Ex2215: Heat capacity of solids
Submitted by: Asaf Szulc
The problem:
Consider a piece of solid whose low laying excitations are bosonic modes that have spectral density
g(ω) = Cω α−1 up to a cutoff frequency ωc , as in the well-known Debye model (items 1-5). Similar
description applies for magnetic materials (item 6). In items 7-8 assume that the solid is a ”glass”,
whose low laying excitations are like two level entities that have a spectral density g(ω).
(1) Write a general expression for the energy E(T ) of the system. This expression may involve a
numerical prefactor that is defined by an α dependent definite integral.
(2) Write a general expression for the heat capacity C(T ).
(3) Write a general expression for the variance Var(r) of an atom that reside inside the solid.
(4) Determine what are α and C and ωc for a piece of solid that consists of N atoms that occupy a
volume Ld in d = 1, 2, 3 dimensions, assuming a dispersion relation ω = c|k|, as for ”phonons”.
(5) Write explicitly what are C(T ) and Var(r) for d = 1, 2, 3. Be careful with the evaluation
of Var(r). In all cases consider both low temperatures (T ωc ), and high temperatures
(T ωc ).
(6) Point out what would be α if the low laying excitations had a dispersion relation ω = a|k|2
as for ”magnons”.
(7) What is the heat capacity of a ”glass” whose two level entities have excitation energies ω = ∆,
where ∆ has a uniform distribution with density C.
(8) What is the heat capacity of a ”glass” whose two level entities have excitation energies ω =
ωc exp(−∆), where the barrier ∆ > 0 has a uniform distribution with density D.
The solution:
(1) Taking the bosonic occupation function with µ = 0 and the spectral density g(ω) = Cω α−1 ,
then integrating up to a cutoff frequency ωc yields:
Z ∞
Z ωc
Z ωc
1
ω α dω
E=
g()df () =
g(ω)ωdω ω/T
=C
(1)
e
−1
eω/T − 1
0
0
0
(2) The heat capacity is the derivative of the energy with respect to the temperature:
dE
C(T ) =
=C
dT
Z
ωc
0
ω ω 2
eω/T
c
α−1
α
ω
dω
≡
CT
F
T
(eω/T − 1)2 T
(2)
(3) The Hamiltonian of our problem is:
N
X
X
p2i
H=
+
Kij (xi − xj )2
2m
i=1
7−→
H=
{ij}(nn)
X P2 X 1
k
+
mωk2 Q2k
2m
2
k
1
k
(3)
where we were lucky enough that the basis in which the Hamiltonian is diagonalized is the Discrete
Fourier Transform of xi , pi given by:
1 X ikl
Qk = √
e xl ;
N l
1 X −ikl
Pk = √
e
pl
N l
(4)
The Hamiltonian was written for a 1D case, however, the generalization to other dimensions is
trivial. Now, as we expressed the position as a sum of independent random variables, we can write:
Z
X 3ωf (ω)
X
3C ωc ω α−2 dω
V ar(Qk ) =
=
(5)
V ar(r) =
mω 2
m 2πc eω/T − 1
L
k
k
where we implemented the equipartition theorem for low temperatures (ωf (ω)/2 for each degree of
freedom), then we used the given density of states for the transformation to an integral and defined
minimum and maximum cutoff frequencies.
(4) Using the connection, E = ω = c|k|, it is convenient to calculate the density of states in k-space.
States with equal energy lie on a d-dimensional spherical shell in k-space, divided by the volume of
a unit cell gives the density of states:
d Ωd k d−1 dk
Ld d
g(k)dk =
=⇒
g(ω)dω
=
Ωd ω d−1 dω
(6)
π d
2d
(2πc)d
L
where Ωd = {2, 2π, 4π} for d = {1, 2, 3} and the pre-factor is due to the consideration of ’d’ different
modes of the lattice (one for each degree of freedom) and only positive values of ’k’.
And now to business, first we will plug eqn.(6) in the number of states equation. For a d-dimensional
crystal with N atoms there are N d normal modes so we can write:
Z ωc
Z
Ωd d d ωc d−1
g(ω)dω =
ω dω
(7)
Nd =
L
(2πc)d
0
0
second, in the equation for the heat capacity to get:
Z ωc
Z
ω 2
d
eω/T
1
Ωd d d ωc
C(T ) =
g(ω)ωdω ω/T
=
L
ω d−1 dω
d
ω/T − 1)2 T
dT
(2πc)
e
−
1
(e
0
0
(8)
Solving eqn.(7) and comparing eqn.(8) to eqn.(2) leads to the final result:
2πc
ωc =
L
Nd
Ωd
1/d
;
C=
Ωd d d N d2
L = d ;
(2πc)d
ωc
α=d
(9)
(5) In this section, throughout our entire calculation for the low temperature limit, T ωc , we
take xc = ωc /T → ∞ and integrating F (xc ) by parts. For the high temperature limit, T ωc , we
take x → 0 ⇒ ex ≈ 1 + x and we keep only the lowest order in x. Thus we get:
For 1D (α = 1, Ωα = 2):
NT
C(T ) =
F (xc ) =
ωc
3L
V ar(r) =
mπc
Z
ωc
2πc
L
π LT
3 c
N
dω
ω(eω/T − 1)
T ωc
T ωc
=
(10)
0
3
2π 2
T
mc2
2
L2
T ωc
T ωc
(11)
For 2D (α = 2, Ωα = 2π):
4N
C(T ) = 2 F (xc ) =
xc
3L2
V ar(r) =
mπc2
Z
ωc
2πc
L
7.212 ·
2N
L2 T 2
πc2
dω
=
ω/T
(e
− 1)
T ωc
T ωc
(12)
T ωc
T ωc
0
3ln(N )
2π
T
mc2
L2
(13)
For 3D (α = 3, Ωα = 4π):
C(T ) =
9N
F (xc ) =
x3c
9L3
V ar(r) =
2mπ 2 c3
Z
2π 2 L3 3
T
5c3
3N
ωc
2πc
L
(Debye T 3 law)
(Dulong P etit law)
T ωc
T ωc
ωdω
=
ω/T
(e
− 1)
(
T ωc
0
9
π
31
3N
4π
T
mc2
L2
T ωc
(14)
(15)
(6) The calculation of the new density of states is similar to the previous problem. Therefore, we
continue by plugging the new dispersion relation ω = a|k|2 into eqn.(6) and get:
g(ω)dω = Ωd Ld
md ω d/2−1 dω
2ma
(2π)d a
g(ω) = Cω d/2−1 ≡ Cω α−1
=⇒
(16)
hence α = d/2.
(7) The partition function for a two level system is:
Z(β) = 1 + e−βω
(17)
leading to the same calculation but with ”+” instead of ”-” sign. Plugging in g(∆) = C to get:
Z
C(T ) =
0
∞
g(∆)e∆/T
(e∆/T + 1)2
∆
T
2
Z
∞
d∆ = −CT
0
x2 e x
π2
dx
=
CT
(ex + 1)2
6
(18)
where we defined x = −∆/T and integrated by parts.
(8) Here we use the given connection, ω = ωc e−∆ and then introduce another change of variables,
x = −ωc e−∆ /T , to get:
(
2
ωc −∆
Z ∞
Z −xc
ln(2)D
T ωc
g(∆)e− T e
ωc e−∆
Dxex dx
2
d∆ =
=
(19)
C(T ) =
ωc −∆
ω
x
2
c
−
e
T
(e + 1)
D
T ωc
+ 1)2
(e T
0
0
8T 2
3