1. science
2. mathematics
3. geometry

How to build a holographic liquid
crystal?
Piotr Surówka
Vrije Universiteit Brussel
Euler Symposium On Theoretical And Mathematical Physics, St. Petersburg 17.07.2013
Motivation
✤
Hydrodynamics is an effective theory that still requires better
understanding
✤
Recent holographic models provided a lot of insight into relativistic
hydrodynamics
✤
Low viscosity over entropy density in the hydrodynamic description
of heavy-ion collision is understood
✤
Progress in the studies of QFT anomalies in hydrodynamics
✤
Superfluidity is investigated via AdS/CFT
✤
What else we can possibly shed some light on using holography?
Solid
Liquid Crystal
Fluid
Solid is characterized by a structural rigidity and resistance to changes of
shape or volume. Described by classical mechanics.
Fluid changes its shape under applied shear stress. Described by
hydrodynamics.
Liquid crystal is a state of matter ”in between”. It shares properties with
fluids (eg. deforms under shear stress) and solids (eg. non-zero elasticity
properties).
Phase transition between isotropic un oriented phase and ordered liquid
crystal phase. There is a number of phenomenological theories (eg. Maier
Saupe mean field theory, Landau-De Gennes theory).
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Parity-odd
IsotropicParity-odd
phasehydrodynamics
-hydrodynamics
hydrodynamics
Relativistic fluid with one conserved charge described by
Relativistic fluid with one conserved charge described by
conservation laws:
conservation laws:
∂µµν
T
µν
∂µ T
=0
=0
µ
∂µ∂j µ j= 0= 0
µν andµ j µ in terms of local
µν
plus
equations
that
express
T
plus equations that express T and j in terms of local
µ:
µ
temperature
T
,
chemical
potential
µ,
and
fluid
velocity
u
temperature T , chemical potential µ, and fluid velocity u :
µνµν = (� + P )u
µ µ
ν uν + P
µνg µν +
µντ µν
T
T = (� + P )u u + P g + τ
µµ
µ µ µ µ
j
=
nu
j = nu + +
ν ν
The
ambiguous
beyond
leading
order.
The definition
definitionofofvelocity
velocityis is
ambiguous
beyond
leading
order.
µνµν
µ =
We
fix
it
by
imposing
u
τ
=
u
ν
µ
µ
We fix it by imposing uµ τ = uµ ν µ 0.
= 0.
Liquid
crystal
Shear
Viscosity
phase (nematic)
Viscosity = Diffusion constant for momentum
In the nematic phase we can measure viscosity
Nematic
Moving plate
v
Force/Area
Viscosity =
Viscosity = [(force/area)]
per unit
velocity gradient Gradient
Unit
Velocity
Viscosity is a
diffusion constant
for momentum
11
We can define a unit vector n called the director to
be the average molecular orientation direction.
Scalar order parameter
�
1�
2
S=
3 cos θ − 1
2
n
ˆ
θ
Viscosity in liquid crystals
2
Letter to the Editor
n
η1
v
(a)
n
n
Universality?
η2
v
n ⊥ v and n � ∇v
(b)
n � v and n ⊥ ∇v
Magnitude?
η3
n ⊥ v and n ⊥ ∇v
Elasticity
Splay
Twist
Bend
igure 10: The three distinct curvature strains of a liquid crystal: (a) splay, (b) twist, and (c) bend.
The question we would like to address here is: how much energy will it
take to deform the director field?
These three curvature strains can also be defined by expanding n(r) in a Taylor series in powers o
The deformation of relative orientations away from equilibrium position
y, z measured from the origin
will manifest itself as curvature strain. The restoring
forces which arise to
2
nx (r) = s1 x + t2 y + b1 z + O(r ),
(20
oppose these deformations will cause curvature stresses
or torques. If
2
ny (r) = −t1 x + s2 y + b2 z + O(r ),
(21
these changes in molecular orientation2 vary slowly in space relative to
nz (r) = 1 + O(r ).
