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Submitted on 1 Jan 1983
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Critical thickness of a hybrid aligned nematic liquid crystal cell
G. Barbero, R. Barberi
To cite this version:
G. Barbero, R. Barberi. Critical thickness of a hybrid aligned nematic liquid crystal cell. Journal de
Physique, 1983, 44 (5), pp.609-616. �10.1051/jphys:01983004405060900�. �jpa-00209638�
Critical thickness of a hybrid aligned nematic liquid crystal cell
G. Barbero (*) and R. Barberi
Unical Liquid Crystal Group, Department of Physics, Calabria University, I-87036 Rende, Italy (Reçu le 5 octobre 1982, révisé le 6 décembre 1982, accepté le 13 janvier 1983)
Résumé. 2014 Nous avons analysé la déformation des cristaux liquides nématiques dans une cellule avec alignement planaire sur une paroi et homéotrope sur l’autre. Nous supposons que l’interaction superficielle est du type (w/2) sin2 (~ - 03A6) et que l’énergie d’ancrage homéotrope est plus grande que celle relative à l’ancrage planaire.
En l’absence du champ électrique, nous démontrons qu’il existe une épaisseur critique dc au-dessous de laquelle
le cristal liquide nématique n’est pas déformé. Nous discutons aussi la dépendance en dc du champ réduit E/Ec (où Ec est un champ de seuil d’instabilité pour transition de type de Freedericks). Ec dépend seulement des énergies d’ancrage Wi sur les parois et du cristal liquide.
La dernière analyse montre qu’il y a deux épaisseurs critiques dc(H) et dc(P) : pour d dc(H) la cellule est homéotrope,
et si d > dc(P) elle est planaire. Enfin nous démontrons que si les énergies d’ancrage homéotrope et planaire sont égales, la cellule hybride est réalisable seulement si E/Ec 1.
Abstract.
2014We discuss the director deformation inside a hybrid aligned nematic (HAN) cell, assuming that the
surface interaction is of the (w/2) sin2 (~ 2014 03A6) kind, and we suppose that the anchoring energy for homeotropic alignment is greater than the planar one.
In the absence of an external electric field, we show that a critical thickness dc, below which the nematic liquid crystal (NLC) is undistorted, exists. Furthermore we investigate theoretically dc vs. the reduced electric field
(E/Ec, where Ec is a threshold field depending upon the anchoring energies and the intrinsic constants of the NLC).
The last analysis shows that there are two critical thicknesses dc(H) and dc(P): for d dc(H) the sample is homeotropic (HOM), while for d > dc(P) it is planar (PLAN). Finally we point out that if the anchoring energies are the same
on the upper and lower plates of the sample, the HAN-configuration is realizable only if E/Ec 1.
Classification
Physics Abstracts
61.30G - 42.65J
-42.1OQ
Introduction.
-The HAN cell has been recently
considered by various authors from a pratical [1-4]
and fundamental [5, 6] point of view. In fact this cell
seems to be particularly suitable for creating coloured displays [1, 2] or optical devices that might use self focusing to get bistability [4]. Furthermore the hybrid alignment gives a volume flexo-electric polarization : by analysing the behaviour of the HAN cell submitted
to an electric field it is possible to determine the bulk flexo-electric constant [6].
M. M. Labes et al. [4] and L. Terminassian-Saraga
et al. [7] show that the NLC film subject to antagonist homeotropic and planar anchoring on two glass plates can be distorted only if the thickness d of the
sample is greater than the critical one dc, and give
an order of magnitude for dr. In the present paper
we will calculate in a rigorous way dc, and will analyse
the influence of the elastic anisotropy K = I - K 11 /K3 3
and of the electric field on dc.
