Model for line tension in three-phase equilibrium

Physica A 173 (1991) 72-110 North-Holland

MODEL FOR LINE TENSION IN THREE-PHASE EQUILIBRIUM B. WIDOM Department of Chemistry, Baker Laboratory, Cornell University, Ithaca, NY 14853, USA

H. WIDOM Department of Mathematics, Applied Sciences Building, University of California at Santa Cruz, Santa Cruz, CA 95064, USA Received 10 September 1990

We propose a model fre.e-energy functional of two order parameters with which to calculate the interracial and line tensions in three-phase equilibrium. The Euler-Lagrange equations for the free-energy minimum are solved exactly, yielding the spatial variation of the order parameters analytically. In terms of a parameter b 2 in the model the three interracial tensions, in dimensionless form, are ~(1 + b2), ~(1 + b2), and 2. When b 2 = 3 the three phases play symmetrical r61es and the line tension, again in the appropriate units, is calculated to be -6/~r + 2/X/-3 = - 0 . 7 5 5 . . . . A wetting transition, where the sum of two of the interfacial tensions becomes equal to the third, occurs as b2---> 1+. A quantity that approximates the line tension is found to vanish proportionally to the first power of the vanishing contact angle as the wetting transition is approached.

1. Introduction and outline

We propose a model free-energy functional for the study of three-phase equilibrium. The intention is to have an exactly soluble model within which to determine the spatial variation of the order parameters near the junction of the ~hree phases, to calculate the interfacial tensions, to determine the sign and ~nagnitude of the line tension, and to study the behavior of the line tension on approach to a wetting transition. Here those goals are largely (but not yet fully) realized. Fig. 1 shows three fluid phases a,/3, and 3, filling the dihedral angles between planar two-phase interfaces. The three-phase line, in which the three interfaces meet, is perpendicular to the plane of the figure. Nothing varies in that direction; the fluid structure varies only in the two directions perpendicular to that line. The dihedral angles are the contact angles in the phase equilibrium. We :Mlaii -'- " denote them by the same symbols c~, /3, and y we have given to the 0378-4371/91/$03.50 (~) 1991 - Elsevier Science Publishers B.V. (North-Holland)

B. Widom, H. Widom / Model ]'or line tension in three-phase equilibrium

,'\/

B

73

,,

Fig. 1. View down the three-phase line of three bulk pha,~ a , / L y Nling the dihedral angle~ between planes. The region A is a circle centered at the three-phase line,

corresponding phases. In stable equilibrium they are related to the interracial tensions o-,~, o'~, and o-,,~ by [ll c o s / 3 = 1 - (or., + % , + ¢ r . , ) ( ~ . o

+ %,

- o'.,)

. etc.

(1.1)

At a wetting transition [ 1,2] one of the tensions, say cr . becomes equal to the sum of the other two, o-,,v = o',8 + o'~

(wetting).

(~2)

and then by (1.1) one of the contact angles, in this instance ~. vanishes. Let "~ be the density of excess free energy associated with the inhomogeneities in the system; i.e., ~ t h the three two-phase interfaces and with the three-phase line. Near the planar interface.s, far from the three-phase line, ~F varies only perpendicularly to the i n t e r f a ~ and vanishes rapidly with increasing distance from them. If z,,g is a coordinate perpendicular to the a/3 interface, say, infinitely far from the three-phase line. the tension cr,,~ of that interface is [3]

or.8 = f ~ dz,,o

(far from three-phase line),

(1.3)

--:x;

and similarly for the tensions cr~v and o-,,~. Near the three-phase line the free-energy density ~ varies in the two directions perpendicular to that line. Integrating ~ through the interior, A, of a circle of radius R centered at the three-phase line (fig. 1), we obtain the line tension z as the next-to-leading term in the asymptotic expansion (R-->~) [1],

B. Widom, H. Widom / Model for line tension in three-phase equilibrium

74

f ~ da = (trot 3 + tr, v + tr~v)R + z + 0 ( 1 / R ) ,

(1.4)

A

where da is an element of area m the plane. Suppose gt is of the form rF = F(p,, P2, . . .) + ½ E mqVpi .Vpj ,

(1.5)

i,j

where the Pi are the density or composition variables (order parameters) in the system, where the function F(p~, P 2 , . . . ) vanishes when the Pi have their bulk-phase values and is positive for all other values of the pi, where the mq are constant parameters, and where V is the two-dimensional gradient operator in any plane perpendicular to the three-phase line. Then the equilibrium p~ satisfy.the Euler-Lagrange equations [1]

OF/Opi= ½ E (m, + mji)V2pj

(i = 1,2,...);

(1.6)

J

and then r, in addition to being given by (1.4), is given also by the KerinsBoiteux formula [4] r = f ( g z - 2F) d a ,

(1.7)

in which the integration is over the whole plane. In the next section we define the model by specifying a function F and coefficients rnq in (1.5), and we write the associated Euler-Lagrange equations (1.6). In section 3 we find the solutions pi of the Euler-Lagrange equations far from the three-phase line, and then obtain the interfacial tensions from (1.3) (and also by a different route). The tensions are found to satisfy (1.2) when a parameter b 2 in the model ~ takes the value 1, thus locating the wetting transition. In section 4 we obtain the solutions p~ of the Euler-Lagrange equations everywhere in the plane of fig. 1, expressing them as integrals through regions of the plane. Those regions have boundaries that we fully determine implicitly and some of the features of which we determine explicitly. We show in section 5 that in this model the integral in (1.4) with the equilibrium p~, and hence the line tension z, may be expressed as the sum of integrals of linear functk~ns of the p~, the integrals extending through the aforementioned regiens. Since by that route the integrands are linear in the p~, it is in this instance an easier route to ~"than that via (1.7).

B. Widom, H. Widom I Model for line tension in three-phase equilibrium

75

We refer to two alternative methods by which the integrals that give r could be evaluated if the regions of integration were sectors with straight-line boundaries; the details of the methods are in appendices B and C. When the parameter b 2 in the model takes the value 3 the three phases a,/3, and y play symmetrical rSles and the regions referred to above become three 12~ sector. In that case the quantity 1-' evaluated by the methods of appendices B and C is the line tension r; the result is in section 5. For any other value of b z (> 1) the quantity ~-' that is calculated by replacing the true regions of integration by sectors with straight-line boundaries is an approximation to r but is not r it~lf. We observe in section 5 that as the wetting transition is app~ached (b~--, rI +, contact angle /3---,0), r' vanishes proportionally to ~l, and we note the possibility that the true r may then also vanish proportionally to ~1, ~ in an earlier, less detailed model [5]. Our results are briefly summarized in the concluding section 6.

2. Model ~P' Here and hereafter we express all quantities (densitics. cncrgics, and lengths) in dimensionless form. There must be at least two order parameters p, in (1.5) and (I.6). or els~ (1.2) would always hold, the contact angles could only be 0 or r:, and there would be no three-phase line [ 1]. For simplicity we assumc only two p,, and wc take their values in the bulk phases to be Pi ,P2

-1,0,

pl/3, p2/Z=O,b,

pi',p2"= 1 0

(2.1)

(These are relative densities, and so may be negative.) We must have

pfl)= e(p,', p,')= o

(2.2)

and F(p~, P2) > 0 for all other p~, P2- We follow Lipowsky [6] (extending his o,-"i g"-~' i . ~ idea for .twO-t,.u~ . . . . '- . . . . e q u m :':b ~- -u-m - to the present "t m. ~. .-. W 'm- -~- - equilibrium) and take F(p~, P2) to consist of three paraboloids, each with its minimum at one of the points ( p ~ , P2"), etc., and joined at seams where they intersect. We take the paraboloids to have equal circular cross sections. The seams, projected onto the p~, P2 plane, are then three straight lines that radiate from the point p~ = 0, P2 = (b2 _ 1 ) / 2 b and divide the plane into three sectors, a,/3. and 3', in which the respective bulk-phase densities lie (fig. 2). Specifically,

B. Widom, H. Widom I Model for line tension in three-phase equilibrium

76

Fig. 2. The (Y,p, y sectors of the p,, p2 plane. The sector boundaries are the solid lines emanating from the point p, = 0, pz = (b* - 1) /2b. The bulk-phase densities ( P,~, p” ) = (-1, 0), ( p,‘. p,“) = (0, b), and (p,“, pzy) = (1,0) are marked with dots and labeled (Y,P, and 7.

‘(PI + 1)2 + &I*2

I

qp,,P,)=

in a! sector,

P,2+(P2-w2

in p sector ,

(P, - 1)” + P2

in y sector.

(23

This F( p,, p2) is continuous but its first derivatives sector boundaries (seams) given by bP2 = ;(b2

-

1) -

p,

bP2 = $(b2

-

1) + p,

(@

boundary)

,

(PY

boundary)

9

(cxy boundary)

PI = 0

are discontinuous

at the

(24

.

