Langmuir 1996, 12, 5221-5230
5221
Static and Dynamic Wetting Properties of Thin Rubber Films Didier Long,*,† Armand Ajdari,† and Ludwik Leibler†,‡ Laboratoire de Physico-Chimie The´ orique, URA CNRS 1382, ESPCI, 10 rue Vauquelin, 75231 Paris Cedex 05, France, and UMR CNRS/Elf-Atochem, Syste` mes Macromole´ culaires He´ te´ roge` nes, 95 rue Danton/B.P. 108, 92303 Levallois-Perret Cedex, France Received May 13, 1996X We model the statics and dynamics of wetting of liquids on thin rubber films. We show that at equilibrium the capillary forces deform soft substrates, leading to undulations of the rubber film surface originating at the triple line. Such deformations propagate when a liquid spreads on the rubber film. This induces the dissipation that may control the spreading dynamics. We derive a general expression for this dissipation as a function of the viscoelastic properties of the substrate. We calculate the spreading velocities for various mechanical relaxation models. Using available mechanical relaxation data, we find that our model may well account for the experimental observations of Carre´ and Shanahan, who demonstrated the importance of viscoelastic braking effects.
I. Introduction Understanding wetting properties is of fundamental importance in a number of practical applications, such as paints formulation, textile dyeing, lubrication, and adhesion. This has been the subject of many theoretical and experimental studies, which demonstrated that both the statics and dynamics of wetting exhibit a great variety of behaviors, depending on the liquid and on the substrate.1-3 Consider first a small liquid droplet at equilibrium on a rigid substrate, in a situation of partial wetting and negligible gravity effects. The corresponding contact angle θe is given by Young’s relation1-3
cos θe )
γS - γSL γL
(1)
(γS, γL, and γSL are respectively the air/substrate, air/liquid, and liquid/substrate surface tensions). Equation 1 expresses the balance of horizontal forces (Figure 1). If at a given instant the contact angle θ differs from its equilibrium value θe, horizontal capillary forces do not balance, so that there is a total pulling force
Fh ) γL(cos θe - cos θ)
(2)
which induces the spreading (or retraction) of the droplet. The dynamics of wetting is determined by the balance between capillary and viscous forces.1,2 On a rigid surface, it is controlled by the viscous dissipation due to the flow inside the droplet. For small contact angles, using the lubrication approximation, one can approximate the flow inside the droplet by a Poiseuille flow (Figure 2). The resulting viscous dissipation inside the droplet is given by1,2
Pdrop )
3V2 ln(r) ηL θ
Figure 1. Liquid droplet at equilibrium on a rigid substrate. The equilibrium is characterized by the contact angle θe.
(3)
†
Laboratoire de Physico-Chimie The´orique, URA CNRS 1382. UMR CNRS/Elf-Atochem, Syste`mes Macromole´culaires He´te´roge`nes. X Abstract published in Advance ACS Abstracts, October 1, 1996. ‡
(1) De Gennes, P.-G. Rev. Mod. Phys. 1985, 57, 827. (2) Le´ger, L.; Joanny, J.-F. Rep. Prog. Phys. 1992, 431. (3) Rowlinson, J. S.; Widom, B. Molecular Theory of Capillarity; Clarendon Press: Oxford, 1982.
S0743-7463(96)00470-2 CCC: $12.00
Figure 2. The instantaneous contact angle θ is larger than θe, and the droplet spreads on the surface. The viscous dissipation due to the flow inside the droplet, approximately a Poiseuille one, controls the spreading dynamics.
where V denotes the spreading velocity, ηL is the viscosity of the liquid, and r is the ratio between the radius R of the droplet and a microscopic length.1,2 Typically, ln(r) is of the order 10. From eqs 2 and 3, the spreading velocity can be calculated using the dissipation law Pdrop ) FhV:
V)
θγL(cos θe - cos θ) 3 ln(r)ηL
(4)
However, the kinetics of wetting can be profoundly modified if the substrate is not rigid. In a beautiful series of experiments, Carre´ and Shanahan4-6 measured the kinetics of the spreading of a liquid droplet on a soft rubber, in the case of partial wetting. Their main observation was that this kinetics can be much slower, by several orders of magnitude, than that of the same droplet on a rigid surface. Moreover, at odds with eq 4, Carre´ and (4) Carre´, A.; Shanahan, M. E. R. C. R. Acad. Sci., Ser. II 1993, 317, 1153. (5) Carre´, A.; Shanahan, M. E. R. Langmuir 1995, 11, 24. (6) Carre´, A.; Shanahan, M. E. R. Langmuir 1995, 11, 1396.
