2Mean field theory
III Theoretical Physics of Soft Condensed Matter
2.2 Nematic liquid crystals
For our purposes, we can imagine liquid crystals as being made of rod-like
molecules
We are interested in the transition between two phases:
– The isotropic phase, where the rods are pointed in random directions.
–
The nematic phase, where the rods all point in the same direction, so that
there is a long-range orientation order, but there is no long range positional
order.
In general, there can be two different sorts of liquid crystals — the rods can
either be symmetric in both ends or have “direction”. Thus, in the first case,
rotating the rod by 180
◦
does not change the configuration, and in the second
case, it does. We shall focus on the first case in this section.
The first problem we have to solve is to pick an order parameter. We want
to take the direction of the rod
n
, but mod it out by the relation
n ∼ −n
. One
way to do so is to consider the second-rank traceless tensor
n
i
n
j
. This has the
property that
A
i
n
i
n
j
is the component of a vector
A
in the direction of
n
, and
is invariant under
n
i
7→ n
j
. Observe that if we normalize
n
to be a unit vector,
then
n
i
n
j
has trace 1. Thus, if we have isotropic rods in
d
dimensions, then we
have
hn
i
n
j
i =
δ
ij
d
.
In general, we can defined a coarse-grained order parameter to be
Q
ij
(r) = hn
i
n
j
i
local
−
1
d
δ
ij
.
This is then a traceless symmetric second-rank tensor that vanishes in the
isotropic phase.
One main difference from the case of the binary fluid is that
Q
ij
is no longer
conserved., i.e. the “total Q”
Z
Q
ij
(r) dr
is not constant in time. This will have consequences for equilibrium statistical
mechanics, but also the dynamics.
We now want to construct the leading-order terms of the “most general” free
energy functional. We start with the local part
f
(
Q
), which has to be a scalar
built on Q. The possible terms are as follows:
(i) There is only one linear one, namely Q
ii
= Tr(Q), but this vanishes.
(ii)
We can construct a quadratic term
Q
ij
Q
ji
=
Tr
(
Q
2
), and this is in general
non-zero.
(iii)
There is a cubic term
Q
ij
Q
jk
Q
ki
=
Tr
(
Q
3
), and is also in general non-zero.
(iv) There are two possible quartic terms, namely Tr(Q
2
)
2
and Tr(Q
4
).
So we can write
f(Q) = a Tr(Q
2
) + c Tr(Q
3
) + b
1
Tr(Q
2
)
2
+ b
2
Tr(Q
4
).
This is the local part of the free energy up to fourth order in
Q
. We can go on,
and in certain conditions we have to, but if these coefficients
b
i
are sufficiently
positive in an appropriate sense, this is enough.
How can we think about this functional? Observe that if all of the rods point
tend to point in a fixed direction, say
z
, and are agnostic about the other two
directions, then Q will be given by
Q
ij
=
−λ/2 0 0
0 −λ/2 0
0 0 λ
, λ > 0.
If the rod is agnostic about the
x
and
y
directions, but instead avoids the
z
-direction, then
Q
ij
takes the same form but with
λ <
0. For the purposes of
f
(
Q
), we can locally diagonalize
Q
, and it should somewhat look like this form.
So this seemingly-special case is actually quite general.
The
λ >
0 and
λ <
0 cases are physically very different scenarios, but the
difference is only detected in the odd terms. Hence the cubic term is extremely
important here. To see this more explicitly, we compute f in terms of λ as
f(Q) = a
3
2
λ
2
+ c
3
4
λ
3
+ b
1
9
4
λ
4
+ b
2
9
8
λ
4
= ¯aλ
2
+ ¯cλ
3
+
¯
bλ
4
.
We can think of this in a way similar to the binary fluid, where
λ
is are sole
order parameter. We fix
¯
b
and
¯c <
0, and then vary
¯a
. In different situations,
we get
λ
f
α < α
c
α = α
c
α > α
c
Here the cubic term gives a discontinuous transition, which is a first-order
transition. If we had ¯c > 0 instead, then the minima are on the other side.
We now move on to the gradient terms. The possible gradient terms up to
order ∇
(2)
and Q
(2)
are
κ
1
∇
i
∇
i
Q
j`
Q
j`
= κ
1
∇
2
Tr(Q
2
)
κ
2
(∇
i
Q
im
)(∇
j
Q
jm
) = κ
2
(∇ · Q)
2
κ
3
(∇
i
Q
jm
)(∇
j
Q
im
) = yuck.
Collectively, these three terms describe the energy costs of the following three
things:
splay
twisting
bend
In general, each of these modes correspond to linear combination of the three
terms, and it is difficult to pin down how exactly these correspondences work.
Assuming these linear combinations are sufficiently generic, a sensible choice is
to set
κ
1
=
κ
3
= 0 (for example), and then the elastic costs of these deformations
will all be comparable.