Liquid Crystals: Lecture 1 Basic properties Support: NSF Oleg D. Lavrentovich Liquid Crystal Institute and Chemical Physics Interdisciplinary Program, Kent State University, Kent, OH 44242 Boulder School for Condensed Matter and Materials Physics, Soft Matter In and Out of Equilibrium, 6-31 July, 2015
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Liquid Crystals: Lecture 1
Basic properties
Support: NSF
Oleg D. Lavrentovich
Liquid Crystal Institute and
Chemical Physics Interdisciplinary Program,
Kent State University, Kent, OH 44242
Boulder School for Condensed Matter and Materials Physics,
Soft Matter In and Out of Equilibrium,
6-31 July, 2015
Acknowledgements
Volodymyr Borshch, Mingxia Gu, Tomohiro Ishikawa, Israel Lazo, Young-Ki Kim, Heung-Shik Park, Chenhui Peng, Oleg Pishnyak, Ivan Smalyukh, Bohdan Senyuk, Jie Xiang, Shuang Zhou (LCI, Kent State University)
Drs. Sergii Shiyanovskii, Antal Jakli (LCI, Kent State University)
Maurice Kleman (Paris), Grigory Volovik (Moscow), Vasyl Nazarenko (Kyiv), Igor Aranson (Argonne NL) Georg Mehl (Hull), Corrie Imrie (Aberdeen)
The lectures are based on the book by
Maurice Kleman and O. D. Lavrentovich
“Soft Matter Physics: An Introduction” (Springer, 2003)
3
Content: Lecture 1
Types of liquid crystalline order Nematic
Cholesteric and blue phases
Twist bend nematic
Smectic A
Lyotropic and Chromonic LCs
Basic Physics Dielectric anisotropy
Surface anchoring
Elasticity
Frederiks effect and modern LCDs
Optics Birefringence
Polarizing microscopy: 2D imaging
Fluorescence Confocal Polarizing microscopy: 3D imaging
4
Content: Lecture 2
Topological defects and droplets Disclinations in uniaxial nematics
Singular
Nonsingular
Homotopy classification
Drops and Conservation laws of topological defects
Cholesteric droplets, Dirac monopole
Chromonic tactoids
Broken chiral symmetry
Wulff construction for liquid crystals
Lamellar phases Free energy density for weak and strong perturbations
Long-range character of deformations; undulations
Focal conic domains
5
Content: Lecture 3
Dynamics of director realignment
Anisotropy of viscosity
Coupling of director reorientation and flow
Statics of colloids in nematic LC
Levitation
Dynamics of colloids in nematic LC
Brownian motion
LC-enabled electrophoresis
LC with patterned orientation as an active medium
Living LC
Swimming bacteria in LC; individual and collective effects
6
Liquid crystal:
a state of matter with long-range orientational order and
complete (nematic)
partial (smectics, columnar phases)
absence of long-range positional order of “building units”
(molecules, viruses, aggregates, etc.)
Our goal: Develop an
intuitive understanding
of what kind of new
physics the orientational
order brings to soft
matter
Crystals, liquid crystals, liquids
pentyl-cyanobiphenyl (5CB):
room temperature nematic
1.2 nm
Temperature
orientational and positional order: Molecular crystal
Elasticity vs Anchoring: Hybrid-Aligned Nematic Film
1. Infinitely strong anchoring
Fixed boundary conditions
Problem: Find the director dependence on z in equilibrium
Assumption #1:
2
3
2
1 cos2
1sin
2
1
=
dz
dK
dz
dKfFO
ˆ , , sin , 0, cosx y zn n n z z = =nn̂
K1 = K3 = KAssumption #2:
2
12FO
df K
dz
=
00z = = dz d = =
The problem reduces to finding (z) that minimizes the integral
2
12
0
z d
FO
z
dF K dz
dz
=
=
=
Euler-Lagrange eq. (next slide gives the outline, home assignment if you do not know it yet):
2
20 0
'
FO FOf fd d
dz dz
= =
0
0
d zz
d
=
2
012
d
FOF Kd
=
eq z
za=
z = eq z a z
z= 0 = z = d = 0
where is such that
z
F z = f eq z a z , 'eq z a' z , z dz0
d
F a
a
a=0
=0 Condition of the extremum:
Euler-Lagrange equation for 1D problem, fixed boundary conditions (leisure time reading)
FFO = f FO ,' ,z dzz =0
z=d
FFO = 12 K '
2dz
z=0
z=d
?
