1. VIBRATIONS OF ONE DIMENSIONALDIATOMIC LATTICE

2. Consider a Diatomic Chain of Two Different Atom Types with nearest-neighbor, Hooke’s Law type forces (F = - kx) between the atoms. This is equivalent to a force-spring model with two different types of atoms of masses, M & m connected by identical springs of spring constant K.
(n-2) (n-1) (n) (n+1) (n+2) K K K K M a) M M m m a b)
Un-1 Un
Un+1
Un+2
Un-2
This is the simplest possible model of a diatomic crystal. a is the repeat distance, so, the nearest-neighbor separation is (½)a 2

3. Equation of Motion for M Equation of Motion for m
This model is complicated due to the presence of 2 different atom types, which, in general, move in opposite directions. M M M m m
Un-1 Un
Un+1
Un+2
Un-2
The GOAL is to find the dispersion relation ω(k) for this model. There are 2 atom types, with masses M & m, so there will be 2 equations of motion, one for M & one for m.
Equation of Motion
for M
Equation of Motion
for m -1 -1

4. Displacement for M Displacement for m
As before, assume harmonic (plane wave) solutions for the atomic displacements Un: M M M m m
Un-1 Un
Un+1
Un+2
Un-2
Displacement
for M
Displacement
for m
α = complex number which determines the relative amplitude & phase of
the vibrational wave. Put all of this into the 2 equations of motion.

5. Equation of Motion for the nth Atom (M)
Cancel Common Terms

6. Equation of Motion for the (n-1)th Atom (m)
Cancel Common Terms

7. Combining (1) & (2) & manipulating:
The Equation for M becomes:
(1)
The Equation for m becomes:
(2)
(1) & (2) are two coupled, homogeneous, linear algebraic equations in the 2 unknowns α & ω as functions of k.
More algebra gives:
Combining (1) & (2) & manipulating:

8. Cross multiplying & manipulating with (1) & (2):
The 2 roots are:

9. Solutions to the Normal Mode Eigenvalue Problem
ω(k) for the Diatomic Chain л / a 2 –л k w A B C
ω+ = “Optic” Modes
ω- = “Acoustic” Modes
There are two solutions for ω2 for each wavenumber k. That is
there are 2 branches to the “Phonon Dispersion Relation” for
each k. From an analysis of the displacements, it can be shown that, at
point A, the two atoms are oscillating 180º out of phase, with their
center of mass at rest. Also, at point B, the lighter mass m is oscillating &
M is at rest, while at point C, M is oscillating & m is at rest.

10. Optic Modes (Optic Branch) Case I
The solution is: K K
Optic Modes (Optic Branch)
Case I
Near BZ Center (qa << 1)
The Optic Mode becomes:
(ω+)2  2K(M1 + M2)/(M1M2)
or ω+  constant
The Acoustic Mode becomes:
(ω-)2  (½) Kq2/(M1 + M2)
or ω-  vsq
vs  sound velocity in the
crystal. Just like an acoustic
wave in air!
Gap qa
Acoustic Modes
(Acoustic Branch)

11. The Diatomic Chain Solution:K K
( = 2a)
Case II
Near BZ edge [q =  (π/a)]
(Assuming M1 > M2)
The Optic Mode becomes:
(ω+)2  2K/M2
The Acoustic Mode becomes:
(ω-)2  2K/M1
So, at the BZ edge, the vibrations of
wavelength  = 2a for the 2 modes
behave as if there were 2 uncoupled
masses M1 & M2, vibrating
independently with identical springs
of constant K.
Optic Modes (Optic Branch)
Gap
Acoustic Modes
(Acoustic Branch)

12. Phonon The linear atom chain can only have
N discrete K  w is also discrete
The energy of a lattice vibration mode at
frequency w was found to be
where ħw can be thought as the energy of a
particle called phonon, as an analogue to photon
n can be thought as the total number of phonons with a frequency w, and follows the Bose-Einstein statistics:
Equilibrium distribution

13. INELASTIC SCATTERING OF PHOTONS BY PHONONS
When a photon interact with phonons in a crystal ,it gets scattered.
Let photon of frequency ν = ω/2π ,
Then wave vector k is given by
K=2 π/ᴧ = 2πν/v
Or k = ω/ν
In case the scattering is inelastic ,the frequency of incident photon is ω and scattered photon is ώ.

