Soft Condesnsed Matter : Polymer Dynamics

Transcription

1 ECNS23 Introductory Course on Neutron Scattering Saint Remy Les Chevreuses 23 Soft Condesnsed Matter : Polymer Dynamics Michael Monkenbusch FZ-Jülich, Institut für Festkörperforschung, IFF, Germany

2 What means SOFT? typical metals 1 rubbery plateau flow In general: high response to weak external perturbations large susceptibilities (linear χ x ) large nonlinear effects caused by small forces

3 Random Flight & Gaussian Chains Random walk = freely jointed chain R 2 = Nb 2 = 6R 2 g Freely rotating chain R 2 = Nb cosθ 1 cosθ = Nb2 C Scaled random walk Φ(R, N) (3/2πNb 2 ) 3/2 exp 3R2 2Nb 2 T S R = k BTΦ(R, N) = 3k BT R Nb 2 R Gaussian Chain U ({R n }) = 3 2b 2k BT N (R n R n 1 ) 2 n=1 [ ] 3/2 [ 3 Φ(R n R m ) = exp 3(R n R m ) 2 ] 2πb 2 n m 2 n m b 2 From M.Doi and S.F.Edwards, The Theory of Polymer Dynamics

4 Good solvents & excluded volume Random walks Self-avoiding walks R 2 N R 2 N 2ν ν = /5 melts Θ-solvent S(Q ) Q 2 good solvents reluctant to interpenetrate S(Q ) Q 1/ν

5 Osmotic pressure, Blobs & Scaling Semidilute Solutions Π k B Tc p = 1 + F Π (c/c ) (1) c p = concentration of polymers c = concentration of segments c = overlap concentration Scaling / dimensional analysis: s = c p R 3 g(n) cn 3ν 1 c/c Local density ρ B within a blob of size R B ξ B ρ B N B R 3 B N 1 3ν B equating this to the overall average density: ρ B = ρ = c p N determines the blob size n B c 1/(3ν 1) ( c 1.39 ) ; R B c ν/(3ν 1) ( c.777 ) For c > c we expect Pi = Π(c) only (for s >> 1): Π k B T c ps 1 3ν 1 c 3ν 3ν 1 c B i.e. F/(blob) k B T!

6 Dilute Semidilute c < c R g R g =1 I(Q).4.2 ξ ξ=1/ Q c > c

7 Why Neutrons? Scattering lengths b H = fm b D = fm b C = fm b N = fm b O = 5.83 fm Contrast ρ = 1 b i v molecule molecule I ρ 2 = (ρ A ρ B ) 2 Labeling & Contrast matching... and dynamics

8 Gaussian coil and Rouse dynamics D CM = k BT Nζ S(Q,t) ( D CM Q 2 t) e τ= (Debye Fktn.) τ = 2.56 τ= (Gauss Fktn.) Q Scaling plot 1. PEE Homopolymer, T=473K.8. S(Q,t)/S(Q) H H PEE: -[-C--C-]- H n HCH HCH H Rouse + Diffusion Rouse (N = ) Q 2 (Wl 4 t) 2

9 Rousemodell Gaussian chain ζ v D R 3kT l 2 x ζ d r n dt = 3k BT ( r l 2 n+1 2 r n + r n 1 ) + f }{{} n random }{{} k 2 r/ n 2 k = entropy spring Solution: relaxation modes of a linear chain τ p = 1 p 2 t/τ p ~ e de Gennes universal scaling function for long chains: S(q, t) = S(q, ) e D Rq 2t g( q 4 Wl 4 t) Wl 4 = 3k B Tl 2 /ζ

10 Neutronenspin Spinrotationen π/2 π π/2 Magnetfeld π π/2 Flipper Präzession 1 Probe Präzession 2 π/2 Detektor Analysator 1. Flipper = Stromblattkasten Counts / 3 msec Phase current (4 turns) / A Echoform S(Q,τ)/S(Q) Meßkurven 2.5% Polyisoprenlösung 1 2 τ/ns Auswertung von 2 Winkelstellungen Principle of NSE first π/2-flipper: start precession main solenoid I precession 2πN +Φ π-flipper rotation: 2πN Φ sample v v+ v main solenoid II precession 2πN + 2πN Φ + Φ +(2πN + Φ) v/v 2nd π/2-flipper: end precession

