, T )i Dc () d
= 2
0
A
=
π
A
=
π
A
=
π
1
~v
2 Z
1
~v
2
1
~v
2
∞
1
(e/kB T
0
3
Z
(kB T )
|0
∞
+ 1)
2 d
x2
dx
(ex + 1)
{z
}
≡ I
(kB T )3 I
f)
CA (T ) = (∂U/∂T )A so CA (T ) will be proportional to T 2 , that is the temperature exponent n = 2.
5
Problem 5 (20 points) BEC
a)
Z
∞
N =
< n > D() d
0
Z
∞
=
0
V
=
4π 2
=
V
4π 2
n =
V
4π 2
1
e(−µ)/kB T
2m
~2
"
V
− 1 (2π)2
3/2 Z
∞
√
2m
~2
3/2
√
#
d
d
e(−µ)/kB T − 1
3/2 Z ∞
√
2mkB T
x
dx
2
(x
−
y)
~
e
−1
0
|
{z
}
≡ I(y)
3/2
2mkB T
I(y)
~2
0
b) Bose-Einstein condensation begins when the above condition is satisfied with µ = 0 which
also means our dimensionless parameter y = 0.
3/2
V
2mkB Tc
nc = 2
I(y = 0)
4π
~2
c)
dU = T dS − P dV + µdN ; change independent variables to T, V, N :
∂S ∂S ∂S dS =
dT +
dV +
dN
∂T T,V
∂V T,N
∂N T,V
So ∂U/∂N |V,T = µ + T ∂S/∂N |T,V .
Now use Maxwell relation derivable from dF = ... on the information sheet: ∂S/∂N |T,V =
−∂µ/∂T |T,N . so
∂U ∂ µ ∂(βµ) 2 ∂ µ
=µ−T
= −T
=
∂ N V,T
∂T N,V
∂T T N,V
∂β N,V
In the Bose condensed phase µ = 0 and is independent of the temperature, so both terms in
∂U/∂N are zero.
6
d)
From answer to part a)
1
n= 2
4π
2mk
~2
3/2
T
3/2
Z
∞
0
√
x dx
e(x−y) − 1
N
C × β 3/2 =
V
Z
0
∞
√
x dx
e(x−y) − 1
where C is a collection of constants.
Differentiate this equation implicitly w.r.t. β and y = µβ,
Z ∞
√
3 N 1/2
e(x−y)
dy
xdx (x−y)
× β =
2
2 V
(e
− 1) dβ
0
Both the term on the left and the term in () are positive definite. Thus dy/dβ > 0.
7
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8.044 Statistical Physics I
Spring 2013
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