Ginzburg-Landau theory of second-order phase transitions Second order= no latent heat (ferromagnetism, superfluidity, superconductivity). Let the transition occur at T=TC Ordered phase below transition (typically) Order parameter h (vanishes in disordered phase): h = macrospopic wave function (superconductivity) h = magnetization (ferro or antiferro) Vitaly L. Ginzburg Example: liquid crystals Lev Landau Order parameter h = excess population at angle Q (or a simple function of Q ) in nematic-isotropic transition in liquid crystals-simplest example (1 real scalar parameter) One of the most common LC phases is the nematic. The word nematic comes from the Greek νημα (nema), which means "thread” Alignment in a nematic phase . 1 (see lectures by A. Salonen) Tfy-0.3252 Soft Matter Physics, Fall 2009 / E. Salonen 2 Equilibrium Condition G(h ) U PV TS Gibbs free energy G 0 h T Gibbs free energy must be minimum at equilibrium at any T dG SdT VdP 2G 2 0 h T How G depends on h near the transition Temperature Order parameter h is 0 or small (since it vanishes in disordered phase) Near the transition G G0 nh n , G0 disordered phase free energy n Gibbs free energy must be minimum at equilibrium in disorderesìd phase G 0 1 2 2h ... 1 for h 0. h T Therefore the expansion around critical point starts from second order G G0 2h 2 3h 3 4h 4 At h=0 there is a minimum or a maximum 3 GL Phenomenological model, with terms up to 4° order: G G0 2h 2 3h 3 4h 4 disordered 2 (T ) 0, T TC G-G0 At the transition T= TC G must be the same for nematic and isotropic (equilibrium). h ordered 2 (T ) 0 Above TC minimum at h=0 (order parameter vanishes at equilibrium in disordered phase 2G 2 0 2 2 6 3h 0 2 (T ) 0 h T 2 (T ) 0 taken linear Below TC maximum at h=0 1 2 2 (T ) (T T * ) A0 parameter T * usually different from TC : T * TC G G0 (T T * ) A0 2 h 3h 3 4h 4 , 4 0 in order to have overall minimum 2 4 1 G G0 (T T * ) A0h 2 3h 3 4h 4 , 2 1 1 1 often written in the form G G0 A0 (T T * )h 2 Bh 3 Ch 4 2 3 4 Discontinuity of Order Parameter h at Tc One obtains the discontinuity of order parameter by imposing equilibrium and minimum at Tc Equilibrium : G G0 1 1 1 1 1 1 A0 (TC T * )h 2 Bh 3 Ch 4 0 A0 (TC T * ) Bh Ch 2 0 2 3 4 2 3 4 G 1 1 1 * 2 3 Minimum : G G0 (T T * ) A0h 2 Bh 3 Ch 4 0 A0 (T T )h Bh Ch 2 3 4 h T G 0 h T GG 0 A0 (TC T * ) Bh Ch 2 0 1 1 1 * A ( T T ) B h Ch 2 0 0 C 3 4 2 Divide the first by 2 and subtract: hC 2B 3C B C h h 2 0 and find hc 6 4 Choosing h 0, B 0 since C 0 (minimum condition) 5 hC T* 2B 3C Choosing h 0, B 0 One can write T* in terms of TC and the other parameters. The order parameter is discontinuous at transition. 1 1 1 G G0 A0 (T T * )h 2 Bh 3 Ch 4 Insert hhc back into G-G0 equation 2 3 4 1 1 1 G G0 at thetransition h 2 ( A0 (TC T * ) Bh Ch 2 ) 0 2 3 4 1 1 2B 1 2B 2 A0 (TC T * ) B( ) C ( ) 0 2 3 3C 4 3C 2 B2 2 B2 * * get A0 (TC T ) TC T 9 C 9 CA0 Order parameter versus T Above, we got the discontinuity hC 2B 3C To obtain h(T) we can start from the equilibrium condition B B 2 4 A0C (T T * ) G * 2 3 0 A0 (T T )h Bh Ch nematic : h (T ) h 2C T 6 To get rid of the double sign, choose the right h at the transition: insert B B 2 4 A0C (T T * ) 2 B2 TC T into h (T ) 9 CA0 2C * 2 B2 B2 B B 4C B 9 C 9 , recalling minimum condition B 0. h 2C 2C |B| B 2 B OK with discontinuity h B B c 3C 3 3 h B 2C 2C wrong 3C 2 we choose h (T ) B B 2 4 A0C (T T * ) 2C : order parameter versusT . 7 Langevin paramagnetism Model: "gas" of ions with magnetic moments g B J N z H g B J .B magnetization M Volume Quantum partition function Z J m J mg B B exp[ ] K BT Paul Langevin (1872-1946) B 1 In the special case J , with g 2 find Z 2cosh( B ) 2 K BT B H N M B tanh V K T B 8 Weiss mean field theory (1907) of Ferromagnetism: Idea:the effective field gets a contribution from the magnetization: H eff H M , B H 1 N J , g 2 M tanh λ to be determined B 2 V K T B B ( H M ) N M B tanh V PIERRE-ERNEST WEISS born March 25, 1865, Mulhouse, France. died Oct. 24, 1940, Lyon, France. K BT The spontaneous magnetization may remain even for H=0 M M N B tanh B V K T B (Ferromagnetism!) One can fix the parameter in terms of Tc as follows: M B M with H 0 since M small, tanh B K BT K BT 2 K BTCV MV B M N B . TC is very useful. 