These boundaries, shown as the solid lines in fig. 2, are the perpendicular bisectors of the lines (not shown) that connect the points cy, p, and y in pairs. We complete the specification of the model by choosing =

ml2

for the

m2,

mij

= 0,

ml1

2

=m22=

(2.5)

in (1.5). Then the Euler-Lagrange

V2P, = p1 + V2P,

= PI 9

V2P,

= p1 -

1

9

V’P,

= P2

V2P2 = P2 -

1 9

V2P2 = P2

b

equations

(a

region)

,

(p

region)

9

( y region)

.

(1.6) are

cw

B. Widom, H. Widom / Model for line tension in three-phase equilibrium

77

The solutions p~, P2 of (2.6) are functions of two spatial variables, say x and y, in any plane perpendicular to the three-phase line, such as the plane of fig. 1. These functions are continuous with continuous first derivatives, but, reflecting the discontinuities in the derivatives of F(p 1, P2), i.e., in the righthand sides of (2.6), they have discontinuities in their second derivatives at the boundaries of three regions in the x, y plane, and as shown schematically in fig. 3a. The location and orientation of the figure in the plane are arbitrary; they are not determined by (2.6). The regions in fig. 3a are mappings of the sectors of fig. 2; in both figs. 2 and 3a the densities #~ and P2 on the boundaries between regions satisfy (2.4). In both figures the boundaries are artifacts of (although they play a crucial rfle in) the model; they are the loci of artificial discontinuities and would have no counterparts in a model with an analytic F(p~, p,). Although the boundaries between regions in fig. 3a are artificial, and the location and orientation of the figure in the x, y plane are arbitrary, the angles between pairs of asymptotes to those boundaries are neither artificial nor arbitrary; they are the contact angles a,/3, and ), of fig. 1, as shown in fig. 3b. That is because the variation of Pt with P2 in the a/3 interface, say, infinitely far from the three-phase line, is given by some trajectory that connects points a and 13 in fig. 2, and such a trajectory necessarily crosses the boundary between the a and/3 sectors. (In section 3 we shall see that the trajectory in question is just the straight line between those points.) Thus, in the x, y plane, far from the three-phase line, the loci of discontinuities in the second derivatives of p~(x, y) and p2(x, y) lie within the two-phase interfaces, to which the asymptotes in fig. 3b must then be parallel. We verify this picture in section 3. where we also calculate the interracial tensions and the contact angles as functions of the parameter b 2.

(o) I

x

(b)

Fig. 3. (a) The a, /3, and 3' regions of the x, y plane. (b) The same as (a), but showing the asymptotes to the boundaries between regions (dashed lines) and the contact angles.

78

B. Widom, H. Widom / Model for line tension in three-phase equilibrium

When b 2 = 3 the three bulk-phase points a,/3, 3' in the p~, 1o2plane of fig. 2 lie at the vertices of an equilateral triangle. The phase equilibrium is then fully symmetric in the three phases and the three regions in the x, y plane of fig. 3a are three 120° sectors with straight-line boundaries. For more general b 2 ( > 1), only the aT boundary in fig. 3a need be a straight line (by symmetry), the a/3 and/3,/boundaries being curves. The form of the curves shown in the figure anticipates the results of section 4.

3. The interfaces

In the interiors of the three regions of fig. 3, far from the boundaries, the densities Pl and P2 have the values p~', P2", etc., that they have in the respective bulk phases, just as in fig. 1. Far from the three-phase junction but near any of the boundaries the gradients Vpl and Vp2 are in the directions z,,a, etc., perpendicular to the interfaces (section 1). There, the solutions of (2.6) that go to the bulk-phase densities (2.1) far from the boundaries, with z,,a, etc. measured from the respective asymptotes in fig. 3b, are

a/3 interface Pl = - 1 + ½ e -Iz-~l

& = ~ h ~-1~o~1

( a side)

P2 = b(1 - ½ e -I~1)

(/3 side),

Pl = ½ e-lZz~l,

Pz = b ( 1 - ½e -I~1)

(/3 side),

Pl = 1 - ½ e -Iz~,l

& = ½b e -Iz~*l

(Y side)

Pl = - 1 + e -Iz~l ,

P2 ~O

(a ~;no~

p~ = 1 - e-t~"'l,

P2 = 0

(31 side) .

P, = -½ e-I~"~l,

,

/_

/3T interface

a,/ interface

(3.1)

As anticipated in section 2, these are continuous and have continuous first derivatives with respect to z,,~, etc., but discontinuous second derivatives at z~t~ = 0, etc. The corresponding interracial trajectories in the p~, p, plane, which specify

B. Widom, H. Widom / Model.for line tension in three-phase equilibrium

79

how Pl and P2 vary with each other within the respective interfaces, are obtained by eliminating z,,o, etc. between p, and P2 in (3.1)"

bPl - P2 = - b

(a/3 trajectory),

bPx + P2

(/3y trajectory),

b

=

(3.2)

( a y trajectory).

P2 -" 0

Here, then, these trajectories are just straight lines connecting the bulk-phase points a,/3, and y [fig. 2 and eqs. (2.1)] in pairs. The sector boundaries in fig. 2 and eqs. (2.4) are their perpendicular bisectors. When each of (3.2) is solved simultaneously with the corresponding one of eqs. (2.4) the results are the coordinates of the intersections of the interracial trajectories with the sector boundaries in the Pi,/12 plane, and are at the same time the values of 01 and P2 in the middle of each two-phase interface and the asymptotic values of p, and 02 along each of the three boundaries in fig. 3, infinitely far from the 1 three-phase line. These are Pl " - ½, 02 "" ½b on the a/3 boundary., p,---~, P2 "" ~b on the fly boundary, and Pl = 0, 02 "" 0 on the a y boundary. The interracial tensions tr,,o, etc. may now be obtained from (1.3) and (1.5), with F and m 0 as in (2.3) and (2.5) and with the densities given as functions of z ~ , etc. in (3.1). Alternatively, we see in (2.3), (2.5), and (2.6) that in this model the two densities O~ and P2 are uncoupled. Each interface, far from the three-phase line, is then like two independent one-variable interfaces, one described by the variation of a variable Pt and the other by that of a variable P 2 ' with 1/¢l = F~(pl ) + lmll[Vpl[ 2 and gr2 = F2(p2 ) + ½maa[Vp2t ~-. The tension of the real interface is the sum of the two one-variable tensions. But each of the latter may be obtained from its F alone (without having first to find 01 and P2 as functions of z,,t3, etc.) as an integral over that density [3]" pi ~

P2B

0"043 : j ~ / 2 m , l F l ( p , ) do, + J ~/2m22F2(p2) dp: , pl a

etc.

(3.3)

p2 a

We thus have two alternative routes to the interracial tensions. By the first route, via (1.3), (1.5), (2.3), and (2.5), 0

tr,,o = f [p, 2 + (P2 - b)2 + p12 + p;21 dz,,o

+ f[(P, + 1) 2 + 0

022 + O;2 + p'22] dz,,t3 ,

(3.4)

80

B. Widom, H. Widom I Model for line tension in three-phase equilibrium

where the primes denote derivatives with respect to z,~,. The integrands are expressed as functions of z,,t3 by (3.1) and the integrals then evaluated. The ter~ions tr~v and tr,~v are found similarly. (By the symmetry of the model it is clear that o-,,~ = trm,.) We find in this way ff,~ = o-or = ~i -{-1 + b 2) ,

(3.5)

o-,,v = 2.

By the alternative route, via (2.3), (2.5), and (3.3), -~

-2f

0

(on+l)do,+2

½b

f

-112

-1

= ½(1+ b 2 ) ,

b

O=dO=+2f(b-o=)do=

0

etc.,

~b

(3.6)

confirming (3.5). From (1.1) and (3.5) we find the cosines of the contact angles (angles between asymptotes in fig. 3b), COS O t - - C O S ~ - -

2 2 1+ b

'

cos/3 =

8 - 1. ( 1 + b 2x2 )

(3.7)

Also from (3.5), we locate the wetting transition, where o'~ = tr~ + o'~, [eq. (1.2)1: b 2= 1

(wetting).

(3.8)

From (3.7) we see that this is consistent with a = y = xr and/3 = 0, as expected. As long as b 2 > 1 the a y interface is not wet by /3; the tensions satisfy the inequality o-~,~+ o'~ > o-~, and the phases meet at a three-phase line with contact angles given by (3.7).

