Response of nematic liquid crystals to Van Der Waals forces

Physrca 66 (1973) 11l-l 30 0 North-Holland Publrshmg Co.

RESPONSE OF NEMATIC

LIQUID CRYSTALS

TO VAN DER WAALS FORCES E R. SMITH*

and B W NINHAM

Department of Apphed Mathematrcs, Research School of Phystcal Sciences, Institute of Advanced Studies, Australian Natronal Umversrty, Canberra, A C T, Australia

Received 18 July 1972

synopis The response of a slab of nematlc hqmd crystal to the Van der Waals forces exerted upon 11 by an amsotroplc crystal 1s studled The slab of hqmd crystal 1s considered as havmg Its molecular alignment fixed on one side, with the other side free to rotate. The amsotroplc crystal exerts a Van der Waals torque which tends to hne up the amsotroplc axis of the liquid crystal with its own anisotropic axls This torque IS opposed by the elastic stresses set up m the hqmd crystal by the induced twist The angle of twist of the amsotroplc axis at a pomt m the hqmd crystal IS calculated as a function of the depth of the pomt below the free surface of the hqmd crystal

1. Introductzon. In this paper we discuss m some detail the role which Van der Waals forces can play m hqmd-crystalhne phenomena. That these forces do play a role m hqmd-crystallme phenomena has been stressed by De Gennes’) m his discussion of the experimental work of Leger and of Rault and Cladls, who have observed the behavlour of droplets of nematlc or cholesterlc liquid crystal when close to the surface of a nematlc or cholesterlc liquid crystal De Gennes also suggests that the arrays of nematlc droplets observed by Meye?) are possibly stabilized by Van der Waals forces. The reason that Van der Waals forces are considered as important m dlscussmg these effects, 1s that there 1s no other available physical mechanism which ~111simply explain the results Expenmentally the interactions are comparatively still strong over distances of 1 or 2 pm, while the,coherence length of the liquid crystal is, charactenstlcally of order 1000 A. This Implies that it 1s not possible for effects observed at distances of 1 or 2 pm to be due to the hqmd-crystallme ordermg of one droplet affectmg that m another droplet There have been some recent developments (Parseglan and Nmham3), Nmham, ParseBan and Welss4) and references therem) m the theory of Van der Waals * Present Australia.

address

Department

of Mathematics,

111

Umverslty

of Newcastle,

N S W 2308,

112

E R. SMITH AND B W NINHAM

forces due to Llfshltz’) which make the apphcatlon of that theory to liquid crystals not very difficult From the point of view of Van der Waals forces, perhaps the most important thmg about nematlc hquld crystals IS that they are optically amsotroplc We may thus expect to see new effects m the theory of Van der Waals forces due to the coupling of this anisotropy to other optically amsotroplc objects (mcludmg other droplets of nematlc hquld crystal) The role of Van der Waals forces m anisotropic media has been investigated by De Gennes’), Kats6) and Parseglan and Welss7) De Gennes used the Hamaker panwise-summation technique to estimate the Van der Waals forces mvolved m his system For reasons we shall discuss further below, this analysis IS not adequate at large distances (z e > 1000 8, a characteristic coherence length for a liquid crystal) The work of Parseglan and Welss7) on amsotroplc crystals and the work of Kats6) on cholesterlc hquld crystals both provide formulations of the Van der Waals energies for the systems consldered at zero temperature However, at room temperatures, Van der Waals forces over distances greater than, say, 1000 8, become dommated by temperature-dependent effects when the media mvolved have slmllar weight den&es3) Parseglan and Welss7) looked at the Van der Waals forces between identical amsotroplc half spaces separated by an lsotroplc medium They found that specifically amsotroplc effects (namely a torque tending to ahgn the two axes of amsotropy) were enhanced when the mtervemng lsotroplc medium tends to look &electrically hke an average of the “transverse” and“paralle1” dielectric properties of the amsotroplc media For a liquid crystalhne medmm, the lsotroplc phase often has a dlelectnc constant which IS the average of the dlelectrlc constants m the nematic phase (though, m this context see footnote 7 of De Gennes’s paper’), where the dlfficultles involved when this does not hold are discussed) Thus we may possibly see comparatively strong effects due to amsotropy m the mteractlon of an amsotroplc crystal or nematlc hquld crystal with a nematlc liquid crystal across a layer of hquld crystalhne matenal m Its lsotroplc phase A further pomt of mterest m the apphcatlon of the theory of Van der Waals forces to hquld crystalline materials anses because nematlc liquid crystals have very small elastic constants This means that the action of Van der Waals forces on a hquld crystallme medmm may well mvolve a distortion of the ordered structure of the liquid crystal This paper IS m fact devoted to this last point, namely the dlstortlon of the ordered structure of a liquid crystal under the actlon of Van der Waals forces exerted by an anisotropic crystal The elastic response of liquid crystals to Van der Waals forces has, to the author’s expenence, not been mvestlgated previously However, De Gennesl) has suggested that the distortion effects observed by Dubols-Vlolette and Parodl’) in arrays of droplets of nematlc hquld crystals may be affected by Van der Waals forces Also, we may consider all membranes to be basically hquld crystalhne in nature9) Most cell-cell mteractlons are m part medlated by Van der Waals mter-