© 1996 American Chemical Society
5222 Langmuir, Vol. 12, No. 21, 1996
Shanahan showed that on soft rubbers the kinetics is independent of the liquid viscosity.5 To account for these effects Carre´ and Shanahan (C&S) introduced a new concept, viscoelastic braking, for which they proposed a phenomenological description,4-6 which we briefly summarize below. At equilibrium the vertical force due to the liquid surface tension, f0 ) γL sin θe, deforms the substrate. The amount of substrate deformation is determined by its elasticity and is of the order of f0/µ0 (µ0 is the static shear modulus).7 Typically, one has f0 = 2 × 10-2 N m-1. For solid substrates, µ0 g 109 Pa; thus, the amplitude of the deformation, ∆h, is negligible and vertical deformations can be altogether forgotten. For soft rubber substrates, with elastic modulus typically smaller than 107 Pa, the deformation ∆h and the corresponding elastic energy E cannot be neglected. When the contact line moves during spreading, the vertical force still deforms the rubber substrate, and a phenomenological fraction ∆ of the energy E is dissipated per unit time, due to viscoelastic effects. The dissipation in the substrate can be much larger than the viscous dissipation in a liquid drop and therefore control the spreading velocity which becomes independent of the solvent viscosity. Our aim in this article is to propose a more precise theoretical description of viscoelastic braking, in the case of spreading on thin rubber films. We start from a simple model of capillary forces and calculate both the dynamical deformation of the substrate and the corresponding dissipation, as functions of the shear relaxation modulus µ(t) of the substrate. This allows us to compare the dynamics of wetting on rubbers of different rheological behaviors. We consider here only the slow velocity regime, when one approaches equilibrium. In a recent paper we have proposed another example of a system in which dissipation in a substrate can control the spreading dynamics: a liquid on a terminally anchored polymer layer (a so called “polymer brush”).8 Due to the very small thickness of such layers, the behavior of a brush is much simpler than that of a thick rubber film: at distances shorter than a characteristic length ξ0, the response of a brush to the vertical force is dominated by the brush surface tension while, at larger distances, the deformation decays exponentially due to the brush incompressibility and finite thickness. When the contact line moves at velocity V, the brush behaves as a viscous fluid set into motion on the scale ξ0. This process involves thus typically a single frequency. The length ξ0 and this frequency depend on elastic brush properties (grafting density) and local monomeric friction. In the present paper we deal with the more complex case of rubber films for which, in contrast to polymer brushes, the elastic response involves a continuum of length scales, as would be the case for a semi-infinite medium, and the motion cannot be described by a single characteristic frequency. This situation can occur even though the film thickness h0 is smaller than the drop radius R. The rubber surface tension provides here mostly a cutoff at short distance. In section II, we describe the basic ingredients of our model: notations, geometry, and the viscoelastic properties of the film. Different models for the linear response functions are then discussed: that of a perfect rubber, the Voigt model9 (which is equivalent to the Rouse model9,10 at low frequency), and the Chasset-Thirion model, which (7) Shanahan, M. E. R.; De Gennes, P.-G. C. R. Acad. Sci., Ser. II 1986, 302, 517. (8) Long, D.; Ajdari, A.; Leibler, L. Langmuir 1996, 12, 1675. (9) Ferry, J. D. Viscoelastic Properties of Polymers; Wiley: New York, 1980. (10) Doi, M.; Edwards, S. F. The theory of Polymer Dynamics; Clarendon Press: Oxford, 1986.
Long et al.
has been found to describe correctly the long time behavior of the shear relaxation modulus of weakly cross-linked rubbers.9,11 In section III the main results are described. We first calculate the static equilibrium deformation induced by capillary forces (section III.A). We also show how to derive the dynamical deformation from the static one using the shear relaxation modulus µ(t) (section III.B). This allows us to give a general expression for the dissipation in the substrate during the spreading of the droplet (section III.C). These general results are then used to calculate the deformation and the dissipation in the cases of a perfect rubber (section III.D), of the Voigt model (section III.E) and of the Chasset-Thirion model (section III.F). The transient regime at the onset of spreading is briefly analyzed (section III.G). Section III is rather technical and can be skipped by a reader interested in conclusions only, which are discussed in section IV. We focus mostly on the comparison between the Rouse model (section IV.A) and the Chasset-Thirion model (section IV.B). Both models are compared with available experimental data. II. Description of the Model Consider a liquid droplet at equilibrium on a soft substrate, obtained by anchoring an incompressible rubber film on a rigid substrate. We use a simple model to calculate the deformation of the film: we take it to be the response to an imposed vertical force f0 along the triple line, with f0 ) γL sin θ (θ is the instantaneous contact angle). Away from equilibrium, the horizontal pulling force is γL(cos θe - cos θ). Note that this approach is not exact: one should in principle minimize the sum of the elastic energy of the substrate and the surface energy of the three interfaces (air/rubber, air/liquid, rubber/liquid). Our description is valid when f0 is small (i.e. f0/γS < 1 and f0/µ0h0 < 1) and amounts then to a first-order perturbation calculation. As stated above, we limit ourselves here to substrates of thickness h0 much smaller than the typical radius R of the droplet. This allows for two simplifications: (1) It will be shown below that deformations due to a localized force are strongly damped beyond h0. Thus, instead of describing the deformation in polar coordinates at the scale of the drop, it is sufficient to express the deformation as a function of a single Cartesian variable x1 in the vicinity of the triple line. Vertical extensions will be measured along the vertical Cartesian coordinate x2 (Figure 3). (2) Inside the droplet, there is an excess pressure Π ∝ γL/R (ref 3) (the Laplace pressure) which exerts on the liquid/film interface a total force which globally compensates the capillary forces f0 applied along the triple line. However, this pressure is uniformly distributed on the liquid/substrate interface (whereas f0 is localized), and its effects can be neglected. Indeed, the effects of a localized force are damped beyond a distance of order h0. Thus, within the liquid/rubber interface (i.e. at a distance larger than h0 from the triple line) the effect of the pressure is that of a uniform pressure on an infinite incompressible film: no deformation occurs. Deformation due to the pressure can occur only near the triple line. The ratio of this Laplace deformation to that due to the localized force f0 is of the order Πh0/f0 ∝ h0/R , 1. The effect of the Laplace pressure is therefore negligible in this context. Within this two-dimensional picture, the Cartesian variable x ) (x1,x2) denotes the position of a point inside (11) Chasset, R.; Thirion, P. In Proceedings of the Conference on Physics of Non-Crystalline Solids; Prins, J. A., Ed.; North-Holland Publishing Co.: Amsterdam, 1965; p 345.
