First order crystal transformation. Consider a crystal that can exist in either of two structures, denoted by α and β. We suppose that the α structure is the stable low temperature form and the β structure is the stable high temperature form of the substance. If the zero of the energy scale is taken as the state of separated atoms at infinity, then the energy density U(0) at τ=0 will be negative. The phase stable at τ=0 will have the lower value of U(0): thus Uα(0)<Uβ(0). If the velocity of sound vβ in the β phase is lower than vα in the α phase, corresponding to lower values of the elastic moduli for β, then the thermal excitations in the β phase will have larger amplitudes than in the α phase. The larger the thermal excitation, the larger the entropy and the lower the free energy. Soft systems tend to be stable at high temperatures, hard systems at low.
(a) Show from Chapter 4 that the free energy density contributed by the phonons in a solid at a temperature much less than the Debye temperature is given by −π2τ4/30v3ℏ3, in the Debye approximation with v taken as the velocity of all phonons.
(b) Show that at the transformation temperature
τc4=(30ℏ3/π2)[Uβ(0)−Uα(0)]/(vβ−3−vα−3).(84)
There will be a finite real solution if vβ<vα. This example is a simplified model of a class of actual phase transformations in solids.
(c) The latent heat of transformation is defined as the thermal energy that must be supplied to carry the system through the transformation. Show that the latent heat for this model is
L=4[Uβ(0)−Uα(0)].(85)
In (84) and (85), U refers to unit volume.