RESPONSE

OF LIQUID

CRYSTALS

TO VAN DER WAALS FORCES

113

actlonslO) and so we may expect that the response of cell membranes to mteractions with other cells or to foreign bodies may involve a drstortlon of the hquld crystalhne order not dlsslmllar to that discussed here An expenment has been devised by Guyonll) to measure the amount of dlstortlon of lrqmd crystalline order caused by the Van der Waals mteractlon of an amsotroplc crystal with a slab of nematlc hqmd crystal separated by a layer of either hquld crystalhne material m its isotropic phase or some other material. The experlmental situation IS sketched m fig. 1 Medmm 1 1s a slab of glass, the upper surface of which 1s carefully scratched m a given dlrectlon Medmm 2 IS a nematlc liquid crystal. The scratches on the glass provide strong anchormg of the molecular ahgnment of the hquld crystal at the hquld crystal-glass Interface

2

Fig 1 A sketch of the experlmental system described Medium 1 IS glass, medmm 2 IS hquld crystal, medium 3 a dlelectnc medmm and medium 4 a strongly amsotroplc crystal with Its amsotroplc axis at an angle 6 to the amsotroplc axls of the hquld crystal at the glass-hquldcrystal interface

The thickness of the liquid crystal 1s d Medium 3 1s a layer of width 6 and may be of lrquld crystalline material m its lsotroplc phase or some other medmm. Medmm 4 1s a strongly amsotroplc crystal whose axis of amsotropy 1s mchned at an angle 8 to the dlrectlon of the scratches on the top surface of the glass. That is, the axis of amsotropy 1s ahgned at an angle 13to the alignment of the amsotroplc axls at the bottom of the nematlc liquid crystal It IS expected that the amsotroplc crystal will exert (as well as a Van der Waals force normal to the surfaces m the system) a torque, attempting to ahgn the upper layers of the hquld crystal with the anisotropIc axis of the amsotroplc crystal. Because of the strong anchormg of the nematlc ordermg at the bottom of the hqmd crystal, we may expect that

114

E R SMITH AND B.W NINHAM

the dlrectlon of the amsotroplc axis m the llqmd crystal WA twist shghtly This will provide an elastic stress energy which will exert a torque tending to oppose the twist so that an eqmhbrmm angle 8, through which the top surface of the hquld crystal 1s rotated will result It 1s this angle which Guyon intends to measure experimentally In this paper we estimate the net change m energy of the system caused by brmgmg up the amsotroplc crystal from a large distance as a functional of 0(z) e(z) 1s the angle of twist of the amsotroplc axis of the nematlc hquld crystal at a distance z from the glass surface The energy 1s made up of a Van der Waals energy of interaction of the amsotroplc crystal with the liquid crystal and a stress energy due to the twisting of the liquid crystal We find the function 0(z) by carrymg out a variational calculation to mmlmlze the energy with respect to the function e(z) The angle B,, 1s then given by 8, = 0(d) The stress energy 1s simply calculated from Frank’s theory12) for the stress energy of a nematlc liquid crystal The distortion mvolved 1s a pure “twist” dlstortlon, so that only the twist elastic modulus occurs m the stress-energy expression Experimental values of the twist elastic modulus may be simply fed mto the theory The calculation of the Van der Waals mteractlons m the system 1s rather more comphcated The first thmg we note 1s that the gap between the nematlc liquid crystal and the amsotroplc crystal 1s to be at least 1000 A For gap widths less than this, the experimental configuration m which the gap 1s filled with liquid crystalline material m its lsotroplc phase would not be possible, as the lsotroplc phase would be rapidly ordered by the mtermolecular forces m the liquid crystal Parseglan and Nmham3) have amply shown that the ultraviolet and infrared contrlbutlons to the Van der Waals energy at these distances are completely negligible due to retardation effects If the media are of similar weight densities This condltlon holds a fortlon for the case of the ordered and amsotroplc phases of nematlc liquid crystals The only contribution to the Van der Waals forces 1s from microwave sources These are strongly temperature dependent and are almost completely nonretarded3) Thus we may calculate the Van der Waals forces m the nonretarded approxlmatlon, which 1s a great slmphficatlon Of course, It 1s not possible to calculate these forces by the pairwise-addition formulation due to Hamaker The next thing to note 1s that Van der Waals forces are small anyway over such large distances and we may expect the angle B0 to be small, so that we may treat the twisting of the hquld crystal as a perturbation The calculation of the Van der Waals energy of a system m the nonretarded approxlmatlon proceeds as follows First we find a dlsperslon relation for the electric surface-wave modes of the system This 1s done by finding general solutions of Laplace’s equation for the system The dispersion relation 1s then found by mslstmg that the solutions correspond to surface waves We do this in section 2 The second step m the calculation 1s then to calculate the Van der Waals energy

RESPONSE

OF LIQUID

CRYSTALS

TO VAN DER WAALS FORCES

115

of the system by the formula4) H(8,d)

= kT2

“g;

i dk,

-co

r dk, [In D (&,6)

- In D (d.,

-co

m)l.

