# Difference between revisions of "Substrate Curvature Resulting from the Capillary Forces of a Liquid Drop"

Entry by Emily Redston, AP 226, Spring 2012

Work in Progress

## Reference

Substrate curvature resulting from the capillary forces of a liquid drop by F. Spaepen. J. Mech. Phys. Solids 44, 675 – 681 (1996)

## Introduction

We typically characterize the surface of solids using two thermodynamic quantities:

• surface (or interface) tension $\gamma$, which is a scalar quantity equal to the work required to create a unit area of new interface at constant strain in the solid
• surface (or interface) stress $f_{ij}$, which is a 2x2 tensor defined such that the surface work required to strain a unit surface elastically by $d {\epsilon}_{ij}$ is $f_{ij} d {\epsilon}_{ij}$

In this paper, Spaepen illustrates the difference between these two quantities by considering a hemispherical liquid drop on a solid surface.

## Geometry of the Droplet

Figure 1 shows a droplet surrounded by a vapor phase on a solid surface. The problem is kept two-dimensional for simplicity. Associated with the three types of interfaces are the tension $\gamma_{lv}$, $\gamma_{sv}$, and $\gamma_{sl}$, as well as the stresses $f_{lv}$ (=$\gamma_{lv}$), $f_{sv}$, $f_{sl}$ (corresponding to the only applicable strain, $\epsilon_{11}$, in the direction of the interface). Spaepen also assumes that the three phases consist of the same single element to avoid complications due to surface segregation.

Fig 1 Diagram of the droplet on the substrate, indicating the relevant interfaces and quantities

## Equilibrium Shape of the Droplet

Consistent with $\gamma_{lv}$ being isotropic, the liquid-vapor interface is considered semi-circular and has a radius of curvature R. The angle between the radii to the end points (OA and OB) is taken to be 2$\theta$. The equilibrium shape of the droplet for a given volume is determined by minimizing the free energy of the system with respect to $\theta$ or R. Placing the droplet on the substrate replaces the solid-vapor interface by solid-liquid interface over an area $A_{sl}$, also creating an area $A_{lv}$ of liquid-vapor interface. The associated changes in free energy are:

$\Delta F = A_{sl}(\gamma_{sl} - \gamma_{sv}) + A_{lv} \gamma_{lv}\ $

Taking the geometry of Fig. 1 into account, we can also write this as:

$\Delta F = 2Rsin\theta(\gamma_{sl} - \gamma_{sv}) + 2\theta R\gamma_{lv}\ $

To minimize this free energy at constant volume $V = R^2(\theta - sin\theta cos\theta)$, a Lagrange multiplier, $\lambda$, is introduced:

$F = 2Rsin\theta(\gamma_{sl} - \gamma_{sv}) + 2\theta R\gamma_{lv} + \lambda R^2(\theta - sin\theta cos\theta)\ $

Minimization gives the conditions:

${\partial F \over \partial R} = 2sin\theta(\gamma_{sl} - \gamma_{sv}) + 2\theta \gamma_{lv} + 2\lambda R(\theta - sin\theta cos\theta) = 0\ [4a]$

${\partial F \over \partial \theta} = 2Rcos\theta(\gamma_{sl} - \gamma_{sv}) + 2R\gamma_{lv} + \lambda R^2(1 - cos^2\theta + sin^2\theta) = 0\ [4b]$

Equation $\lambda R$ found from both equations and simplifyling the trigonometric functions gives:

$\theta cos\theta({\gamma_{sl} - \gamma_{sv} \over \gamma_{lv}} + cos\theta) = sin\theta({\gamma_{sl} - \gamma_{sv} \over \gamma_{lv}} + cos\theta)\ $

Equation 5 can only be satisfied in two ways: $\theta$ = 0 (complete wetting), which requires $\gamma_{sl} + \gamma_{lv} < \gamma_{sv}$, or, more interestingly

$cos\theta = - {\gamma_{sl} - \gamma_{sv} \over \gamma_{lv}}\ $

which is the well-known Young equation for the equilibrium wetting angle $\theta$. This scalar equation has an equally well-known vector representation, shown in Fig. 2. Although the interfacial tensions are formally represented as vectors in this diagram, it is important to remember that these vectors are not forces.

Fig. 2. Vector diagram showing the horizontal balance of the interfacial tensions that yields the wetting angle $\theta$

Solving for the Lagrange multiplier in equilibrium:

$\lambda = -{\gamma_{lv} \over R}\ $

This is the pressure difference between the liquid and vapor across the curved interface.

## Curvature of the Substrate

After establishing the shape of the droplet from the relation between the tensions, Spaepen considered the strains in the substrate from the forces exerted by the droplet and its interfaces. Since the displacements under consideration are elastic, the interfacial stresses are the relevant quantities.

The stress inside the droplet is hydrostatic. The pressure exceeds that in the vapor by $\Delta p$, which is found by the well known Laplace force equilibrium, illustrated in Fig. 3.

Fig. 3. Portion of the liquid-vapor interface, indicating the relevant quantities for Laplace's calculation of the pressure difference between liquid and vapor

The component of the force normal to the surface is balanced by the components of $f_{lv}$ in that direction:

$2Rd\theta \Delta p = 2f_{lv}d\theta\ $

For the liquid $f_{lv} = \gamma_{lv}$, so we can write:

$\Delta p = {\gamma_{lv} \over R}\ $

A free body diagram of the vertical forces on the substrate is shown in Fig. 4.

Fig. 4. Free body diagram of the substrate with the vertical capillary forces

The vertical components of the capillary forces, $2\gamma_{lv}sin\theta$, spaced a distance $L= 2Rsin\theta$ apart, are balanced by the force from the hydrostatic pressure $\Delta p$. The substrate curvature resulting from this load was estimated using simple beam bending with a strain that varies linearly though the thickness. This approximation is reasonable if the thickness of the substrate, t, is less than L. There is no curvature to the left and right of the droplet. Under the droplet, the curvature varies, being maximum in the middle and going to zero at the ends. Spaepen focused on calculating an average curvature, since that's what one measures in condensation experiments. Standard balancing of forces and moments gives for the strain in the top surface:

$\epsilon_0(x) = {6Fx \over Et^2}({x \over L} - 1)\ $

where E is the Young's modulus of the substrate. The total elongation of the top fiber is

$\Delta L = \int_0^L \epsilon_0(x)dx = -{FL^2 \over Et^2}\ $

This translates into an average curvature of

$\kappa_1 = {2\Delta L \over tL} = -{2FL \over t^3E} = -{4 \gamma_{lv} R sin^2\theta \over t^3E}$