(22
the molecular distance scale, we may describe the response of the liquid
We crystal
now postulate
the Gibbs
energy density
g oftheory.
a liquid crystal, relative to its free energy
with athat
version
of a free
continuum
elastic
nsity in the state of uniform orientation can be expanded in terms of six curvature strains
Phase transitions
A phenomenological theory of phase transitions
was established by Landau. He suggested that
Pressure
phase transitions were manifestation of a broken
symmetry. In the simplest cases through the
definition of an appropriate order parameter, Q,
the macroscopic behaviour of a phase may be
followed. Typically Q = 0 in the more symmetric
(less ordered) phases and Q �= 0 in the less
Science 12, 207, vol. 315 no. 5809 196-197
symmetric (more ordered) phases.
The theory, though originally introduced to describe continuous phase
transitions in solids, appeared (as we know today) to correctly account for
symmetry change observed at majority of continuous and first order phase
transitions.
The Landau theory generaly fails in the temperature adjacent to the
transition, in which the behaviour of a system is dominated by fluctuations
Landau thery tailored to describe liquid
crystals
The LT appears as a necessary intermediate step (and also as a tool) in
constructing generalized theory that include fluctuations - it is known as
Landau-Ginzburg-Wilson theory:
Majority of insight into physics of liquid crystals relies on the
implementation of Landau’s ideas (physics of defects, elasticity,
hydrodynamics, relaxation, etc. ), and the approach has been rather
successful.
A major breakthrough in including liquid crystals in the Landau
reasoning is due to de Gennes. He realized that instead of the scalar
order parameter one should use a tensorial quantity. For the nematic
phase it reads:
Qαβ
S
= (3ˆ
nα n
ˆ β − δαβ )
2
Landau-de Gennes theory
Once the appropriate tensor order parameter of the system is identified we can
assume, in a spirit of Landau theories, that the free energy density F is an
analytic function of the order parameter. The Landau-de Gennes theory of the
nematic-isotropic transition starts by assuming that a spatially invariant dimensionless, order parameter is small in the nematic phase close to the transition
point. The difference in free energy density (per unit volume) of the two phases
it thus expanded in powers of the order parameter. Since the free energy must
be invariant under rigid rotations, all terms of the expansion must be scalar
functions of the tensor.
FLdG (P, T, Qαβ ) = F0 + αF Qαβ Qβα + βF Qαβ Qβγ Qγα + γF Qαβ Qβα Qγρ Qργ
The most general form capturing the uniaxial phase is typically truncated at
fourth order.
Holography=spherical cow approx.
3 AdS/CFT correspondence
Strongly coupled QFTs are difficult to study
arting point for construction of some more realistic models. We will review some
Use
a playground
perties
of N = 4 SYM. as
The
lagrangian is uniquely fixed by supersymmetry and
following schematic form,
�
Apply holography,
treat as a toy-model (spherical cow) �
�
1
1 µν
L = 2 TrNc − F Fµν + Dµ Φa Dµ Φa +
[Φa , Φb ]2 + fermions .
(3.1)
gY M at universal
4
Aim
properties, understand
qualitative features
a,b
angle
zero.
The field successful
content of N
= 4context
gauge supermultiplet
includes the
It to
was
particularly
in the
of
hydrodynamics.
ds Aµ , four Weyl fermions and six real scalars Φa . Supersymmetry requires that
Quantitative understanding why viscosity over entropy density is small at
must transform in the same representation of the gauge group, namely the adjoint
Useful
thethe
context
anomalies
due
to their coupling
ation,RHIC.
and all
must in
have
same of
mass.
By gauge
invariance,
a mass for the
nature.
Natural
language
models
lds isindependent
zero, hence, the
fermion
and scalar
fields to
areinvestigate
massless as two-fluid
well. Moreover,
Maybe
same controls
tool canall
be interactions
used to better
understand
only (superfluidity).
one coupling constant
gYthe
in the
theory.
M which
2
one combines
the
two
parameters
g
and
N
into
a
combination
λ
=
g
liquid crystals.
YM
c
Y M Nc
’t Hooft coupling constant.