4. pa(r),

p~(r) in

the whole plane

What we do here in deriving p~(r) and P2(r) from (2.6) (and much of what we do later in appendix B) will be formal but can, no doubt, be made completely rigorous. In particular we will be working with Fourier transforms of functions which do not have Fourier transforms in the usual sense; they must be interpreted as distributions (generalized functions). Let r be any point in the x, y plane of fig. 3. We denote the Fourier

B. Widom, H. Widom / Model for line tension in three-phase equilibrium

81

transform by a circumflex, so that for an integrable function f(r) (i.e., a function in the class L ~)

z< = f

e-i-~"f(r)

da,,

with da, an element of area in the plane. If we denote the characteristic function of a set E by XE, then eqs. (2.6) yield P,(~)=

~ ( ~ ) - ~',,(~) ~2+1 '

b~a(~) P2(~) = ~ 2 + 1

(4.1)

(interpreted as distributions), where ~: = [~ [. Since the inverse Fourier transform of (~2 + 1)-1 is (2"rr)-iK0(r), where r = and K 0` in standard notation [7], is the Bessel function of order 0 that decays exponentially rapidly for large values of its argument, and since the Fourier transform of a convolution is the product of the Fourier transforms, we deduce

[r[

(4.2) 2~rP2(r ) = b

f K,,(ls- rl)da

,

where the integrals are through the interiors of the indicated regions. Equivalently, that the p~ and P2 given by (4.2) are the solutions of (2.6) that reduce to (2.1) in the interiors of the regions may be verified by substituting (4.2) into (2.6), first noting that (2xr)-~K0(r) is the potential function in two dimensions, satisfying V2K0(r)= Ko(r)- 2rrr(r). As expected, the Pi and P2 of (4.2) are continuous with continuous first derivatives, but discontinuous second derivatives, at the boundaries between regions. The solution (4.2) does not become fully explicit until the boundaries of the regions a,/3, and 7 are specified. The requirement that the pl(r) and PE(r) of (4.2) satisfy (2.4) identically, for all points r on any of the boundaries, implicitly determines those boundaries. The aY boundary, by symmetry, is always just a straight line, as in fig. 3. We are not able to determine the whole of the a/3 and/33, boundaries explicitly for general b, but we determine here their qualitative shapes including their form near their point of intersection. We already know the asymptotes [fig. 3b and eqs. (3.7)]. Let 0 be the point of common intersection of the three boundaries in the x, y plane of fig. 3. From (2.4),

B. Widom, H. Widom / Model for line tension in three-phase equilibrium

82

p,(0) = 0 ,

p2(0) -= ½ ( b - b - l ) .

(4.3)

The first of these is satisfied by the first of (4.2) just by symmetry, and so yields no new information, but from the second of (4.2) and (4.3), and (3.7), f

Ko(s ) das = rr(1

-

lib 2) =

1 - cos(/3 / 2) 2at 2 - cos(/3/2) '

(4.4)

/3

where s is the distance of s from O. For a sector /3' with straight-line boundaries and vertex at 0, and sector angle that we also call/3',

Ko(s ) da s = fl' f Ko(s ) s ds = / 3 ' . sector/3'

(4.5)

0

The second equality follows from the more general [8]

f Ko(s) s n-1 ds = 2"-2[F(n/2)l 2 ,

(4.6)

o

which we shall need later (appendices A and C). Since 2"tr[1-cos(/3/2)]/ [ 2 - c o s ( / 3 / 2 ) ] < / 3 for /3<120 °, and K0(s)>0 for all real, positive s, the /3 region when /3 < 120° [i.e., when b 2 < 3 , from (3.7)] is contained within a sector of angle/3 whose vertex is also at 0, as indicated schematically in fig. 4a. Since the/3 region (when b 2 < 3) thus fits inside a sector of angle/3 that has the same vertex, and the angle between asymptotes is also/3, the opening angle

p

V

/ /

U

(o)

(b)

Fig. 4, (a) Region /3, with vertex at 0 and angle /3 (<120 °) between the asymptotes to its boundaries, and a .sector of angle/3 with vertex at O. (b) Angle/3" between the tangents to the boundaries of the/3 region at O, and tht angle/3 between asymptotes.

B. Widom, H. Widom / Model for line tension in three-phase equilibrium

83

between the a/3 and fly boundaries at 0, i.e., the angle 13" between the tangents to those boundaries at 0 (fig. 4b), must be less than /3 (when 13 < 120°). Indeed, since Ko(s) is positive and decreases exponentially rapidly with increasing s for large s, and since the integral of Xo(s) over a sector is the sector angle, by (4.5), the integral over the /3 region in (4.4) is a rough measure of that initial angle/3*. For small/3 the right-hand side of (4.4) is ~/32/4, from which we conclude that for small 13 the opening angle/3* is of order 132 We now obtain a formula for the opening angle /3* from which we may confirm some of the features of fig. 4. Let r, 0 be the polar coordinates of r, with origin at 0 and with the angle 0 measured counterclockwise from the a y boundary (the vertical line in figs. 3 and 4), and let s, 4, be the coordinate of s. Then from dKo(s)/ds =-Kl(s ) [7], we have for small r,

K 0 ( l $ - r l ) = Ko(s ) + KI($ ) r c o s ( 0 - t~) + - -

(4.7)

(r--~0) .

Then from (4.2),

27rpl(r ) = 2r sin

0f

KI(S ) sin ~b da, + - - -

(r---, 0),

(4.8) 2-rrP2(r) = "tr(b - 1/b)

+ br cos 0 f Kl(s) cos & da~ + - - .

(r--* 0) .

/3 For the first of (4.8) we have made use of obvious symmetries between integrals over the cz and ), regions, while for the second we have used, in addition to (4.2), the first of the equalities (4.4) and the fact that f~ K~(s)sin 4~ d a ~ - 0 by symmetry. Points on the /37 boundary near 0 have angular coordinate 0---ax - / 3 * / 2 (fig. 4b). Therefore, from (4.8) and the second of (2.4),

f K (s) cos & da, tan(B*/2) = - ½ b 2 ~

f Kl(s

.

(4.9)

sin ~b da,

The integrals in (4.9) are convergent. That in the numerator is negative because K, > 0 and cos O < 0 in the/3 region, and that in the denominator is positive because sin ~b > 0 in the 3, region. Thus, tan(/3*/2) is finite and greater than 0. Therefore/3" # 0, so the tq3 and/33, boundaries do not meet in a cusp (except asymptotically, as/3 ~ 0, when, as we saw,/3" vanishes proportionally

84

B. Widom, H. Widom I Model for line tension in three-phase equilibrium

to/]2); and/3* # ~r, ,.;o neither do those two boundaries meet at 0 tangentially with vanishing slope. This confirms the qualitative features of figs. 3 and 4. When bE __ 3 the raodel is symmetric in the three phases a, fl, and % as one sees in (2.3) and fig. 2. The t~,/3, and ~/regions in figs. 3 and 4 (as well as those in fig. 2) are then 120° sectors, and fl* =/3. That is also consistent with (4.4) and (4.5). In that fully symmetric case, (4.2) is an explicit solution for the densities p l ( r ) and pE(r). As bE-->1+ (the wetting transition), the angle/3 --> 0 while a = y--> xr, again consistently with (4.4). The aft and fly boundaries approach each other and become extensions of the line of the a~, boundary. This is the only other case for which the boundaries in figs. 3 and 4 are known explicitly.

5. Line tension

Moving the center of the circle A of radius R from where it is in fig. 1 to any other point becomes equivalent, as R---->~, to changing the nominal location of the three-phase line, to which the line tension r is known to be invariant [1]. In evaluating r from (1.4) we may therefore choose the center to be anywhere in the plane, provided only that we fix it before letting R---->~. Choosing it at the intersection of the asymptotes (dashed lines) in fig. 3b would correspond to fig. 1, but we find it slightly more convenient to choose it at the point of intersection of the boundaries; i.e., at 0 in fig. 4. For now, then, the center is at 0. This question arises only when evaluating r by (1.4), not by the KerinsBoiteux formula (1.7), which makes no reference to any center; but in the present model, as will soon be apparent, the route via (1.4) is the more convenient. From (1.5), (2.5), and Green's theorem, the integral (1.4) in the present model is

f '~" da = f [F(Pl, P 2 ) - (pI~72pl "~- p2~72p2)] da + f n " ( p , V p I + p2VP2) d l , A

A

C

(5.1) where C is the circumference of the circle, dl is an element of length along it, and n is the unit outward normal to it. The gradients Vpl and Vp2 vanish in the interiors of the regions a,/3, and y in fig. 3, far from the boundaries, and they do so exponentially rapidly with increasing distance from the boundaries, as seen in (3.1). Not far from the boundaries, but far from their junction at O, those gradients are perpendicular to the boundaries, while as R--->~ the boundaries themselves become orthogonai to the circle circumference C.