11 Monomerdimensionen nichtuniverselle (lokale) Eigenschaften PIB bei T=47K, Q=.3.4 Å 1 1. S(Q,t) / S(Q).5 Q=.15 ROUSE Q= t / ns S(Q,t) / S(Q).5 Q=.15 INT VISCOSITY Q= t / ns Daten: IN11, Zusammenarbeit mit: A. Arbe, J. Colmenero, Universidad del Pais Vasco, San Sebastian, Spanien

12 REPTATION: virtual tube=topological constraint 1 log[r 2 (t)] 1 2 τ d = 77 µs τ e = 6 ns τ R = 1.6 µs log(t)

13 REPTATION: virtual tube=topological constraint Coherent scattering from h-labeled chains in a melt of d-labeled chains: S(Q,t > τ e ) S(Q) = (1 F(Q) ) exp t erfc( t/τ ) τ } {{} density fluctuations along the tube + F(Q) [Creep(t/τ d ) const.] Form factor of the virtual tube: F (Q) = exp[ (Qd/6) 2 ]

14 Polyethylene (PE) 19 g/mol Single chain structure factor from chains in a melt at T=59 K 1. F(Q 1 ) F(Q 2 ) S(Q,t)/S(Q) Q 1 =.3 A 1 Q 2 =.5 A 1 Q 3 =.77 A 1 Q 4 =.96 A 1 Q 5 =.115 A 1 F(Q 3 ) F(Q 4 ) F(Q 5 ) t / ns tube dimension d = 49Å follows from exploitation of: F(Q) = exp[ (Qd/6) 2 ]

15 Incoherent scattering: h-pe self-correlation function S inc (Q,t) 1 S inc (Q, t) = exp 1 6 Q2 r 2 (t) r 2 (t) = 6 log [S inc (Q,t)]/Q 2 (?)

16 τ e 1 ns Free Rouse segmental diffusion (t < τ e ): r 2 (t) = 2 Wl 4 /π t 1 2 (t > τ e ) r 2 (t) locrep = d 2/3(Wl 4 t /π) 1 4 log <r 2 > 1/2 τ e =6 ns 1/4 log t Diffusion along a contorted tube (Fatkullin & Kimmich PRE52,3273(1995)) (t >> τ e ) S inc (Q,t > τ e ) = exp Q4 d 2 72 r 2 (t) 3 erfc Q2 d 6 2 r 2 (t) 3

17 Incoherent Scattering from PE Resolve coherent/incoherent inconsistency of τ e and d Rouse 1,,75 S inc (Q,t),5,25 Q = 1. nm -1 Q = 1.5 nm -1 Fatk.&Kim., Fourier time t / ns 1 1 Q = 1. nm -1 Doi&Edw. <r 2 (t)> / nm 2 1 Q = 1.5 nm -1 t 1/4 t 1/ Fourier time t / ns Plot in terms of PSEUDO-mean-squared-distance

18 Modulus due to Rouse chains G(t) = c N k BT p=1 exp( t 2p 2 /τ R ) Re[G(ω)] Im[G(ω)] η ω J e ()η ω 2 ω 1/ ω τ R

19 Rouse Dynamics: Single Chain Scattering I ζ v D R 3kT 2 x l ξ d R n dt k BT b 2 ( R n+1 2 R n + R n 1 ) = f n (t) with f n (t) with f nα (t 1 )f mβ (t 2 ) = 2k B Tξδ nm δ αβ δ(t 1 t 2 ) assuming a continuous index variable n: ξ R n t 3k BT b 2 2 Rn 2 n = f n (t) boundary condition: R n / n n=,n =. X p (t) = 1 N N dn cos( pπn N ) R n (t) yields: 2Nξ X p t + 6k BTπ 2 p 2 Xp = f Nb 2 p (t)