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Langmuir, Vol. 12, No. 21, 1996 5223
an elastic contribution due to its bulk deformation. The equilibrium deformation field can be computed by minimizing the total energy, including the driving term -f0∆h(0) (∆h(0) is the amplitude of the deformation at the point where the force is applied). Within the frame of linear response, the free energy is the sum of the free energy of each deformation mode. Thus, following Fredrickson et al.,12 let us consider the increase of the film energy for a deformation of the free surface
∆h(x1) ) cos(qx1)
(7)
The surface energy contribution per unit surface is12 Figure 3. Side picture of the rubber film, of thickness h0. We aim to calculate the response to a force f0 (per unit length in the third dimension), applied along a line perpendicular to the plane (x1,x2). The force can be static (V ) 0), or it can move at velocity V.
the rubber film. We will describe both static and dynamic deformations within the frame of linear elasticity and will verify at the end that the deformations indeed remain within its range of validity. To solve this problem we must specify the boundary conditions imposed on the rubber film. In particular the droplet exerts capillary forces which we represent as a vertical force f0 per unit length of the triple line. The stress imposed on the free surface of the rubber film, i.e. at x2 ) h0, is σ22(x1) ) f0δ(x1)
(6)
In this dynamical situation, the response of the film will depend on the bulk relaxation shear modulus µ(t) of the rubber.9 To illustrate the physics involved in this problem, we will consider different situations: (a) A perfectly elastic medium with a shear modulus described by µ(t) ) µ0Y(t), where Y is the Heaviside function. (b) A substrate described by one of the simplest viscoelastic models, the Voigt model, which involves a single time scale τ: µ(t) ) µ0Y(t) + ηδ(t), where η is a viscosity and τ ) η/µ0. At low frequencies this model is equivalent to the Rouse model of rubbers.9,10 The time τ corresponds to the Rouse time of the strands between crosslinks. (c) The Chasset-Thirion model9,11 µ(t) ) µ0(1 + (t/τm)-m), where m is a number between 0 and 1 and τm is a characteristic time. Both parameters depend on the considered material. Chasset and Thirion found that the relaxation of weakly cross-linked rubbers, which are of interest here, is better described by such a model at long times than by a Rouse one. III. Deformation and Dissipation. General Results (A) Static Deformation. We calculate here explicitly only the free surface deformation ∆h(x1). However, though cumbersome, the expression of the bulk deformation ui(x) can be obtained straightforwardly.12 Consider the static deformation of the rubber film submitted to a localized force f0. It is limited by the free energy penalty arising from two different contributions: a surface energy contribution due to the increase of the film free surface and
(8)
To calculate the elastic contribution, one needs to solve the static elasticity equations in the rubber film, which describe the equilibrium of an elastic incompressible medium:
∂jσij(x) ) 0
(
σij(x) ) µ0
(5)
in the static case (δ is the Dirac function). In the dynamical situation, we consider that the applied force is still punctual and moves at a given velocity V (Figure 3): σ22(x1) ) f0δ(x1-Vt)
∆Fs ) 1/4γSq22
(9)
∂ui ∂uj + (x) - p(x)δij ∂xj ∂xi
)
(10)
∂ui (x) ) 0 ∂xi
(11)
The indices i and j take the values 1 and 2, and summation over repeated indices is implicit. The fields ui(x), p(x), and σij(x) denote the displacement, pressure, and stress tensor, respectively. δij is the Kronecker symbol. Equation 9 expresses the local equilibrium of forces, and eq 11, the incompressibility of the material. The constitutive eq 10 relates the stress field tensor to the displacement field in the linear regime. To these equations, one has to add the boundary conditions. Here, the vertical displacement applied at x2 ) h0 is imposed u2(x1,h0) ) ∆h(x1) ) cos(qx1), the tangential force has to vanish on the free surface, and there is no displacement at the lower surface. By solving these equations, and then by calculating the corresponding elastic energy, Fredrickson et al.12 obtained a simple expression for the elastic energy penalty per unit surface in two different limits:
∆FEL ) 1/2µ0q2 for qh0 > 1
(12)
µ02 ∆FEL ) 3/4 2 3 for qh0 < 1 q h0
(13)
The deformation free energy F[∆h] corresponding to a surface profile ∆h(x1) can formally be written12
F[∆h] )
1 -1 χ (q) hˆ (q) hˆ (-q) dq ∫-∞+∞2π
1 2
(14)
where hˆ (q) is the spatial Fourier transform of ∆h(x1):
hˆ (q) )
∫-∞+∞∆h(x1) exp(iqx1) dx1
(15)
(12) Fredrickson, G. H.; Ajdari, A.; Leibler, L.; Carton, J.-P. Macromolecules 1992, 25, 2882.
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Long et al.
where ξ and ξ′ are characteristic lengths of order h0. The deformation exhibits a trough at x1 = h0, as can be expected from the incompressibility and the finite thickness of the film, but we furthermore predict a spatial oscillation of the surface. (2) At intermediate distances Rh0 ) γS/µ0 < |x1| < h0 from the contact line, the deformation is given by
∆h(x1) = Figure 4. Schematic response of the film to the static force. The profile is symmetric, so only half of it is represented.
