1. LATTICE VIBRATIONS

2. VIBRATIONS OD ONE DIMENSIONAL MONOATOMIC LATTICE
Consider a
Monatomic Chain of
Identical Atoms
with nearest-neighbor, “Hooke’s Law” type forces (F = - Kx) between the atoms.
This is equivalent to a force-spring model, with masses m & spring constants K.

3. This is the simplest possible solid
This is the simplest possible solid. Assume that the chain contains a large number (N  ) of atoms with identical masses m. Let the atomic separation be a distance a.
Assume that the atoms move only in a direction parallel to the chain. Assume that only nearest-neighbors interact with each other (the forces are short-ranged). a a a a a a
Un-2
Un-1 Un
Un+1
Un+2

4. Consider the simple case of a monatomic linear chain with only nearest-neighbor interactions.
Expand the energy near the equilibrium point for the nth atom. Then, the
Newton’s 2nd Law
equation of motion is: a a
Un-1 Un
Un+1 l ..

5. Total Force = Force to right – Force to left
This can be seen as follows. The total force on the
nth atom is the sum of 2 forces.
The force to the right is: a a
The force to the left is:
Un-1 Un
Un+1
Total Force = Force to right – Force to left
Diatomic benzet ..
The Equation of Motion of each atom is of this form.Only the value of ‘n’ changes. 5

6. Put all of this into the equation of motion: ..
Assume that all atoms oscillate with the same amplitude A & the same frequency ω. Assume harmonic solutions for the displacements un of the form:
Displaced
Position:
Undisplaced
Position:
Put all of this into the equation of motion: ..
Now, carry out some simple math manipulation:

7. Equation of Motion for the nth Atom Cancel Common Terms & Get:

8. This is the solution to the normal
Mathematical Manipulation finally gives:
After more manipulation, this simplifies to
This is the solution to the normal
mode eigenvalue problem
for the monatomic chain.

9. Let C be the stiffness and ρ be linear mass density ,where C = force /strain =F/(u/a) C/a=F/u=force /displacment=K where K = spring constant therefore , C=Ka and ρ=m/a hence ω= 2/a (C/ρ)1/2 sin (ka/2)
ω= 2/a V0 (sin (ka/2))

10. “Phonon Dispersion Relations” or Normal Mode Frequencies or ω versus k relation for the monatomic chain. w A C B k k –л / a л / a 2 л / a
Because of BZ periodicity with a period of 2π/a, only the
first BZ is needed. Points A, B & C correspond to the
same frequency. They all have the same instantaneous
atomic displacements.

11. At low frequency When k→0 , sin(ka/2) →ka/2 ω = V0 k
Phase velocity is ,Vp = ω/k
Group velocity is ,Vg = dω/dk
Here both the velocities are same ,so medium behaves as homogeneous elastic medium for long wavelength.

12. Vp = ω/k = (2V0 /ka) sin (ka/2)
At higher frequencies
The phase velocity is not same as group velocity these can be obtained as follow
Phase velocity is
Vp = ω/k = (2V0 /ka) sin (ka/2)
For k→0 , vp → V0
And group velocity is
Vg = dω/dk
Vg = V0 cos(ka/2) ,
For k→0 , Vg = V0

13. the  for Bragg reflection!k
At the Brillouin Zone edge: l
The maximum frequency occurs at this k. In that mode, every atom is oscillating π radians out of phase with it’s 2 nearest neighbors. That is, a wave at this value of k is A STANDING WAVE.
Black: k = π/a
or  = 2a
Note! This is also
the  for Bragg reflection! x

14. The Monatomic Chain
k = (/a) = (2/);  = 2a
k  0;   

15. vg  (dω/dk) = a(K/m)½cos(½ka)
Group Velocity, vg in the 1st BZ
vg  (dω/dk) = a(K/m)½cos(½ka)
At the 1st BZ Edge:
vg = 0
This means that a wave with λ corresponding to a zone edge wavenumber
k =  (π/a)
Will Not Propagate!
That is, it must be a
Standing Wave!