- Email: [email protected]
Torque balance at a line of contact
Available online at www.sciencedirect.com
Physica A 319 (2003) 151 – 162
www.elsevier.com/locate/physa
Torque balance at a line of contact Dirk Jan Bukman Department of Chemistry and Chemical Biology, Cornell University, Ithaca, NY 14853, USA Received 25 July 2002
Abstract Kerins and Boiteux (Physica A 117 (1983) 575) were the .rst to apply Noether’s theorem to the van der Waals theory of non-uniform 3uids. In particular, for a three-phase line of contact, they showed that the translation invariance of the variational integral for the excess free energy implies a balance of forces at the three-phase line. In this paper we consider the implications of the rotation invariance of the variational integral for the excess free energy. Intuitively, one would expect this invariance to lead to an equation of torque balance, and this is indeed the case—the total moment of the forces around the line of contact is zero. In the course of the calculation it will be necessary to .nd an expression for the surface of tension for a model with a multi-component density, which is a simple extension of earlier work by Fisher and Wortis (Phys. Rev. 29 (1984) 6252). At the same time, we extend to multi-component densities those authors’ results for the Tolman length and the equimolar surface, making contact with a more general calculation by Groenewold and Bedeaux (Physica A 214 (1995) 356). c 2002 Elsevier Science B.V. All rights reserved. PACS: 68.03.Cd; 05.70.Np; 68.05.−n; 82.60.−s Keywords: Three-phase line; Noether’s theorem; Inhomogeneous 3uids
1. Introduction In the van der Waals theory of inhomogeneous 3uids, the equilibrium density pro.les are determined by minimizing the total free energy of a system. The total free energy is given by an integral over space of a local free-energy density, which in the present version of the theory is a functional of the density pro.les and their gradients. Thus we have a variational problem in which the quantity to be minimized is an integral, the integrand of which (the free-energy density) does not explicitly contain E-mail address: [email protected] (D.J. Bukman). c 2002 Elsevier Science B.V. All rights reserved. 0378-4371/03/$ - see front matter PII: S 0 3 7 8 - 4 3 7 1 ( 0 2 ) 0 1 4 5 6 - 5
152
D.J. Bukman / Physica A 319 (2003) 151 – 162
the coordinates that are integrated over. This invariance, or symmetry, under coordinate transformations leads, with the use of Noether’s theorem, to the construction of conserved (divergenceless) currents. In 1983, Kerins and Boiteux [1] explored the consequences of Noether’s theorem for, among others, a system with a three-phase line of contact. They showed that the translational invariance leads to equations that express the balance of forces at the three-phase line. In addition, they derived an expression for the line tension as an integral over a “line tension density” that is localized near the three-phase line. In this paper, we will show that the rotational invariance leads to an equation of balance for the torque around the line of contact. In order to identify the torque, it is necessary to derive an expression for the surface of tension for a multi-component van der Waals theory. For a one-component theory this was done by Fisher and Wortis [2], and in Appendix A we extend their work to the multi-component case. There we also give a generalized version of their expression for the Tolman length, and compare the results with microscopic integral expressions obtained by Groenewold and Bedeaux [3]. The equation of torque balance is an interesting result in itself, since it is a macroscopic result that follows from the microscopic theory. On a practical level, it has some implications for numerical studies of lines of contact, which will be discussed in the conclusion. 2. The line of contact The system we will consider consists of m equilibrium phases separated by m planar interfaces that all meet at a straight line of contact. The usual case would be that of a three-phase contact line, as shown in Fig. 1, but the arguments presented here are valid for an arbitrary number of phases meeting at a “multi-phase” contact line. The system has translational symmetry in the direction parallel to the line of contact. The excess free-energy density is given by a functional , and the total excess free energy, for a cylindrical box of cross-section A and unit length, parallel to the contact line, is F = d x dy ; (2.1) A
where the x–y plane is perpendicular to the line of contact. This total excess free energy includes contributions from the surface tensions of the interfaces and from the line tension of the line of contact [4]. The line tension of the line of contact is given by m d x dy −
R ; (2.2) = D
=1
where the excess free-energy functional is integrated over a circular disk D of radius R (with R → ∞), centered near the line of contact (the line tension is independent of the exact location of the center), and is the surface tension of the th interface. (The line tension is not necessarily de.ned only in terms of a circular disk of radius
D.J. Bukman / Physica A 319 (2003) 151 – 162
153
Fig. 1. The geometry of the area where the three phases ; , and meet at the three-phase line of contact (black dot), which is perpendicular to the plane of the .gure.
R, but for simplicity we will use only that de.nition, even though a more general one is available.) We work in a van der Waals theory with n density components i . The equilibrium density pro.les i (x; y) are those that minimize the free energy F, which follow from the Euler–Lagrange equations F = 0 (i = 1 : : : n) : (2.3) i (˜r) We use an excess free-energy functional that is given by n 1 ˜ i · ∇ ˜ j mij ∇ = F(i ) + 2
(2.4)
i; j=1
with mij = mji . The function F(i ), the bulk excess free energy, has minima F(i ) = 0 ( = 1 : : : m), for the m stable equilibrium phases; elsewhere F(i ) is positive. The Euler–Lagrange equations for the functional given by Eq. (2.4) are n 9F − mij ∇2 j = 0 (i = 1 : : : n) : (2.5) 9i j=1
Together with the appropriate boundary conditions, these equations can be solved to give the equilibrium density pro.les.