With this definition, the inverse Fourier transform is
∆h(x1) )
1 2π
∫-∞+∞hˆ (q) exp(-iqx1) dq
(16)
According to eqs 8, 12, and 13 and definition 15, one has -1 χ-1(q) ) γSq2 + χEL (q) with
χ-1 EL(q) = 2µ0q for qh0 > 1
(17)
-3 -2 χ-1 for qh0 < 1 EL(q) = 3µ0h0 q
(18)
Let us compare the contributions of the surface tension and elastic terms at short distances (qh0 > 1). We define the corresponding ratio R at q ) 2h0-1, namely, R ) γS/ µ0h0. The typical case of a rubber film corresponds to R much smaller than 1, even for thin films. Indeed, for µ0 ) 106 Pa, h0 ) 10 µm, and γS ) 10-1 J m-2, one has R = 10-2. Hence, there are three different regions in Fourier space: (1) At low wavenumbers, i.e. qh0 < 1, the deformation energy is dominated by the elastic term 3µ0h0-3q-2. The incompressibility of the rubber and the finite thickness of the film play an important role. (2) At intermediate wavenumbers, i.e. 1 < qh0 < R-1, the elastic term 2µ0q dominates. The elastic response corresponds to that of a semi-infinite medium. Surface perturbations at such wavenumbers relax in a layer of thickness = q-1 so that the finite thickness of the film has little influence. (3) At high wavenumbers (qh0 > R-1) the deformation energy is dominated by the substrate surface tension. Let us now calculate the deformation ∆h(x1) of the free surface due to the force f0. It is obtained by minimizing the functional ∆F ) F[∆h] - f0∆h(0) with respect to ∆h. Using ∆h(0) ) 1/2π∫+∞ ˆ (q) dq and eq 14, we obtain -∞ h
∫
1 ∆F ) 2
1 -1 χ (q) hˆ (q) hˆ (-q) dq -∞ 2π 1 f 2π 0 +∞
∫-∞+∞hˆ (q) dq
∆h(x1) )
∫-∞ χ(q)f0 exp(-iqx1) dq
1 2π
( ) ( )
|x1| -|x1| -f0 sin exp µ0 ξ ξ′
The same results for the deformation are obtained for a semi-infinite elastic medium submitted to a line force.12 (3) The logarithmic divergence of eq 21.2 is suppressed at short distances (x1 < Rh0), where the substrate surface tension dominates, and the deformation is limited to a finite value
∆h(0) =
-1 f0 ln(R) 2π µ0
(21.3)
With the values just quoted above and f0 ) 2 × 10-2 N m-1, one has Rh0 ) 10-7 m, h0 ) 10-5 m, and ∆h(0) = 10-7 m. In particular, the microscopic cutoff Rh0 is much larger than the molecular size for such a soft substrate. The surface tension cutoff may be much smaller for a rigid substrate (µ0 = 109 Pa). In this latter case, the logarithmic behavior is suppressed at the molecular level.7 We will see below that the role of the cutoff Rh0 is crucial for understanding the dissipation in the film and the dynamics of spreading. Finally let us note that, due to the cutoff Rh0, the strain gradient ∂1∆h(x1) is smaller than f0/γS, which is smaller than 1, so that the rubber deformation remains in the linear elasticity regime. There is a simple way to take into account the surface energy effects. If there was no surface tension, the deformation of the rubber film would be equal to the response of a 2-D elastic medium to a force applied in a point (i.e. along a line in the 3-D geometry). For |x1| < h0 such a response is given by eq 21.2 even in the limit x1 f 0: eq 21.2 is the Green function12 G(x1,0) for the free surface deformation of the pure elastic problem:
∂jσij(x) ) 0
(
σij(x) ) µ0
(22.1)
∂ui ∂uj + (x) - p(x)δij ∂xj ∂xi
)
(22.2)
∂ui (x) ) 0 ∂xi
(22.3)
with boundary conditions
(20)
As expected from the analysis of the deformation in Fourier space, the free surface deformation exhibits three regions of qualitatively distinct behaviors (Figure 4) (for more details, see the Appendix): (1) At distances larger than the thickness h0, the deformation decays exponentially:
∆h(x1) ∝
(21.2)
(19)
The minimization yields hˆ (q) ) χ(q)f0 and +∞
( )
|x1| -f0 ln 2πµ0 h0
(21.1)
σ22(x1) ) f0δ(x1)
(22.4)
σ12(x1) ) 0
(22.5)
on the free surface of the film (x2 ) h0) and
ui(x1) ) 0 (i ) 1, 2)
(22.6)
on the bottom surface which is fixed on a rigid substrate (x2 ) 0). As discussed above, the rubber surface tension yields the cutoff at |x1| ) Rh0 below which eq 21.2 is no longer valid. However, according to eq 21.3, we can replace the singular stress f0δ(x1) by a stress σ22(x1) ) f0/2Rh0 spread uniformly between x1 ) -Rh0 and Rh0 (so that the total force is still f0) and consider only the elastic problem.
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Langmuir, Vol. 12, No. 21, 1996 5225
Indeed, in this problem, the deformation is given by ∆h(x1) Rh0 ) ∫-Rh G(x1,x′1)σ22(x′1), which is practically the same 0 deformation as that calculated by taking the surface tension into account. Another equivalent point of view consists in considering eqs 22.1-22.6 and the solution (eq 21.2) only for |x1| > Rh0, which is then a real space cutoff. In conclusion, in Fourier space, the surface tension can be accounted for by a large wavenumber cutoff at quv ) (Rh0)-1, while the finite thickness provides a small wavenumber cutoff at qir ) h0-1. (B) Dynamical Deformation. Following the preceding discussion, we consider in this section explicitly the purely elastic response of the film. The rubber surface tension is taken into account by introducing the appropriate cutoff, either in real or in Fourier space. The static elastic problem is described by eqs 22.1-22.6. This set of equations, which we call problem 1, is complete in the sense that there is a unique solution ui(x) (i ) 1,2).13 Consider now the dynamical situation for which a punctual force moves on the top side of the rubber film (Figure 3). The imposed stress at the free surface is then
Thus the equations of elasticity in time Fourier transform are
σ22(x1) ) f0δ(x1 - Vt)
ui(x1,ω) ) 0 (i ) 1, 2)
(24)
(25)
(26) where µ(t) is the shear relaxation modulus and ∂t and ∂i denote the time derivative ∂/∂t and the spatial derivative ∂/∂xi, respectively. Using the time Fourier transform, eq 26 becomes
(
)
∂ui ∂uj + (x,ω) - p(x,ω)δij ∂xj ∂xi
∫-∞+∞f(t) exp(-iωt) dt
(28)
and with the particular definition of µ(ω)
∫0+∞µ(t) exp(-iωt) dt
µ(ω) ) iω
(30.3)
)
-ix1ω 1 σ22(x1,ω) ) f0 exp V V
(30.4)
σ12(x1,ω) ) 0
(30.5)
on the free surface and
(29)
(13) Landau, L. D.; Lifshitz, E. M. Elasticity Theory; Pergamon Press: New York, 1986.