2 N B K BTC N B VK B For T TC 9 Set B M N M B tanh can be solved graphically. V K BT M N x B M B tanh x . Graphical solution: K BT V tanh x B M B T K K T M M B B N N TC B B B V V KB N B2 xT , TC TC VK B x=y 1 solution if T>Tc, 2 otherwise B M N M B tanh can be solved graphically. V K BT From the graphical solution we gain the trend with T of the M(T) curve. Increasing T at H=0 M must vanish, but how? PIERRE-ERNEST WEISS Spontaneous magnetization for H=0 born March 25, 1865, Mulhouse, France. died Oct. 24, 1940, Lyon, France. T Tc from below: M is very small x M xT x3 xT tanh x . We need the third power: x TC 3 TC M x K BT B M B K BT x 3(1 1 T ) TC TC T . Power law!Critical exponent 1/2. 11 A different experiment: fix T =Tc and measure M dependence on H at for H0 . Since M is small we can expand B ( H M ) N M B tanh V K T B C 3 B ( H M ) 1 B ( H M ) N B [ ...] V K BTC K BT 3 N B2 N M ( H M ) B B VK BTC V K BT N B2 Using TC , the VK B We are left with: 3 3 2 2 2 3 3 ( H 3H M 3H M M ) N B2 l.h.s simplifies with M . VK BTC 3 B N N 3 2 2 2 3 3 0 H B ( H 3H M 3H M M ). VK BTC V K BT 2 B Setting M H 1 for H 0, neglecting H 2 and H 3 3 1 B 3 3 N B2 N 3 3 0 H B M M H 3. M H . VK BTC V K BT Then, neglect of H 2 and H 3 justified a posteriori. Power law!Critical exponent 1/3. 12 Behavior for T >Tc , in paramagnetic phase M small B ( H M ) N M B tanh V K T B N B 2 (H M ) V K BT This gives N M V B 2 B 2 N B2 N H H , TC 2 N B V K B (T TC ) VK B K BT V which is the Curie law. Critical exponent=-1. Of course this holds in para phase while M is small. How do these critical exponents arise? Widom in 1965 put forth the scaling hypothesis for magnetic materials in field B: G Gibb's free energy: r , p : G( r ( T T ), p B) G(( ), B) G homogeneous. TC TC f homogeneous function a,b:f( a x, b y) f ( x, y) This produces relations between various critical exponents. The dependence on approximation scheme and dimensionality is dramatic. (Weiss theory) (Weiss theory) 14 Ising Model in 1d , defined on a closed ring (model invented in 1920 by Wilhelm Lenz as a thesis for Ernst Ising) H Ising Jsn sn1 H sn n H H Simplify notation : n H Ising Jsn sn 1 H sn n n where the sums over sites and sn is a classical spin taking the values +1,-1. The Partition function is defined by: Z Tre H Ising From Z one derives thermodynamics: F KT log Z . Z e s1 s2 J ( s1s2 s2 s3 sN s1 ) H ( s1 s2 sN ) e . sN 15 Z e J ( s1s2 s2s3 sN s1 )e H ( s1 s2 sN ) . s1 s2 Rearrange: sN Z [e Js1s2 e Hs2 ][e Js2s3 e Hs3 ] [e JsM s1 e Hs1 ]. s1 s2 sN Vs1s2 [e Js1s2 e Hs2 ] 1 2 There is a [ ] factor for each bond; each factor is a matrix product Jsn sn1 Hsn1 (VV ) ( V ) ( V ) [ e ] 1 2 sn , sn1 1 sn , s 2 s, sn1 where V VV 1 2 called transfer matrix (V1 ) s , s e Jss (V2 ) s ,s e Hs ( s, s) e J V J e e J e H e J 0 0 e ( J H ) ( J H ) H e e Z [VV 1 2 ]s1s2 [VV 1 2 ]s2 s3 s1 s2 sN e ( J H ) e ( J H ) [VV 1 2 ]sM s1 . 16 Z [VV 1 2 ]s1s2 [VV 1 2 ]s2 s3 s1 s2 sN [VV 1 2 ]sM s1 . But pbcimply a Tr : M M Z [(VV ) ] TrV , V VV 1 2 s1s1 1 2 called transfer matrix s1 V V V V V recall : Z [e Js1s2 e Hs2 ] [e Js2s3 e Hs3 ] [e JsM s1 e Hs1 ]. s1 s2 e J V J e sN e J e H e J 0 0 e ( J H ) ( J H ) H e e e ( J H ) e ( J H ) e ( J H ) e ( J H ) Det[V I ] ( J H ) 0 ( J H ) e e 2 e J (e H e H ) e2 J e2 J 0 2 2 e J cosh( H ) 2sinh(2 J ) 0 17 2 2 e J cosh( H ) 2sinh(2 J ) 0 e J cosh( H ) e2 J cosh( H ) 2 exp(2 J ) exp(2 J ) e J cosh( H ) e J cosh( H )2 1 exp(4 J ) e J cosh( H ) e J sinh( H )2 exp(4 J ) For H0, e e J J exp(4 J ) e TrV TrV N N N J J 2 J e e e J e J 2cosh( J ) 2sinh( J ) Z N N . 