B. Widom, H. W i d o m / Model for line tension in three-phase equilibrium

85

Therefore, as R--> ~ the gradients become orthogonal to n, the normal to C, where C crosses the boundaries. Hence, as R--> ~ the iategrand in the integral over C in (5.1) vanishes on the whole of C, in such a way that the integral itself vanishes in that limit. Then from (2.3), (2.6), and (5.1), fgtda= A

[l+p,(r)lda r - b

f a(r
+

f

f

[ - b + p2(r)l da,

[3(r
(5.2)

[1-pl(r)]da,+°(1),

v(r
where the notation a ( r < R) means that the integration is extended through that part of the a region that is within the distance R of 0, etc., and o(1) is a correction that vanishes as R---->~. It is because the integrands in (5.2) are linear in p~ and P2 that the route to z via (1.4) is more convenient than that via (1.7), by which the corresponding integrands are quadratic in p~ and Pz. From (5.2) and (4.2), f

1 tit da = 2a R + b213R + -- 1,,r

1

'IT

A

where a R is the area of the region

'IT

1~,,

b2 ,~_ 1,[3 + o( 1 ) , ~

(5.3)

a(r < R), etc., and where

f y o, lS rl, a ar

etc

a ( r < R ) [3

For (5.3) use was made of the first three of the symmetry relations

(5.5) By (4.5), the integral of K 0 over the whole plane is 2rr, so we have the identities 2"rran = I,~o, + l,~t3 + I,~,t ,

(5 o~' J J 2'n'/3R = 2113,, + ! t ~ , for the second of which use was also made ui" the fourth of (3.3). TiLe corresponding formula for 2~ryn yields no new information, because of the symmetries (5.5).

B. Widom, H. Widom / Model .for line tension in three-phase equilibrium

86

From (5.3) and (5.6) we find

f

~ da = 1 (21~ v + l~tj + b21t3~) + o ( 1 ) .

(5.7)

'It'

A

We show in appendix A that l,,v --- I~, --- la~ ---arR as R---->0o. Then from (1.4), (3.5), and (5.7), z = lim 1 [2(I,,, - "rrR) + (I,~ - ,rR) + b (la~ - "rrR)]. i

2

R---, ~ 'iT

(5.8)

Thus. to calculate ~"we have in principle to find the O ( 1 ) corrections to "rrR in the asymptotic (R---> oo) expansions of the integrals I ~ , etc., of (5.4), and then combine them as in (5.8). For b 2 = 3, we have seen, the regio6s a,/3, and 3, are three 120 ° sectors with straight-line boundaries. The sector boundaries in figs. 3 and 4 then coincide with their asymptotes, the point of intersection of the asymptotes is the same as 0, and/3* is the same as 13 (= 120°). For that case, as will be seen, we are able to calculate ~- exactly. By the same methods by which one calculates ~- when /3 = 120 ° one may, more generally, calculate the constant term (i.e., the R ° term) in (5.2) for large R when the regions a,/3, and 3' in (5.2) and (4.2) are arbitrary sectors. We shall carry out the computation only in the case where a = 3 ' and 2"rr = a +/3 + 3' but with/3 now any angle less than 180 °. The center of the circle of radius R is at the common vertex of the sectors. We continue to take the angles a , / 3 , and 3' to be related to the p a r a m e t e r b 2 by (3.7); now b 2 may have any value greater than 1. Then that constant term in (5.2) is ~-for b 2 = 3 and is a certain approximation to z, which we call z', for all b 2. In appendices B and C we show that z'= _1

[ - 2 / 3 cot(½/3) + (b 2 - 1)(xr - / 3 ) cot/3 + b 2 + 3].

(5.9)

'IT

The derivation in appendix B is by Mellin-transform methods and starts from (5.2) [with (1.4)]. In appendix C we outline a more elementary (but equally lengthy) derivation starting from (5.8). Both methods are instructive, and the coincidence of their results is an important check of the correctness of (5.9). In the case of three equal sectors (/3 = 2,r/3, b 2= 3), for which ~" =~', we find from (5.9) ~"= - 6 / ' r r + 2/X/3 = - 0 . 7 5 5 . . .

(b z = 3 ) .

(5.10)

B. Widom, H. Widom / Model .for line tension in three-phase equilibrium

87

We note that this line tension is negative, as it often is in other models [1, 5]. The behavior of z' as /3-*0 (bE---> 1+) is also easily found; (3.7) gives b 2 - 1--- ¼/32, so it follows readily from (5.9) that ~-' --- -/3/4

(/3-->0).

(5.11)

This is no longer ~-, but it is possible and perhaps even likely that as/3-~, 0 the real ~-, like ~-', vanishes through a range of negative values and does so proportionally to the first power of /3. What makes this believable is the resemblance between the sector geometry for which r' is calculated and the geometry of fig. 3. In particular, for given b 2 the angle/3 between asymptotes in fig. 3b is the same as the sector angle/3 in the calculation of r'. Also, the difference between ~- and r' is expressible as integrals through regions which, although unbounded, have finite areas. If these areas tend to zero a s / 3 - * 0 (even somewhat less suffices) then it would follow that r - * 0. Unfortunately we have not been able to show this. Nevertheless it seems likely that r does tend to zero as/3 -* 0 and that the sign and magnitude of ~"are the same as that of r'. It is possible that even the coefficient 1/4 is the same; but we stress that none of this is yet proven. The vanishing of ~-proportionally to /3, if it does so vanish, would be as in an earlier model [5].

6. Summary and conclusions

We have defined a model free-energy density gr as a functional of two spatially varfing order parameters Pl and P2, from which to calculate interracial and line tensions in three-phase equilibrium. The interracial tensions are displayed in eqs. (3.5), the contact angles in (3.7), and the spatial variations of pl and P2 in (4.2), all as functions of a parameter b E in the model ~. The line tension r is the R-independent term in the asymptotic expansion of (5.2) for largc R. For the c a s e b E - 3, in which the three phases play symmetrical r61es, the regions of integration a,/3, and 3' in (4.2) and (5.2) are three 120° sectors. The methods of appendices B and C then yield the line tension r given in (5.10). More generally, what is calculated in the appendices B and C is an approximation to ~', called ~-', obtained by replacing the regions of integration t~, /3, and 3' by sectors whose ~ngles are the contact angles in the original problem. This r' is given by eq. (5.9). In the symmetric case referred to above, and only there, it is the same as ~'. It is found that ~ " - - - f l / 4 as the contact a n g l e / 3 - , 0 (i.e., at a wetting transition, which occurs when b E - * 1+), and it is plausible that z, too, vanishes proportionally to the first power of/3, through

88

B. Widom, H. Widom / Model for fine tension in three-phase eqvilibrium

negative values, as/3 --> 0. If this were so it would agree with what was found in an earlier, less detailed model; and it would strengthen the supposition that the vanishing of ~-at a wetting transition may be universal, and that its vanishing with the first power of the vanishing contact angle may be characteristic of all mean-field theories. Although the calculation of the line tension in this model has proved unexpectedly difficult and challenging, and we have succeeded only partly, the model's seductive simplicity makes us hopeful of future progress.

Acknowledgements We acknowledge the collaboration of A.S. Clarke in an early stage of this work, and continuing helpful discussions with I. Szleifer. The work of B.W. was supported by the US National Science Foundation and the Cornell University Materials Science Center. The work of H.W. was supported by the US National Science Foundation.

Appendix A Here we show that the integrals I,,,, I,~, and I ~ defined as in (5.4) are all asymptotic to "trR as R---~, as required in (5.8). More generally, I~----rrR, where tt and v are any two of the regions a,/3, and % Because the integrand in (5.4) is short ranged the important contributions to it come when s and r are both close to the boundary between the two regions of integration. When R is large most of even those contributions come from the distant parts of the boundary where the boundary is close to its asymptote. Then in the asymptotic limit R---~, as we see from the definition (5.4) and from fig. 5, ~

,,~-~ff 0

lr-arcsin(z/h)

f z

Ko(h,hdg, dhdz,

(A.1)

arcsin(z/h)

k. Fig. 5. Geometry for the integral (A. 1). L is the asymptote to the boundary between regions tt and v; r and s are points on opposite sides of L: the distance between r and s is h, the distance between r and L is z, and the angle between L and the line joining r and s is g~.

B. Widom, H. Widom / Model for line tension in three-phase equilibrium

where 0 ~<

arcsin(z/h)

89

~< ~r/2. Thus,

Rf J(z)dz

(A.2)

0

where

J(z) = f [~r - 2 arcsin(zlh)]Ko(h ) h dh .

(A.3)

From (A.3), and with the prime meaning derivative with respect to the indicated argument,

(x 2 - 1)-"2xK0(zx)

J'(z) = - 2 z f

dx ;

(A.4)

1

so from (4.6),

f

z"J'(z) dz =

-~rn!

(n = 0, 1, 2 , . . . ) .

(A.5)

0

Thus, all the moments n = 0, 1, 2 , . . . of J'(z) are - r r times the corresponding moments of e x p ( - z ) . Therefore J'(z) = - r r e x p ( - z ) and J(z) = ~r e x p ( - z ) + const. But Ko(h ) vanishes rapidly as h---~c, so J(0c) = 0, by (A.3). Therefore const = 0 and

J(z)

= "rr e -z .

(A.6)

Then from (A.2), I.~ --- rrR

(R--~ ~ ) ,

(A.7)

as we wished to show.

