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.