(1)

D (CII,6) = 0 IS the dlsperslon relation for the system when the gap between the liquid crystal surface and the amsotroplc crystal 1s 6 D (IE,, 6) IS the dispersion

relation D (0,6) evaluated at the rmagmary frequency w = I&,, where 5, = 2xnkTltt andn =0,1,2, The prime on the sum mdxates that the term n = 0 should be given half wetght k, and k,, are the wave numbers for the surface-wave modes, parallel to the surfaces, so that we must mtegrate over both variables to get the contrlbutlon to the energy from all modes The factor Lz In eq. (1) 1s the area of the surfaces m the system We carry out this evaluation m section 3 The vanatlonal calculation to find e(z) 1s carried out rn section 4 2 Duperson relatzon for the system We construct a set of coordmates for the system as shown m fig. 2 We write the dlelectrlc permlttlvlty m repon I as q(z), that is, the dlelectrlc permlttlvlty 1s a 3 x 3 tensor. It 1s only a function of z m regon 2 (the hqmd crystal), but It also varies from region to re@on Laplace’s equation for the electnc potential @ m region z may now be written

v * E,(Z) V@ =

0

(2)

x

Glass

Anisotropic crystal

Llquld crystal

1

jl_ Y

1

d

Qt."21

Fig 2 A diagram of the coordmate

3

2

:

E3

4 E4t,E41

system used m denvmg the dlsperslon relation

2 1 Region

1 The glass medium of regon 1 1s lsotroplc so cl(z) 1s a constant diagonal tensor with the property

(3)

E R SMITH AND B W NINHAM

116

Thus we may write Laplace’s equation m regon 1 as El VV,

= 0.

(4)

We may then wnte down the solution @ Immediately !ijl (x, y, z) =

rdk, -*

idk,, ei(kXx+PyY) a, (k,, k,) epZ, -m

(5)

where e2 = k: + k;

(6)

A general solution to eq (4) would include m the mtegrand a term bI (k,, k,,) e-O’, but since we require the solution to correspond to a surface mode, so that hm,, _ m @ (x, y, z) = 0, such a term 1s not present m our solution 2 2 Region 2 The amsotroplc hqmd crystal which IS the medium of region 2 has a dlelectnc tensor given by the formula ‘521

E2(z) =

-

-

(E21

+ (c2, -

sin2 19(z)

e2t)

~2~)

sin

33

+ (e2, - c2J sm 26 (z) ant

(4

+

and

0

0

where 13(z)1s the angle through which the has been rotated at a point distant z from the m the mtroductlon, we make the assumption e(z) IS a monotomcally increasing function except that

e(o) =0

sin2 f&z)

(E~,-E~~)

0

i

0

e(z) -3 1

E2t

9 (7)

1

amsotroplc axis of the hquld crystal scratched glass surface As mentioned that e(z) 1s very small for 0 4 z -< d of z, but IS otherwise undetermined,

for

z E [O,4

(9

Because of the size of e(z) we may write e(z)

EL1 ~~(4

=

e(z)

(~2,

-

e2t

+)

0

&2t

c2t)

0

0

0

(

-

(9

,

E2t i

to first-order accuracy m e(z) Laplace’s equation m regon 2 may then be WrItten m the form

aw2

E21 F

+

E2t -

aw2

aw2

+

aY2

E2t

F

+

28

(4

(E2,

-

e2t)

820, = axay

-

o

.

(10)

We may dlvlde both sides of eq. (10) by &2tand transform the x coordinate to x* where x* = (E2JE2,)*X = yzx,

(11)

RESPONSE

OF LIQUID

CRYSTALS

TO

VAN DER WAALS FORCES

117

so that Laplace’s equation becomes P@, + -

ax*2

a2Q2 82~~ + ay2 az2

2

-+-

a’@, 44 -=. ax* ay (&2&m)*

&21 -

Egt

o

(12)

If for the moment we write 0(z) = 0,,8, (z), then we may write eq. (12) m the form (L, + B&) o2 = 0. Tlns suggests that we may develop a perturbation scheme for Q2 m the form Q2 = c.“=,, IX-$&,. Usmg thts form, the equation becomes [email protected] + 0oL,@2,0 + f&Lo@2,1 = 0. Then, by equating the coeffictent of 0: to zero we have L,@, o E !?!!z

ax*2

a2@2,0 + [email protected]

+

w

o

F=

(13)

’

while by equating the coefficient of &, to zero we have

LoQ2,1

=

a2Qi2., [email protected] : a2@2pl I

ax*2

=-

2 y;

ay2

; 21

y

e,(z)

2t

az2

!S!L

ax*ay

z

(14)

[email protected].