Appendix B Here we derive (5.9) from (5.2) [with (1.4)], By definition, z' is the quantity thus calculated when the regions a, /3, and ), are taken to be sectors with straight-line boundaries, with the center of the circle of radius R at the vertex

90

B. W i d o m , H. W i d o m I M o d e l f o r line tension in three-phase equifibrium

of the sectors. The sector angles t~,/3, and y are still taken to be related t o b 2 by (3.7). For this appendix alone we adopt a simplified notation: x and y in place of the former r and s, and dx and dy in place of the former da, and da S. Also, DR is the disc of radius R centered at the vertex of the sectors, and the intersection D R f'l p for some region p is the restricted region previously denoted p(r < R). Finally, in the subsequent calculations we denote by 6 half of the sector angle /3, so 28 =/3. For arbitrary sectors p and q we write

Ipq = Ipq(R)=

2~rl f f Ko(]x _ y])dy dx

(B.I)

DRN p q

and denote by Tpq the constant term in a presumed asymptotic expansion

Ipq(R) 32R2 + a,R + Tpq+ O(R -I) =

(R--> ~).

(B.2)

Then it follows from (5.2) [with (1.4)] and (4.2) that T' = T~,~, -- "raa -- b2Tot3 -- T~,.r + "rva --

2T,,r -- 2%,~ -

b2Tot3 ;

(B.3)

the last follows by symmetry. The formula for Tpq is rather complicated and before stating it we introduce some more notation. Suppose the sectors p and q are given by inequalities Pl ~
ql~
respectively. Then we define

d~pq(O)=length of [q,, q2] N [p~ - 0, P 2 - 0 ] . The graph of ~)pq is given in fig. 6 in the case Pl - ql < P2 -- q2" Whichever of P ~ - ql, P 2 - q2 is smaller is on the left; if they are equal then the graph is

/,,,, ,,I

,

Pl-q2 P!-ql

Pz-qz P2-qw

Fig. 6. Graph of ~t,,"

B. Widom, H. Widom I Model for line tension in three-phase equilibrium

91

triangular rather than trapezoidal. The sides of the graph have slopes +_1. The usefulness of this function lies in the fact that

ff

f (~pq'O'

q P

d0

,B.4,

-~

for any function ~. We denote by II011 th~ ,it-periodic extension of the function ]O[ ([t91<~ Ir/2) and by 1[[01[[ the 2~r-periodic extension of 101 (101 ~ ) . Alternatively,

Iioli=

dist(O, "trT/),

IIio111= dist(O, 2"n'7/),

where 7/denotes the set of integers. Let 4, denote any function with the qualitative properties of 4)pq: it is continuous, piecewise linear, and vanishes outside a finite interval. We associate with ~b two quantities which we denote by I(~b) and II(~b). The first is simple, I(~b) = -

(B.5)

2--~1 ~ Iloll cotlloll [,~'(o + ) - ,t,'(o-)]. ,

where the sum is taken over all points of discontinuity 0 of 0'. To define the second we look at the piecewise constant function

IIIolll'x[o,=+~j(lllolll)~'(o)

(IIiolii' = dillolll/dO)

(B.6)

(XE is again, as in section 4, the characteristic function of a set E), denote by J any of its intervals of constancy where the function is nonzero, and denote by vj the value of the function on J. Write aj and bj (0 ~ aj < bj ~<"tr/2) for the values of II10111at the end-points of J. Then II(~b) =

21~ vj(cot b s - cot as) ,

(B.7)

J

v~ where we make the ,.onven.v,, " " ":"" cot 0 = O tx,j. After all this our formula for rpq can be stated simply as (B.8)

Zpq = I(~bpq)+ II(~bpq). Hence, from (B.3), r ' = I(~b) + II(~b),

where 4) = 24),,~ - 2~b.~ -

b2~/3~

.

(B.9)

92

B. W i d o m , H . W i d o m I M o d e l f o r line tension in three-phase equilibrium

-T+8

-28

28

0

,

,

,

i . f / -

/

-2-2~ " " ' - , , , . . . / ' 4 4

~ "n'- 8

"n'+8

2"n"

Fig. 7. Graph of ~b for 8 < ~r/3.

Before giving the rather lengthy derivation of (B.8) we shall show how (5.9) follows from (B.9). Fig. 6 helps us draw the graph of ~b which is given in fig. 7 for the case 8 <~r/3. (When 8 = ~ / 3 some of the points coincide.) The (self-explanatory) tables I and II list the ingredients necessary for the computation of I(~b) and II(4~) in the case 8 <-rr/4. We find that in this case I(4~) = - ~ 1 [(1

-

b2)28 cot 28 - 48 cot 8 + 3 +

b 2]

II(~b) = (1 - b 2) cot 2 6 ,

(B.IO)

and so (5.9) holds (with 26 = / 3 ) . Tables III and IV are the analogues of tables I and II, respectively, when rr/3 I> 8 ~>~r/4. (When 8 = Tr/3 rows 1 and 2 of table III are to be combined Table I

Table III

0

~ ' ( 0 + ) - ~'(0-)

II01l

0

~ ' ( 0 + ) - ~'(0-)

II0ll

-'tr+8 -28 0 28 rt - 6 "tr + ,3 2~r

-2 -b z 4 + 2b 2 2 - b2 -2 -4 2

6 23 0 26 6 6 0

-~r+6 -28 0 28 "tr - 6 "tr + 8 27

-2 -b z 4 + 2b 2 2 - b2 -2 -4 2

8 28 0 28 - rr 6 8 0

Table II J

vs

as, b s

(-'tr/2, - 2 ~ ) ( - 2 & 0) (0, 26 ) (28, ~/2) (3"rr/2, 27r)

2 2 + b: 2 + b" 4 2

26, rr/2 0, 26 0, 26 26, ~/2 O, "rt/2

Table IV J

Os

as, bs

(-~r/2, 0) (0, 7r/2) (3w/2, 2~r)

2 + b" 2 + b2 2

0, Tr/2 0, rr/2 0, rr/2

B. Widom, H. Widom I Model for line tension in three-phase equilibrium

into the row -2~r/3, - 2 ~r/3.) We find that

93

b 2, ~r/3 and rows 4 and 5 combined into 2~r/3, - b 2,

I(~b) = -~1 [(1 - b2)(28 - ~r) cot(28 - "rr) - 48 cot 8 + 3 + b 2] II(~b) = O, and so (5.9) holds in this case also. It is certainly curious that these formulas for I(~) and II(4~) and those given by (B.10) are different, yet the formula for their sum is the same in the two cases. This is not a coincidence; (5.9) holds for all values of 8 ( = ~ / 2 ) and the industrious reader could, from the results derived here, check that this is so. (It is seen to be so by the alternative m e t h o d of appendix C.) O f course it would be nice to be able to show a priori that r is analytic in & We proceed now to the derivation of (B.8), which rests on the study of the Mellin transform of Ipq(R),

f

R-S-'Ipq(R) dR.

(B.I1)

o

It is not hard to see that as R - + 0

lpq(R) = cR 2 + O(R 3) for some constant c and so J'~ R-S-llpq(R)dR is analytic for R e s < 2 and continuous for Re s <~ 2, s ~ 2; in fact it extends analytically to Re s < 3 except for a simple pole at s = 2. The expansion (B.3) implies that j'~ R-S-llpq(R)dR is analytic for Re s > 2, continuous for Re s = 2 (s ~ 2), and continues analytically to Re s > - 1 except for simple poles at s = 2, 1, 0 where it has residues a 2, a~ and rpq respectively. Thus we consider (B.11) first for Re s = 2 (s ~ 2) and extend it analytically into Re s < 2; the quantity Tpq is the residue of this continuation at its pole at s = 0. Integration by parts and (B.1) give

f R-S-~lpq(R) dR = - s 1 f R -sl;q(R) , dR 0

,0

o

-

2"n'slf

f [xl-SKo(ix-

p q

Yl)dy d x .

94

B. Widom, H. Widom / Model for line tension in three-phase equilibrium

Thus ~-pqequals the value at s = 0 of

Ixl-SKo(lx yl) dy d x ,

2"a"

-

(B.12)

P q

where of course we mean the value of the continuation from Re s = 2. Now /-

1

2'n"

j Ko(lx- yl) dy

is the convolution of the functions (2~r)-lK0(lx]) a n d Xq(X) and so its Fourier transform equals

I¢12 +

1 "

It follows then from Parseval's relation in two dimensions that if we write es(X ) = [xl -s then (B.12) is equal to 1 f~ / ~ ' q ( - - ~ ) dse . 4"u2 esXp(¢) I¢1 + 1

(B.13)

Introducing polar coordinates shows that P2

esx"~'p(Se)-

ff

r -s+l

e -ira'n°

dr dO

Pl 0

where n o is the unit vector in the direction 0. For ~. n o ~ 0 and Re s = 2 (s ~ 2) the inner integral e q u a l s / - ' ( 2 - s) (i~. n o)s-2 where the exponential is defined by specifying that arg(i~:, n o ) = +-~r/2 .

