PHASE TRANSITIONS IN SOFT MATTER SYSTEMS

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1 OUTLINE: Topic D. PHASE TRANSITIONS IN SOFT MATTER SYSTEMS Definition of a phase Classification of phase transitions Thermodynamics of mixing (gases, polymers, etc.) Mean-field approaches in the spirit of Landau See, for example, the book by R. A. L. Jones

2 How to define a phase State (form) of matter. Liquid Solid Gas Plasma Condensate Order parameter. The order parameter characterizes any given phase. Phase transitions are characterized by abrupt changes in the order parameter. Phase. A distinct state of matter in a system, characterized by some well-defined physical properties. In liquid-crystals, e.g., Nematic Smectic Cholesteric phases, etc.

3 Examples of phase transitions Water: Transition from liquid to gas. 1st order transition Order parameter: density difference between the fluid and gas phases density Order parameter: Overall, it is the parameter, scalar or vector, characterizing the transition from one phase to another. It is zero in one of the phases and non-zero in the other. temperature

4 Examples of phase transitions Phases found in liquid crystals: smectic, nematic,, and including the isotropic phase at high T crystalline smectic nematic θ 2nd order Legendre polynomial. Often used in studies of liquid crystals, lipids, through techniques such as NMR, etc. Order parameter characterizing the ordering of a molecule with respect to a fixed reference direction.

5 Examples of phase transitions crystalline smectic nematic disordered at high T θ Order parameter characterizing the ordering of a molecule with respect to a fixed reference direction. S Nematic-disordered transition temperature

6 Examples of phase transitions smectic nematic Phases found in liquid crystals: smectic, nematic,, and including the isotropic phase at high T In biological membranes reminiscent, in part, to liquid crystals: Fluid lamellar, high T Solid-like lamellar, low T

7 Phase equilibrium For simplicity, consider a closed system comprised of α partitions, which include different particle types j = 1,2,. Then U, V, and N remain constant, though intensive variables (p, T, etc.) as well as N may fluctuate from one partition to another. Extensive variables are additive: S = Σ α S α, V = Σ α V α, U = Σ α U α Let also N jα to be the number of particles of type j in a partition α. Then the total number of particles of type j is T α, p α, V α N j = Σ α N jα, Since each partition is in equilibrium, S α = S(U α, V α, {N jα })

8 Phase equilibrium Since each partition is in equilibrium, S α = S(U α, V α, {N jα }) The differential for internal energy is du = TdS pdv + μdn, T α, p α, V α and hence ΔS α = (ΔU α / T α ) + (p α /T α ) ΔV α Σ j (μ jα /T α ) ΔN jα. For simplicity consider α = A, B. Then for an isolated system ΔU A = ΔU B, and so on, and hence ΔS = Σ α ΔS α = ( (1/T A ) (1/T B ) ) ΔU A + ( (p A /T A ) (p B /T B ) ) ΔV A Σ j ( (μ ja /T A ) (μ jb /T B ) ) ΔN ja. In equilibrium ΔS = 0, and we have conditions for equilibrium: T A = T B, p A = p B, μ ja = μ jb.

9 Phase transitions In equilibrium for two phases the above conditions for p and T have to hold. Also, the chemical potential of each particle type in each phase must be the same: μ ja = μ jb (j = 1,2,.). The phase transition between two phases is characterized by an order parameter. E.g., for a gas-liquid transition (water) it is essentially the density difference of the two phases. Examples of phase diagrams

10 Phase transitions of 1st and 2nd order 1st order phase transition Ψ Involves a finite and non-zero change in enthalpy. The first derivative of a thermodynamic potential Ψ is discontinuous. Hence the order parameter changes discontinuously at the phase transition (melting of some crystals, and boiling of water, for example) T c T 2nd order (continuous) phase transition Ψ Involves no change in enthalpy. The second derivative of a thermodynamic potential Ψ is discontinuous. Thus, the order parameter changes continuously at the phase transition (change from a liquid to a gas at a critical point, for example) T c T