18 We need to note that another reasoning is possible. Instead of Vs1s2 [e Js1s2 e Hs2 ] 1 2 one can decide to distribute the interactions with H symmetrically among the bonds and define a different transfer matrix, which is also common in textbooks, with elements: 1 Vsl ,sm e 1 2 Vs2 ,s2 [e Js1s2 e 2 Jsl sm H ( sl sm ) , e ( J H ) V J e 1 H ( s1 s2 ) 2 ] e J e ( J H ) This is transfer matrix has the same trace and the same determinant as the previous one, hence the eigenvalues are the same. 19 Thermodynamic limit: only large eigenvalue matters Z N N N e J cosh( H ) e J sinh( H ) 2 exp(4 J ) F NKT ln NJ NK BT ln cosh( H ) sinh( H ) 2 e 4 J . F sinh( H ) Magnetization (after simplification) . 2 4 J H sinh( H ) e No spontaneous magnetization. The peak in specific heat is due to some short-range order (increasing T at low temperatures spins are reversed and this costs energy). 20 no ferromagnetism, no phase transition, no long-range order! 1d theory fails to explain magnetism. Why? Ground state : 1 broken bond : Breaking a single bond breaks long range order. It costs energy 2J. S=K B logW. The free energy change is F 2 J K BT ln( N ) For large N there is a gain in F at any T. In 2d is different. 21 Rudolf Peierls Centre for Theoretical Physics 1 Keble Road Oxford, OX1 3NP England 22 Percolation problem: Renormalization group approach to the onset of conductivity A.P.Young and R.B. Stinchcombe J. Phys. C 8 L535 (1975) Consider a regular lattice (sites connected by bonds, with Z bonds per site) Remove a fraction 1-p of bonds at random. Conductivity is a function of p. There is a critical value pc such that for p< pc the lattice is not conducting and consists of clusters of connected sites in a sea of isolated sites. If we pick two sites far apart the probability of finding a continuous path joining them gets small. Conversely, for p> pc the film as a whole is conducting , even if it contains nonconducting islands. The macroscopic conductivity S of the lattice is a function SS(p, s, s bond conductivity . Let us consider a square lattice: How to find an approximation to pc ? By scaling! For exampe, discard half of the x and half of the y and consider a new lattice where the nodes can be connected or not with some probability. The new problem is similar. We can exploit that! 23 Decimation Rescaled length (rescaling factor b ) Rescaled probability p pc p1 ( p, b) 0 p1 p1 ( p, b) for b p pc nonconducting islands disappear p pc system unchanged for b for b p1 p1 ( pc , b) 24 A similar problem can be posed on the lattice with vertices 1,2, .. or with the rescaled lattice A,B,…on a larger square mesh. p1= probability in rescaled network , s1 bond conductivity in rescaled network SS(p1, s1 25 in this square lattice case: rescaling factor b 2, Renormalization of bond probability : p1 p at each scaling p changes by a factor ( p ). Rescaled length (rescaling factor b 2 ), the correlation length index b 2 1 is defined by b c ln(b) ln( 2) 0.8547 3 ln( ) ln( ) 2 By comparison with the exact results one can evaluate approximations. To proceed find a simple approximation to p1 p1 ( p, b) Consider only paths involving ,,c,. What is the probability p1 that nodes A and B of new lattice are connected? 1 A B 2 C D P( , , , ) probability that , , , Let are connected, P( , , ) probability that α,β,γ are conected but δ is not, and so on. p1 P( , , , ) P( , , ) P( , , ) P( , , ) P( , , ) P( , ) P( , ) renormalization of p : p1 p4 4 p3 (1 p) 2 p2 (1 p)2 2 p2 p4 27 Renormalization of bond probability : p1 4 p(1 p 2 ) p p1 2 p 2 p 4 p1 4 p(1 p 2 ) p at each scaling p changes by a factor ( p). p1 2 p2 p4 Fixed point equation: pc 2 pc 2 pc 4 Fixed points x 2 x2 x4 x 0 or x3 2 x 1 0 which yields x 1 or Roots : x 0, x 1, x x2 x 1 0 5 1 5 1 0.618, x 1.618 2 2 pc 0 Trivial Fixed point : repeated scaling no conductance pc 1 Trivial Fixed point : repeated scaling perfect conductance pc 1.6 nonsense fixed point nontrivial solution : pc 0.618 (Exact result pc 0.5) 28