(B.14)

Thus P2

e~p(~:) = F(2 - s) f (i~ • no)S- 2 d 0 . Pl

This can be extended to other values of s. Althou~h~ ~,.~(i~"n , ) '-2 i,:.~ not a locally integrable function of ~: when Re s < 1 it can be analytically continued,

B. Widom, H. Widom I Model for line tension in three-phase equilibrium

as a distribution, the distributional lim (is

l

E-O+

to all values of s [9]. It is denoted by (it limit

l

95

it, + O)‘-’ since it is

n, + 15)‘~~,

where we take ]arg(i[

l

in conformity

n, + &)I6 ~12

(B.15)

with (B.14).

So the analytic continuation

of ezP is given by

PZ eTx,C 6) = r(2 - s) 1 (ie* n, + O)‘-’ d6

.

PI

Taking s = 0 gives 42 = I (-it*

i,(-4)

n, + 0)02 de.

41

Hence (B. 13) is equal to 92

P2

- r(2 - s) 111 4n2 91 Pl

(is

l

n,, + 0)“-2(-i&

l

n, + 0)-2

dt de, de,

IslZ+l

L

and the next step is the evaluation of the inner integral. If we rotate the e-plane so that ns, is the direction of the new horizontal axis we see that the inner integral equals

I

(it 9n, + 0)“~‘(-it, IsI’+ 1

+ 0)-2 d5

(5 = (6,

e2), 8 = 0, - e2).

If we make the change of variables q:=

5

17

r7=5-f?=5*

we have%l,T‘I)ms

9

cos 8 + t2 sin

8,

82)= sin 0 and an easy computation gives

so the last integral is equal to

96

B. Widom, H. Widom t Model for line tension in three-phase equilibrium f f Isin 01

(in + o y - 2 ( - i n , + 0) -2 sin20 --1-'i~i" "l- n 2 - - 2r/i 77 cos 0 dr/l dr/ "9

•

(B.16)

We evaluate S---dT/l by going into the upper half-plane. Note that ( - i t h + 0) -2 is analytic on (I) and above the real axis and so we only pick up the residue at the pole of the integrand in the upper half-plane. This occurs at 71cos 0 + ilsin 0l~ / 1 + n 2 and we find that (B.16) is equal to

(in O)s-2 drl +

--71"

_

~/1 + r / 2 ( r / c o s 0 + i J s i n 0 l ~ l

-q-n2) 2

Recapitulating (and using the fact that F ( 2 - s) equals 1 at s - 0 ) , we have shown that ~-pq is the value at s = 0 of q2 P2

~c

lfff 4~ ql P l

-~

(i,.q + O)s-2

~/1 + 172117 COS(01 -- 02) -I-ilsin(0, - 0z)IV1 + ¢12 dn dO, dOE 0¢

:4---~

~bpq(O)dO_ Xffl+n2(ncosO+ilsinOl~/l +n2) 2

(Recall (B.4).) We now study the continuation of the inner integral from Re s = 2 to the neighborhood of s = 0. We write the inner integral as f (it/) s-2 dn _~ ~/i + ,/2 (77 cos 0 + ilsin 0]~/1 + r/2) 2 '

(B.17)

where we shall think of r/as a complex variable, Im r/<~ 0, with the exponential determined by [arg(ir/)[ ~< zr/2 in conformity with (B.15). We a~sume first that 0 < 0 < ~r. The exponential in the integrand is analytic for Im 77~<0 and so we may deform the contour by making a little indentation below and around 7/= 0, but above the pole of the integrand at r / - - s i n 0. With the change of variable 7/= sinh y this integral becomes _

!

(i sinh y)'-2

.....~ sinh2(y + i0) d y ,

(B.18)

B. Widom, H. Widom / Model for line tension in three-phase equilibrium

97

° -i8

I' r

• -iT

• - i ' t r - i0 Fig. 8. Integration path F in complex y plane for case 0 < ~r/2.

the contour being indented as before, this time above the pole at y = - i 0 . We consider the cases 0 < xr/2, 0 > ar/2 separately. The case 0 < rr/2" With F as indicated in fig. 8 we see that (B. 18) is equal to

d

_ f (isinh y)S-2 dy + 2ari ~yy (i sinh r sinh2(y + i0) = _f

(i sinh

y)S-2 dy -

, srnh-5-y7?;)

y)S_21.v--i0

2ar(s - 2) sin s- 30 cos 0 ,

the terms on the right arising from the double pole at y = - i 0 . The integral term is an entire function of s (uniformly for 0 < 0 < ~/2) and its value at s = 0 is f

sinh_2y sinh2(y + i0) d y .

(B.19)

F

If we integrate 1 y sinh-2y "rri sinh2(y + i0) over the rectangle pictured in fig. 9 and use the fact that the integrand in

° iw'-i0

1"+ i'n"

°0 ° -i0

Fig. 9. Rectangular contour in complex y plane for case 0 <'rr/2.

98

B. Widom, H. Widom I Model for line tension in three-phase equilibrium

°0

17' •

-

iTl2

• -it9 Fig. 10. Integration path F in complex y plane for case ~r/2 < 0 < ~r.

(B.19) has period ~i we find that (B.19) itself is equal to - 2 x (sum of residues of

y sinh-2y sinh2(y + i0)

aty=0and

y=-i0)

0 cos 0 - sin 0 = -4

sin30

(B.20)

Hence we have shown that in this case 0 < 0 < x r/2 the expression (B.18) equals -2xr(s - 2) sinS-30 cos 0

(B.21)

plus a function analytic near s = 0, uniformly in 0, whose value at s = 0 is given by (B.20). The case ,rr/2 < 0 < xr: We now replace the contour in (B.18) by the F pictured in fig. 10. The resulting expression is an entire function of s, uniformly for xr/2 < 0 < xr. Its value at s = 0 is f sinh_2y r sinhe(y + i 0 ) d y ,

(B.22)

which can be evaluated by integrating 1 y sinh- 2y rri sinh2(y + i0) over the rectangle indicated in fig. 11. We find that in this case (B.22) is equal to

- 4 (~ - O) cos(rr - 0) - sin(rr - 0) sin 3(~ _ O) " We combine the two cases by recalling that

(B.23)

99

B. Widom, H. Widom / Model for line tension in three-phase equilibrium ['+iT • i"n'-iO

°O • -i#

Fig. 11. Rectangular contour in complex y plane for case ~/2 < 0 < ~r.

ii011__{° ~'-0

if 0~< 0 < ~ / 2 ' if ~/2~< O~
Thus for 0 < 0 < ~r both (B. 20) and (B.23) are equal to

Iloll coslloll - sinlloll sin~llol I

-4

.

(B.24)

The term (B.21) enters only if 0 < 0 < "tr/2, and so for all values of 0 in (0, rr) it is equal to -2~r(s - 2) sinS-30 cos 0 x(O), where we write X here as short-hand for X[o,=/21- Finally, if we observe that (B.17) is an even function of 0 and has period 2~r we deduce that for all 0 it is equal to -2,rr(s - 2)sinS-3lll0lll coslll0111 x(lll0111) plus a function analytic at s = 0 and equal to (B.24) there. Recall that "rpqis obtained by multiplying (B.18) by dppq/4'rr, integrating over 0, and evaluating the result at s = 0. The last step in our derivation of (B.8) is to show that for any 4, which is continuous, piecewise linear, and vanishes outside a finite interval

t" / "tr a 1

Ilollcos[[O[[

sin[lOll

sin3[[0[[

_ 1 f ( s - 2)sin s 2

d)(0) dO = I(d~) + A

lllolll coglllolll x(lllolll)

dOl =o = l I ( $ ) i

(B.25)

A, (8.26)

100

B. Widom, H. Widom I Model for line tension in three-phase equilibrium

where I(~b) and I I ( ~ ) are given by (B.5) and (B.7), and where A is a constant which will drop out when we add. We shall use a variant of the formula for integration by parts,

f F'(O) g(O)dO = - f F(O) g'(O) dO -

~ [F(0+) -

F(O-)]g(O),

(B.27)

which holds when the function F has jump discontinuities. The sum on the right is taken over all points of discontinuity 0 of F. We shall prove (B.25) for a wider class of functions th, those for which the condition of piecewise linearity is replaced by piecewise C 2 This widening will allow us to assume first that ~ vanishes to high order at 0 = 0, -_+'r , . . . (when the two parts of the integral in (B.25) can be manipulated separately) and then derive the result for our particular ~b by an approximation argument. So assume ~b does vanish to high order at 0 = 0 , _-_xr,... and apply (B.27) with F(O) -

1

II011'

2 sin:ll011

'

g(O) = II011~(0).

"[he singularities of F occur at 0 = 0, ---rr,..., where they do not matter since g vanishes to high order there, and at 0 = +- 1aT, _ 23r- r , . . , where it has jumps of 1 and where g(O)= ½"rr~b(0). We obtain

f II011cosll011 sin3ll0ll d~(0) d0 1

1

II01l' sin~ll011 [11011~(0)1' dO- ½~ Y, ~((k + ½)~)

cotll011 [11011~(0)1"dO - ½~ E ~b((k + ½),rr),

(B.28)

the last by ordinary integration by parts, valid since 4' vanishes to high order at 0, +--xr,. . . . The sums on the right are taken over all integers k but, of course, are finite. Again, ordinary integration by parts gives

sin211oll 4,(o)dO- -

cotll011 ~'(0)II011' dO.