11 Examples of phase transitions Specific heat Fluid lamellar, high T T c temperature Solid-like lamellar, low T Behavior close to the phase transition point is referred to as critical phenomena. Universality-classes 1st order transitions (enthalpy change associated with the transition) 2nd order (continuous) transitions (no enthalpy change)

12 Entropy of mixing As an example of phase transitions, consider the situation where two liquids are miscible at all proportions at high temperatures, while at low temperatures they separate into two distinct phases. For simplicity, let us consider a binary system (only two different species A and B). At high T, the species A and B mix to form a single-phase homogeneous liquid. Assume that there is no volume change on mixing: V = V A + V B. The volume fractions are T Two phases Single phase φ A The above description is the so-called regular solution model.

13 Entropy of mixing Assume each particle to occupy one lattice site. Then there are n = V A + V B sites. The entropy S is determined by the number of ways Ω to arrange molecules on the lattice (the number of states) S = k B ln Ω. The number of translational states of a given particle is the number of independent positions that a particle can have on a lattice. Thus in a homogeneous mixture of A and B, each particle has Ω AB = n (mixed system) possible states. In a similar manner, the number of states Ω A of each particle of species A before mixing (in a pure A state) is equal to the number of lattice sites occupied by A Ω A = n φ A. (non-mixed)

14 Entropy of mixing For a single molecule of species A, the entropy change on mixing is ΔS A = k B ln Ω AB k B ln Ω A = k B ln (Ω AB / Ω A ) = k B ln (1 / φ A ) = k B ln φ A > 0. To calculate the total entropy of mixing, the entropy contributions from each particle in the system need to be summed: ΔS mix = n A ΔS A + n B ΔS B = k B ( n A ln φ A + n B ln φ B ). This is an extensive quantity. To find an intensive counterpart, we note that n A = n φ A. The entropy of mixing per lattice site is then ΔS mix = ΔS mix / n, Note: does not depend ΔS mix = k B (φ A ln φ A + φ B ln φ B ). on energetics in any This holds for the regular solution model. manner!

15 Energy of mixing Next we need to find the energy of mixing, U mix. The energy of mixing can be either negative (promoting mixing) or positive (opposing mixing). Regular solution theory writes the energy of mixing in terms of three pairwise interaction energies (u AA, u AB, u BB ) between nearest neighbors on a lattice. It further uses a mean field to determine the average pairwise interaction U A of a particle of species A interacting with its nearest neighbors. The probability of a neighbor being a monomer of species A is then φ A (mean field approximation), while the probability of a neighbor being of type B is φ B = 1 φ A. Then U A = u AA φ A + u AB φ B and U B = u AB φ A + u BB φ B. On a lattice, the coordination number z gives the number of nearest neighbors (on a 2D square lattice z = 4). Then the total interaction energy of the mixture is U = [ U A φ A + U B φ B ] z n/2.

16 Energy of mixing The total interaction energy of the mixture is U = [ U A φ A + U B φ B ] z n/2. Let us denote the volume fraction of species A by φ = φ A = 1 φ B. Then combining the above descriptions we find the total energy of the system after mixing: U = [ [ u AA φ + u AB (1 φ) ] φ + [ u AB φ + u BB (1 φ)](1 φ) ] z n/2 U = [ u AA φ u AB φ(1 φ) + u BB (1 φ) 2 ] z n/2. The energy before mixing: the number of A particles is nφ and the interaction energy per site in a pure A component before mixing is zu AA /2. Hence U 0 = [u AA φ + u BB (1 φ) ] z n/2.