F a
a=
f
af
'
'
a
dz
0
d
=f
z
f
'
d z
dz
dz
0
d
d z
dz
f
'dz
0
d
= z f
' z=0
z=d
0
z d
dz
f
'dz
0
d
f
d
dz
f
'
z dz = 0
0
d
f
d
dz
f
'
ddz
0
d
= aF a
a
a =0
= dF = 0or
f
d
dz
f
'= 0
Euler-Lagrange equation for 1D problem; its solution
has 2 constants of integration defined
from the boundary conditions, for example, = z,c1,c2
z = 0 = 0 z = d = dand
F = f ,' ,z dz0
d
f s0 0 0 f sd d d
zzz eq ad ==
z is not necessarily 0 at the boundaries!
dfs 0 0 da
=d
dafs 0 0,eq a z = 0 = 0
dfs0
d0
F a
a=
f
af
'
'
a
dz
0
d
dfs 0
d0
0 dfsd
dd
d
d z
dz
f
'dz
0
d
= z f
' z=0
z=d
0
z d
dz
f
'dz
0
d
f
d
dz
f
'
z dz
f
'
dfs0
d
z=0
0 f
'
dfsd
d
0
d
z=d
d = 0
f
d
dz
f
'= 0
Euler-Lagrange equation for 1D problem; its solution
has 2 constants of integration defined
from the boundary conditions = z,c1,c2
f
'
dfs0
d
z = 0
= 0f
'
dfsd
d
z= d
= 0and
Euler-Lagrange equation for 1D problem, soft boundary conditions (leisure time reading)
Elasticity vs Anchoring: Hybrid-Aligned Nematic Film
2. Finite (weak) anchoring
Problem: Find the director vs z in equilibrium with new
boundary conditions:
0
0
0'
FO s
z
f df
d =
=
n̂
2
12FO
df K
dz
=
21
0 0 0 02sf W = 2
12sd d d df W =
n̂
0'
FO sd
z d
f df
d =
=
0
0
d zz
d
=
0 0
0 0
0
d
d
L
d L L
=
2
012
0
d
FO
d
F Kd L L
=
0 0 0 0 0dK W d = 0 0d d d dK W d =
0
0
d d
d d
d
L
d L L
=
Finite anchoring makes the director gradients weaker
and the energy smaller, effectively increasing the cell
thickness
0 0/L K W=
/d dL K W=
0
0
d zz
d
=
Summary of hybrid aligned N film
Ld = K /Wd
0
0
d zz
d
=
2
012
d
FOF Kd
=
0
0
d zz
d
=
0 0
0 0
0
d
d
L
d L L
=
0
0
d d
d d
d
L
d L L
=
2
012
0
d
FO
d
F Kd L L
=
0 dd d L L
L: anchoring extrapolation length
Infinitely strong Finite anchoring
NB: at large scales, surface anchoring takes over, enslaving the director field, the director follows the “easy axis” prescribed by surface anchoring potential
~FOF Kl2~anchoringF Wl
00z = =
dz d= =
End of story for hybrid aligned films?
No. At submicron thicknesses, the director is unstable
w.r.t. in-plane deformations (similar to buckling)
2
2
1 1
1 2
1 1 1 1ˆdiv ~
2 2K K
R R
n
Int. J. Mod. Phys 9 2389 (1995)
1 2
1 1ˆ ˆ ˆ ˆdiv div curl
2R R= n n n n 24
24
1 2
2ˆ ˆ ˆ ˆdiv div curl
KK
R R =n n n n
Free Energy of a Nematic in an External Field
Dielectric case (no ions, no flexoelectric/surface polarization)
21
02ˆ
FO af f e e= n E
211
02ˆ
FO af f m = n B m0 = 4 107
Henry / m
Energy density should depend on two vectors, the field and the director
ˆ ˆ= n nE
E D= e0eE2 e0ea n E
2
a = e e e
0i ij jD E= e e
||
0 0
0 0
0 0
e
e e
e
=
Electric displacement:
Energy density (x 2):
FFO = f FO dV
U=const
FE = f E dV =12 E D dV
dFE = 12 E dD dV
dDz
The energy of the electric field
and its change:
Which leads to a change in the
surface charge density by
When n reorients, surface charge density changes;
to keep the voltage constant at the electrodes, one
needs to supply energy from the electric source:
dFG = dDz
A
dA= div dD dVElectric potential
div dD =divdDdD
E =
dFG = E dD dV = 2dFE
The minimum of the total bulk free energy is achieved when dFFO dFE dFG = d FFO FE = 0
D = e0eE e0e| |E| | = e0eE e0ea E n n E D= e0eE2 e0ea n E
2
21
02ˆ
FO af f e e= n E
211
02ˆ
FO af f m = n B m0 = 4 107
Henry / m
Free Energy of a Nematic in an External Field
Dielectric case (no ions, no flexoelectric/surface polarization)
Splay Frederiks Transitions
2 21
1 0
1 1ˆ ˆdiv