1. HAN cell without external field
-The sample geometry is shown in figure 1. A NLC is placed between
two glass plates (at Z = - d/2 and Z = d/2) coated
to induce homeotropic and planar orientations [8]
respectively. In this condition the sample has under-
gone a mixed splay-bend distortion. We can obtain the director deformation of the NLC by the conti-
nuum theory, supposing that the anisotropic inter-
action between the liquid crystal and the substrate is not strong. In this paper the surface energy is modelled as (wi/2) sin2 (Qi
-oi), where wi is the anchoring strength coefficient, tP¡ the easy axis and (Pi the director angle at the surface substrate [9, 10].
In the case of the HAN cell we have tP 1 = 0 and tP2 = n/2 on the lower and upper walls respectively,
and we suppose wl > w2 [11].
Article published online by EDP Sciences and available at http://dx.doi.org/10.1051/jphys:01983004405060900
610
Fig. 1.
-Schematic representation of a HAN cell in the
case of finite anchoring energy. T, and T2 represent the tilt angles at the interfaces (in the strong anchoring hypo-
thesis they are given by (p, = 0 and CP2 = n/2).
1.1 ONE CONSTANT APPROXIMATION.
-In the one
constant approximation (K11 1 = K33 = K, i.e. K = 0) by minimizing the total free energy we obtain
with the boundary conditions
where Li = K/wi are the de Gennes-Kleman [12]
extrapolation distances, the Z-axis is normal to the substrate and 9 is the tilt angle formed between the director n and the Z-axis.
The subscript Z means the derivative with respect to Z.
By solving equation ( 1.1 ) we obtain :
where the constants Q2 = Q(d/2) and 91 1 = Q(2013 d/2)
are determined by equations (1.2). In the strong anchoring hypothesis wi -> oo and hence cp 1 = 0,
Q2 = n/2; then 9 = (n/2 d) (Z + d/2) [3, 6]. When
the anchoring energies are finite, but L d, it is possible to linearize the boundary conditions (1.2)
and after trivial calculations we obtain :
Consequently, from equation ( 1. 3)
In these cases (i.e. wi -> oo or Li d) for any sample-
thickness d the tilt angle Q is linearly increasing with
Z : this implies that a critical thickness dc, below which the sample is undistorted, does _not exist.
In the opposite situation, i.e. Li ~ d, the boundary
conditions give, for example,
By solving this transcendent equation for 2 Q 1 we find that
where h is an integer and
With the aim of determining the critical thickness dc,
it is necessary to find the situation where QZ -> 0,
e.g. 92 - 9i
Then we can restrict ourselves to the case in which qJ¡ E (0, n/2). Using this assumption in equation (2 .1 )
we have h = 0 and h = 1 only.
Furthermore, by taking into account that 0, 1 = 0, ø2 = n/2 and w1 > w2, we get Q2 > qJt’ and hence sin (P2 > sin Q 1. By using equation (2. .1 ) the latter condition can be rewritten as
These relations are equivalent to
Since L, L2, we deduce that h = I is not accep- table.
In consequence we have
and hence
The function Q2 = Q2(d) is monotonically increasing
with d : this implies that equation (2.3) has only
one solution.
From equations (2.2), (2. 3) it is possible to deter-
mine Qi, and then, by means of ( 1. 3), the tilt angle Q.
The transcendent equation (2.3) has a non-trivial
solution only if
which implies d > dc(H) = L2 - [1. Then the equa- tions (2.2), (2.3) have the unique solution Qi; = 0
if d dc(H) : consequently the sample is undistorted.
For d > dc(H), Qi # 0, i.e. it is possible to obtain the
HAN cell. We point out that for d dcH) the NLC subject to conflicting boundary conditions destroys
the smaller anchoring energy, reacting to the increasing distortion ; thus the result for the total free energy is
w2/2.
_ _The previous result, dc(H) = L2 - Li, is in agree- ment with that given by M. M. Labes [4].
It is possible to rewrite equation (2.2) as
Now it is necessary to test the stability of the solu- tion Q2 = 0 (i.e. (Q = 0 : the sample is in the HOM-
configuration) for d d, .
