Order Parameters and Defects in Liquid Crystals - e-lc.org · Liquid Crystal-substance with a...

Order Parameters and Defects in Liquid Crystals

Ferroelectric phenomena in liquid crystalsJune, 2007

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Liquid Crystal-substance with a degree of crystalline order that remains in a liquid state

• Simple picture- long thin molecules

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Energy to describe static stable configurations

Made up of competing energies• Molecular packing (molecular alignment)• Thermal effects (randomize alignment)• External stresses (mechanical, electrical..)

Temperature dependent-different terms become dominant at different temperatureselectr

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Isotropic

Nematic

SmecticT

• Nematic phase-energy seeks local alignment of long axes.

• n(x) is the “local average”of long axes.• |n(x)|=1

n(x)

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n(x) is a unit vector field

• Molecules do not have a head and tail as n(x) does.• In problems where this does not lead to inconsistencies n(x) is the

simplist way to describe a nematic liquid crystal.• n(x)-director field

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Oseen Frank Energy

• Energy records the cost of distortions away from n(x)=const.

• Written in terms of pure splay, twist, and bend. These have K1 ,K2, K3 as coefficients respectively

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Pure Distortions

Splay Bend Twist

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ionapproximatconstant one 0,K, case, Special 4321 ==== KKKK

vector identity

=⇒ )(nFFelectronic-

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Boundary Conditions

Ω∂= on 0nn1)

2) 2))(( xencFb ⋅= ∫ Ω∂

1) Strong anchoring problem2) F=FF+Fb weak anchoring

If c<0 then n(x) tends to be parallel to e(x). e(x) is the easy axis.

If n(x) or e(x) are parallel to the boundary normal

we are promoting “homeotropic” boundary values.

1, 0 =n

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If c>0 then n(x) tends to be e(x) on the boundary. If e(x) is parallel to the boundary normal then n(x) tends

towards tangential boundary values.

⊥

ThmAssume is a smooth surface, n0 is smooth, K1 ,K2, K3 >0.Then there is a minimizer for F(n) in the class

}||,|,:|{ 20

2 ∞<∇=→Ω= ∫ ΩΩ∂ unuSuuA

Ω∂

.nonemptyisAprovided

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n(x) may have singularities, “defects”

• ex . n(x)=x/|x|“Hedgehog”. n(x) has homeotropic b.v. and

a defect at one point. Minimizers from the theorem can not be singular on a curve. Line singularities in liquid crystals do exist, “disclination lines”.

Try n(r, ,z)=er and get

RB=Ω

θ

∞=∇∫ 2|| nelectr

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Cladis and Kleman consider :

withzr ererrn )(sin)(cos)( ϕϕ += 0)( =Rϕ

R

rz

r

zn(r)

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rdrr

dxdynF RBR

)cos)'((2|| 2

22

02 ϕϕπ +=∇= ∫∫

They prove the minimizer is )(tan22

)( 1

Rrr −−=

πϕ

R

Solution is not singular. It escapes to the third dimension.

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f( ) probability density that the director of molecules near x are within an angle

Of n(x).f0( ) =s(x) is the the isotropic order parameter .Measures how concentrated f is near =0.

π41

ϕ

))((cos23)(

20

2∫∫ −=S

ffxs ϕ

ϕ

ϕ

Introduce an order parameter that allows n to “melt’ near r=0

s=1 strongly nematic, s=0 isotropicelectronic-

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Ericksen considers a nematic described by the pair (s,n)

2s

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}0)(,0)(,)(sin)(cos),({ 0 >==+ sRsRererrs zr ϕϕϕ

Mizel, Roccato, and Virga prove that this has a minimizer

( ) in the family

of the form 0)0(~,00)(~,0~ =>>≡ sforrsϕ

•

ns ~,~

The solution is allowed to melt into the isotropicn phase at r=0,

has finite energy, and has a disclination line along the z axis.

1≤kIf

.2

)0(~0)0(~1 πϕ =>> andshassolutionthethenkIf

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uniaxial

biaxial

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Smectics

∫=π

ρ2

)(x

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Molecules locally align and in addition form layers

ν

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• Layer normal • Density modulation

• , = layer thickness• If then layers exist - smectic phase• If then no layers - nematic phase

01 >ρ

)2cos()( 10 xd

x ⋅+= νπρρρ

dq π2=

ν

d

01 =ρelectr

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• If - smectic A• If not - smectic C• Complex order parameter

• nematic, smectic

n||ν

)()( xiex ϕρψ =

0=ψ

*1 cos2 ψψϕρ +=

0≠ψ

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Energy

III II I

A>0

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• I prefers smectic phase• II locally the level sets of are

the layers,

• III penalizes phase transitions

11 =ρ

,|| nϕ∇ q≈∇ || ϕ

ϕ

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0

r

Relectronic-

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Rffrdr surfR +∫ 0

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reasy

iiii

eekSK

== ,2

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Cholesteric Phases

τπ

=

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Chirality introduces a stress onto the liquid crystal that affects its phase transition.

τ lowers the transition temperature

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Smectic C

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Set )(xie ωψ =

Then

Now let xk ⋅=ω and let n= const , then

k| k

n⋅

n⋅electronic-

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This minimizes for =2|| k

and =⋅

=||

cosknkθ

θ

||2kπ

||2kπ

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dxFF FSm )(F +∑

= ∫

dxFF FSm )(F +∑

= ∫electr

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Boundary value problem for F.M. Calderer and J.Park

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N. Clark et al Jakli et al Bauman et al

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