1. Vibrational & Thermal Properties of Crystal Lattices
Heat Capacity—The Einstein Model
The Debye Model — Introduction
A Continuous Elastic Solid
Counting Modes and Finding N()
The Debye Model — Calculation

2. A. Heat Capacity—Einstein Model (1907)
Properties of solids that come from collective vibrations of the atoms about their
equilibrium positions.
A. Heat Capacity—Einstein Model (1907)
Model for a vibrating atom: m ky kx kz
Classical Statistical Mechanics
The Equipartition theorem:
In thermal equilibrium each quadratic term in the E has an average energy
so:

3. Classical Heat Capacity
For a solid with N atomic oscillators:
Total energy per mole:
Heat capacity at constant volume per mole:
This Law of Dulong and Petit (1819) is approximately obeyed by most solids at high T ( > 300 K). But by the middle of the 19th century it was clear that
CV  0 as T  0 for solids.
So…what was happening?

4. Einstein Uses Planck’s Work
Planck (1900): vibrating oscillators (atoms) in a solid have quantized energies
[later QM showed is actually correct]
Einstein (1907): Model solid as collection of 3N independent 1-D oscillators, all with same , & use Planck equation for energy levels
Occupation of energy level n:
(probability of oscillator being in level n):
Classical Boltzmann factor)
Thermal Average Total Energy
of The Solid:

5. Some “Nifty” Summing!! So finally: Now let Using Planck’s equation:
This can be rewritten:
To give:
Now use the infinite sum:
So finally:

6. The Heat Capacity! So finally: Using the definition: Differentiating:
Now define an “Einstein temperature”:
So finally:

7. Limiting Behavior of CV(T)
High T limit:
Low T limit: 3R CV
T/E
These predictions are qualitatively correct: CV  3R for large T and CV  0 as T  0:

8. High T behavior: Reasonable agreement with experiment
Now, Take a Closer Look:
High T behavior: Reasonable agreement with experiment
Low T behavior:
CV  0 too quickly as T  0 !

9. B. The Debye Model (1912)
Despite its success in reproducing the approach of CV  0 as T  0, the Einstein model is clearly deficient at very low T. What might be wrong with the assumptions it makes?
• 3N independent oscillators, with the same frequency 
• Discrete allowed energies:

10. Details of the Debye Model
Pieter Debye succeeded Einstein as professor of physics in Zürich, and soon developed a more sophisticated (but still approximate) treatment of atomic vibrations in solids.
Debye’s model of a solid:
3N normal modes (patterns) of oscillations
• Spectrum of frequencies from  = 0 to max
• Treat the solid as a continuous elastic medium (ignore
details of the crystal structure)
This changes the expression for CV because each mode of oscillation contributes a frequency-dependent heat capacity and we now have to integrate over all :
Einstein function for one oscillator
# of oscillators per unit 

11. C. The Continuous Elastic Solid
We can describe a propagating vibration of amplitude u along a rod of material with Young’s modulus E and density  with the wave equation:
for wave propagation along the x-direction
By comparison to the general form of the 1-D wave equation:
we find that
So the wave speed is independent of wavelength for an elastic medium!
is called the dispersion relation of the solid, and here it is linear (no dispersion!)
group velocity

12. D. Counting Modes and Finding N()
A vibrational mode is a vibration of a given wave vector (and thus ), frequency , and energy How many modes are found in the interval between and ?
# modes
We will first find N(k) by examining allowed values of k. Then we will be able to calculate N() and evaluate CV in the Debye model.
First step: simplify problem by using periodic boundary conditions for the linear chain of atoms:
We assume atoms s and s+N have the same displacement—the lattice has periodic behavior, where N is very large. s
s+N-1
s+1
s+2
x = sa
x = (s+N)a
L = Na

13. First: finding N(k)
Since atoms s and s+N have the same displacement, we can write:
This sets a condition on allowed k values:
independent of k, so the density of modes in k-space is uniform
So the separation between allowed solutions (k values) is:
Thus, in 1-D:

14. Next: finding N()
Now for a 3-D lattice we can apply periodic boundary conditions to a sample of N1 x N2 x N3 atoms:
N1a
N2b
N3c
Now we know from before that we can write the differential # of modes as:
We carry out the integration in k-space by using a “volume” element made up of a constant  surface with thickness dk:

15. N() at last!
Rewriting the differential number of modes in an interval:
We get the result:
A very similar result holds for N(E) using constant energy surfaces for the density of electron states in a periodic lattice!
This equation gives the prescription for calculating the density of modes N() if we know the dispersion relation (k).
We can now set up the Debye’s calculation of the heat capacity of a solid.

16. F. The Debye Model Calculation
We know that we need to evaluate an upper limit for the heat capacity integral:
If the dispersion relation is known, the upper limit will be the maximum  value. But Debye made several simple assumptions, consistent with a uniform, isotropic, elastic solid:
• 3 independent polarizations (L, T1, T2) with equal propagation speeds vg
• continuous, elastic solid:  = vgk
• max given by the value that gives the correct number of modes per polarization (N)

17. N() in the Debye Model First we can evaluate the density of modes:
Since the solid is isotropic, all directions in k-space are the same, so the constant  surface is a sphere of radius k, and the integral reduces to:
Giving:
for one polarization
Next we need to find the upper limit for the integral over the allowed range of frequencies.

18. max in the Debye Model
Since there are N atoms in the solid, there are N unique modes of vibration for each polarization. This requires:
Giving:
The Debye cutoff frequency
Now the pieces are in place to evaluate the heat capacity using the Debye model! This is the subject of problem 5.2 in Myers’ book. Remember that there are three polarizations, so you should add a factor of 3 in the expression for CV. If you follow the instructions in the problem, you should obtain:
And you should evaluate this expression in the limits of low T (T << D) and high T (T >> D).

19. Debye Model: Theory vs. Expt.
Better agreement than Einstein model at low T
Universal behavior for all solids!
Debye temperature is related to “stiffness” of solid, as expected

20. Debye Model at low T: Theory vs. Expt.
Quite impressive agreement with predicted CV  T3 dependence for Ar! (noble gas solid)
(See SSS program debye to make a similar comparison for Al, Cu and Pb)

21. Debye Model Heat Capacity
・Low temperature T<<θ
・High temperature T>>θ
This is consistent with the Equipartition Theorem:
The average energy per freedom is (½)kBT

22. Heat capacity CV in the Debye Approximation or the Debye model
kB=1.38x10-23JK-1
kBmol=7.70JK-1
3kBmol=23.1JK-1

23. Heat capacity of Si, Ge & solid Ar
in the Debye model
Solid Ar
Si and Ge
1 cal/mol K=4.185J/mol K
3kB mol=5.52cal K-1

24. Heat capacity of Diamond
and it is even worse for diamond which is an insulator.
this does indeed go to zero at zero temperature it seems
double log scale: T increases one order of magnitude, C increases three orders of magnitude
maybe on blackboard: log(C)=log(const*T^3)=log(const)+log(T^3)=log(const)+3*log(T)
So: any microscopic theory should at least give two things: DP at high T, zero at low T, possibly also T^3 law