(30.6)
on the bottom surface. The set of equations 30.1-30.6 constitutes problem 2. One can write the imposed stress of problem 2 (eq 30.4) as a sum of punctual forces:
∫-∞+∞ exp(
)
-iy1ω δ(x1-y1) dy1 (31) V
1 σ22(x1,ω) ) f0 V
Thus we are led to consider problem 3 constituted by eqs 30.1-30.3 and the following boundary conditions:
(32.1)
on the bottom surface and
(
)
-iy1ω 1 δ(x1-y1) σ22(x1,ω) ) f0 exp V V
(32.2)
σ12(x1,ω) ) 0
(32.3)
on the free surface, y1 being a given point along the free surface. Let us denote by ui,y1 the solution of problem 3. Then the displacement field
νi,y1(x,ω) )
( )
iy1ω Vµ(ω) exp ui,y1(x,ω) µ0 V
(33)
satisfies the equations of problem 1, but with the force being applied at x1 ) y1 instead of x1 ) 0. Thus one has
(27)
where the time Fourier transform of a function f(t) is
f(ω) )
∂ui (x,ω) ) 0 ∂xi
ui(x1,ω) ) 0
∫-∞t µ(t-t′) ∂t′(∂iuj+∂jui)(x,t′) dt′ - p(x,t)δij
σij(x,ω) ) µ(ω)
(30.2)
(
Within the linear response theory, the stress tensor can be expressed as a function of the displacement field at earlier times:9
σij(x,t) )
)
∂ui ∂uj + (x,ω) - p(x,ω)δij ∂xj ∂xi
while the incompressibility condition reads
∂ui (x,t) ) 0 ∂xi
(30.1)
with the following boundary conditions:
(23)
The deformation of the rubber film is now characterized by a displacement field which depends on time, as well as the stress field tensor. As the inertial effects are negligible (we consider a velocity V much smaller than the sound velocity of the rubber, and the Reynolds number is much smaller than 1), the stress field tensor still satisfies the force balance:
∂jσij(x,t) ) 0
(
σij(x,ω) ) µ(ω)
∂jσij(x,ω) ) 0
νi,y1(x,ω) ) uSi (x-y1)
(34)
where uSi (x-y1) is the static deformation when the force is applied on the free surface (x2 ) h0) at x1 ) y1. (x - y1 is a shorthand notation for (x1 - y1,x2).) As the equations of elasticity are linear, the solution ui(x,ω) of problem 2 (i.e. the time Fourier transform of the actual displacement field) is a linear superposition of the solutions of problem 3 and thus is a linear superposition of the solutions of the static problem:
ui(x,ω) )
1 V
0 exp( ∫-∞+∞µ(ω)
µ
)
-iy1ω S ui (x-y1) dy1 V
(35)
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Long et al.
Then, by calculating the inverse Fourier transform, we obtain the actual displacement field
1 1 2π V
ui(x,t) )
0 exp( ∫-∞+∞µ(ω)
µ
)
-iy1ω × V
exp(iωt)uSi (x-y1) dy1 dω (36) which we can write equivalently:
1 1 2π V
ui(x,t) )
0 exp( ∫-∞+∞µ(ω)
µ
)
-ix1ω × V exp(iωt)uˆ Si
( )
ω ,x dω (37) V 2
uˆ Si (ω/V,x2)
is the spatial Fourier transform of the where static deformation along the x1 direction for the value of the wavenumber q ) ω/V:
uˆ Si (q,x2) )
∫-∞+∞exp(iqx1)uSi (x1,x2) dx1
(38)
Thus the displacement field can be calculated from the knowledge of both the static deformation and the shear relaxation modulus. Such a simple result holds when one considers homogeneous boundary conditions (stress or zero displacement field imposed at the surface) and when the inertial effects are negligible so that superimposing solutions is allowed. In particular we deduce from eq 37 that, at a given velocity V, the response of the material is determined by the behavior of µ(ω) for frequencies ω such that the wavenumber q ) ω/V lies in the spectrum of the static deformation, i.e. ω1 ) V/h0 < ω < ω2 ) V/Rh0. The knowledge of the dynamical shape allows us now to calculate the dissipated power in the rubber film. (C) Dissipation. Let us calculate the dissipation Pfilm (Pfilm > 0) which occurs in the rubber film when the punctual force moves at velocity V. We start from the general result14
Pfilm )
∫Bσij(x,t) ∂i∂tuj(x,t) d2x
(39)
where the integration is made over the bulk B of the rubber film (as we are looking at a stationary problem, the value of the integral does not depend on time). Let us express σij(x,t) and ∂i∂tuj(x,t) in terms of their Fourier transform. We obtain
∫Bd x ∫-∞ dω dω′ ×
2 Pfilm ) (2π)2
2
ui(x,t) ) uSi (x1-Vt,x2)
(43)
Let us now calculate the dissipated power from eq 41. Due to the symmetry x1 f -x1 of the static deformation, the quantity ∫uSij(x) uSij(x - y1) d2x is invariant when one changes y1 to -y1. Thus by changing y1 f -y1 and ω f -ω, one obtains Pfilm ) -Pfilm and thus Pfilm ) 0. This result is obvious for a system in which there are no dissipative losses. (E) Rouse Model. The Voigt model, although very simple, provides a useful physical insight into the behavior of rubber films. In this model the viscoelastic response is given by