Now, as L,, 1s Laplace’s operator, we can find the general solution of eq (13) and substitute it mto the right-hand side of eq. (14) Since we may also wrote down the Green function for the operator L,,, we may then solve eq. (14), thus generating the solutton to Laplace’s equation m region 2. The general solution to eq. (13) may be written down immedtately It 1s @2,0 (x*, y, z) =

[dk: rdki -00 -00

ei(kx’x*+kyy) [& (k:, ki) e@’+ bz (k:, k;) e-“‘1. (1%

The Green function for Laplace’s operator m region 2 is given by13)

-00

-a

x exp(1[k (x* - x0>+ ky (Y -

yo)

+ (244

(z - zdll.

(16)

E R SMITH

118

AND

B W NINHAM

Using this form for the green function, we may write down an Integral form for @2,1 immediately It IS Qs,,, (x”.

y, z)

-2 (Q’- &2J_-!-

=

(E~~F~J+ 45c2d

x exp (1 [k,x* + k,y + (2xm/d) (z - zo)]j

ss

dk; 13,(z) k:k;

dk:

X

-a:

-co

x expl1ho (kk - k> + y. (kl -

[a:(k:,

x

k,)l)

k:) e“‘, + b: (k:, kh) eBpzo]

(17)

We can do several of the mtegrals m eq (17) and after performing transform x* back to x and find the form

Q2.1(x7Y, z>=

them, we

-2(&Z’ - E2t)

4, d

ss m

X

cc

dk, e i(k,x+kyY) dz,

dk

-CC

s

-CC

5

In=-m

0

b2 (kx, k,Jeozro + b2 (k,, k,) e-0*20]

x

@:

+

8,(zo) k,k,

e2rrimfz-z0)/d 3

(2xm/d)z

W-0

where Q: = (k:/y:) + kz We may now write the general solution to Laplace’s equation m region 2 to first-order accuracy m 8, The solution 1s aj CC @2

(x,

Y,

4

=

dk, s

dk,

exp

b (kx

+

k,.dl

s

--ac -m a2 (k,, k,) e”*’

x

1 - 2 “’ - E2t k,k, d&z,

d

x

s

dzo

0

m

Wo) L

In=-m

exp [(e2 - 2xlm/d) (z. - z)] ~‘2 + (2xm/d)’

>

RESPONSE OF LIQUID CRYSTALS TO VAN DER WAALS FORCES

+ bz (k,, k,) e-o2z

1 - 2 “Id;

119

“’ k,k,, 21

d

x

s dzo

e(zo)

ff

I

exp K~z + 2x144 (z - ~011 .

VI=-cc

Q: + (2xm/d)’

0

(19)

In future, we shall wnte the two round brackets (multlplymg a2 eQzz and b2 e-‘*‘, respectively) as [l - f+(z)] and [l - f-(z)], for slmphclty The result (19) is, as can be seen, extremely comphcated, but It should be noted that It 1s a sum (or integral) over wave modes of wave vectors k, and k,, m the x and y duectlons, Just as was the solution to Laplace’s equator m region 1 and as the solution will be shown to be m regions 3 and 4 This fact 1s useful as It allows us, later, to separate the solutlon m all regions mto separate surface modes m the x-y plane 2 3 Region 3 The lsotroplc liquid crystal or other lsotroplc dielectric which forms the medium of region 3 has an lsotroplc dlelectnc permlttlvlty tensor, that IS

(20)

Thus Laplace’s equation has the form E3V2G3

(21)

=o,

with the solution G3 (x, y, z) =

Tdk, idk,, ei(Lxx+kyy) -m --a,

x 1~3(k, k,>eaz + b3(k, k,) e-‘T ,

(22)

where Q IS defined as m eq. (6) 2.4 Region 4 In regon 4 we have as medium an amsotroplc crystal with one axis of amsotropy at an angle 8 to the x axrs of fig 2. In the coordinate system we are using, Q(Z) takes the form

Q(Z) =

cos2 f&, + sm2 0~~~

+ sm 28 (E.+,- Q,)

0

+ sm 20 (Q, - cqt)

sin2 eE4, + ~0s~ eE4,

0

0

0

E4t

.

