LATTICE VIBRATIONS

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.

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

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 ..

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

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:

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

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.

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))

“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.

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.

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

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

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

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!