2 2af K = m n B n
Assuming deviations are small:
m
221
0
2
2
1 sin2
1cos
2
1B
dz
dKf a
=
221
0
2
12
1
2
1m
B
dz
dKf a
=
=0 at z=0, z=d
E-L equation: 02
22 =
dz
d
za
za sincos 21 =General solution: Boundary conditions yield and 01 =a nd =/
, the critical field for the Frederiks tra nsition Non-trivial solution at a
c
K
dBB
m
1
0
1
=
a
K
B m
1
0
11
=
, , cos ,0, sinx y zn n n z z=
Three basic geometries of Frederiks effect
a
c
K
dB
m
1
0
1
=
a
c
K
dB
m
1
0
2
=
a
c
K
dB
m
1
0
3
=
Electric field case can be treated similarly
Splay deformation
Twist deformation
Bend deformation
Heliconical director in electric field
0cE
p PE
=
Bimesogens: Ideal for electrically induced twist bend, since the bend constant is
very small
A cholesteric with a small bend-twist ratio K3 /K2 =1 and e>0 adopts an
oblique helicoidal shape in an electric field with the field-dependent pitch:
EE
0e
Color tuning: Vertical field
E
J. Xiang et al, Advanced Materials, 3014 (2015)
E
Liquid Crystals. Lecture 1.2
Optics
Support: NSF
Oleg D. Lavrentovich Liquid Crystal Institute and
Chemical Physics Interdisciplinary Program,
Kent State University, Kent, OH 44242
Boulder School for Condensed Matter and Materials Physics,
Soft Matter In and Out of Equilibrium,
6-31 July, 2015
LCs: Birefringent materials
1.7
extraordinary wave
en
Director n
=Optic axis
1.5
ordinary wave
on
Polarization E of light
Birefringence: Double refraction of light in an ordered material,
manifested through dependence of refractive indices on
polarization of light
0.2e on n n =
Birefringence revealed through pair of
polarizers: textures and LCDs
LCs: Ordinary and Extraordinary waves
/ t = E B / t = H D 0 =B 0 =D
0i ij jD Ee e= 0i ij ij jB Hm d =
||
0 0
0 0
0 0
e
e e
e
=
Light propagation in a homogeneous medium
El. displacement Mag. field induction El. field Mag.field strength
0, exp , , ...t i i t t= =E r E k r H r
=k E B =k H D 0 =k H0 =k D
Consider a plane monochromatic wave:
Eliminating mag. field H: 2 2
0 k m = D E k Ek
2 0ij i j ij jN N N Ed e =0 0
1 c
v e m= N kRefractive index “vector”
and using constitutive eq. for D:
2 2
0 0 ij j i i j jE k E k E k= m e e
Fresnel equation; Propagation of Light in LC
The three homogeneous equations have a nontrivial solution only if the
determinant of coefficients vanishes (Fresnel equation):
2 2 2 2
|| ||Det , 0z x yN N N N e e e e e = =
k
Two waves: ordinary, with
and extraordinary, with refractive index that
depends on the angle between the wave-vector
and the optic axis:
oN n e= =
,2 2 2 2cos sin
o ee eff
e o
n nN n
n n = =
on e=
en e=
optic axis
2 0ij i j ij jN N N Ed e =
Polarizing microscopy: Principle
sin
cos:Entry
A
A
Phase difference =2d
0
ne no
dntA
dntA
e
o
0
0
2cossin
2coscos
:Exit
nematic slab
n X
Y d
polarizer
aryextraordinc
dn
ordinaryc
dn
e
o
,
,:propagate toTime
Projected onto analyzer’ direction:
=
=
dntAA
dntAA
e
o
0
2
0
1
2coscossin
2coscossin
Acos t Resulting vibration:
with amplitude 2
2 2
1 2 1 22 cosA A A A A =
= oe nn
dII
0
22
0 sin2sin
nematic slab
n X
Y d
polarizer
Y
X
analyzer
Polarizing microscopy: Principle
= oe nn
dII
0
22
0 sin2sin
Dark Brushes:
regions where n is || or to polarizers
0; / 2; ... =
P A
Polarizing microscopy: Principle
Polarizing microscopy:
Degeneracy of two director fields
= oe nn
dII
0
22
0 sin2sin
P A
Polarizing microscopy: Degeneracy of two
director fields: Radial or Circular?
