A slope-dependent disjoining pressure for non-zero contact angles

Abstract

A thin liquid film experiences additional intermolecular forces when the film thickness $h$ is less than roughly 100 nm. The effect of these intermolecular forces at the continuum level is captured by the disjoining pressure $\Pi $. Since $\Pi $ dominates at small film thicknesses, it determines the stability and wettability of thin films. To leading order, $\Pi = \Pi (h)$ because thin films are generally uniform. This form, however, cannot be applied to films that end at the substrate with non-zero contact angles. A recent ad hoc derivation including the slope $h_x$ leads to $\Pi = \Pi (h, h_x )$, which allows non-zero contact angles, but it permits a contact line to move without slip. This work derives a new disjoining-pressure expression by minimizing the total energy of a drop on a solid substrate. The minimization yields an equilibrium equation that relates $\Pi $ to an excess interaction energy $E = E(h, h_x )$. By considering a fluid wedge on a solid substrate, $E(h,h_x )$ is found by pairwise summation of van der Waals potentials. This gives in the small-slope limit $$\Pi = \frac{B}{h^3}\big(\alpha ^4 - h_x^4 + 2hh_x^2 h_{xx}\big),$$ where $\alpha $ is the contact angle and $B$ is a material constant. The term containing the curvature $h_{xx} $ is new; it prevents a contact line from moving without slip. Equilibrium drop and meniscus profiles are calculated for both positive and negative disjoining pressure. The evolution of a film step is solved by a finite-difference method with the new disjoining pressure included; it is found that $h_{xx} = 0$ at the contact line is sufficient to specify the contact angle.