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). ! "#$%%&!%'!()*$+,)*-#.! /01234"156!78!"7/59:(;570! Texture - distribution !"#$%&#'()"(*'+,-'.,"+)of crystallographic /'0.$()1(,$2$#)34556) orientations Liquid crystals 7$8'.,%)9:'-$) ! "#$%%&!%'!()*$+,)*-#.! /01234"156!78!"7/59:(;570! ;8$%.,%)<)/#'=$($26)9:'-$) "#$%&#'()"(*'+,-'.,"+) /'0.$()1(,$2$#)34556) ;8$%.,%)<)/#'=$($26)9:'-$) !"#!$#%!$&' ()**+,-).'/012-34,056'467'8-7+-'94-4*+,+-:.' ;4*<-07=+'%!$&' >' 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

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