µ(t) ) µ0Y(t) + ηδ(t)
(44)
More microscopically, this model corresponds to a Rouse model of a regularly cross-linked rubber network. The characteristic time τ ) η/µ0 corresponds to the Rouse relaxation time of the network strands.9,10 This description is valid for sufficiently long time scales for which consideration of high-frequency internal modes of relaxation of strands is not necessary. For very long time scales the model is not adequate when some stress relaxation may still occur,11 and their effects will be discussed below. From definition 29, eq 44 is equivalent to µ(ω) ) µ0 + iηω. The dynamical deformation is given by eq 36, so that
ui(x,t) )
dt′ ∫0+∞uSi (x1-V(t-t′),x2) exp(-t′ τ )
(45)
∫-∞0 uSi ((x1-Vt-y1),x2) exp(Vτ1 ) dy1
(46)
1 τ
or equivalently
ui(x,t) )
1 τ
y
+∞
i(ω+ω′)t
(iω′)e
µ(ω) uij(x,ω) uij(x,ω′) (40)
Then, using eq 35, one has the general formula
Pfilm )
expressed as a function of the static deformation. Let us now apply these general results to different models for the viscoelastic behavior of the film, as listed at the end of section II. (D) Response of a Perfectly Elastic Rubber. Let us first look at the limit case of a perfectly elastic rubber, the viscosity of which can be neglected. The shear relaxation modulus is then given by µ(t) ) µ0 and is frequency independent: µ(ω) ) µ0. Consider first the dynamical deformation. From eq 36 we deduce that it is the one we would have if a punctual static force were applied at x1 ) Vt, which is natural, as no memory effect is present:
∫Bd2x ∫-∞+∞dω dy1 (iω)V1 ×
1 π
exp
(
)
-iy1ω µ02 S u (x) uijS(x-y1) (41) V µ(ω) ij
where we have used the usual notation
uij )
(
)
1 ∂ui ∂uj + 2 ∂xj ∂xi
(42)
As for the deformation, the dissipated power can be (14) Landau, L. D.; Lifshitz, E. M. Fluid Mechanics; Pergamon Press: New York, 1959.
The factor exp(-t′/τ) in eq 45 arises from the pole at ω ) iτ-1 of the function 1/µ(ω). More generally, if one considers a shear relaxation modulus with different relaxation times, each of them will provide a contribution similar to eq 45. The shape of the rubber film appears as the superposition of deformations produced at earlier times by a punctual force, with an exponentially decreasing weight. From eq 46 we deduce in particular that, contrary to the static shape, the dynamical deformation is not symmetric (the front is sharper than the back). Furthermore, for velocities of order h0/τ, the maximum of the amplitude of the deformation lay behind the application point, at a distance of order h0, due to negative contributions arising from the trough of the static deformation (Figure 5). Consider now the shape of the deformed film at low velocities. According to eq 46, for the differences between the static shape and the dynamic one to be small, one must verify V|∂1uS2 (0,h0)|t′ < uS2 (0,h0) for t′ < τ. As |∂1uS2 (0,h0)| e f0/γS, this amounts to Vτ < (γS/µ0)|ln R| )
Wetting Properties of Thin Rubber Films
Langmuir, Vol. 12, No. 21, 1996 5227
the deformation penetrates to a depth q-1; thus,
∫qq
Pfilm = ηV2
uv
ir
hˆ (q) hˆ (-q)q3 dq
(50)
As hˆ (q) = f0/µ0q, we obtain
()
Pfilm = ηV2
Figure 5. Example of a possible response of the film when the applied force moves at an intermediate velocity.
Rh0|ln R|. Thus, forgetting the logarithmic factor, we define the low-velocity regime by
h0 τ
V
3ηL ln(r) θ
(66)
For θ ) 0.2 rad, ηL ) 10-3 Pa‚s, and the values quoted above for γS and f0, eq 66 is satisfied when η is larger than about 4 Pa‚s. Consider a typical case of a polyisobutylene rubber substrate for which good viscoelastic data are available. Suppose that it is regularly cross-linked with a number of monomers Nc between cross-links of about 100. It has a shear modulus µ0 = kBT/Ncb3 = 3 × 105 Pa (with b = 5 Å). The Rouse time of the portions of chains between cross-links is10
τ=
ζb2 Nc2 3π2kBT
(67)
IV. Results and Discussion
where ζ is the monomeric friction coefficient, which is about 4 × 10-8 N‚m-1‚s at room temperature.15 Then τ = 10-3 s, we get η ) µ0τ = 3 × 102 Pa‚s, and from eq 66 the dissipation in this substrate dominates that in the droplet. This is even more the case for softer rubbers because µ0 scales as Nc-1 while the time τ scales as Nc2.
In the preceding section, we have derived general results for the deformation of the rubber substrate and the
(15) Ferry, J. D. Viscoelastic Properties of Polymers; Wiley: New York, 1980; p 330.