AdS/CFT correspondence
The AdS/CFT correspondence
The dictionary of gauge/gravity duality
N = 4 supersymmetric
Yang-Mills theory in R4
Equivalence means:
Strongly coupled gauge
� 4
−Ssugra
d xφ
0 (x)O(x)
fields
equivalent
to
weakly
e
≈
Z
=
Z
≡
�e
�
string
CF T
AdS5 × S 5 10D spacetime
coupled strings
IIB superstrings in curved
Identification of some corresponding quantities:
The dictionary
Strongly coupled gauge fields ⇔ Weakly coupled strings
Potentially very useful tool but
Gauge side
we one
need
practical
How
can some
relate physical
observables in these two theories?
µν
TrF
F
µν
Theimplementation
dictionary ofofgauge/gravity
duality
the above
Tµν
equivalence:
ence means:
e
−Ssugra
≈ Zstring = ZCF T ≡ �e
�
dilaton
graviton gµν
dimension of operator
mass of the field
...
...
d4 xφ0 (x)O(x)
cation of some corresponding quantities:
String side
�
z =�
reg
z 2 =�
where SSU GRA represents the dynamics of closed string modes and SDB
where
ν is a open
positive
number
thattoonly
depends
on the
of this talk is the implications
of
brane
system
with
attached
strings
coupled
gravity.
Expandin
where ν is a positiveand
number
(2k) that only depends on the scale dimension
(0) of t
a
are
local
functions
of
the
sources
f
.mea
L
or gauge theories
(2k) around flat background and taking the (0)
action
low-energy
limit
(this
and a
are localdivergent
functionspart
of the
sources
f� limit
. Last
thing
to counte
do is
and
take
the
�
→
0.
The
dimensionless
parameters
fixed
while
taking
α counterterms
→ 0), all interaction
terms
divergent
part and
take
the
limit
�
→
0.
The
depend
onmt
2
2
on
the
surface
z
=
�,
characterized
by
the
induced
ations are huge due to oneNformula
of
4 SYMztheory,
with gY M = gsby
and
supergravity.
2
on the=surface
=reads
�, characterized
thefree
induced
metric
γµν /�. The ren
�
N = 4 SYM is not the only low-energy limit (α → 0)� of D3-brane sys
onjecture:
reads
(0)�
previously D3-brane is a (0)
solution of� supergravity
To[fsee
Sren(0)
[f (0) ] =(3.12).
lim Sreg
; that
�] + Si
Mapping of parameters
Sren
[f ] such
= limthat
Sregthe
[f branes
; �] + are
S�→0
�);at
�] the
. ori
ct [F(x,
to shift a coordinate
system
located
radius of curvature
�→0
coordinate system.
yD3explicit
= 0 andform
weSitter
introduce
distance
from the
To Then
get the
of Sctgeometry
[F the
(x, �);
�] we
Anti-de
(0)have to e
SYM
on
a
stack
of
To The
get the
explicit
form by
of aSctstack
[F(x, �); �] we have
tobe
express
f as
as a fun
� �4
metric
generated
may
rewritten
calculate
a(2k) (fof(0)D3-branes
), and determine
the divergent
par
AdS/CFT at finite temperature
ctical
holography
L D3-branes
calculate a
(f (0) ), and determine the divergent part. We show how th
in practice
in4 �Chapter
5, regularizing
�
�
� 12 the D7-brane a
− 12
4
in practice
5, regularizing
the
D7-brane Laction. 2
L
R
2
µ
2
2
implications
oflin Chapter
�s this talk is the
cnamics
of
T=0
or gauge theories
(2k)
ds =
s
S719
1+
r4
dx dxµ + 1 +
r4
(dr + r dΩ5 ).
3.6
Gauge/gravity
duality
at
finite
For
an
observer
at
infinity,
far
away
from
the
branes,
at
r
�
L,
the
space
3.6
Gauge/gravity
duality
at
finite
temperature
one
formula of Minkowski space-time. Close to the branes, for r � L,
a ten-dimensional
So far we have discussed a system of extremal D3the
‘1’
in
the
above
metric
So far we have discussed a system of extremal D3-branes which is dua
cations
arelength
huge due to
string
conjecture:
R
theory at zero temperature.