154
D.J. Bukman / Physica A 319 (2003) 151 – 162
3. The conserved currents As Kerins and Boiteux [1] pointed out, because the integrand in Eq. (2.1) is independent of the origin and orientation of the coordinate axes, it follows from Noether’s ˜ associated with theorem [5] that there exist conserved (i.e., divergenceless) currents H this independence. More precisely, if the integrand in Eq. (2.1) is invariant under an ˜ ·H ˜ = 0 holds in.nitesimal coordinate transformation (x ; y ) = (x + x; y + y), then ∇ for a current de.ned by n 9 (9x i x + 9y i y) x − 9(9x i ) i=1 ˜ = H (3.1) n 9 y − (9x i x + 9y i y) 9(9y i ) i=1
with the i (x; y) determined by the Euler–Lagrange equations (2.3). Two transformations that were previously studied by Kerins and Boiteux [1] are the two translations along the coordinate axes, (x; y) = (; 0) and (x; y) = (0; ). We will here also look at the rotation around the origin, (x; y) = (y; −x). The conserved currents corresponding to these three transformations are, respectively, 9 F − 12 mij (9x i 9x j − 9y i 9y j ) − 9 x i 9(9x i ) ij i = ˜1= H ; 9 − m 9 9 − ij x i y j 9 x i 9(9 ) ij y i i 9 9 y i − 9(9x i ) i = ˜2= H F + 9 − 9 y i 9(9y i ) i
˜ 3 = yH ˜ 1 − xH ˜2 : H
−
mij 9x i 9y j
; mij (9x i 9x j − 9y i 9y j ) ij
1 2
ij
(3.2)
˜ ·H ˜ i = 0 over an area A perpendicular to the When we integrate the expression ∇ line of contact, we .nd with Gauss’ theorem that ˜ ·H ˜i= ˜i=0 ; d x dy ∇ d‘nˆ · H (3.3) A
9A
where d‘ is a line element on the boundary 9A of A, and nˆ is the outward normal to 9A. If we take 9A to be a circle C of radius R, then in polar coordinates around the center of C, cos nˆ = ; sin
D.J. Bukman / Physica A 319 (2003) 151 – 162
155
˜ i is and Eq. (3.3) for the three currents H
2 1 1 mij 9r i 9r j − 2 9 i 9 j R d cos F − R 2 ij 0 sin +2 9r i 9 j R
2
0
1 1 R d sin F − mij 9r i 9r j − 2 9 i 9 j 2 ij R
cos 9 r i 9 j −2 R
2
0
=0 ;
R d
=0 ;
mij 9r i 9 j = 0 :
(3.4)
ij
We now choose the center of C near the line of contact, and choose the radius R large enough so that the m interfaces that meet at the line of contact cross C perpendicularly (up to order 1=R), and at the crossing point have the structure of the equilibrium two-phase interface, so that their structure is not aNected by the presence of the line of contact. The angle between the th interface and the positive x-axis will be designated by . Now at the large distance R from the line of contact, the th interface is then (up to order 1=R) a planar interface whose density pro.le only depends on the coordinate z = R sin( − ), the distance from the line through the origin under an angle . Writing i as a function of this local coordinate, and de.ning i = 9i =9z , we have 9r i (z ) = i sin( −
)
9 i (z ) = i R cos( −
; )
(3.5) :
(3.6) i
Assuming that each interface has a .nite width, say " ; will be non-zero only for values of z ∼ " . Since R" , this means that the relevant values of for a particular interface must satisfy " ∼ R sin( − ), so that sin( − ) ∼ " =R1, and cos( − ) ∼ 1 + O("2 =R2 ). Near the th interface, the integrals in (3.4) are then to leading order in 1=R 1 R d cos F + mij i j = 0 ; 2 ij
R d sin
R d
ij
1 F+ mij i j 2 ij
mij i j z = 0 :
=0 ; (3.7)
156
D.J. Bukman / Physica A 319 (2003) 151 – 162
For large R, the diNerential R d ≈ d z, and since the separations between the m interfaces are much larger than their widths, we can rewrite the integral around the circle C as a sum of integrals perpendicularly across the individual interfaces, over all values of z : 2 m ∞ R d · · · ≈ d z · · · : (3.8) 0
−∞
=1
For a planar interface, we know from (2.1) and (2.4) that the surface tension is given by ∞ 1 (3.9) dz F + mij i j ;
= 2 ij −∞ so that the .rst two equations of (3.7) read m
cos
=0 ;
sin
=0 :
=1 m
(3.10)
=1
These two equations express force balance in the x- and y-directions, and for m = 3 are equivalent to Neumann’s triangle [4]. This result was already found by Kerins and Boiteux [1], who used a triangular contour instead of a circle. The third equation is m
zs; = 0 ;
(3.11)
=1
where 1 zs; =
∞
−∞
d z
mij i j z :
(3.12)
ij
It is shown in Appendix A that zs; de.ned in this way is the location of the surface of tension of the th interface. Then, if we imagine the surface tension of an interface to be acting perpendicular to the line of contact, along the surface of tension of that interface, then zs; is the perpendicular distance between the line of action of the force
and the origin, i.e., it is the lever arm of the force. The product zs; is then the torque due to the surface tension of the th interface, and Eq. (3.11) means that the total torque around the origin is zero. Since the forces are balanced, see Eq. (3.10), the total torque is the same around any point and thus independent of the choice of origin. In the case of three interfaces meeting at a three-phase line, one can choose the origin of the x- and y-coordinates so that it coincides with the intersection of two of the surfaces of tension. Two of the zs; are then zero, so due to Eq. (3.11) the third zs; must also be zero. In other words, for three interfaces meeting at a three-phase
D.J. Bukman / Physica A 319 (2003) 151 – 162
157
line, the three surfaces of tension intersect in a single line. Also in the case of three interfaces, one might de.ne dividing surfaces within and parallel to the three physical interfaces. If these three dividing surfaces are chosen so that they intersect in a single line, the total moment of the surface tensions acting along the dividing surfaces is zero. Similarly, one can think of a system of two coexisting surface phases, such as e.g. exists at a prewetting transition, as two planar interfaces meeting at a two-phase line of contact, the boundary line. Then Eq. (3.11) for m = 2 shows that the surfaces of tension of the two interfaces must coincide. 4. Conclusion In this paper we have examined a new application of Noether’s theorem to the van der Waals theory of inhomogeneous 3uids. From the rotation symmetry of the variational integral (2.1), which describes the microscopic structure of the density pro.les, it follows that the total torque due to the interfaces meeting at a line of contact is zero. Like the equations of force balance, this is a statement about the macroscopic properties of a system, derived from microscopic considerations. The equation of torque balance can be important in numerical calculations of the density pro.les around a line of contact [6]. In such calculations, one numerically solves the Euler–Lagrange equations (2.3) in a large region around the line of contact, while imposing Dirichlet boundary conditions along the edges of the region, i.e., prescribing certain values for the density pro.les on the boundary. These prescribed values are the bulk densities, and the density pro.les of the two-phase interfaces, to which the density pro.le reduces far away from the line of contact. The boundary conditions are chosen such that the two-phase interfaces are oriented under angles that satisfy the balance of forces, Eq. (3.10). As we have seen here, the balance of torque, Eq. (3.11), is a property satis.ed by the minimizing density pro.le, and hence the boundary conditions must also be chosen to satisfy this equation. If this is not the case, the numerical solution of the Euler–Lagrange equations will not lead to the true minimum of the excess free energy. In previous work on three-phase lines and boundary lines between two-dimensional phases [6], the bulk excess free-energy F(i ) has usually been chosen such that two of the bulk phases are related by symmetry; in such a case the symmetry ensures that the torque balance is satis.ed. In the general case, in the absence of such a symmetry, it is an extra requirement that must be satis.ed by any boundary conditions that are imposed. Acknowledgements The author is grateful to B. Widom for useful conversations and comments, and for a reading of the manuscript, and to J. Groenewold for useful comments. This work was done in the group of B. Widom, and was supported by the National Science Foundation and the Cornell Center for Materials Research.