In this case the total free energy is given by :
where
’Ps represents the nematic-substrate interaction.
From equations (2. 2) and (1. 3) we find that :
and hence, by using Q2 as parameter, we obtain, from (5)
Consequently
From these relations we deduce that T2 = 0 is a stable
configuration only if d dc(H’), while for d > dc (H)
the total free energy corresponding to HOM-situation is a maximum (see appendix 1).
Finally we point out that if the surface interaction is of the parabolic form the critical thickness does not exist.
_ _
In fact if ’YS = (1/L1) Q2 1 + (1/£2) (7t/2 CfJ2)2 the boundary conditions (1.2) become :
Taking into account equation (1.3) we find that the constants qJi are given by (1.4) and the tilt angle by (1. 5) : hence the sample is always distorted
1. 2 GENERAL CASE.
-In the general case (K 11 K33, i.e. K # 0) equations (1.1), (1.2) become :
where Li = K33/Wi and C2 is an integration constant,
fixed by the boundary conditions (6.2). From (6.1), (6.2) we obtain
being
By solving the simultaneous equations (7) we can
deduce the unknowns Q 1, Q2, Q and C (see Figs. 2).
Figures 2 show the behaviour of Q2 and Q 1 vs.
the sample-thickness d for some elastic anisotropies.
From figures 2 we see that if d >> d c (H) the HAN cell
assumes a configuration near the ideal one, with strong anchoring energies.
In order to determine dc(H) , from (7.1) written for Z = ± d/2 we determine, after some algebra,
(Pl (Pl((P2) and
From (8) it is easy to show that ðd( ({J2)la({J2 > 0 still,
for any Q2 E (0, nI2), and hence dIHI = lim d(qJ2).
for any T2 c
CP2 -+ 0
In the case of small Q2, routine calculations give
Therefore, from (8), we have d((P2) = L2 - L1 + 0(Q2) and then dc,(H’) = L2 - L1, as in the case of
elastic isotropy; but Li = K33Iwi if we consider the situation L2 > L1(Wl > w2). In the opposite situation
an analogous calculation (where Q1 - n/2 if d - dc(p))
shows that the sample is planar if d d,(P) = L1 L2,
where Li = K 11 /wi. In order to examine the stability
of the HOM-configuration for d dc (H) we proceed
as in the previous case.
Now the total free energy W is still given by equa-
612
Fig. 2.
-a) Plot of Q2 and b) of l{JI vs. the sample thickness d (J.1m), for K = 0, K = 0.3 and K = 0.6, which are reasonable
values [ 12]. We note that K > 1 only next to the transition
N - SA, and in this range the HAN cell is not used. The
case L1 = 1 J.1m and L2 = 5 J.1m is considered, giving dcH) = 4 um. We point out that lim CP2 = 7r/2 and
d-+ 00
lim Q1 1 = 0.
d-+ 00
tion (5), but f(Q, cpz) = ( 1 - K sin2 Q) cpi. Using (6. .1 ) and (7 . 2) it is possible to rewrite as
In the limit of a small deformation close to the
HOM-configuration (e.g. rp2 - 0) we have from the
previous discussion
as was the case when K11 = K33. Hence the HOM-
configuration is stable for d dv(H).
We note that the present case, in the absence of an
external field, is very important for the analysis of the
non-linear birefringence in a HAN cell [5]. Here for
E Et, E, being the threshold field for the Freedericks transition, it is possible to expand the tilt angle (p as a
power series in e = E/Et. This way the zero approxi-
mation gives equation (7), while higher orders give
linear equations integrable only if the zero order is
known [14] (see appendix 2).
2. HAN cell submitted to external field.
-Now let
us analyse in greater depth the case of a HAN cell
submitted to an electric field parallel to the X-axis.
For a small positive dielectric anisotropy, ea = s ||
-