(23)

120

E R SMITH AND B W NINHAM

This IS a very mconvement form to use m the solution of eq. (2), so we express eq (2) m terms of a new set of coordmates (x’, y’, z) formed by rotatmg the old x and y axes so that the new x axis corresponds to the axis of amsotropy of the amsotroplc crystal For this set of coordmates, G+(Z)takes the form

(24)

,

and m terms of the new coordmates, eq. (2) becomes t4]

--

84t

a2Q4+ a2G4+ a2G4 ai2 ap az2 -

o

-=

(25)

We now make the substltutlon

x* = (Eqt/E4,)* x’ =

y4x’,

(26)

and Laplace’s equation becomes

aw,

-+

ax*2

aw4

aw4

ayf2

F

-+

o

=

(27)

’

which has the general solution Q4

(x*,

y',

z)

r&-L rdk; ei(kx’x*+ky’y’)

= -m

x

-00

[bi (k:, k:)

e-“* +

ai(k: , k:) e”‘],

(28)

where @I2= kL2 + ki2. Note that this solution IS general and does not correspond to a surface-wave mode as It diverges as z + + co We now transform the coordinates (x*, y’, z) back to the orlgtnal (x, y, z) system to obtain Q4 (x, y, z) We find Qi, (x, y, z) =

fdk,

-00

rdk,, ei(k~x+k~y) -03

x [b4 (k,, k,) e-04’ + a4 (k,, k,) ep4’],

(29)

where pi = k: [(co? O/y:) + sin2 131+ k,’ [cos2 8 + (sin2 O/r:)]

+ k,k, sin 28 [(l/y:) - 11

(30)

RESPONSE

OF LIQUID

CRYSTALS

TO VAN DER WAALS FORCES

121

Thus solutron IS also not a surface wave and we obtam the drspersron relatron for the system by forcmg rt to be a surface wave, that IS by msrstmg that a, (kX, k,) = 0 The boundary condrtrons are that the transverse component of the electnc field and the normal component of the drsplacement vector be contmuous across any surface m the system. For the electric potentral m our system, these condmons reduce to (a) @ IS contmuous everywhere, (b) et (a@/&) IS contmuous everywhere From the appearance of the solutron in eqs (5) (19) (22) and (29) we can see that the solutrons may be separated mto separate surface-mode solutrons corresponding to each value of the wave vectors k, and k,, m the x and y dn-ectrons We may now demand that the boundary condrtrons be satisfied separately for each such surface-wave mode. We obtain the drspersron relation from the surface boundary condrtrons by calculatmg the a, (k,, k,,) of eq (29) m terms of the behavrour of the electric potentral at the Interfaces and the value of a, (k,, k,) We then set a4 (k,, k,) = 0 to obtain the drspersron relation. We illustrate this process by calculating a4 (k,, k,), b4 (k,, k,.) m terms of u3 (k,, k,,), b3 (k,, k,) The contmmty condrtron gives a3

(k,, k,,) ep(d+d) + b3 (k,, k,) e-p(d+d) = a4 (k,, k,) eoatd+‘) + b4 (k,, k,) e-‘Qtdfd),

(314

while the condrtron for the contmmty of E, (&D/&) gives @s3a3

(k,, k,) eq(d+d) - p3b3 (k,, k,) e-“(d+d)

= ~4~4~~4(k,, k,) e“4(d+d) - e4E4,b4 (k,, k,) e-“4(d+d’

Wb)

We may write both of these equations together m the form

X

(31c)

We obtain similar (but more complicated) matrix expressrons for [a3 (k,, k,,), b3 (k,,

kJ1 in terms of [a2 (k, k,J, bz (k, kJ1 and for [a2 (k, k,), b2 (k,, k,Jl In terms of [aI (k,, k,), 01. Multrplymg all the matrices together we find an expressron of the form (32)

122

E.R

SMITH AND B W NINHAM

The dispersion relation for the system is then D(fIo,6)=

M,,,(o,d)

=o.

(33)

If we carry out this program we find

D (CO,6) =

W.M-1cE4te4 + E3ej fbtw4,ee2e4

F(O) Y(O) + G(O)

wF(d) E3eFW

- Ezte24 (4 + Ezte2 9 (4

+(:::z:g>[( xe -

2Q2d

> (34)

where we have used the definitions

w = 1 -f+(z),

g(z) = 1 -f+(z) -

G(z) = 1 -f-(z),

s(z) = 1 -f-(z)

+

(lle2)fM, (lle2)f.l.(z>,

(35)

andf+(z) andf_(z) are as defined m the discussion immediately followmg eq. (19) By substitutmg the dispersion relation (34) mto eq (1) we are now readv to calculate the change m the Van der Waals energy caused by brmgmg the amsotropic crystal from an infinite distance to a distance 6 from the top surface (at z = d) of the liquid crystal

3. Evaluation of the Van der Waals energy for the system. The dispersion relatron (34) is very comphcated Indeed To obtain meaningful results from it, it IS necessary to make some approximations For the first of these we note that d s 6 Physically we would have d 2 1 mm while 6 I 50 pm Thus we may make the approximations 1 + 6/d z 1 and e-2Qd/e-2Qdx 0 We simplify the dispersion relation (34) usmg the second approximation and then substitute the result into