158
D.J. Bukman / Physica A 319 (2003) 151 – 162
Appendix A. The surface of tension In this appendix, we derive expressions for the surface of tension and the Tolman length of an interface, for the excess free-energy functional (2.4). The surface of tension is a property of the planar interface; even though the calculation involves a spherical droplet of radius R, the surface of tension is well de.ned in the planar limit R → ∞. The calculation is a straightforward extension to a multi-component system of the work of Fisher and Wortis in Ref. [2]; that reference contains more details and a more thorough discussion. For the interface between two of the equilibrium phases, say and , we consider a spherical drop of radius R of phase surrounded by phase . The bulk excess free-energy F(i ) in the functional (2.4) has equal minima (F =F =0) corresponding to the two phases and , which therefore coexist in bulk, separated by a planar interface. To describe an equilibrium spherical droplet of phase, the system is taken slightly oN coexistence by including a change in chemical potential − i O$i i in F. (If one of the density components is the entropy density, the corresponding O$i is the temperature.) The calculation consists of two parts. First, we calculate the excess free-energy F of the droplet, which is related to the (radius-dependent) surface tension [R] by F = 4 R2 [R] :
(A.1)
In the limit R → ∞; [R] approaches , the surface tension for the planar interface. We are interested in the O(1=R) corrections to [R]. Second we calculate the pressure diNerence Op between the inside and the outside of the droplet. For large R; Op ∼ 2 =R, and we are again interested in the next order in 1=R. The surface of tension is at that radius Rs for which Op = 2 [Rs ]=Rs exactly, and the two previous results can then be combined to .nd an expression for Rs . This expression is independent of the radius R of the droplet, and thus remains valid for a planar interface. In the following, we will let integrals over r run from −∞ to ∞; since the integrands go to zero exponentially far away from the interface at r = R, the errors can be neglected. We can thus equally well think of r as the coordinate perpendicular to a planar interface. Since we are interested in the limit of large radius R, we expand the (spherically symmetric) density pro.les i (r) and the chemical potentials O$i in 1=R: i1 (r) i2 (r) i (r) = i0 (r) + + ··· ; + R R2 O$i1 O$i2 (A.2) + 2 + ··· : R R The two phases that coexist in bulk, corresponding to the minima of F, have densities
; i0 . Due to the chemical potentials O$i , the densities of the two phases and in the
system with a drop, ; i , are at the minima of ! = F − i O$i i , and their excess free energies are ! ; = −p ; . From the minimization 9!=9i = 0 we .nd, by expanding to order 1=R, 92 F
j1 = O$i1 at i = ; : (A.3) i 9 9 j0 i0 j O$i =
D.J. Bukman / Physica A 319 (2003) 151 – 162
159
We now derive some necessary intermediate results from the Euler–Lagrange equation (2.3). In spherical coordinates, it is 2 9F − O$i − mij 92r + 9r j = 0 : (A.4) r 9i j Expanding r −1 = R−1 (1 − (r − R)=R + · · ·), the leading orders in powers of 1=R of Eq. (A.4) are 9F mij j0 = ; (A.5) 9 i0 j 92 F j1 − O$i1 ; (A.6) mij (j1 + 2j0 ) = 9i0 9j0 j j 92 F mij (j2 + 2j1 − 2(r − R)j0 ) = j2 9i0 9j0 j j +
93 F 1 j1 k1 − O$i2 ; 2 9i0 9j0 9k0
(A.7)
jk
where here a prime indicates diNerentiation with respect to r. If we multiply Eq. (A.5) by i0 , sum over i, and integrate with respect to r, we .nd that 1 (A.8) mij i0 j0 = F(0 ) 2 ij and hence, with (3.9), that ∞
= dr mij i0 j0 : −∞
(A.9)
ij
As was mentioned before, we have here extended the integration range of r to (−∞; ∞), which should only lead to exponentially small errors. Similarly, multiplying Eq. (A.6) by i0 , summing over i, and integrating over r, we .nd, after partially integrating twice, and making use of Eqs. (A.5) and (A.9) ∞ dr O$i1 i0 : (A.10) − 2 = −∞
i
Another intermediate result follows when we multiply Eq. (A.6) by i1 , sum over i, and integrate: ∞ ∞ 92 F dr 2mij i0 j1 + O$i1 i1 = dr i1 j1 : 9 9 i0 j0 −∞ −∞ ij i ij (A.11) Next we turn to Eq. (A.7), again multiplying it by i0 , summing over i, partially integrating twice, and using Eq. (A.5) ∞ dr 2mij (i0 j1 − (r − R)i0 j0 ) −∞
ij
160
D.J. Bukman / Physica A 319 (2003) 151 – 162
1 − i1 j1 9r 2 ij
92 F 9i0 9j0
+
O$i2 i0
=0 :
(A.12)
i
If we partially integrate the term containing F, and de.ne Kij = 92 F=9i0 9j0 , we can write this ∞ 1 O$i2 (i0 (∞) − i0 (−∞)) = Kij i1 j1 2 i
−
∞
−∞
dr
ij
ij
9F i1 j1 + 2mij (i0 j1 − (r − R)i0 j0 ) : 9i0 9j0
Using Eq. (A.11) and Eq. (A.3), this is O$i2 (i0 (∞) − i0 (−∞)) =
∞
−∞
i
−
−∞
2
dr
(A.13)
2mij ((r − R)i0 j0 − 2i0 j1 )
ij
1 O$i1 (i1 (∞) − i1 (−∞)) : 2 i
(A.14)
The excess free energy of the drop of radius R is R 1 2 2 F = 4 R [R] = 4 dr r mij i j + F() − O$i i − ! 2 ij −∞ i ∞ 1
2 : (A.15) mij i j + F() − O$i i − ! + 4 dr r 2 ij R i Now, expanding in 1=R, writing r 2 = R2 (1 + 2(r − R)=R + · · ·), using Eqs. (A.8) and (A.9), and, after partially integrating once more, Eq. (A.5), we .nd 4 R2 [R] = 4 R2 + 4 R ∞ dr 2(r − R) mij i0 j0 − O$i1 (i0 − bulk ; i0 ) −∞
ij
i
(A.16) where bulk is de.ned to be equal to for r ¡ R, and for r ¿ R. After a .nal partial integration, we have 2 4 R2 [R] = 4 R2 1 − + · · · = 4 R2
R ∞ + 4 R (A.17) dr 2(r − R) mij i0 j0 + O$i1 (r − R)i0 ; −∞
ij
where the Tolman length has been introduced.
