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Next: Landau model of surface Up: Surface melting Previous: Preface

Phenomenological thermodynamics model

Surface melting can be regarded as a case of wetting [27], namely a wetting of the solid by its own melt. As in the case of adsorption of a gas onto a hard wall one may observe complete or incomplete wetting, depending on whether the quasiliquid thickness diverges or remains finite as $T \rightarrow T_m$. By the same analogy, the case of non-wetting corresponds to an absence of surface melting, e.g. the surface remains dry up to $T_m$. What makes the surface premelt or remain dry? If a single crystal is cleaved along it $\{hkp\}$ plane, where $h,k,p \in Z$ are the crystallographic indices, then the surface free energy per unit of area $\gamma_{sv}^{\{ hkp\}}$ is defined as the work needed to create a unit area of dry surface (the subscript 'sv' refers to the solid-vapor interface). On the other hand, the free energy of a surface that at $T_m$ is covered with a thick melt layer, is given by $\gamma_{sl}^{\{ hkp\}}+\gamma_{lv} $, where the indices 'sl' and 'sv' refer to the solid-liquid and liquid-vapor interfaces, respectively (See Fig. [*]).

Figure: The solid-liquid and liquid-vapor interfaces
\begin{figure}\centerline{\epsfxsize=6.8cm \epsfbox{/home/phsorkin/Diploma/Theory/Pict/slv.eps } }\end{figure}
Surface melting will only occur if there is a gain in the free energy, that is, if
\Delta \gamma^{\{hkl\}}\equiv\gamma_{sv}^{\{hkl\}}-\left(\gamma_{sl}^{\{hkl\}} +\gamma_{lv} \right) > 0
\end{displaymath} (3.1)

On the other hand, if $ \Delta \gamma^{\{hkp\}} <0$, then the surface will remain dry up to $T_m$. The sign and magnitude of $ \Delta \gamma^{\{hkp\}} $ depend not only on the material, but also on the surface orientation. In general the most open crystal faces, e.g. the (011) face for fcc crystals and (111) for bcc crystals, are most likely to exhibit surface melting. For a system exhibiting complete wetting there is a unique relation between the temperature $T$ and the equilibrium thickness of the quasiliquid layer $\bar {l}(T)$. The fact that there is a finite equilibrium thickness of the film at a given temperature $T$ is a result of balance between two opposite thermodynamic forces. On one hand, the quasiliquid becomes more liquid-like for increasing layer thickness, which results in some gain in the free energy. This corresponds to an effective repulsive force between the solid liquid and liquid-vapor interface. The effective interaction energy between the interfaces at either side of the quasiliquid layer is given by $\Delta \gamma^{\{hkp\}}exp{\{-\frac{2l}{\zeta_b}\}}+Wl^{-2} $, where $l$ is the thickness of the film, $\zeta_b $ is a characteristic length scale over which the crystalline order decays [23,26] as it is measured from the crystal-quasiliquid interface, and $W$ is a positive constant which is called the Hadamar constant. On the other hand there is the free energy cost associated with supercooling of the quasiliquid layer. For a layer of thickness $l$ the energy cost per unit of area is $L_ml(1-T/T_m)$, where $L_m$ is the latent heat of melting per unit volume. This yields an attractive force between the two interfaces.

The total free energy $F(l)$ of the surface covered with a melt layer of thickness $l$ is:

F(l)=\gamma_{sv}^{\{hkp\}}+\gamma_{lv}+\Delta \gamma^{\{hkp\...
\end{displaymath} (3.2)

The equilibrium thickness $\bar {l}(T)$ is the value of $l$ for which $F(l)$ is minimal, i.e. $dF(l)/dl=0$. Let us define define a crossover thickness $l_c$ as a thickness for which the long-range contribution $Wl^{-2}$ to $F(l)$ is equal to the short-contribution $\Delta \gamma^{\{hkp\}}exp{\{-\frac{2l}{\zeta_b}\}}$. Consequently two different regime are considered. The first regime is for $\bar {l}(T) << l_c$, where the system is governed by the short-range exponentially decaying interactions and the equilibrium thickness is given by:
\bar {l}(T)=\frac{\zeta_b }{2} ln\left [ \frac{2\Delta \gamma^{\{hkp\}}T_m }{L_m(T-T_m)\zeta_b} \right] \simeq -ln(t)
\end{displaymath} (3.3)

where the reduced temperature is given by $t=(1-T/T_m)$. This kind of logarithmic divergence of $\bar {l}(T)$ is characteristic for metals and semiconductors. But for rare-gas crystals (or if $T_m$ is approached very-very closely [23]) the long-range force must eventually dominate the melting behavior in the second regime $\bar {l}(T) >> l_c$. The short-ranged force will dampen out and one is left with van der Waals type dispersion forces. Thus minimizing $F(l)$ with respect to $l$ yields:
\bar {l}(T)=\left[\frac{(T-T_m)}{2T_mW} \right]^{-1/3}
\end{displaymath} (3.4)

Here it is implied that $W >0$ and the liquid is less dense than the solid. This power law was tested in numerous experiments carried out with rare-gas crystals. The agreement between the theoretical prediction and the experimental results is very good.

next up previous
Next: Landau model of surface Up: Surface melting Previous: Preface