123

RESPONSE OF LIQUID CRYSTALS TO VAN DER WAALS FORCES

eq. (1) We find, on expanding the drspersion relatron to first order m &, that H(6, 6) z kT- L2 5’ r 4x2 n=O

d4 { e de

0

0

(36)

where we have transformed the sz dk, jz dk, of eq (1) to an integration over polar coordmates. jp d$ Jc e de The varrables k, and k,, are then grven by k, = .gcos cj~and k, = e sm C# If we substttute these formulae mto the expresstons for e2 and e4, we find @2 =

e (1 +

2cos2

4 K&21

@4 =

@ (1 +

2cos2

(4

-

-

~21)/2~2tl}+ = P2@

e> ME41

-

~4W4t113=

Wa)

P4@

Wb)

Eqs. (37a) and (37b) show that the expressrons [(et@,- e,ej)/(erer + e,e,)] which occur m eq (36) for the Van der Waals energy are independent of e. This provrdes a considerable srmphficatlon of the problem of performing the integrals m eq (36) We now expand the mtegrand of eq (36) using the expansron In (1 + P + f3,Q) x P + O,Q Thus IS perhaps not accurate smce when O. IS small, P2 may well be of the same order of magmtude as (or even larger than) 0,Q However, the object of our study IS to be able to predict the way the hqmd crystal twists. For this we require a mmrmtzatron of the energy of the whole system wrth respect to the function O(z) This energy IS a sum of the Van der Waals energy calculated here and the stress energy of the hqmd crystal Thus, for our purposes the Important quantity m thts calculation IS that part of the Van der Waals energy mvolvmg O(z) and rt IS certainly adequate for our purposes to use the above expansion Using this expansion of the logarithm, the expression (36) for the Van der Waals energy becomes

124

E R SMITH

AND

B W NINHAM

where we have already performed the mtegratlon over Q m the first term. We now perform the 4 mtegratlon for this first term In domg the 4 integration m the first term on the right--hand side of eq (38) we may consider two different physical sltuatlons The treatment of the second term IS the same m both cases In fact, our mam concern 1s with the second term, so we do the two cases of the first term now to get them out of the way. The first case IS that when medium 3 1s the lsotroplc phase of the liquid crystal formmg medium 2 The second 1s when we have some other dielectric as medium 3 3 1 Medium 3 the lsotroplc phase of medium 2 First we make the assumption that the dielectric constant Ed of the lsotroplc hqmd crystal 1s the mean of the dlelectnc constants of the ordered liquid crystal’), that 1s 3E3 = E21+ 2E2t

(39)

Next we note that because (Ed, - E~J/B~~1s m general small for a hquld crystal14), we may write p2 Z 1 + cos2 C#J (&21- &2t)/2~2t Thus, to first order m (Ed, - Ebb)/ &2t, we may write E3

-

EztP2

E3

+

E2tP2

&21

=

-

3 (83

E2t +

(1 - 3 co? 4)

(40)

E2t)

We now write the Van der Waals energy (38) m the form N(S, 19)= HO (S, f3) - H1 [6, 0, O(z)] and obtain Ho

(d,e)

kTL2

_

fr

~21 -

167~~6~n=o 3 (~3

+

E2t E2t)

2n

X

s

d$ (1 - 4 cos2 4)

1-

2E3 &4tP4

). +

(41)

E3

0

Because p4 > 1 and because we may expect t4 B c3 ‘I), it 1s useful to write We note that jp d4 (1 - 3 cos2 9) = 3x ExpresUk4tP4 + E3) = UE4tP4 slon (41) for Ho (6, 0) then becomes

3cos2e

[Z+(&)

_ 2(1AfA)

++)]}),

+ (1 + A)+

(42) where A = (c4, - E~J/E~~ and K(x) and E(x) are the complete elliptic mtegrals of the first and second kinds, wrltten m the notation of Abramowitz and Stegun”)

RESPONSE OF LIQUID CRYSTALS TO VAN DER WAALS FORCES

125

3 2 Medium 3 not similar to medium 2 In descrlbmg medium 3 as not similar to medium 2, we mean that I(F~, - E~~)/(Q + 2&J 1s small We make the expansion 83 -

&2tP2

F3 +

&ZIP2

=(~~J:::)-(~~cE),.,~.,,,cos2~,

(43)

which 1s correct to first order m (c2, - E~,)/(E~ + &2t) We use this expansion m eq (38) and performmg the integration on C$as before (that is, using the approxlmatlon E~/E~~4 1) we obtain kTL2 m Ho (6, 0) x ~ “[4(::I:::)-‘(~:‘+:::)(~3fl~2~)] 16x2d2 n=o

X

12”

-

E4t

X

F3 cos &a (1

+

2% (1 + 4’