Wetting Properties of Thin Rubber Films
Langmuir, Vol. 12, No. 21, 1996 5229
In this regime where the dissipation in the film dominates, the spreading velocity is given by a simple relation
V)
γL(cos θe - cos θ) (f0/γS)2η
(68)
For a rigid substrate, one has
V)
θγL(cos θe - cos θ) 3 ln(r)ηL
(69)
With γL ) 10-1 N m-1 and cos θe - cos θ = 0.1, the triple line spreads thus at a velocity V = 0.8 mm s-1 on the considered soft rubber instead of the velocity V = 60 mm s-1 one would have on a rigid substrate. A different situation may occur for rubbers with lower dissipation losses, for example a hevea rubber weakly vulcanized with dicumyl peroxide.15 The monomeric friction coefficient at room temperature of this rubber is much less than that of the polyisobutylene rubber formerly considered; namely, one has ζ = 4 × 10-10 N‚m-1‚s. With Nc ) 100 and b ) 6.8 Å, the shear modulus of such a rubber is µ0 = 1.3 × 105 Pa. The corresponding Rouse time is τ = 2 × 10-5 s, and the viscosity η = 2.6 Pa‚s. In that case, the dissipation in the substrate does not dominate the dissipation in the droplet, though this rubber is very soft. In conclusion, we stress that the Rouse model predicts that the slowdown of droplet spreading should be particularly important when one reduces the temperature just above the glass transition temperature. In this temperature region we expect a dramatic increase of monomer friction coefficient and a comparatively small increase in modulus µ0. It is important to note that in practice many soft rubbers exhibit slow relaxation processes. It might be that the slow spreading of droplets such as that observed in Carre´ and Shanahan’s experiments corresponds to these long time relaxations rather than to “fast effects” which might be described by simple Rouse-like models. To get an insight into the problem of the applicability of the Rouse model to the interpretation of these experiments, we have to make some indirect estimates because in ref 5 the mechanical relaxation data on the silicone rubber substrates studied (RTV 630) are not reported. On this rubber, C&S measured very small velocities indeed.5 For example, with the conditions ∆ cos θ = 0.1, f0 = 3 × 10-2 N m-1, and γS ) γSL = 10-1 N m-1, the velocity is about V = 0.03 µm s-1, while our prediction on polyisobutylene rubber would be here larger by a factor of 104, i.e. V = 0.3 mm s-1. Note that the experiment has been realized with a tougher rubber of elastic modulus µ0 = 1.5 × 106 Pa, which is a priori less favorable to slow velocities than our example. A higher monomeric friction coefficient of the sample used in this experiment, which we do not know actually, could be an explanation of such slow velocities. This is not very probable for a silicone rubber. Moreover, within the framework of the Rouse model, one predicts the substrate-dissipated power Pfilm to scale as V2, while Carre´ and Shanahan measured Pfilm ∝ V1+n with n = 0.23 (ref 5) in the experiment discussed here and n = 0.5-0.6 in some other experiments with modified materials.6,7 We conclude therefore that the Rouse model cannot explain the scaling law for the dissipated power as well as the slow velocities observed in these experiments. Probably long time relaxations play an important role. We are led to consider the more realistic Chasset-Thirion model which should be better suited to account for the C&S
results. We will show indeed that this model leads to very slow spreading velocities, even with rubbers of low monomeric friction coefficient. (B) Chasset-Thirion Model. We have found, according to the Rouse model, that the dissipation in the rubber film can be indeed the dominant braking effect. However, for viscoelastic braking to be efficient, one needs weakly cross-linked rubbers, and their relaxation is not adequately described by the Rouse model for long characteristic times. Chasset and Thirion11 indeed observed much slower relaxations. In particular, the time required to achieve equilibrium increases dramatically when the cross-link density decreases. They found that the shear relaxation modulus is better represented by
( ( ) )
µ(t) ) µ0 1 +
t τm
-m
(70)
the parameters m and τm depending on the material, temperature, and cross-link density. A theoretical justification could be that a weakly cross-linked rubber has many dangling chains. Its slow relaxation has thus been interpreted by Ferry9 and Curro and Pincus16 as the consequence of the exponentially long times required by such chains to reach their equilibrium configuration. Indeed, the relaxation is due to the star-polymer-like retraction of dangling chains in their tubes. The entropic barriers slow such a motion considerably.17 For such a rubber, described by eq 70, we have shown that the dissipated power is
Pfilm )
( )
f02 Vτmµ0 m V γS γS
(71)
When the substrate dissipation dominates (Pfilm > Pdrop), the spreading velocity obtained from Pfilm ) FhV is
V)
(
)
γLγS∆cos θ f0
2
1/mRh
τm
(
)
γLγS∆cos θ
0
)
f0
2
1/m
γS µ0τm
(72)
where ∆cos θ ) cos θe - cos θ. First, we note that, with m ) 0.23, we recover the experimental power law of C&S.5 Secondly, in rubber systems analyzed in the literature,9,11 the exponent m is smaller than 1. Thus eq 72 shows that the spreading can be much slower than that predicted by the Rouse model (eq 68). To give an estimation of the spreading velocity, one needs to know the value of τm. As already stated, m and of course τm are not universal but depend on the rubber (cross-link density, chemical nature). Moreover, according to eq 72, the spreading dynamics at a given value of ∆cos θ depends strongly on the specific values of the surface tensions, particularly when the parameter m is small. As we do not know the parameters m and τm and the surface tension of the rubbers used by Carre´ and Shanahan, a precise comparison with our theoretical model is therefore not possible. However, to illustrate how the Chasset-Thirion response leads to a very slow spreading velocity, compatible with those observed in C&S experiments, let us consider what our models predicts for natural rubber samples. Good relaxation measurements for these systems are available.18 Dickie and Ferry18 have found an empirical relation between m, τm, and the creep compliance Je which describes well the effect of the variation of cross-link density. At room temperature and for a shear modulus µ0 = 2.5 × 105 (16) Curro, J. G.; Pincus, P. Macromolecules 1983, 16, 559. (17) De Gennes, P.-G. J. Phys. (Paris) 1975, 36, 1199. (18) Dickie, R. A.; Ferry, J. D. J. Phys. Chem. 1966, 70, 2594.