i
D.J. Bukman / Physica A 319 (2003) 151 – 162
161
Next we calculate the pressure diNerence across the surface of the drop of phase surrounded by phase , expanded to second order in 1=R: O$j1 O$j2 1 O$j1
+ 2 j0 − j1 : Op = ! − ! = − 2 R R 2 R j j
Using Eqs. (A.10) and (A.14), this gives ∞ 1 2
− 2 dr 2mij ((r − R)i0 j0 − 2i0 j1 ) : Op = R R −∞ ij The last term in this equation can be rewritten by partial integration, ∞ ∞ dr mij r(i0 j1 + i0 j1 ) ; dr mij i0 j1 = − −∞
ij
ij
−∞
(A.18)
(A.19)
(A.20)
and hence, using Eqs. (A.5) and (A.6) and partially integrating twice, ∞ ∞ ∞ 2 dr mij i0 j1 = 2 dr mij ri0 j0 + drO$i1 ri0 : −∞
ij
ij
−∞
i
−∞
(A.21) On using this, and Eqs. (A.9) and (A.10), we .nally obtain for Op ∞ 1 2
dr 2mij (r − R)i0 j0 + 2 Op = R R −∞ ij + 2
∞
−∞
dr
O$i1 (r −
R)i0
:
(A.22)
i
Now, the surface of tension is given by the value R = Rs that satis.es the relation Op = 2 [Rs ]=Rs , so we must have ∞ dr(r − Rs ) mij i0 j0 = 0 ; (A.23) −∞
or
ij
Rs =
∞
−∞
dr r
mij i0 j0 = :
(A.24)
ij
Since we are in the limit of a very large droplet, we can think of r as the coordinate perpendicular to a planar interface, and then, because of Eq. (A.9), the location of the surface of tension is seen to be independent of the choice of the origin of r. From Eq. (A.17) we also have an expression for the Tolman length, ∞ 1 =− dr O$i1 ri0 − Rs : (A.25) 2 −∞ i
162
D.J. Bukman / Physica A 319 (2003) 151 – 162
The .rst term in Eq. (A.25) can be viewed as an equimolar radius Re for the densities weighted with the chemical potentials O$i1 , Re ∞ dr O$i1 (i0 − bulk ) = dr O$i1 (i0 − bulk ): (A.26) i i −∞
Re
i
i
Then the Tolman length is given by = Re − Rs . Groenewold and Bedeaux [3] derived microscopic integral expressions for the quantities Re and Rs . The present results for the van der Waals theory can be obtained by replacing the pair density in their expressions by the product of the densities, (2) r 1 ;˜r 2 ) = i (˜r 1 )j (˜r 2 ), and making a Taylor expansion of j (˜r 2 ) around ˜r 1 [7]. ij (˜ Appendix B. The Kerins–Boiteux expression for the line tension Kerins and Boiteux also derived an alternative expression for the line tension [1]. In the present context, this expression can easily be rederived by considering the ˜ ·V ˜ = xH ˜ 1 + yH ˜ 2 . On the one hand, we have that ∇ ˜ = 2F, so that quantity V ˜ ·V ˜ = ˜ : dA∇ dA 2F = d‘nˆ · V (B.1) D
D
C
˜ around the circle C is, in the limit of large R, On the other hand, this integral of nˆ · V 2 1 1 R d F − mij 9r i 9r j − 2 9 i 9 j 2 ij R 0 m =
R :
(B.2)
=1
Using this in the de.nition of , Eq. (2.2), gives n 1 ˜ i · ∇ ˜ j − F ; = dA mij ∇ 2 D
(B.3)
i; j=1
the expression originally derived by Kerins and Boiteux [1]. References [1] [2] [3] [4] [5] [6]
J. Kerins, M. Boiteux, Physica A 117 (1983) 575. M.P.A. Fisher, M. Wortis, Phys. Rev. B 29 (1984) 6252. J. Groenewold, D. Bedeaux, Physica A 214 (1995) 356. J.S. Rowlinson, B. Widom, Molecular Theory of Capillarity, Clarendon Press, Oxford, 1989 (Chapter 8). E.L. Hill, Rev. Mod. Phys. 23 (1951) 253. I. Szleifer, B. Widom, Mol. Phys. 75 (1992) 925; D.J. Bukman, B. Widom, Physica A 251 (1998) 27; S. Perkovic, E.M. Blokhuis, E. Tessler, B. Widom, J. Chem. Phys. 102 (1995) 7584. [7] E.M. Blokhuis, D. Bedeaux, Mol. Phys. 80 (1993) 705.