= oe nn
dII
0
22
0 sin2sin
P A
PM problem: Two orthogonal n fields
Solution: optical compensator (quartz wedge, Red plate, etc.)
en
on
en
on
en
on
en
on
en
on
Lower retardation,
yellow of the
first order of
interference
Higher retardation,
blue of the
second order of
interference
Radial and circular patterns produce different pattern
of interference colors when the compensator is added
Ultimate Compensator: LC PolScope
R. Oldenburg, G. Mei, US Patent 5,521,705
YK Kim et al, J Phys Cond Phys 25, 404202 (2013)
LC PolScope image of chromonic: shows both
the in-plane director and retardance
(a) T = 32.9 oC (c) T = 30.6 oC, t = 0 s (d) T = 30.6 oC, t = 156 s (b) T = 30.7 oC
0 nm
136 nm
50 μm
1
2
1 2
1
2
1
2
1 2 1 2 Retardance:
; e o
eff
d n planar n n n
d n titled director
= =
=
PolScope creates a map of the orientation of the optic axis in the sample and of the local value of the
optical phase retardation;
Nematic
Isotropic
Polarizing microscopy
= oe nn
dII
0
22
0 sin2sin
Interference colors: 0I I =
Visible when 0~ 1 3e od n n
Limitation: 2D image
Why the interference colors change?
= oe nn
dII
0
22
0 sin2sin
Limitation: 2D image
= oe nn
dII
0
22
0 sin2sin
Derived in approximation of planar director,
and constant thickness d ˆ , ,0x yn n=n
BTW! When , how does the texture look like? , , 0, 0,1x y zn n n =
Limitation: 2D image
2 2
0
0
sin 2 sin , , , , , , ,
d
effI I f x y z n x y z d x y dz
= n
When and , one deals with: , ,x y z=n n ,d d x y=
Limitation: 2D image
??????????????
Confocal Microscopy: 3D image of 3D
Usual
microscope
objective Confocal
microscope
Voxel=3D pixel
~1 mm 3
objective
Pinhole #1 Pinhole #2
(Minsky, 1957)
Computer 3D image Scanning
PMT PMT
Light
source
Light
source
Confocal Microscopy: Principle
Pinzhole #1 Pinhole #2
Fluorescence Confocal Microscopy
3D image of concentration (positional distribution) of
fluorescent dopant….
Fluorescent tag increases contrast of tissues
Living cell
PMT PMT
Light
source
Light
source
1980 M. Petran and A. Boyde
Two distinctive features:
1. Anisometric fluorescent dye aligned by LC
2. Polarized light
….but we are interested in orientations
rather than in concentrations…
Fluorescence Confocal Polarizing Microscopy
FCPM: Fluorescent anisometric dyes
N,N'-Bis(2,5-di-tert-butylphenyl)-3,4,9,10-
perylenedicarboximide)
BTBP
t
BTBP
350 400 450 500 550 600 650
Wavelength, nm
Ab
sorp
tion
, arb
.u
Ar Laser Excitation at
=488nm
Flu
ore
scen
ce E
mis
sion
, arb
.u.
A | |
A |
F | |
F |
0.01 % of BTBP in ZLI-3412
FCPM: Anisotropic Fluorescence
Fluorescence signal = f (orientation of dye)
n
n
P
minimum signal
n
n
P P
a
I ~ cos4a
n
n
P P
maximum signal
I. Smalyukh et al, Chem Phys Lett 336, 88 (2001)
FCPM: Anisotropic Fluorescence
Objective:
40X NA0.6;
60x NA1.4
Sample
Focal plane
PMT
Light source
Illuminating aperture
Confocal aperture
Beam splitter
filter
488 nm,
100 nW
510nm-550 nm
z
Polarizer/Analyzer
mznnn oe m10;1.0/||
mz m1~
/e oz n n z n
I. Smalyukh et al Chem Phys Lett 336, 88 (2001)
FCPM: Frederiks Effect
Relative Intensity scale:
Imin Imax
I ~ cos4a
P 0V
a=0
d=15mm
3V
a= f(z)
P
d=15mm V
I. Smalyukh et al Chem Phys Lett 336, 88 (2001)
FCPM: Planar Cholesteric
P
8 mm
Pn/2 <1 (to avoid the Mauguin regime)
I. Smalyukh et al, Phys. Rev. Lett. 90, 085503 (2003)
Summary: What have you learned Liquid crystals: Orientationally ordered media