Part III. Polymer Dynamics molecular models

Transcription

1 Part III. Polymer Dynamics molecular models I. Unentangled polymer dynamics I.1 Diffusion of a small colloidal particle I.2 Diffusion of an unentangled polymer chain II. Entangled polymer dynamics II.1. Introduction to Tube models II.2. «Equilibrium state» in a polymer melt or in a concentrated solution: II.3. Relaxation processes in linear chains Source: Polymer physics Rubinstein,Colby 1

2 I. Unentangled polymer dynamics I.1 Diffusion of a small colloidal particle: random walk Simple diffusive motion: Brownian motion Friction coefficient: Diffusion coefficient A constant force applied to the particle leads to a constant velocity: Einstein relationship: Friction coefficient Stokes law: (Sphere of radius R in a Newtonian liquid with a viscosity η) Stokes-Einstein relation: Determination of 2

3 I.2 Diffusion of an unentangled polymer chain: I.2.1. Rouse model: R - Each bead has its own friction ζ - Total friction: N beads of length b - Rouse time: t < τ Rouse : viscoelastic modes t > τ Rouse : diffusive motion - Stokes law: - Kuhn monomer relaxation time: 3

4 Rouse relaxation modes: The longest relaxation time: For relaxing a chain section of N/p monomers: (p = 1,2,3,,N) N relaxation modes 4

5 - At time t = τ p, number of unrelaxed modes = p. - Each unrelaxed mode contributes energy of order kt to the stress relaxation modulus: - Mode index at time t = τ p : (density of sections with N/p monomers: Volume fraction, for τ 0 < t < τ R Approximation:, for t > τ 0 5

6 , for τ 0 < t < τ R 6

7 Viscosity of the Rouse model: Or: 7

8 Exact solution of the Rouse model: with Each mode relaxes as a Maxwell element The Rouse relaxation time is the half of the end-to-end correlation time. Or: The Rouse relaxation time is the end-to-end correlation time. 8

9 Rouse time determined from the free energy: Free energy U of a Gaussian chain (see Part I): Forces equilibrium after a displacement of ΔL: After the disorientation spring force from the free energy 9

10 Example: Frequency sweep test: For 1/τ R < ω < 1/τ 0 For ω < 1/τ R 10

11 Rouse model: also valid for unentangled chains in polymer melts: Relaxation times τ rel M c M For M < M c : τ rel M 2 For M > M c : τ rel M

12 I.2.2. Zimm model: Viscous resistance from the solvent: the particle must drag some of the surrounding solvent with it. Hydrodynamic interaction: long-range force acting on solvent (and other particles) that arises from motion of one particle. When one bead moves: interaction force on the other beads (Rouse: only through the springs) The chain moves as a solid object: Zimm time: (In a theta solvent) 12

13 Rouse time versus Zimm time: - τ Zimm has a weaker M-dependence than τ Rouse. - τ Zimm < τ Rouse in dilute solution - real case: often a combination of both 13

14 Slope: 5/9 to 2/3 (theta condition) 14

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16 Rheological Regimes: (Graessley, 1980) Dilute solutions: Zimm model H.I. no entanglement Unentangled chains (Non diluted): Rouse model no H.I. no entanglement Entangled chains: Reptation model no H.I. entanglement 16

17 II. Entangled polymer dynamics II.1. Introduction to Tube models The test chain is entangled with its neighbouring polymers entanglement Molecules cannot cross each other 17

18 The tube model Doi & Edwards (1967), de Gennes (1971) The molecular environment of a chain is represented by a tube in which the chain is confined entanglement M e : molecular weight between two entanglements 18

19 Tube model and entanglements: 19

20 Concentrated solutions or polymer melts: G(t) Glassy region Pseudo equilibrium state Relaxation processes Pseudo equilibrium state: t The chains are not in their stable shape. The other chains prevent their relaxation. The tube picture becomes active. M e : M w of the longest sub-chain relaxed by a Rouse process 20

21 II.2. «Equilibrium state» in a polymer melt or in a concentrated solution: Segment between two entanglements 1 l Khun segment N e 2N e Z = segments number =4 tube 3N e 4n e a = tube diam. L eq. Each segment between two entanglements is relaxed by the Rouse process The primitive path of the chain is defined by its segments between the entanglements. Its length = L eq. Coarse grained model 21

22 Primitive path - Simpler description of the chain - Coarse grained model: subchains of mass M e are considered as a segment of length l - relaxation time of subchain M e : Rouse relaxation 22