Physica A 319 (2003) 151 – 162
www.elsevier.com/locate/physa
Torque balance at a line of contact Dirk Jan Bukman Department of Chemistry and Chemical Biology, Cornell University, Ithaca, NY 14853, USA Received 25 July 2002
Abstract Kerins and Boiteux (Physica A 117 (1983) 575) were the .rst to apply Noether’s theorem to the van der Waals theory of non-uniform 3uids. In particular, for a three-phase line of contact, they showed that the translation invariance of the variational integral for the excess free energy implies a balance of forces at the three-phase line. In this paper we consider the implications of the rotation invariance of the variational integral for the excess free energy. Intuitively, one would expect this invariance to lead to an equation of torque balance, and this is indeed the case—the total moment of the forces around the line of contact is zero. In the course of the calculation it will be necessary to .nd an expression for the surface of tension for a model with a multi-component density, which is a simple extension of earlier work by Fisher and Wortis (Phys. Rev. 29 (1984) 6252). At the same time, we extend to multi-component densities those authors’ results for the Tolman length and the equimolar surface, making contact with a more general calculation by Groenewold and Bedeaux (Physica A 214 (1995) 356). c 2002 Elsevier Science B.V. All rights reserved. PACS: 68.03.Cd; 05.70.Np; 68.05.−n; 82.60.−s Keywords: Three-phase line; Noether’s theorem; Inhomogeneous 3uids
1. Introduction In the van der Waals theory of inhomogeneous 3uids, the equilibrium density pro.les are determined by minimizing the total free energy of a system. The total free energy is given by an integral over space of a local free-energy density, which in the present version of the theory is a functional of the density pro.les and their gradients. Thus we have a variational problem in which the quantity to be minimized is an integral, the integrand of which (the free-energy density) does not explicitly contain E-mail address: [email protected] (D.J. Bukman). c 2002 Elsevier Science B.V. All rights reserved. 0378-4371/03/$ - see front matter PII: S 0 3 7 8 - 4 3 7 1 ( 0 2 ) 0 1 4 5 6 - 5
152
D.J. Bukman / Physica A 319 (2003) 151 – 162
the coordinates that are integrated over. This invariance, or symmetry, under coordinate transformations leads, with the use of Noether’s theorem, to the construction of conserved (divergenceless) currents. In 1983, Kerins and Boiteux [1] explored the consequences of Noether’s theorem for, among others, a system with a three-phase line of contact. They showed that the translational invariance leads to equations that express the balance of forces at the three-phase line. In addition, they derived an expression for the line tension as an integral over a “line tension density” that is localized near the three-phase line. In this paper, we will show that the rotational invariance leads to an equation of balance for the torque around the line of contact. In order to identify the torque, it is necessary to derive an expression for the surface of tension for a multi-component van der Waals theory. For a one-component theory this was done by Fisher and Wortis [2], and in Appendix A we extend their work to the multi-component case. There we also give a generalized version of their expression for the Tolman length, and compare the results with microscopic integral expressions obtained by Groenewold and Bedeaux [3]. The equation of torque balance is an interesting result in itself, since it is a macroscopic result that follows from the microscopic theory. On a practical level, it has some implications for numerical studies of lines of contact, which will be discussed in the conclusion. 2. The line of contact The system we will consider consists of m equilibrium phases separated by m planar interfaces that all meet at a straight line of contact. The usual case would be that of a three-phase contact line, as shown in Fig. 1, but the arguments presented here are valid for an arbitrary number of phases meeting at a “multi-phase” contact line. The system has translational symmetry in the direction parallel to the line of contact. The excess free-energy density is given by a functional , and the total excess free energy, for a cylindrical box of cross-section A and unit length, parallel to the contact line, is F = d x dy ; (2.1) A
where the x–y plane is perpendicular to the line of contact. This total excess free energy includes contributions from the surface tensions of the interfaces and from the line tension of the line of contact [4]. The line tension of the line of contact is given by m d x dy −
R ; (2.2) = D
=1
where the excess free-energy functional is integrated over a circular disk D of radius R (with R → ∞), centered near the line of contact (the line tension is independent of the exact location of the center), and is the surface tension of the th interface. (The line tension is not necessarily de.ned only in terms of a circular disk of radius
D.J. Bukman / Physica A 319 (2003) 151 – 162
153
Fig. 1. The geometry of the area where the three phases ; , and meet at the three-phase line of contact (black dot), which is perpendicular to the plane of the .gure.
R, but for simplicity we will use only that de.nition, even though a more general one is available.) We work in a van der Waals theory with n density components i . The equilibrium density pro.les i (x; y) are those that minimize the free energy F, which follow from the Euler–Lagrange equations F = 0 (i = 1 : : : n) : (2.3) i (˜r) We use an excess free-energy functional that is given by n 1 ˜ i · ∇ ˜ j mij ∇ = F(i ) + 2
(2.4)
i; j=1
with mij = mji . The function F(i ), the bulk excess free energy, has minima F(i ) = 0 ( = 1 : : : m), for the m stable equilibrium phases; elsewhere F(i ) is positive. The Euler–Lagrange equations for the functional given by Eq. (2.4) are n 9F − mij ∇2 j = 0 (i = 1 : : : n) : (2.5) 9i j=1
Together with the appropriate boundary conditions, these equations can be solved to give the equilibrium density pro.les.