14. Conservation law of energy and momentum gives ħω = ħώ + ħΩ ( 1)
ħω = ħώ + ħΩ ( 1)
ħk = ħќ + ħK (2)
Now for long wavelength ω=ώ ,k=ќ

15. If velocity of sound is V0 , Then frequencyof emitted phonon is Ω=V0K
from fig. K2 =(k-ќ).(k-ќ)
For k = ќ
K2 =
or K =
If velocity of sound is V0 ,
Then frequencyof emitted phonon is
Ω=V0K
Ω =V0

16. Classical Theory of Heat Capacity of Solids
The solid is one in which each atom is bound to its side by a harmonic force. When the solid is heated, the atoms vibrate around their sites like a set of harmonic oscillators. The average energy for a 1D oscillator is kT. Therefore, the averaga energy per atom, regarded as a 3D oscillator, is 3kT, and consequently the energy per mole is =
where N is Avagadro’s number, kB is Boltzmann constant and R is the gas constant. The differentiation wrt temperature gives;

17. Thermal Energy & Heat Capacity Einstein Model
The Einstein solid is a model of a solid based on assumptions:
Each atom in the lattice is an independent 3D quantum harmonic
oscillatorrAll atoms oscillate with the same frequency (contrast with the Debye model)
Average energy of a harmonic oscillator and hence of a lattice mode of angular frequency at temperature T
The probability of the oscillator being in this level as given by the Boltzman factor

18. According to the Binomial expansion for x«1 where
(*)
According to the Binomial expansion for x«1 where

19. Eqn (*) can be written
Finally, the
result is

20. The heat capacity:

21. Case 1: when T >> θE
This is specific heat at constant volume

22. Case 2: As discussed earlier, the increase of is out powered by the increase of As a result, when
T << θE

23. Cv vs T
Points:
Experiment
Curve:
Einstein Model
Prediction

24. Vibrational modes of a continuous medium
In 1 D crystal
In 3 D crystal wave in an arbitrary direction
As with ideal gas we can count number of all waves with wavelength greater than , G() is given by

25. THE DEBYE APPROXIMATION
Acc. to Debye , atoms of crystals are considered as harmonic oscillator .They are assumed to be coupled together ,so they are capable of propagating elastic waves whose frequency might vary over a wide range.
Let no. of modes per unit frequency range of frequency = G(ν)
Over a range ν and ν+dν = G(v)dν

26. In terms of frequency
But there are 2 transverse waves and one longitudinal
At maximum frequency, total # of waves = # degree of freedom

27. Density of states and partition function
Partition function, independent oscillators with various frequencies
Where as in the Einstein model
Vibrational part

28. Partition function and F
Replacing summation with integration
Thus the free energy

29. Energy E
Energy

30. Energy and heat capacity
With x=hv/kT and u=hmaxv/kT
And after lengthy derivation
All can be expressed in terms of the Debye function, and Debye temperature

31. Debye Temperature
High Debye temperature for solids with large atomic density (N/V) and high speed of sound (high modulus and low density)

32. High - Temperature Limit
With x=hv/kT and u=hvmax/kT
T ∞
Thus heat capacity

33. Low - Temperature Limit
With x=hv/kT and u=hvmax/kT
For T 0, u  ∞
with
The heat capacity is

34. LIMITATIONS OF DEBYE MODEL
This is valid only for long wavelengths.
The assumption that are 3N modes of vibration is also not justifiable since an elastic continuum is supposed to possess infinite frequencies.
Since velocities of transverse and longitudinal waves are different ,hence the Debye cut off frequency should also be different for these two.