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236                           Dielectric materials

                                   10.11  Acoustic waves
                                   Atoms can vibrate. They can change their positions relative to each other.
                                   When these motions are regular we can talk about acoustic waves. This is a
                                   purely classical phenomenon but there are lots of people who are fond of using
                                   terms which smell of quantum mechanics. Instead of acoustic waves, they talk
                                   of acoustic phonons. Our own view is to give quantum mechanics the respect
                                   it deserves, but when a phenomenon is classical we prefer to use the classical
                                   terminology.
                                     The simplest way to describe the propagation of acoustic waves is by
                                   nearest-neighbour interaction. Let us look at Fig. 10.11, which shows three
                                   atoms, n –1, n, and n + 1. When quiescent they are at a distance a from each
                                   other. Under wave motion each one of them is displaced from its equilibrium
                                   position by the amounts x n–1 , x n , and x n+1 .The forceonthe nth atom depends
                                   on the displacement of atom n relative to that of atoms n–1 and n+1. If x n+1 –x n
                                   is larger than x n – x n–1 then the force is in the positive direction. Taking force
                                   proportional to displacement, we may write for the net force on atom n,

                                                     F n = β(x n+1 – x n )– β(x n – x n–1 ),  (10.47)

                                   where β is a force constant. The equation of motion may then be written as

                                                        2
                                                       d x n
                                                    m        = β(x n+1 + x n–1 –2x n ),    (10.48)
                                                        dt 2
                                   where m is the mass of the atom. Assuming a wave solution in the form
                                                        x n = x 0 exp[–i(ωt – kna)],       (10.49)


                                   eqn (10.48) reduces to

                                                                      ka
                                                          2
                                                                   2
                                                         ω m =4βsin       .                (10.50)
                                                                       2
                                   Hence the dispersion equation is
                                                                 1/2
                                                             4β        ka

                                                        ω =        sin     .               (10.51)
                                                             m          2
                                   The frequency range of acoustic waves is clearly from 0 to (4β/m) 1/2 .


                                                                     a

                                                     n – 1     n           n + 1



     Fig. 10.11
     Displacement of elements in a
     one-dimensional lattice.                          x n – 1   x n       x n + 1
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