154
D.J. Bukman / Physica A 319 (2003) 151 – 162
3. The conserved currents As Kerins and Boiteux [1] pointed out, because the integrand in Eq. (2.1) is independent of the origin and orientation of the coordinate axes, it follows from Noether’s ˜ associated with theorem [5] that there exist conserved (i.e., divergenceless) currents H this independence. More precisely, if the integrand in Eq. (2.1) is invariant under an ˜ ·H ˜ = 0 holds in.nitesimal coordinate transformation (x ; y ) = (x + x; y + y), then ∇ for a current de.ned by n 9 (9x i x + 9y i y) x − 9(9x i ) i=1 ˜ = H (3.1) n 9 y − (9x i x + 9y i y) 9(9y i ) i=1
with the i (x; y) determined by the Euler–Lagrange equations (2.3). Two transformations that were previously studied by Kerins and Boiteux [1] are the two translations along the coordinate axes, (x; y) = (; 0) and (x; y) = (0; ). We will here also look at the rotation around the origin, (x; y) = (y; −x). The conserved currents corresponding to these three transformations are, respectively, 9 F − 12 mij (9x i 9x j − 9y i 9y j ) − 9 x i 9(9x i ) ij i = ˜1= H ; 9 − m 9 9 − ij x i y j 9 x i 9(9 ) ij y i i 9 9 y i − 9(9x i ) i = ˜2= H F + 9 − 9 y i 9(9y i ) i
˜ 3 = yH ˜ 1 − xH ˜2 : H
−
mij 9x i 9y j
; mij (9x i 9x j − 9y i 9y j ) ij
1 2
ij
(3.2)
˜ ·H ˜ i = 0 over an area A perpendicular to the When we integrate the expression ∇ line of contact, we .nd with Gauss’ theorem that ˜ ·H ˜i= ˜i=0 ; d x dy ∇ d‘nˆ · H (3.3) A
9A
where d‘ is a line element on the boundary 9A of A, and nˆ is the outward normal to 9A. If we take 9A to be a circle C of radius R, then in polar coordinates around the center of C, cos nˆ = ; sin
D.J. Bukman / Physica A 319 (2003) 151 – 162
155
˜ i is and Eq. (3.3) for the three currents H
2 1 1 mij 9r i 9r j − 2 9 i 9 j R d cos F − R 2 ij 0 sin +2 9r i 9 j R
2
0
1 1 R d sin F − mij 9r i 9r j − 2 9 i 9 j 2 ij R
cos 9 r i 9 j −2 R
2
0
=0 ;
R d
=0 ;
mij 9r i 9 j = 0 :
(3.4)
ij
We now choose the center of C near the line of contact, and choose the radius R large enough so that the m interfaces that meet at the line of contact cross C perpendicularly (up to order 1=R), and at the crossing point have the structure of the equilibrium two-phase interface, so that their structure is not aNected by the presence of the line of contact. The angle between the th interface and the positive x-axis will be designated by . Now at the large distance R from the line of contact, the th interface is then (up to order 1=R) a planar interface whose density pro.le only depends on the coordinate z = R sin( − ), the distance from the line through the origin under an angle . Writing i as a function of this local coordinate, and de.ning i = 9i =9z , we have 9r i (z ) = i sin( −
)
9 i (z ) = i R cos( −
; )
(3.5) :
(3.6) i
Assuming that each interface has a .nite width, say " ; will be non-zero only for values of z ∼ " . Since R" , this means that the relevant values of for a particular interface must satisfy " ∼ R sin( − ), so that sin( − ) ∼ " =R1, and cos( − ) ∼ 1 + O("2 =R2 ). Near the th interface, the integrals in (3.4) are then to leading order in 1=R 1 R d cos F + mij i j = 0 ; 2 ij
R d sin
R d
ij
1 F+ mij i j 2 ij
mij i j z = 0 :
=0 ; (3.7)
156
D.J. Bukman / Physica A 319 (2003) 151 – 162
For large R, the diNerential R d ≈ d z, and since the separations between the m interfaces are much larger than their widths, we can rewrite the integral around the circle C as a sum of integrals perpendicularly across the individual interfaces, over all values of z : 2 m ∞ R d · · · ≈ d z · · · : (3.8) 0
−∞
=1
For a planar interface, we know from (2.1) and (2.4) that the surface tension is given by ∞ 1 (3.9) dz F + mij i j ;
= 2 ij −∞ so that the .rst two equations of (3.7) read m
cos
=0 ;
sin
=0 :
=1 m
(3.10)
=1
These two equations express force balance in the x- and y-directions, and for m = 3 are equivalent to Neumann’s triangle [4]. This result was already found by Kerins and Boiteux [1], who used a triangular contour instead of a circle. The third equation is m
zs; = 0 ;
(3.11)
=1
where 1 zs; =
∞
−∞
d z
mij i j z :
(3.12)
ij
It is shown in Appendix A that zs; de.ned in this way is the location of the surface of tension of the th interface. Then, if we imagine the surface tension of an interface to be acting perpendicular to the line of contact, along the surface of tension of that interface, then zs; is the perpendicular distance between the line of action of the force
and the origin, i.e., it is the lever arm of the force. The product zs; is then the torque due to the surface tension of the th interface, and Eq. (3.11) means that the total torque around the origin is zero. Since the forces are balanced, see Eq. (3.10), the total torque is the same around any point and thus independent of the choice of origin. In the case of three interfaces meeting at a three-phase line, one can choose the origin of the x- and y-coordinates so that it coincides with the intersection of two of the surfaces of tension. Two of the zs; are then zero, so due to Eq. (3.11) the third zs; must also be zero. In other words, for three interfaces meeting at a three-phase
D.J. Bukman / Physica A 319 (2003) 151 – 162
157
line, the three surfaces of tension intersect in a single line. Also in the case of three interfaces, one might de.ne dividing surfaces within and parallel to the three physical interfaces. If these three dividing surfaces are chosen so that they intersect in a single line, the total moment of the surface tensions acting along the dividing surfaces is zero. Similarly, one can think of a system of two coexisting surface phases, such as e.g. exists at a prewetting transition, as two planar interfaces meeting at a two-phase line of contact, the boundary line. Then Eq. (3.11) for m = 2 shows that the surfaces of tension of the two interfaces must coincide. 4. Conclusion In this paper we have examined a new application of Noether’s theorem to the van der Waals theory of inhomogeneous 3uids. From the rotation symmetry of the variational integral (2.1), which describes the microscopic structure of the density pro.les, it follows that the total torque due to the interfaces meeting at a line of contact is zero. Like the equations of force balance, this is a statement about the macroscopic properties of a system, derived from microscopic considerations. The equation of torque balance can be important in numerical calculations of the density pro.les around a line of contact [6]. In such calculations, one numerically solves the Euler–Lagrange equations (2.3) in a large region around the line of contact, while imposing Dirichlet boundary conditions along the edges of the region, i.e., prescribing certain values for the density pro.les on the boundary. These prescribed values are the bulk densities, and the density pro.les of the two-phase interfaces, to which the density pro.le reduces far away from the line of contact. The boundary conditions are chosen such that the two-phase interfaces are oriented under angles that satisfy the balance of forces, Eq. (3.10). As we have seen here, the balance of torque, Eq. (3.11), is a property satis.ed by the minimizing density pro.le, and hence the boundary conditions must also be chosen to satisfy this equation. If this is not the case, the numerical solution of the Euler–Lagrange equations will not lead to the true minimum of the excess free energy. In previous work on three-phase lines and boundary lines between two-dimensional phases [6], the bulk excess free-energy F(i ) has usually been chosen such that two of the bulk phases are related by symmetry; in such a case the symmetry ensures that the torque balance is satis.ed. In the general case, in the absence of such a symmetry, it is an extra requirement that must be satis.ed by any boundary conditions that are imposed. Acknowledgements The author is grateful to B. Widom for useful conversations and comments, and for a reading of the manuscript, and to J. Groenewold for useful comments. This work was done in the group of B. Widom, and was supported by the National Science Foundation and the Cornell Center for Materials Research.