In
order to generalize tha
Mapping
g N =of parameters
2
2
theory gravity
at zero temperature. In2order
to2 the2 case with fin
! coupled
r to µgeneralize
L that
2
pled QFT
we needAdS
to
find
a
gravitational
solution
with
a scale t
radius weakly
of curvature
Schwarzschild
geometry
ds
=
dx
dx
+
dr
+
L
dΩ
,
µ
5
2
2
we need to find a gravitational solution
with a rscale
that may correspond
L
and some notion of entropy. This solution is known t
solutions
of
a
class
of
theories
and
some
notion
of
entropy. This solution is known to be AdS-Schwarzc
� �4
and identify the emergent geometry as a product of two spaces, five dime
L
2
heory cannot be solved by perturbation theory
r
Sitter space AdS5 , with 2a five-dimensional
sphere
S 5 2, both
with
radius
2
2
i
L
r
ds
=
(−f
(r)dt
+
dx
dx
)
theory dual =⇒ Einstein’s
general
relativity!
T>0
2
2
i 2
2
2i +
2
�s
have two distinctdssets
those
propagating
in
the
Minkowski
= of2 modes,
(−f (r)dt
+ dx
dx
)
+
dr
+
L
dΩ
Li
(
5 ,fsp
2
4
2
c
4
s
ls
f Nc D3-branes.
R
L region where the geometry
f (r)ris AdS × S 5 . Th
propagating in the ‘throat’
5
4
R
modes decouple from
each
other
in
the
limit.(3.50)
Far away
from t
where
f (r) = 1 − r4 .low-energy
The solution
describes
R4
where
f (r) modes
= 1 −survive,
solution
(3.50)
describes
the near-horizon
ge
4 . The
massless
while
in
the
throat
there
is
a
whole
tower
of
massiv
r
extremal D3-branes. It can be depicted as a black h
extremal
D3-branes.
can be depicted
as aBecause
black hole
located
in
the r
cannot climb up with
theIt
gravitational
potential.
we
have
two
distinct
bottom-up
the horizon at R. We interpret this black hole
4
with
the
horizon
at
R.
We interpret
holeequivalent,
as a thermodynam
the same D3-brane system
we expectthis
thatblack
they are
meaning t
GSI 2009 – p.23/42
string length
Top-down vs.
Spin 2 Lagrangian
Complicated plus various consistency issues
S=
�
d
d+1
x
√
�
1
1
1 2
2
µα 2
2
µν
−g − (∇µ ϕαβ ) + (∇µ ϕ ) + (∇µ ϕ) − ∇µ ϕ ∇ν ϕ − mϕ (ϕµν ϕµν − ϕ2 )
2
2
2
�
1
1 2 2
1
α 2
µν
µα νβ
2
µν
2
− Fµν F − mA Aµ + Rµναβ ϕ ϕ −
Rϕ − A (ϕµν ϕ − ϕ ) ,
4
2
2(d + 1)
2
We need a quadratic coupling between the fields. A cubic coupling of the
form a1 Aµ Aν ϕµν + a2 A2 ϕ would not allow for spontaneous hairy black holes.
The spin 1 field would just act as a source for the massive spin 2 field and all
charged black holes would have a secondary hair. Instead we want to have a
hair appearing below a critical temperature. The situation would be similar to
trying to build a superconductor using a dilatonic coupling eφ F 2 . It would not
work since the electric field would source the dilaton.
Equations of motion
Our ansatz:
ϕµν = diag(0, 0, ϕx1 x1 (z), ϕx2 x2 (z), ϕx3 x3 (z))
µ
Cµ dx = φ(z)dt
where
2
l
ϕij = 2 ψ(z)
z
We get the following EOMs
��
ψ +
�
�
f
d−1
−
f
z
�
�
d−1
d−2
ψ� +
�
�
�
δi1 δj1
2 2
m
αφ
ϕl
+ 2 − 2
f
z f
2
2
d−3 �
αl 2
φ −
φ + − 2 ψ −
z
z f
��
1
−
d−1
m2A l2
z2f
�
�
�
ψ = 0,
φ = 0.