28 A>+

(44)

As we discussed m the mtroductlon, the size of the gap 6 across which the Van der Waals mteractlons take place 1s so great that we need only consider microwave contrlbutlons to the Van der Waals energy In practice this means consldermg only the II = 0 terms m the sums c:Zo Examining expressions (42) and (44) for Ho (8, 0) m the two cases, we see that m the case of 3 1 the energy is positive (since &21> E~J and the force repulsive. In the case of 3 2, however, the force may be repulsive or attractive dependmg on the sign of (Ed - ~~0 The term dependent on cos 20 (which gives the torque exerted on the amsotroplc crystal by the hquld crystal) m both cases shows that the Van der Waals force tends to align the amsotroplc axes of medium 4 and medium 2 We now return to the part of the Van der Waals energy dependent on the twisting of the liquid crystal Substltutmg the expression forf;(d) mto the relevant part of eq (38), and remembering to take only the first term m the sum xA’Zo, we find

k = (1b3 Hi kt

e, e(4

E R SMITH AND B W NINHAM

126

X

s

dec2 e-‘06 ;

exp Ke - 2xlmld)

t In=-ul

e2 +

GO

-

(2xm/d)2

211

Cd

(45)

0

We have used the approxlmatlon ez w Q, so that eq (45) IS only correct to first order m (.z2, - &21)/~2t We can perform the sum on m fairly easily by makmg use of the Poisson summation formula16) and we obtam

= -d(l

+ 1/2x) e-P(d-zo)(l+ax’ [l + 0 (6/d)]

(46)

We Ignore the terms proportional to 6/d as small The Integration over Qm eq (45) IS then straightforward Again usmg the approxlmatlon E~/E~~+ 1, we find, for the integral over C/J ZIF

d$ sm 4 cos 4 E4tP4 - ” E4tP4 +

s 0

=

2+A

25 sm 28

K

(1 + A)”

A

E4t

83

2(1+& A

E (~E~/E~~)sin 28 5‘(A)

A ( l+A

>I (47)

Substltutmg eqs (46) and (47) mto eq. (45) and performing the mtegratlon over Q we find

x sm 28 5(A) f dz, 8(2,)/[26 + (d - zo) (1 + 1/2x)13

(4ga)

0

E

-e"(e)

j,,

F(y)/[26

+ y (1 + 1/2x)13,

(48b)

0

where, to simplify the appearance of h, for the variational calculation, we have defined Y(y) = 8 (d - y) We note that the Van der Waals torque on the hquld

RESPONSE

OF LIQUID

CRYSTALS

TO VAN

DER

WAALS

FORCES

127

crystal tends to twist the amsotroplc axis of the hqmd crystal round to be parallel with that of the amsotroplc crystal The factor sm 28 indicates that the torque 1s a maximum when the angle 8 IS at x/4, a result which might have been expected on physical grounds. The function [(A) IS plotted m fig 3.

Fig 3 A plot of the function A z 04

C(A) for 0 I A I 10 [(A) reaches a maxlmum of about 0 222 at Note that at A = 1, [(A) IS approximately half Its maximum value

At large A, c(A) % log A/A *

4 Varlatlonal calculation of the twist of the kquld crystal We may write down the stress energy of the hqmd crystal correspondmg to the twist described by the function 19(z)lmmedlately from the Frank theory12) We have hT = HT/L2 =

jfk,,[W(z)]” dz = j:k,,

[Y’(y)12 dy,

(49)

0

where HT IS the stress energy of a slab of hquld crystal of size L x L x d twisted accordmg to the function e(z) and kz2 IS the twist elastic modulus. The total energy per unit area of surface for the system IS then d

h = hT - h, =

3kx 0

[~‘(Y)I~

-

‘ce)

[26 + y(l

vy)

+ 1/2x)13 >

dy

(50)

We now mmlmlze h with respect to Y(y) SubJect to the condltlon that Y(d) = 0. This does not completely specify h To do this we have to write h m terms of Y(0) and then further mlmnuze h with respect to Y(0). To carry out the mmlmlzatlon of h over the functions Y(y), we wnte Y(y) = Ye(y) + 6YI (y), where Ye(y) IS the function we are endeavounng to find. Substituting this form for Y mto

128

E R SMITH AND B W NINHAM

eq (50), we find

d

=

!A2

V'XY)I'

W)

-

[26

+

WY) (l +

y

k22cm Y,(O)

1,2x)13

0 d

3((e) [26 + y(l

0

+

1/2x)13 >

Yl(Y)

(51) 1

For h to be a mmlmum, the coefficient of 6 has to be zero for all functions ul,, including those for which Y1(0) = 0 Thus we must have Y;(y) Integratmg we find ug(Y)

= -W3)/k22

[26 + y(1 + 1/2x)13

eq (52) subJect to the constramt

=

00+ -f

(52) that Ye(O) = B. and Yo(d> = 0,

q”(e) 4k,,6 (1 + 1/2x)’

