PHYS430 STAT. THERMODYNAMICS
DEBYE MODEL FOR HEAT CAPACITY IN
SOLIDS
Eren ÇANGA
1274927
Submitted to: Doç. Dr. Altuğ ÖZPİNECİ
DEBYE MODEL FOR HEAT CAPACITY IN SOLIDS
1. INTRODUCTION
The amount of energy required to raise the temperature of one kilogram of the substance
by one kelvin. The SI unit for specific heat capacity is the joule per kilogram kelvin, J·kg1
·K-1.By heat capacity, it is often referred that heat capacity at constant volume, which is
more fundamental than the heat capacity at constant pressure.The heat capacity at
constant volume is defined as
Eqn. 1
Cv= T  (S / T )V = ( U / T )V
Where S is the entropy, U is the energy, and T is temperature.
The experimental facts about the heat capacity of solids are these:
1. In room temperature range the value of the heat capacity of nearly all monoatomic
solids is close to 3Nk, or 25 J mol-1 deg -1.
2. At lower temperatures the heat capacity drops rapidly and approaches zero as T3
in insulators and as T in metals.If metal becomes semiconductor, the drop is faster
than T. [1]
Figure 1.Heat capacity vs. temperature of Mg2SiO4, Mg (diamonds), Si (circles), O
(triangles). [2]
The Debye model is developed by Peter Debye in 1912.He estimated the phonon
contribution to the heat capacity in solids. The Debye model treats the vibration of the
lattice as phonons in a box, in contrast to Einstein model, which treats the solid as noninteracting harmonic oscillators. The Debye model predicts the low temperature depence
of heat capacity T3 that confirms the experimental results. Moreover, it covers the high
temperature limits like the Einstein model. [2]
2. EINSTEIN MODEL
The average energy of an oscillator of frequency  , is n  . For N oscillators in one
dimension, all having same frequency, the thermal energy is [1]
Eqn. 2
U  N  n  
N 
e
1
 / 
Where   kT , k boltzman constant and n is the thermal avarage of the number of
phonons in an elastic wave of given frequency.
Then the heat capacity of oscillator is
Eqn. 3
  2  e  / 
 U 
  / 
CV  
  Nk 
 1) 2
 T V
   (e

  kT and e kT  1   kT
(a) At high temperatures
Eqn. 4
U
3 N

 3 NkT  3RT
kT
independent of . Also C = dU/dT = 3R, in agreement with experiment.

(b) At low temperatures
Eqn 5.
  kT and e kT  1
U  3 Ne


kT
And C can be found as follows,
    kT
C  3 Nk 
 e
 kT 
2
Eqn. 6

   E
 3R E  e T
T
2
Therefore, it can be seen that Einstein model succesfully predicts that C decreases with
decreasing T. However, exponential decrease is not observed; if low frequencies are
present, then  will be small, much smaller than kT even at low temperatures; C will
remain at 3kT to much lower frequencies and the fall off is not as dramatic as predicted
by the Einstein model.
3.DEBYE MODEL
Debye model uses wide spectrum of frequencies to describe the complicated pattern of
lattice vibrations. It is assumed that hypothetical oscillators generate simple sine waves
throughout the crystal and these will displace the atoms away from their equilibrium
positions by an amount equal to the amplitude of the sine wave at that point. If whole set
of such oscillators generates sine waves of certain frequencies and amplitudes, then it can
be predicted that the superposition of such waves will simulate the complicated pattern of
the actual atomic vibrations. The distribution of oscillators is quasi-continuous in  hence
integration can be used instead of summation. [1]
Eqn. 7 E ( )    k (e
k
kT
 1)
k
U   N ( ) E ( )d
In order to go further there are two problems to solve:
1. Density of states function is required
2. Need to set the range of frequencies over which the integration is to be performed,
i.e. the cut-off or limiting frequency needs to be determined
Density of States
For standing waves, and considering a cube of material of side L, the appropriate
boundary condition for vibrational waves reflected from mechanically free surfaces is
that an antinode of the vibration amplitude should exist at each surface. This corresponds
to there being an integral number of half-wavelengths of the standing wave along the
length of the cube[3]. The allowed values of the standing wave vectors are given by
Eqn. 8

k si  n i   ,
 L
i  x, y, z
n i  1, 2, 3, .......
0
2
u x, k
1
u x, k
1
3
2
=2L, k=2(/L), n=2
3
2
4
u x, k
3
u x, k
 = 2L, k=/L, n=1
0
u x, k
2
u x, k
L
6
6
3
=2/3 L, k=3 /L, n=3
6
8
0
0.5
1
x

Figure 1
Schematic 1D illustration of a standing wave set up between the free
surfaces of a cube of an elastic continuum with antinodes at the free surfaces.[3]
Each allowed standing-wave solution of the wave equation consistent with the
boundary conditions is represented by a point in the reciprocal space containing the kvectors. The spacing between allowed k-values is ks=/L, and so the volume of k-space
corresponding to the one k-value (standing-wave state) is
Eqn. 9