If we expand [lloll~(o)i" by Leibniz's formula and use the fact that see that adding the right sides of (B.28) and (B.29) results in

(B.29)

iloll" = O w e

B. Widom, H. Widom I Model for line tension in three-phase equilibrium

101

if II011cotll011 ~b"(0) d0 - ½~r~'~ ~b((k + ½)~r). Thus we have shown that the left side of (B.25) is equal to

if

2~r Ilollcotlloll ~"(0)dO + ½ ~ ~((k + ½)~) if ~b vanishes to high order at 0 = 0, ---'tr, . . . . Now take a sequence of 4~'s approaching our true 4~, which is continuous and piecewise linear. Then the corresponding sequence of second derivatives ~" approaches (in the distributional sense) the true 4~", which is equal to

[4,'(o + ) - ,t,'(o-)]~o. Here 8o is the Dirac distribution at 0 and the sum is taken over Thus (B.25) follows, with

all corners of 4,.

A=½~((k+~)rr). Finally, we prove (B.26). The left side of (B.26) is continuous for Re s i> 2 (s :~ 2) and analytic for Re s > 2. We assume first that Re s > 2 and manipulate the integral so that the value of its continuation to s = 0 is revealed. We apply (B.27) with

F(O)- -½ Ill0111'sin'-~lll0111 x(lll0111),

g(O) = 6 0 ) .

Note that since Re s > 2 , F is continuous where III0111-0. At 0 = +--~av, ---3~r,... it has jumps of ½. Hence the left side of (B.26) equals

if III0111'sin~-~lll0111 x(lll0111) ~'(0)dOThe constant term on the right equals - A I z.

r

J III0111'sinS-=lll0111 x(lll0111) ~'(0)d0]

½ E 4'((k+ ½)ax). and so it remains to show that

I

= II(4'). s=0

With the notation of (B.7) the above integral, with its factor, caiJ be written bj

!Eo,

sinS-20 d 0 . aj

(B.30)

102

B. Widom, H. Widom I Model for line tension in three-phase equilibrium

Now clearly, if O ~< a < b <~~ / 2 then b

f (sinS-20 - 0 ~-2) dO a

is analytic for Re s > - 1 and its value at s = 0 equals b

f (sin-20 - 0 -2) dO = c o t a - a -1 - c o t b + b-1

(B.31)

a

with cot a - a -i interpreted as 0 if a = 0. But for Re s > 1 b

{bS-I - a

f Os-2 dO = a

1

s-I

x

bS_l

s-1

if a ~ 0 , if a = 0 ,

whose continuation to s = 0 has the value a l - b -!

if a ~ 0 ,

- b -1

if a = 0 .

Adding this to the right side of (B.31) shows that the value at s = 0 of the analytic continuation of (B.30) is II(th), as claimed.

Appendix C

Here we outline an alternative route to the quantity r' calculated in appendix B and we verify (5.9). r ' is the quantity calculated from (5.8) and (5.4) when the regions t~,/3, and y are replaced by sectors with straight-line boundaries, with sector angles (which we also call a. 8. and y ) still related tn t h o n a r ~ a m o t o r h 2 k,, ("1 "/~ and with the center of the circle of radius R taken to be at the common vertex of the sectors. It is the same as the r ' calculated in appendix B because the passage from (5.2) [with (1.4)], which was the starting point for the calculation in appendix B, to (5.8), which is the present starting point, is the same whether a , / 3 , and y are the original regions or the present sectors. Let # and v be any two of the sectors a, fl, and y and let L in fig. 5 now be the boundary between/~ and v and t'le extension of that boundary through the

B. Widom, H. Widom / Model for line tension in three-phase equilibrium

103

vertex, so that L divides the plane into two half planes; and let z be the distance to L from a point r in sector It, again as in fig. 5. If S is the half plane on the side of L opposite r, we have from (A.3), (A.6), and fig. 5,

f

Ko(lS -

rl) das = J(z) = ~re -z .

(C.1)

S

If A is the third sector, then L, as the extension of the Itv boundary, divides A into two parts, A' and A", the former, say, bordering the sector v and the latter bordering It; see fig. 12. The half plane S is the union of v and A'. Then from (5.4) and (C.1),

lt`~ = - l t ` ~ , + ~r

f

e -z dar,

(C.2)

t`(r
with It, A, defined by the analog of (5.4). But the sectors It and A' have only their vertex in common, not a boundary (fig. 12); so from the definition of It`^,, as R--~ oo, It`u=-Ht`A,+~r

/

e -zdar+o(1 ),

(c.3)

t`(r
where

H"A'= R--=liml.^, = f f Ko(IS - rl) da. da r ,

(C.4)

t`A'

this double integral being over the full sectors/z and A'. This integral is O(1), while the integral of e x p ( - z) over the restricted sector It(r < R) is R + O ( l ) , L

Fig. 12. Sectors/z, u, A, the last divided into two parts, A' and A", by the extension of the /.tu boundary.

104

B. Widom, H. Widom / Model for line tension in three-phase equifibrium

in agreement with (A.7). We need to evaluate both O(1) contributions to (C.3) and we shall then have lim(l,~ - "rrR), with/.t and v two of the sectors a, /3, and y, as required in (5.8). We tint evaluate the double integral H,a, in (C.4). Let r, 0 be the polar coordinates of r and s, $ those of s, with the origin at the vertex of the sectors and with angles measured from an arbitrary, fixed polar axis through that origin. Replace s as a variable of integration by h = I s - r I = ( r 2 + s 2 2rs cos ~,)1/2 with ~/, = ~b(0, 4,) the angle between r and s. We specify 0 < ~, ~< xr so that sin $ i> 0. Integrate first over r for fixed h, 0, and ~b. Two matters must be carefully attended to in this integration. The first is to recognize that

s ds =

r cos ~b + 1)h d h , ---~/h2 _ r 2 sin2$

(C.5)

where the top sign applies when s > r cos $ and the bottom sign when s < r cos $; so that when ~ < xr/2 the integral over r consists of two integrals with overlapping ranges, one from r = 0 to r = h/sin ~, with the top sign from (C.5) and one from r = h/sin ~, to r = h with the bottom sign; while when ~, > "tr/2 the r integration is simply over 0<~ r~< h with the top sign from (C.5). The second matter that requires care is that one must take arcsin(sin ~/,)= when ~, < ~r/2 and arcsin(sin ~,) = xr - $ when ~, > "rr/2, if the arcsin function that arises in the integrations is not to exceed xr/2. The form of the result then proves to be the same for both ranges of ~. The r integration as just outlined leaves for the integratio~as over h, 0, and 4, an integrand that is Ko(h ) h 3 times a function of ~b(0, 4,) ~ione. The integral over h, which is from 0 to 0% then follows from (4.6). The double integral on the right-hand side of (C.4) has now become 02 ~b2

f H,a, = 2

f

1 + ('rr- ~,)cot ~, .4.,..~,~ sin2$ u,~, ,,,,

(C.6)

°1 ~l

with ~, = ~,(0, 4,) (0 < ~, <~rr) s, the magnitudes of r and s, sectors/x and A' ' ed by respectively. With O and ~b measured 0<~0<2"tr, 0 < ~ b < 2 x r , we 2 ~ - iO - 4,1, q~ = min(lO - 4,], 2rr - ] 0

the angle between the original r and s (only r and having so far been integrated over), and with the the angular ranges 01 ~< 0 ~< 02 and 4', ~ 4' <~ ~b2, from the same (arbitrary) polar axis, and with have ~, equal to the smaller of 10- 1 and

- 4,[).

(C.7)

B. Widom, H. Widom I Model for line tension in three-phase equilibrium

105

Let

x=0-+,

x’=9+ip

(C.8)

be new variables of integration replacing 6 and 4. Then d0 d4 = ;clX’ dx. The integration, which was over the rectangle Ors 0 s 6!, 41 G 4 s & in the 0, 4 plane, is now over the corresponding rectangle in the x, x’ plane. Since by (C.7) the integrand in (C.6) is a function of x alone we may immediately integrate over x’. For definiteness, we choose the polar axis to lie in the A” sector and measure angles from it in the direction of the A’ sector (fig. 13). We continue to use the same symbols p, V,A (=A’ + A”)for the sector angles as for the sectors themselves. In doing the x’ integration we must carefully distinguish the two cases p < A’ (i.e., Oz- & < 6, --.$3,j knd p > A’ (i.e., Oz- & > O1- 4,). In the former case, as seen from (C.8) and the rectangle in the 0, 4 plane, the x’ integration yields the factor ?(& - e1 + x) for x in the range O1- & s x s Oz- &, the factor 2(8, - OJ = 2~ for x in the range Oz- 4z s xq - 419 and the factor 2(8, - & - x) for x in the range 8, - 4,~ x s Oz- & ; whereas in the latter case the x’ integration yields the factor 2(& O1+ x) for x in the range O1- & s x s t+ - +l, the factor 2(rb, - 4,) = 2X for x in the range O1- & s x s & - &, and the factor 2(4 - 4, - x) for x in the range Oz- & s x s t$ - &. Then (with O1- & = V, Oz- O1= p, 4z - 4, = A’, 01 - 4 = n, and 2n + 41 - ez = A”,from fig. 13) the result of the x’ integration 1s

Fig. 13. Sectors p, v, A (=A’ + A”), with polar axis (dotted line) in sector A’! Angles are measured from that axis in the direction of the A’ sector, as indicated by the arrow. The angle between the axis and the A”,A’ boundary is then d,, etc., with 0 < 4, c c#+c 8, < ez c 2~ and 0, - 4, = 7~.