23 Equilibrium state in a polymer melt: parameters L eq = length of the primitive path a Z = number of segments between two ent. n e tube l = length of a segment between two ent. 1 l a = tube diameter b = length of a Khun segment L eq. N e = number of Khun segments between two entanglements N = total number of Khun segments in the chain Gaussian chain in solution: the end-to-end distance R 0 2 = Nb 2 Gaussian chain in a melt or concentrated solution: l 2 N e b 2 23

24 Constitutive equation of a polymer melt For a Rouse chain (unentangled chain): Entropic material: U = -TS (free energy) U(R 0 ) S = k lnp In a fixed direction: (Gaussian) (free energy) 0 R 0 24

25 For a polymer melt: Constitutive equation of a polymer melt Additional boundary conditions: ψ(r) = 0 on the tube surface U(L chain ) 0 L eq L chain At equilibrium (U min.): Nb 2 a.l eq 25

26 1 l Equilibrium state in a polymer melt: parameters N e tube L eq. Gaussian chain in solution: R 0 2 = Nb 2 a L eq = length of the primitive path Z = number of segments between two ent. l = length of a segment between two ent. a = tube diameter b = length of a Khun segment N e = number of Khun segments between two entanglements N = total number of Khun segments in the chain Gaussian chain in a melt or concentrated solution: l 2 N e b 2 l a Nb 2 a.l eq 26

27 Material parameters in the tube model Defined at the equilibrium state of the polymer The molecular weight between two entanglements: M e The plateau modulus: (x 4/5) The relaxation time of a segment between two entanglements: Very simplified coarse-grained model 27

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29 II.3. Relaxation processes in linear chains G(t) Glassy region Rouse process Plateau modulus : tube effect t Relaxation of the polymer (as Reptation) Φ = 1 Φ = 0 Plateau modulus Unrelaxed fraction of the polymer melt Not anymore memory of the initial deformation 29

30 II.3.1. Main relaxation process in linear chain: Reptation 30

31 Reptation D curvilinear diffusion As the primitive chain diffuses out of the tube, a new tube is continuously being formed, starting from the ends. This new tube portion is randomly oriented even if the starting tube is not. Part of the initial tube + Movie 31

32 Reptation Curvilinear diffusion coefficient: Reptation time: Total friction of the chain Monomeric friction In solution: D e : along L eq <R 02 > D 32

33 Reptation PBD, 130 kg/mol. M (Kuhn segment): 105 g/mol N=1240 Kuhn segments τ 0 = 0.3 ns. τ e = 0.1 µs. M e =1900g/mol. N e =18. M/M e = 68 entanglements 33

34 Reptation: Doi and Edwards model - First passage problem - diffusion of a chain along the tube axis 0 s L eq p(s,t): survival probability of a initial tube segment localized at a distance s. Variables separability 34

35 Doi and Edwards model p(x,t) time Using the usual coordinate system x 0 1 x 0 x 35

36 Doi and Edwards model Relaxation modulus: Related viscosity: for M < M c for M > M c in the real case 36

37 Zero-shear viscosity vs. log (M) Slope of 3.4 Discrepancy between theory and experiments: inclusion of the contour length fluctuations process 37

38 Initial D&E model (with CR) 38

39 II.3.2. Contour length fluctuations Equilibrium length of an arm: Entropical force: Topological force: L eq = the most probable length Equilibrium length L eq real length This part of the initial tube is lost 39

40 II.3.2. Contour length fluctuations Symmetric star: Fixed point x=1 x=0 No Reptation Only fluctuations (see the next lesson) 40

41 II.3.2. Contour length fluctuations + movie 41

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43 Linear chains: Doi and Edwards model including the contour length fluctuations (CLF): - the environment of the molecule is considered as fixed. Rouse-like motion of the chain ends: Displacement along the curvilinear tube: For t < τ R Displacement of a monomer in the 3D space: Rem/ Before t = τ Rouse, the end-chain does not know that it belongs to the whole molecule 43

44 L eq,0 L(t) Partial relaxation of the stress: At time t = τ R : For τ e < t < τ R With µ close to 1 44

45 In the same way: Rescale of the reptation time in order to account for the fluctuations process Slope: 3.4 Log(M) 45

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47 D&E model with contour length fluctuations (with CR) 47

48 Linear chains: model of des Cloizeaux: Based on a time-dependent coefficient diffusion: D e 48

49 Des Cloizeaux model (with CR) 49