5230 Langmuir, Vol. 12, No. 21, 1996
Pa, the relation gives m log τm ) -0.5. For m = 0.23 it corresponds to a time τm = 7 × 10-3 s. Note that if such a natural rubber was perfectly cross-linked, it would have a viscosity η ) µ0τ = 1 Pa‚s (according to eq 67), which is relatively small and would be insufficient to control the droplet spreading dynamics. On the contrary with the values ∆cos θ ) 0.02, γS ) 10-1 N m-1, γL ) 10-1 N m-1, and f0 ) 3 × 10-2 N m-1, we predict here a velocity V = 0.08 µm s-1. This velocity is of the same order of magnitude as velocities observed by C&S under analogous conditions (V = 0.03 µm s-1 with ∆cos θ ) 0.1 instead of 0.02). Although our example does not correspond exactly to the C&S experiment (as the numerical results depend on data specific for the considered rubber), it clearly shows that one can obtain much slower velocities than one could expect with the Rouse model. Moreover, the calculated velocity is of the order of magnitude of that observed in spreading experiments. We conclude here that the slow relaxation modes of a real rubber lead to a spectacular slowing down of the spreading dynamics, even though their contribution to the storage modulus is small. The spreading velocity is reduced by many orders of magnitude compared to the case of a supposedly regularly reticulated rubber, and this reduction is comparatively more important when one is closer to equilibrium. Indeed, when the frequencies involved in the dynamics are smaller, processes with longer relaxation times participate in the dissipation. The spreading motion toward equilibrium can thus take a very long time!
Long et al.
Appendix We aim here to calculate ∆h(x1) ) 1/2π∫+∞ -∞ f0χ(q) exp-1 -1 (q) with χEL (q) = (-iqx1) dq. One has χ-1(q) ) γSq2 + χEL -1 2µ0q at short wavelengths (qh0 > 1) and χEL(q) = 3µ0h0-3q-2 at large wavelengths (qh0 < 1). Formally, the integral is given by a summation over the poles of χ(q). In particular, the behavior of the deformation at long distances (i.e. |x1| > h0) is given by the contribution of the pole having the smallest imaginary part. This pole arises from the competition between the short distance elasticity and the large distance elasticity and can thus be roughly estimated by solving the equation (for Re(q) > 0)
3µ0h0-3q-2 + 2µ0q ) 0
This equation has two solutions with Re(q) > 0. If one considers x1 > 0, the relevant one is qc1 ) (3/2)1/3 exp(iπ/3)h0-1. For Re(q) < 0, the pole is given by
3µ0h0-3q-2 - 2µ0q ) 0
(19) Carre´, A.; Shanahan, M. E. R. Langmuir 1995, 11, 3572.
(74)
whose relevant solution is qc2 ) (3/2)1/3 exp(i4π/3)h0-1 and has the same imaginary part as qc1. As already stated, eqs 73 and 74 provide only estimates of the poles qc1 and qc2, and we retain here that both poles have an imaginary part and a real part of order h0-1. Thus, at distances larger than h0, one has
V. Conclusion We have described the statics and dynamics of wetting on a soft rubber film. We have derived a general expression for the dissipation in the substrate as a function of the shear relaxation modulus of the material. For time scales for which the relaxation can be described by the Rouse model, we predict the dissipation in the film to scale as V2. It can control the kinetics of wetting. However, we expect that, in rubbers with dangling chains for which long time relaxation effects are important, the spreading toward an equilibrium angle can be a very long process. By describing the relaxation by the ChassetThirion model, we have shown that the dissipated power scales as V1+m. The value of m depends on the material and can be measured independently in mechanical relaxation experiments. These predictions are consistent with experiments recently performed by Carre´ and Shanahan. Our approach could be extended to the case of dewetting19 or forced wetting experiments. Measuring the long time mechanical relaxations in rubber films deposited on substrates is not easy. Observing the spreading of liquid droplets on such films may provide an interesting new tool to characterize their viscoelastic properties.
(73)
∆h(x1) =
( ) ( )
|x1| -|x1| -f0 sin exp µ0 ξ ξ′
where both ξ and ξ′ are of the order h0 (an unknown prefactor of order unity has been omitted). To calculate the deformation at short distance (|x1| < h0), let us write ∆h(x1) ) 1/π∫0+∞f0χ(q) cos(qx1) dq. Then one can approximate the deformation by
∆h(x1) =
f0 πµ0
∫0q 3q-21h -3 cos(qx1) dq +
f0 πµ0
ir
0
∫q
f0 1 cos(qx1) dq + 2q πγS
quv ir
∫q+∞q12 cos(qx1) dq uv
where qir ) (h0)-1 and quv ) (Rh0)-1 ) µ0/γS. The contribution of the first term and that of the third term are never larger respectively than f0/9πµ0 and f0/πµ0. The contribution of the second term, for Rh0 , x1 , h0, can be approximated by f0/2πµ0∫q+∞ cos(u)/u du = -f0/2πµ0 irx1 ln(x1/h0) and is the dominant contribution at short distance. Finally, when x1 e Rh0, the second term, which is still the dominant factor, is f0/2πµ0∫qqiruvxx11 cos(u)/u du = -f0/2πµ0 ln(R). LA9604700