158
D.J. Bukman / Physica A 319 (2003) 151 – 162
Appendix A. The surface of tension In this appendix, we derive expressions for the surface of tension and the Tolman length of an interface, for the excess free-energy functional (2.4). The surface of tension is a property of the planar interface; even though the calculation involves a spherical droplet of radius R, the surface of tension is well de.ned in the planar limit R → ∞. The calculation is a straightforward extension to a multi-component system of the work of Fisher and Wortis in Ref. [2]; that reference contains more details and a more thorough discussion. For the interface between two of the equilibrium phases, say and , we consider a spherical drop of radius R of phase surrounded by phase . The bulk excess free-energy F(i ) in the functional (2.4) has equal minima (F =F =0) corresponding to the two phases and , which therefore coexist in bulk, separated by a planar interface. To describe an equilibrium spherical droplet of phase, the system is taken slightly oN coexistence by including a change in chemical potential − i O$i i in F. (If one of the density components is the entropy density, the corresponding O$i is the temperature.) The calculation consists of two parts. First, we calculate the excess free-energy F of the droplet, which is related to the (radius-dependent) surface tension [R] by F = 4 R2 [R] :
(A.1)
In the limit R → ∞; [R] approaches , the surface tension for the planar interface. We are interested in the O(1=R) corrections to [R]. Second we calculate the pressure diNerence Op between the inside and the outside of the droplet. For large R; Op ∼ 2 =R, and we are again interested in the next order in 1=R. The surface of tension is at that radius Rs for which Op = 2 [Rs ]=Rs exactly, and the two previous results can then be combined to .nd an expression for Rs . This expression is independent of the radius R of the droplet, and thus remains valid for a planar interface. In the following, we will let integrals over r run from −∞ to ∞; since the integrands go to zero exponentially far away from the interface at r = R, the errors can be neglected. We can thus equally well think of r as the coordinate perpendicular to a planar interface. Since we are interested in the limit of large radius R, we expand the (spherically symmetric) density pro.les i (r) and the chemical potentials O$i in 1=R: i1 (r) i2 (r) i (r) = i0 (r) + + ··· ; + R R2 O$i1 O$i2 (A.2) + 2 + ··· : R R The two phases that coexist in bulk, corresponding to the minima of F, have densities
; i0 . Due to the chemical potentials O$i , the densities of the two phases and in the
system with a drop, ; i , are at the minima of ! = F − i O$i i , and their excess free energies are ! ; = −p ; . From the minimization 9!=9i = 0 we .nd, by expanding to order 1=R, 92 F
j1 = O$i1 at i = ; : (A.3) i 9 9 j0 i0 j O$i =
D.J. Bukman / Physica A 319 (2003) 151 – 162
159
We now derive some necessary intermediate results from the Euler–Lagrange equation (2.3). In spherical coordinates, it is 2 9F − O$i − mij 92r + 9r j = 0 : (A.4) r 9i j Expanding r −1 = R−1 (1 − (r − R)=R + · · ·), the leading orders in powers of 1=R of Eq. (A.4) are 9F mij j0 = ; (A.5) 9 i0 j 92 F j1 − O$i1 ; (A.6) mij (j1 + 2j0 ) = 9i0 9j0 j j 92 F mij (j2 + 2j1 − 2(r − R)j0 ) = j2 9i0 9j0 j j +
93 F 1 j1 k1 − O$i2 ; 2 9i0 9j0 9k0
(A.7)
jk
where here a prime indicates diNerentiation with respect to r. If we multiply Eq. (A.5) by i0 , sum over i, and integrate with respect to r, we .nd that 1 (A.8) mij i0 j0 = F(0 ) 2 ij and hence, with (3.9), that ∞
= dr mij i0 j0 : −∞
(A.9)
ij
As was mentioned before, we have here extended the integration range of r to (−∞; ∞), which should only lead to exponentially small errors. Similarly, multiplying Eq. (A.6) by i0 , summing over i, and integrating over r, we .nd, after partially integrating twice, and making use of Eqs. (A.5) and (A.9) ∞ dr O$i1 i0 : (A.10) − 2 = −∞
i
Another intermediate result follows when we multiply Eq. (A.6) by i1 , sum over i, and integrate: ∞ ∞ 92 F dr 2mij i0 j1 + O$i1 i1 = dr i1 j1 : 9 9 i0 j0 −∞ −∞ ij i ij (A.11) Next we turn to Eq. (A.7), again multiplying it by i0 , summing over i, partially integrating twice, and using Eq. (A.5) ∞ dr 2mij (i0 j1 − (r − R)i0 j0 ) −∞
ij
160
D.J. Bukman / Physica A 319 (2003) 151 – 162
1 − i1 j1 9r 2 ij
92 F 9i0 9j0
+
O$i2 i0
=0 :
(A.12)
i
If we partially integrate the term containing F, and de.ne Kij = 92 F=9i0 9j0 , we can write this ∞ 1 O$i2 (i0 (∞) − i0 (−∞)) = Kij i1 j1 2 i
−
∞
−∞
dr
ij
ij
9F i1 j1 + 2mij (i0 j1 − (r − R)i0 j0 ) : 9i0 9j0
Using Eq. (A.11) and Eq. (A.3), this is O$i2 (i0 (∞) − i0 (−∞)) =
∞
−∞
i
−
−∞
2
dr
(A.13)
2mij ((r − R)i0 j0 − 2i0 j1 )
ij
1 O$i1 (i1 (∞) − i1 (−∞)) : 2 i
(A.14)
The excess free energy of the drop of radius R is R 1 2 2 F = 4 R [R] = 4 dr r mij i j + F() − O$i i − ! 2 ij −∞ i ∞ 1
2 : (A.15) mij i j + F() − O$i i − ! + 4 dr r 2 ij R i Now, expanding in 1=R, writing r 2 = R2 (1 + 2(r − R)=R + · · ·), using Eqs. (A.8) and (A.9), and, after partially integrating once more, Eq. (A.5), we .nd 4 R2 [R] = 4 R2 + 4 R ∞ dr 2(r − R) mij i0 j0 − O$i1 (i0 − bulk ; i0 ) −∞
ij
i
(A.16) where bulk is de.ned to be equal to for r ¡ R, and for r ¿ R. After a .nal partial integration, we have 2 4 R2 [R] = 4 R2 1 − + · · · = 4 R2
R ∞ + 4 R (A.17) dr 2(r − R) mij i0 j0 + O$i1 (r − R)i0 ; −∞
ij
where the Tolman length has been introduced.