Looking for an instability
The dynamical fields ψ, A1 obey hypergeometric equations when there is no
coupling α = 0. Their fall-off is
ψ(z) = ψD z
φ(z) = µz
d−2
2 −
�
d
2−
(d−2)2
4
�
d2
4
+m2ϕ l2
+m2A l2
+ ψN z
+ φN z
�
d
d2
2 l2
+
+m
ϕ
2
4
�
(d−2)2
d−2
2 l2
+
+m
A
2
4
+ ...
+ ...
We impose regularity at the horizon and Dirichlet condition at the boundary
1.5
Instability!
!ΨN#
1.0
We have spin-2
condensate
Tc
0.5
0.0
0.80
0.85
0.90
T!Tc
0.95
1.00
Partition function
The Euclideanduality
black hole
solution
is interpreted as a saddle-point in the path
Gauge/gravity
at finite
temperature
integral corresponding to the thermal partition function. The supergravity
action evaluated for this solution is interpreted as the leading contribution to
the freeasenergy.
Free energy
for a stack
D3-branes
erpreted
the leading
contribution
toofthe
free energy. For the D3-brane sys
π2 2 4
F = T Ssugra = − Nc T .
8
Eq. (3.53) we can compute other thermodynamic variables, for example, the e
ty To
reads
prove the instability for the spin-2 system we need to calculate the
2
∂F
π
partition functions for the sisotropic
phase
and
the
phase
with
the
2 for
3
=−
= Nc T .
∂T lowers
2
condensate and show that the system
the free energy by
picture
becomes
for studying a finite temperature field theory at stro
developing
thefeasible
condensate.
. We recall that there are indications that real QGP investigated at RHIC
d = 4, using dimensional
analysis, we have the dimensions [ψ] = [φ] = [µ] = [φN
2
2
0 for
≤m
1 = 3, which is the allowed range for the mass (exce
Al < d −
Mat−1z. =
We
find0the
following
plots.
which is special and that we do not consider here).
w
the
action
can
be
written
as
Free energy
�
�
1 . Our outgoing unit
z l/(z
�
�
e c is a parameter
to
be
determined
normal
to
z-slices
is
n
=
−δ
√
1
1
α 2
µ
µ
d+1
µν
µ
µν
2
µα
ν
mplicity,
we
will
only
look
at
d
=
4
but
we
will
keep
the
masses
general.
We
conside
=
d
x
−g
ϕ
E
+
A
E
+
A
ϕ
ϕ
−
ϕ
−
αA
A
(ϕ
ϕ
2
µν The variation
µ
µν
µ
νansatz to
α
h of course satisfies n =
+1.
of
the
total
action
reduces
for
our
2
2
2
withFor
counter-terms
�
� keep the masses
�
�simplicity,
� we will
we
will
look
at
d
=
4
but
general.
2
3
�
�
√
�
l �
2c l
l f �
µ
2 µ
5 d√
µν
3
cA
Stot =+ d dx x −g−γ
(Eµνnδϕ
dtd
x
φ δφ µ+ 2 √ φ δφ − 3 ψ δψ
.
µΨ + E
µ δA, ) −
d √
z
Stot = S + c d x −γz Aµ A , z f
z=0�
Free energy analysis
where c is a parameter to be determined 1 . Our outgoing unit normal to z-slices is nµ = −δµz l/(z
mpose the Dirichlet condition
0. The terms involving ψ therefore vanish. Us
D =
2 = +1. ψ
which
of
course
satisfies
n
The
variation
of the
total
action
reduces
for our ansatz to
The
variation
of
the
total
action
reduces
for
our
ansatz
to
ptotic fall-off
for
φ (2.21), we find that we
need
5 to tune
�1
�
�
�
1
1
1
1
2
3
µ
µ
αβ
µν
µ
µν
µ
µν
√ ∇ ϕ +
l ϕ ϕ −l fF � Aν .