[

e”(e)

e,+ 2k,,

(1 + 1/2x)’

qe) __ 2kz2 (1 + l/2+

1

--1 26

>I

26 + d(1 + 1/2X)

’ [2d + y(1 + 1/2x)]-’

(53)

We can now substitute this expression mto eq (50) The angle B. IS thenglven by the equation i%/8Bo = 0 Usmg expresslon (53) this equation reduces to d

ah

ae,=-

s(

k

dv -E-

Y;(y)

d

+

w3 (1 [2S + y(l

~14

+ 1/27t)j3> = O

(54)

0

By substltutmg the expresslon from eq (53) for Y:(y) mto eq (54) and performmg the mtegratlon over y, we find

80%

s(e) d 8k,, (1 + 1/2x) 6’

(55)

Usmg the value x/4 for 0 and 0 2 for &A), and some charactenstlc values for the ddectnc-constant ratios (for the case when medmm 3 1s the lsotroplc phase of the hquld crystal) we find, as a rough estimate 8 (x/4) x 2 x lo- l7 erg A

RESPONSE OF LIQUID CRYSTALS TO VAN DER WAALS FORCES

129

characterlstlc value for kzz might be 2 x 10m7 dyne 17) An estimate for 8, would then be 80 x (lo-l”/8) (d/a2) For 6 x 1000 A and d = 1 cm, we would then For distances 6 greater have 8o w _Brad z 7”, while for 6 = 1 pm, 8, z 2 than 1 or 2 pm, we would expect the dlstortlon of the liquid crystal to be small. For distances 6 less than 1000 A, we would expect large distortion of the hquld crystal, so that our perturbation expansion might break down Of course, to have a gap 6 < 1000 A, we would need some other isotropic medium than the lsotroplc phase of the hqmd crystal, as such a thm layer of tsotroptc hquld crystal would be unstable The material would have to be chosen to have a similar weight density to the hqmd crystal, or there might be some difficulty with the approxlmatlon that we need only consider the nonretarded microwave contrlbutlons to the Van der Waals mteractlons3) For distances considerably less than 1000 A, It would be necessary to take the retarded infrared and ultraviolet contrlbutlons into account Finally we note that the deformation of the liquid crystal has been assumed to be a contmuous twist It would seem unlikely that the liquid crystal would form two domains twlsted with respect to each other because of the large energy required to form a domam boundary However, it 1s possible that only the first few layers of molecules are twlsted round by the Van der Waals torque The stress energy associated with this sort of &stortlon would be small as It would correspond to a very mcomplete domain boundary If we let the mtermolecular distance be u and model this sort of distortion by the distortion fun&on Olylna na I y I d,

where n 1s the number of molecular layers twisted round (assumed small) then the net energy of the system 1s given by h z E,, (n, 6,) - R(O) t90na/863,

(57)

where E,, (n, 0,) IS the stress energy per unit area associated with the dlstortlon described m eq (56) As we have not been able to develop a theory of thrs stress energy yet, we have been unable to pursue this posslblhty further Acknowledgement The authors would hke to thank P G De Gennes and E Guyon for mtroducmg them to this problem.

REFERENCES 1) De Genres, P G , CR Acad SCI 271 (1970) 469 2) Meyer, R B , Phys Rev Letters 22 (1969) 918 3) Parseglan, V A and Nmham, B W , J Collold Interface SC] 37 (1971) 332

130 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17)

E R SMITH AND B W NINHAM

Nmham, B W , Parseglan, V A and Weiss, G H , J statist Phys 2 (1970) 323 Dzyaloshmsku, I E , Llfshltz, E M and Pltaevsku, L P , Advances m Phys 10 (1961) 165 Kats, E I , Soviet Physics-JETP 33 (1971) 634 Parseglan, V A and Weiss, G H , J Adhesion 3 (1972) 259 Dubols-Vlolette, E and Parodl, 0 , J Phys 30, C4 (1969) 57 Ambrose, E J , Paper presented at the Faraday Symposmm on Llquld Crystals, London, November 1971 Parseglan, V A and Gmgell, D , to be pubhshed Guyon, E , private commumcatlon Frank, F C , Disc Faraday Sot 25 (1958) 19 Morse, P M and Feshbach, H , Methods of TheoretIcal Physics, McGraw-HI11 (New York, 1953) p 1255 Pelzl, G and Sackmann, H , Paper presented at the Faraday Symposmm on Llquld Crystals, London, November 1971 Abramowltz, M and Stegun, I A , Handbook of Mathematical Functions, Nat Bur Stand (Washmgton, 1970) ch 17 LIghthIll, M J , Fourier Analysis and Generahzed Functions (CambrIdge, 1970) p 69 Saupe, A , Z Naturforsch 15a (1960) 815