i
D.J. Bukman / Physica A 319 (2003) 151 – 162
161
Next we calculate the pressure diNerence across the surface of the drop of phase surrounded by phase , expanded to second order in 1=R: O$j1 O$j2 1 O$j1
+ 2 j0 − j1 : Op = ! − ! = − 2 R R 2 R j j
Using Eqs. (A.10) and (A.14), this gives ∞ 1 2
− 2 dr 2mij ((r − R)i0 j0 − 2i0 j1 ) : Op = R R −∞ ij The last term in this equation can be rewritten by partial integration, ∞ ∞ dr mij r(i0 j1 + i0 j1 ) ; dr mij i0 j1 = − −∞
ij
ij
−∞
(A.18)
(A.19)
(A.20)
and hence, using Eqs. (A.5) and (A.6) and partially integrating twice, ∞ ∞ ∞ 2 dr mij i0 j1 = 2 dr mij ri0 j0 + drO$i1 ri0 : −∞
ij
ij
−∞
i
−∞
(A.21) On using this, and Eqs. (A.9) and (A.10), we .nally obtain for Op ∞ 1 2
dr 2mij (r − R)i0 j0 + 2 Op = R R −∞ ij + 2
∞
−∞
dr
O$i1 (r −
R)i0
:
(A.22)
i
Now, the surface of tension is given by the value R = Rs that satis.es the relation Op = 2 [Rs ]=Rs , so we must have ∞ dr(r − Rs ) mij i0 j0 = 0 ; (A.23) −∞
or
ij
Rs =
∞
−∞
dr r
mij i0 j0 = :
(A.24)
ij
Since we are in the limit of a very large droplet, we can think of r as the coordinate perpendicular to a planar interface, and then, because of Eq. (A.9), the location of the surface of tension is seen to be independent of the choice of the origin of r. From Eq. (A.17) we also have an expression for the Tolman length, ∞ 1 =− dr O$i1 ri0 − Rs : (A.25) 2 −∞ i
162
D.J. Bukman / Physica A 319 (2003) 151 – 162
The .rst term in Eq. (A.25) can be viewed as an equimolar radius Re for the densities weighted with the chemical potentials O$i1 , Re ∞ dr O$i1 (i0 − bulk ) = dr O$i1 (i0 − bulk ): (A.26) i i −∞
Re
i
i
Then the Tolman length is given by = Re − Rs . Groenewold and Bedeaux [3] derived microscopic integral expressions for the quantities Re and Rs . The present results for the van der Waals theory can be obtained by replacing the pair density in their expressions by the product of the densities, (2) r 1 ;˜r 2 ) = i (˜r 1 )j (˜r 2 ), and making a Taylor expansion of j (˜r 2 ) around ˜r 1 [7]. ij (˜ Appendix B. The Kerins–Boiteux expression for the line tension Kerins and Boiteux also derived an alternative expression for the line tension [1]. In the present context, this expression can easily be rederived by considering the ˜ ·V ˜ = xH ˜ 1 + yH ˜ 2 . On the one hand, we have that ∇ ˜ = 2F, so that quantity V ˜ ·V ˜ = ˜ : dA∇ dA 2F = d‘nˆ · V (B.1) D
D
C
˜ around the circle C is, in the limit of large R, On the other hand, this integral of nˆ · V 2 1 1 R d F − mij 9r i 9r j − 2 9 i 9 j 2 ij R 0 m =
R :
(B.2)
=1
Using this in the de.nition of , Eq. (2.2), gives n 1 ˜ i · ∇ ˜ j − F ; = dA mij ∇ 2 D
(B.3)
i; j=1
the expression originally derived by Kerins and Boiteux [1]. References [1] [2] [3] [4] [5] [6]
J. Kerins, M. Boiteux, Physica A 117 (1983) 575. M.P.A. Fisher, M. Wortis, Phys. Rev. B 29 (1984) 6252. J. Groenewold, D. Bedeaux, Physica A 214 (1995) 356. J.S. Rowlinson, B. Widom, Molecular Theory of Capillarity, Clarendon Press, Oxford, 1989 (Chapter 8). E.L. Hill, Rev. Mod. Phys. 23 (1951) 253. I. Szleifer, B. Widom, Mol. Phys. 75 (1992) 925; D.J. Bukman, B. Widom, Physica A 251 (1998) 27; S. Perkovic, E.M. Blokhuis, E. Tessler, B. Widom, J. Chem. Phys. 102 (1995) 7584. [7] E.M. Blokhuis, D. Bedeaux, Mol. Phys. 80 (1993) 705.