17 Energy of mixing The total energy of the system after mixing: U = [ u AA φ u AB φ(1 φ) + u BB (1 φ) 2 ] z n/2. The energy before mixing: U 0 = [u AA φ + u BB (1 φ) ] z n/2. We find U U 0 = φ(1 φ) (2u AB u AA u BB ) z n/2 and as an intensive property (divided by n), the energy change on mixing is ΔU mix = φ(1 φ) (2u AB u AA u BB ) z/2. Defining the Flory-Huggins interaction parameter χ = (z/2) (2u AB u AA u BB ) / k B T, we can write the energy of mixing per lattice site as ΔU mix = χ φ(1 φ) k B T.

18 Free energy of mixing The entropy of mixing per lattice site ΔS mix = k B (φ ln φ + (1 φ) ln (1 φ). The energy of mixing per lattice site ΔU mix = χ φ(1 φ) k B T. Combining these, we arrive at the Helmholtz free energy of mixing per lattice site: ΔF mix = ΔU mix T ΔS mix Flory-Huggins equation = k B T [φ ln φ + (1 φ) ln (1 φ) + χ φ(1 φ) ] Sign? Promotes or opposes χ = 2.5 mixing? χ = 2.0 χ = 0 (ideal mixture) χ = 1.0

19 Mixing: stability Free energy as a function of angle θ The definition of thermodynamic equilibrium is the state of the system with minimum free energy. How about the stability of a homogeneous mixture of composition φ 0 with free energy F mix (φ 0 ). To be found: local stability is determined by the sign of the second derivative of free energy with respect to composition.

20 Mixing: stability Unstable case Locally stable case System with overall composition φ 0 is in a state with two phases, with volume fraction of A species in the α phase φ α, and the fraction of A component in the β phase φ β. Use lever rule. With the fraction f α of the volume of the material having composition φ α (and fraction f β = 1 f α having composition φ β ), the total volume fraction of A component in the system is the sum of contributions from the two phases: φ 0 = f α φ α + f β φ β. For example: Homogeneous phase Nucleated phase

21 Mixing: stability One finds f α = (φ β φ 0 )/(φ β φ α ) and f β = (φ 0 φ α )/(φ β φ α ). The free energy of the demixed state is the weighted average of the free energies of the material in each of the two states (neglecting the surface tension between the two phases), F αβ (φ 0 ) = f α F α + f β F β = (φ β φ 0 )F α + (φ 0 φ α ) F β φ β φ α This linear composition dependence of the free energy results in straight lines in the above figure that connect the free energies F α and F β of the two compositions φ α and φ β.

22 Mixing: stability If the dependence of the free energy is concave, the system can spontaneously lower its free energy by phase separating into two phases, since F αβ (φ 0 ) < F mix (φ 0 ). Similarly in opposite order for the convex case, where F mix (φ 0 ) < F αβ (φ 0 ), making the mixed state locally stable. Criterion for local stability: ( 2 F mix / φ 2 ) < 0, unstable. ( 2 F mix / φ 2 ) > 0, locally stable. If it is zero, then the system is metastable; the locus of these points is known as the spinodal.

23 Mixing: phase diagram Composition φ To be considered as an exercise. Finally, the above analysis for the entropy of mixing can easily be extended to model polymer systems (to be discussed later).

24 OUTLINE: Topic E. POLYMERS Polymers, monomers, polymerization Classes of polymers Homo, hetero, random polymers etc. Individual polymers vs concentrated solutions Single polymer properties (ideal chains) See, for example, the book by R. A. L. Jones. Also M. Rubinstein and R. H. Colby Polymer Physics

25 Polymers Polymerization The word (poly)-(mer) means (many)-(parts) and refers to molecules consisting of many repeating units, called monomers. Monomers are structural repeating units of a polymer connected by covalent bonds (during polymerization). The number of monomers in a polymer is called its degree of polymerization N. The molar mass M of a polymer is equal to N timesthe molarmassof itsmonomer: M = N M mon. Two sequence isomers of polypropylene

26 Polymers The three structural isomers of polybutadiene. Different polymer architectures. Star H Linear Ring Comb Ladder Dendrimer Randomly branched