106

B. Widom, H. Widom ! Model ]'or line tension in three-phase equilibrium ~+v

w

f

H,,~,. = 2 f (X - v)j(q,) dx + 2 min( A',/.t)

j(~b) d x

,.g.

it/

2"n-A"

+2

f

(2~r - i t " -

x)j(~b) d x ,

(C.9)

~t+v

where J(q0 is the integrand in (C.6), j(O) = 1 + ('n"- q,)cot sin2q~

,

(C.IO)

now with q, = m i n ( x , 2,n" - X)

(C.11)

from (C.7) and (C.8). Next, we replace ;r by qJ as the integration variable in (C.9), taking account of (C.11) and/x + v + it = 2rr, with the result '71"

H,,~, = 2

'IT

f (~-

v)j(e)de + 2 min(it', ~ ) f j ( ~ , ) d e A

A

+2f

(O - it")j(~b) dqs.

(C.12)

These integrals, with j(qJ) given by (C.10), are now elementary. Taking account of it = A' + it", we thus find H.^, = 1 + (-tr - u) cot u + (rr - it") cot h " - ('rr - it) cot it

-(x'-

sin2it + cot it ,

(C.13)

with O(x) the unit step function, equal to 1 for x > 0 and 0 for x < 0. From (5.8), we see that we require I~,~ for the three cases in which/xv is ay, t~/3, and fla. For these, the sector A is/3, y, and y, respectively; the angle A' is /3/2, "tr-/3, and fl/2, respectively; and the angle it" is ill2, ill2, and "tr-/3, respectively (fig. i4; cf. fig. 12). We have also c~ = y = r r - fl/2, and the sector angles are all less than -n, so i t ' - / x < 0 in all three cases. In these cases of interest, then, the last terms in (C. 13) are absent, and we have

B. Widom, H. Widom / Model for line tension in three-phase equilibrium

107

B

a

I

Fig. 14. Sectors ct,/3, and y with the sector boundaries extended (dashed lines).

H~,^, = 1 - ( , r - / 3 ) [ c o t

1 3 - cot(/3/2)]

(when /xv is c~y),

H~,^. = 1 + (at - / 3 ) cot/3 + ~r cot(/3 / 2)

(when /zv is c~/3),

H~,a, = 1 - / 3 cot/3

(when /zv is /3or).

(C.14)

Next we turn to the evaluation of the integral of e x p ( - z ) over the restricted sector/.~(r < R) in (C.3); we call the integral E~(R), E~,(R)=

f

(c.15)

e-Zdar.

~(r
We recall that z is the distance of the variable point r in sector/x from the line L (fig. 12) and r is the distance of r from the vertex of the sectors. We again take 0 to be the angular coordinate of r, which we now measure from the extension of the/xv boundary (dashed line in lag. 12), through the sector A" and into the sector/z, so that the range of 0 in the sector/z is xr - tt <~ 0 <~at. Then ,r

E~,(R)=

R

f f e-rsin° rdrdO *t-it

0

,IT

- f [ s,nO

1) e -RSin° +

t

-I

sin20 dO.

(C.16)

Except near 0 = ,r the R-dependent terms are exponentially small in R and their contribution to E~,(R) is therefore only o(1). We then distinguish two cases,/x < ,r/2 and/x > ,r/2. In the former case 0 exceeds ,tr/2 over the whole range "rr - / x ~< 0 <~ "rr while in the latter case we split the integration over 0 into

108

B. Widom, H. Widom I Model for line tension in three-phase equilibrium

the two ranges ~r-/~ ~<0~<~r/2 and ~r/2~<0<~r. In this latter case, in the integral over the lower range, which does not include 0 = at, we drop the terms in e x p ( - R sin 0). Thus, with O(x) again the unit step function, E~(R) = - O ( t t -

+

~/2) cot/x

i[

1

sin 0

~)

e-R

sin 0 +

-,

d0 + o(1)

sin

9

(C.17)

where m is the larger of ~ - tt and ~/2, m = max(-rr-/x, -rr/2).

(C.18)

Replace 0 as integration variable by y = R sin 0" E~,(R) = - O ( / z -

"rr/2) cot/.t

R sin m

+R

f

1 - ( l + y2) e Y

o

-y

1 ~/1

-

yE/R2

dy + o(1)

(C.19) "

Expand the factor ( 1 - y 2 / g 2 ) - l / 2 in the integrand in powers of y2/R2 and perform the integration with the leading term: E,~(R) = - ¢9(p.- ~/2) cot ~ + R - (sin m)-' R sin m

"k

+ E

c n R -(2n-l)

f [1-(l+y) e-Y]y2n-2dy+o(1)

(C.20)

n=l

0

with c,, the coefficient in ( l - x 2) - -

c,,x"~ t /

_....~

(x z < 1 ).

(C.21)

rt ~ O

But for large R the leading contribution to each of the remaining integrals in (C.20) comes from large y, where the exponential in the integrand is negligible. Each of those integrals is then asymptotic to (2n - 1)-~(R sin rn) 2n- 1, so R ~2"-I~ times the integral is ( 2 n - l)-t(sin rn) 2"-~ + o(t). Therefore,

B. Widom, H. Widom / Model for line tension in three-phase equilibrium

109

E~, (R) = - 19(/x - xr / 2) cot tx + R - (sin m ) + ~

(2n - 1)-~c,(sin m) 2"-1 + o ( 1 ) .

(C.22)

n=l

But from the definition of c, in (C.21) we have identically, for any x ( x 2 < 1), x

f

(2n-1)-'c.x2"-' = y-2[(1- yZ)-'/z-1]dy n-- 1

0

=x

-'[1 - (1 - x2)~/21.

(C.23)

Therefore E~,(R) = - O ( / z -

"rr/2) c o t / z + R + c o t m

+ o(1),

(C.24)

since "rr/2 ~< m < xr, by (C.18). But cot rn = - 19(xr/2-/s,) cot ~, by (C.18), so we have finally

E~,(R) =

R-cot/x

+ o(1).

(C.25)

We require the two special cases of (C.25) in which ~ = c~ ( = ~ - / 3 / 2 ) /.t =/3. T h e n from (C.25),

E~(R) =

and

R + cot(/3/2) + o(1), (C.26)

E~(R) = R-cot

13 + o ( 1 ) .

In (C. 14) and (C.26) we have the pieces we need to assemble r ' from (5.8). From (C.3), (C.14), (C.15), and (C.26), I ~ = "rrR - 1 +/3 cot(/3 / 2) + (~ - / 3 ) cot/3 + o ( 1 ) , l~0 = "rrR - 1 - (~r - 13) cot 13 + o ( 1 ) ,

(C.27)

l ~ = ~ r R - 1 - (~r-/3) cot/3 + o(1). Then from (5.8),

r'=-l

'11"

[-2~cot(~/2)+(b2-1)Or-~)cot~+b2+3l .

This is (5.9) as tound in appendix B, now verified.

(C.28)

110

B. Widom, H. Widom ! Model for line tension in three-phase equilibrium

References [1] J.S. Rowlinson and B. Widom, Molecular Theory of Capillarity (Oxford Univ. Press, Oxford, 1982) ch. 8. [2] J.W. Cahn, J. Chem. Phys. 66 (1977) 3667. [3] J.S. Rowlinson and B. Widom, Molecular Theory of Capillarity (Oxford Univ. Press, Oxford, 1982) ch. 3. [4] J. Kerins and M. Boiteux, Physica A 117 (1983) 575. [5] B. Widom and A.S. Clarke, Physica A 168 (1990) 149. [6] R. Lipowsky, unpublished. M. Nilges, Diplomarbeit, Universit~it Miinchen, 1984. [7] G.N. Watson, A Treatise on the Theory of Bessel Functions, 2nd ed. (Cambridge Univ. Press, Cambridge, 1952) pp. 77-80. [8] G.N. Watson, A Treatise on the Theory of Bessel Functions, 2nd ed. (Cambridge Univ. Press, Cambridge, 1952) p. 388. [9] I.M. Geifand and G.E. Shilov, Generalized Functions, vol. 1 (Academic Press, New York, 1964) §4.4.