Molecular Theories of Linear Viscoelasticity THE ROUSE MODEL (P. 1)

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1 THE ROUSE ODEL (P. 1) odel polymer dynamics by a system of N + 1 beads connected by N springs. Figure 1: apping the Polymer Chain onto a Chain of Beads Connected by Springs. ROUSE SCALING Recall that a polymer chain (or part of a chain) acts like a linear spring when stretched. Spring Length a Spring Constant kt a 2 Bead Friction ζ

2 ROUSE SCALING (P. 2) Einstein Relation: Chain Friction ζn Diffusion Coefficient D = kt ζn 1 N The relaxation time λ is the time the chain takes to diffuse a distance equal to its size R. λ = R2 D NR2 N 2 since R = an 1/2. The terminal modulus G(λ) is kt per chain G(λ) = νkt = ρrt 1 N ν = ρn 0 / is the number density of chains ρ is the density R = N 0 k is the ideal gas constant Viscosity η = λg(λ) N Thus the Rouse model correctly predicts the viscosity proportional to chain length for an unentangled polymer.

3 THE ROUSE ODEL (P. 3) Rouse solved the system of coupled differential equations that describe the chain of springs, and got G(t) = ρrt e t/λp (2-94) with Rouse relaxation times (eigenvalues) λ p = a2 N 2 ζ 6π 2 p 2 kt text has typo! P N (2-95) Equation (2-94) is in the form of a generalized axwell model with N modes of equal weight. From equations (2-94) and (2-95) we can calculate all viscoelastic functions. Viscosity η 0 = G(t)dt = ζρa2 N N 0 is Avogadro s Number 0 = /N is the molecular weight per spring Rearranging (2-96) (2-96) Substitute into (2-95) a 2 ζ = η 0 ρn 0 λ p = η 0 N 2 6π 2 p 2 kt ρn 0 Canceling terms and using = 0 N λ p = 6η 0 π 2 p 2 ρrt The longest relaxation time (p = 1) is λ R = 6η 0 π 2 ρrt = 0.608η 0 ρrt (2-97) (2-98)

4 THE ROUSE ODEL (P. 4) There are N relaxation modes in the Rouse model λ p = 6η 0 π 2 p 2 ρrt (2-97) The longest relaxation time is with p = 1, 2, 3,..., N λ R = 6η 0 π 2 ρrt = 0.608η 0 ρrt whereas in our scaling argument we concluded (2-98) λ R = η 0 ρrt Scaling always fails to give a prefactor of order unity (in this case 6/π 2 ). SIGNIFICANCE OF THE LONGEST RELAXATION TIE Relaxation of internal stresses occurs on this time scale Determines the diffusion coefficient D = R 2 /λ 1 Determines the viscosity (equation 2-98) For unentangled polymers, λ R 2 The longest mode involves all N springs Shorter modes involve smaller numbers of springs, with The shortest mode λ p = λ R p 2 λ N = λ R N 2 = a2 ζ 6π 2 kt

5 THE ROUSE ODEL (P. 5) Storage odulus G(t) = ρrt e t/λp (2-94) OSCILLATORY SHEAR G (ω) ω 0 G(s) sin(ωs)ds (2-65) and Loss odulus G (ω) = ρrt G (ω) ω 0 ω 2 λ 2 p 1 + ω 2 λ 2 p (2-99) G(s) cos(ωs)ds (2-66) G (ω) = ρrt ωλ p 1 + ω 2 λ 2 p (2-100) Equations (2-94), (2-99) and (2-100) are special cases of the generalized axwell model G(t) = G (ω) = G i exp( t/λ i ) (2-25) i=1 G (ω) = with all G i = ρrt G i (ωλ p ) (ωλ p ) 2 (2-67) G i ωλ p 1 + (ωλ p ) 2 (2-68)

6 THE ROUSE ODEL (P. 6) Dimensionless oduli [G ] R G ρrt = N [G ] R G ρrt = N ω 2 λ 2 p 1 + ω 2 λ 2 p ωλ p 1 + ω 2 λ 2 p These moduli are UNIVERSAL for the Rouse model in the large N limit. Figure 2: Rouse odel Predictions of Storage and Loss oduli. Note that τ 1 λ 1.

7 THE ROUSE ODEL TERINAL RESPONSE [ ] 1.08ρRT G = ω 2 λ 2R = 0.4η2 0ω 2 ρrt (2-102) G = 1.645ρRT ωλ R = ωη 0 (2-103) Recall the steady state compliance ( ) G JS 0 = lim ω 0 (G ) 2 J 0 S = 0.4 ρrt HIGH FREQUENCY (SHORT TIE) RESPONSE Rouse model is a POWER LAW at short times (2-101) G(t) = Ct 1/2 (2-104a) 3ρRT η0 C = 2π (2-104b) π G (ω) = G (ω) = 2 Cω1/2 (2-106) tan(δ) = G G = 1 Whenever G(t) is a power law G(t) = ct m (2-26) The storage and loss moduli are parallel power laws G (ω) G (ω) ω m δ = mπ 2

8 APPLICATIONS OF THE ROUSE ODEL 1) UNENTANGLED ELTS The Rouse odel correctly describes the full viscoelastic character of short polymers. However, there is another complication at low molecular weights because the monomeric friction coefficient depends on chain length. This is reflected in the glass transition temperature. T g = T g K n (D-2) This is understood as a chain end effect chain ends have more free volume v f than segments in the middle of the chain. v f = v f + C n The free volume theory of the glass transition relies on a constant amount of free volume at T g, so an increase in free volume lowers T g. It also lowers the monomeric friction coefficient and hence the viscosity via the Doolittle equation. 2) ENTANGLED ELTS ln η = ln A + B(v v f) v f (10-11) The Rouse model describes the short time (or high frequency) relaxation of high molecular weight polymers. The molecular theory for terminal relaxation of high molecular weight polymers, REPTATION, is simply Rouse motion of a chain confined in a tube.

9 Chain End Effects THE GLASS TRANSITION Chain Ends 1 n T g = T g K n (D-2) Figure 3: Glass Transition Temperature of Low olecular Weight Polybutadiene.

10 Chain End Effects SPECIFIC VOLUE 1 ρ = 1 + C ρ n Figure 4: Density of Low olecular Weight Polybutadiene.

11 Chain End Effects KINEATIC VISCOSITY ν η ρ units Stokes = cm2 s Figure 5: Isothermal Kinematic Viscosity of Low olecular Weight Polybutadiene.

12 Chain End Effects CORRECTED KINEATIC VISCOSITY Figure 6: Kinematic Viscosity at Constant Friction Coefficient for Low olecular Weight Polybutadiene.