27 Polymers Polymer networks

28 Polymers A-A-A-A-A-A-A-A-A-A-A Homopolymer Types of copolymers (two types of monomers) Types of tripolymers (three types of monomers)

29 Polymer liquids (melts and solutions) Polymer melts, a bulk liquid state in the absence of solvent solutions (Silly Putty) Polymer solutions by dissolving a polymer in a solvent, see on the left. Mass concentration c, the ratio of the total mass of polymer dissolved in a solution and the volume of the solution. Volume fraction φ, the ratio of occupied volume of the polymer in the solution and the volume of the solution. For polymer density ρ, φ = c / ρ. Overlap volume fraction φ* = N v mon / V Volume of solution spanned by the polymer chain, V = R 3 for the size of the chain R. Occupied volume of a single monomer, in a polymer chain comprised of N monomers.

30 Ideal chain A flexible polymer of n+1 monomers (n bonds). The bond vector r i goes from monomer A i-1 to monomer A i. Ideal state corresponds to a situation where there are no interactions between A i and A j for which i j >> 1. l i = r i. For a freely jointed chain model l = constant and

31 Freely rotating chain model Model which assumes all 1. torsion angles to be equally probable 2. all bond lengths are fixed (constant) 3. all bond angles are also fixed (θ ) We want to find (prove it)

32 Freely rotating chain model We have a term describing the decay rate of correlations along the chain, Assuming decay to be rapid, Number of monomers in a persistence segment. Persistence length. Again, the end-to-end distance scales as n.

33 Radius of gyration Polymer of N monomers whose positions are given by We assume all monomers to have the same mass m. Center of mass position As a measure of size, one can use the end-to-end distance for linear chains, or the radius of gyration for any polymer. For an ideal chain one can derive R g 2 = R ee2 / 6 (prove it)

34 Real chains Then do the real chains really behave like ideal ones? Yes and no. Scaling of polymer size: R g ~ N x In 2D the exponent x is 0.75 for real chains (good solvent) 1/2 given by the ideal chain models Temperature T In 3D for the exponent x 0.59 for real chains (good solvent) 1/2 given by the ideal chain models 1/3 for real chains (poor solvent) Molecular weight characterizing the number of monomers in a polymer Radius of gyration Overall, the value of x depends on monomer-monomer and monomersolvent interactions, and the conservation laws for momentum

35 Polymers and random walks Random walk in 2 dimensions Constructing polymers can be interpreted as an unbiased / biased random walk depending on the model considered. For an ideal polymer without any correlations, one has a pure random walk. Then what is the distribution of the end-to-end distance? For a one-dimensional case, it turns out to be Please find the Gaussian nature!

36 Free energy and entropic force The polymer is subject to thermal fluctuations which change its configuration, the entropy of the polymer being Number of conformations of a freely jointed chain of N monomers with endto-end distance R Probability distribution function for the end-toend vector of an ideal linear chain of N monomers in 3 dimensions (known to Gaussian)

37 Free energy and entropic force Helmholtz free energy For an ideal chain, U does not depend on the end-to-end vector since there are no explicit monomer-monomer interactions. The free energy increases quadratically with the magnitude of the end-to-end distance, thus the entropic force increases linearly with the distance. The entropic elasticity follows Hooke s law. The force is proportional to temperature T!

38 Force-extension relations in DNA How about experiments? Interpolation from the low-force to highforce cases. Low force: f = ( k B T/ ξ ) ( 3 x / 2 L ). High force: f = ( k B T/ ξ ) ( 1 / 4 ) ( 1 (x / L)) 2. Interpolation formula: f ( k B T/ ξ ) [ ( 1 / 4 ) ( 1 (x / L)) 2 + ( x / L ) ( 1 / 4 ) ]. Ahsan et al., Biophysical Journal vol. 74, 132 (1998); Bustamante et al. (1994).

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