V   .
 L
The number of k-values contained in a unit volume of k-space is
3
s
k
Eqn. 10

s
k

V

3
where V=L3 is the sample volume. The density of k states is uniform in k space and
depends only on the sample size. For large samples, k can be taken as a continuous
variable rather than as a discrete quantity.
Eqn. 11
1
V
Vk 2 dk
2
g i (k)dk = 4 k dk 3 =
.
8

2 2
The density of states for a mode i in terms of frequency, gi(), may be obtained using the
linear dispersion relation valid for long-wavelength acoustic modes, giving[3]
Eqn. 12
g i ( )d =
V 2
d ,
2 2 v 3i
where vi is an appropriate sound velocity for mode i. However, three acoustic modes, one
longitudinal (LA) mode and two degenerate transverse (TA) modes, can propagate in
continuous elastic media, and so the total vibrational density of states is given by
Eqn. 13
3V 2
g( )d =
d
2 2 v 3o
where vo is an appropriate average of the LA- and TA-mode velocities. This quadratic
frequency dependence is the so called Debye density of vibrational states.
Eqn. 15

D
3V
2
,
3N =
2 3   d
2 v o 0
the Debye frequency is given by
Eqn. 16
 6 2 N 
D  

 V 
1
3
vo .
Now N can be found that,
Eqn. 17
N ( ) 
V
8 3
 dS  4k
dS
V
    8
k
3
vg
 dS
2
  vg k
Eqn. 18
N ( ) 
V
8 3 v g
4 k 2 
V 2
, (see Eqn. 12)
2 2 v g 3
Eqn. 19
 max
3
V  max
0 N ( )d  N  6 2 v 3g
3
  max

6 N 2 3
vg
V
Eqn. 20
U   N ( )E( )d
V 2
 2 3
2 v g
V 

2 2 v 3g
Z

e

kT
 max
1
3

0
d
e

kT
d
1

kT
; d =
dZ
kT

Eqn. 21
V kT  kT 
U 
 
2 2 v3g   
3 Z max

0
Z3
dZ
eZ  1
Eqn. 22

3
max

6 2 Nv 3g
V
Eqn. 23
4
3N  kT
U  3
max 3
Z max

0
Z3
dZ
eZ  1
Eqn. 24
D 
max
k
This is called as Debye temperature
Eqn. 25
 T
U  3N kT  
 D 
3 Z max

0
Z3
dZ
eZ  1
x3 for three branches, i.e. 2 transverse acoustical and 1 longitudinal acoustic
Eqn. 26
 T
U  9 N kT  
 D 
3 Z max

0
Z3
dZ
eZ  1
Debye Specific Heat Capacity
Eqn. 27
V 
U
2 2 v 3g
 max
3

0
e

kT
d
1
Eqn. 28
dU
V 

dT 2 2 v 3g
- 
kT 2
V 1

2 2 v 3g
2
kT 2

3N
3
 max
 max

0
e
3
 kT

 e  1


4
 2  kT  kT
 
kT 2    
Z max
3 Z max
Z4 eZ
 e
0
Z
2
d
 1
2
eZ Z 4
 e
0
Eqn. 29
 T
CVD  3Nk 
 D 
d
e kT  4
 kT

 e  1


0
2

 max


kT
dZ
Z
 1
2
dZ
Debye specific heat capacity for a single acoustic branch. However, there are three
branches, i.e. two transverse acoustic and one longitudinal acoustic branch. The total
specific heat is therefore [3]:
Eqn. 30
3 Z
 T  max Z4e Z
CVD  9Nk   Z
2 dZ
 D  0 e  1
At low temperature
Eqn. 31
T << D
Let Zmax  

Z3
4
dZ

0 eZ  1
15
Eqn. 32
3NkT 4  T 
 
U 
5
 D 

3
3Nk 4 4
T
5D3
Eqn. 33
CVD
dU 12 4  T 

  Nk  
dT
5
 D 
3
Debye T3 approximation at low temperature as in the experimental results
At high temperature
Eqn. 34
Z0
eZ  1  Z and / or 1
Eqn. 35
CV D
 T
 9Nk  
 D 
3 Z max
 T
 9Nk  
 D 
3 Z max
 T
 9Nk  
 D 
3

0
Z4
dZ
Z2
 Z dZ
2
0
Z3max
3
 3Nk
Figure 3.Comparision between Debye and Einstein model [4]
CONCLUSION
For sufficiently low temperatures, the Debye approximation should be quite good, as here
only long wavelength acoustic modes are excited. These are just the modes that may be
treated as in elastic continuum with macroscopic elastic constants. The energy of short
wavelength modes is too high to be populated at low temperatures.
REFERENCES
[1] Charles Kittel, Solid State Physics, 1976, John Wiley & Sons Inc.
[2] http://images.google.com.tr/images?svnum=10&hl=tr&lr=&q=heat+capacity+vs+temperature
[3] Introduction to Lattice Dynamics, Martin T. Dove, Cambridge topics in mineral physics and
chemistry lecture notes
[4] http://en.wikipedia.org/wiki/Debye_model