�
Ψ
=
−
+
ϕ∇
ϕ
−
5 ϕ
µνϕ ϕν�
µ
3− l ϕ
� ∇ν ϕ2c
αβ
�
δStot = d 2x −g (Eµν δϕ + E1µ δA )2− dtd x 2 φ δφ + 2 √ 2 φ δφ − 23 ψ δψ
.
z
z f
z=0
c=
1 + m2 l2 − z1 ,
A
2l
nowimpose
go to the
Euclidean
Euclidean action is given by
We
Dirichlet signature.
condition ψThe
D = 0. The terms involving ψ therefore vanish. Usin
fixthe
c invariation
order
afind
well
define
Dirichlet
problem
der We
to set
ofhave
the we
action
in
thewefinite
form
suitable
for the Dirichlet problem,
asymptotic
fall-off
for φ to
(2.21),
that
need
to tune
E = Stot
�
��
�S�
1 µν
5 √
c µν
= δϕ +1E
+µm
δStot = d x −g (E
δA2Aµl2) −
+ 1 ,dtd3 x (J + �O�δµ) ,
Stot is evaluated in Euclidean signature
with the Euclidean time τ = it runnin
2l
znh .order
In the
case
of
the uniaxial
nematic
phase
in d suitable
= 4, defining
the volume
V =n
to
set
the
variation
of
the
action
in
the
finite
form
for
the
Dirichlet
problem,
e the
vacuum
expectation
value
dual toand
µ isevaluate the action
We
pass
to
Euclidean
signature
the free energy in the
ensemble (T,
� grand canonical�
� µ fixed)
5 √ �
2lµν + E δA2µ )2+� dtd3 x (J + �O�δµ) ,
�
δStot = d x −g
(E
δϕ
µν
2
2
3
�
�O�zh= − αl2 3 φ12 ψ+2µ mA l φNlφφ
, �
cl φ
l f ψψ
z
F (ψ �= 0) = SE /β = V
dz h
+V
+ 2√ +
3
2z
2z 3 z=0
z f
where the vacuum expectation value
to (z)z
µ is
0 dual 2f
F (ψ �= 0) = V
�
β/π
0
αl3 φ2 ψ 2 1
dz
− �O�µ .
3
2f (z)z
2
An instability confirmed
where we used β = πzh . When there is no condensate, the free energy will simply be given b
econd term in the above expression with �O� evaluated in the normal state,
�
2
2π l
c+
�O�n = − 2
1+
µ.
β
c− difference of the free
Finally we can write down an expression
for the
m2A l2
energies in the normal and condensed phase
Therefore,
��
�
β/π
αl3 φ2 ψ 2 1
F (ψ �= 0) − F (ψ = 0) = V
dz
− (�O� − �O�n )µ .
3
2f (z)z
2
0
The free energy can be evaluated numerically. The result is displayed in Figure 4.1. The un
ematic phase is (dis)?favored.
0
Difference of free energy between
�4000
the uniaxial nematic phase and the
Let us compute
the stress-tensor of the action (1.1) under the assumption that ϕ = 0 = ∇µ ϕµ
�6000
normal phase as a function of the
ave
�8000
temperature T = 1/β. The uniaxial
γ
�10 000
δΓ
2
δL
α µβ
ν
favored.
T µν = √
= g µν L + ∇µ ϕαβ ∇ν ϕαβ + 2∇nematic
ϕ ∇α ϕphase
+ 4 is δα
[ϕγβ ∇δ ϕαβ ]
Bulk stress-tensor
F�T�
5
�2000
�12 000
−g δgµν
0.65
0.70
β
0.75
0.80
0.85
0.90
0.95
1.00
δRαβδγ
δgµν
Future directions
Check if the model gives the equations of
nematodynamics
Use an anisotropic ansatz for the metric and study
backreaction
Calculate viscosity coefficients
Investigate the non-relativistic limit