20 Rouse Dynamics: Single Chain Scattering II Solution: Relaxation modes for p > : X pα (t)x qβ (t) = δ αβ δ pq k B T τ R = ξn 2 b 2 /(3π 2 k B T). Nb 2 p2 exp( t ) 3k B T2πp2 τ R Mode p = center-of-mass diffusion: X α (t)x β () = δ αβ 2k B T Nξ t S(Q, t) = N Scattering Function: n,m=1 exp[i Q ( R n (t) R m (t))] Gaussian chain! averages analytically exp[iq ( R n (t) R m (t))] = exp[ 1 2 α=x,y,z Q2 α (R nα (t) R mα (t)) 2 ] = exp[ Q2 ( R n (t) R m (t)) 2 ] }{{} Φ nm (t) S(Q, t) = 1 N exp( 1 n,m 6 Q2 Φ nm (t)) (2)

21 Rouse Dynamics: Single Chain Scattering III S(Q, t) = 1 N n,m exp( 1 6 Q2 Φ nm (t)) (1) the rms distance: Φ nm (t) = Φ D (t) + Φ nm + Φ 1 nm(t) Φ D (t) = 6Dt Φ nm = n m b 2 the center-of-mass diffusion Gaussian chain structure Φ 1 nm = 4Nb2 π 2 p=1 1 p 2 cos(pπm N ) cos(pπn N )[1 exp( tp2 /τ R )] dynamic part, Φ 1 nm (t = ) =. The center-of-mass diffusion is as for any diffusing object represented by a common factor exp( Q 2 Dt): S(Q, t) = exp( Q 2 Dt)S intrachain (Q, t) For t = is S(Q, t) = S(Q), i.e. the structure factor corresponds to a snap-shot of the chain structure: S(Q) = 1 N exp( 1 n,m 6 Q2 n m b 2 )

22 Rouse Dynamics: Single Chain Scattering IV S(Q) = 1 N n,m exp( 1 6 Q2 n m b 2 ) summations integration R 2 g = 1 6 Nb2 Debye-function: S(Q) = Nf Debye (Q 2 R 2 g) f Debye (x) = 2 x 2(e x 1 + x) 1 1 dσ/dω / cm incoherent incoherent / 3 d incoherent Q / nm 1

23 Rouse Dynamics: Single Chain Scattering V S(Q, t) = 1 N n,m exp( 1 6 Q2 Φ nm (t)) F(Q, t) = S(Q, t)/s(q) 1..8 S(Q,t).6.4 t=.2 t= Q R g h(u) = 2 π high-q behavior of S intrachain (Q, t) [degennes] S intern (Q, t) = 12 du exp[ u Γ Q 2 b 2 q t h(u/ Γ q t)] characteristic relaxation rate: Γ q = k BT 12ξ Q4 b 2 with dx cos(xu)(1 e x2 )/x 2 = 2 π e u2 /4 + u [erf(u/2) 1]

24 Rouse Dynamics: Single Chain Scattering VI Rouse scaling: 1. PEE Homopolymer, T=473K.8 S(Q,t)/S(Q) Rouse + Diffusion Rouse (N = ) Q 2 (Wl 4 t) Kumulante (PEE) Γ(q) / (s 1 ) T=533K T=473K Wl 4 q 4 q / A 1.1

25 Zimm Dynamics: Single Chain Scattering in Solution I ζ v D R f 3kT l 2 x H( r) = 1 8πηr 1 + r r r 2 R n t = m H nm k 2 m 2 R m + f m (t) preaveraging... H nm = 1 6πη 1 R n R m 1 (6π 3 n m ) 1/2 ηb) 1 = 1 h(n m).

26 Zimm Dynamics: Single Chain Scattering in Solution I Fourier transform (mode representation): X p / t = q h pq (k q Xq + f q ) FT[h(m n)] h pq δ pq 1/(2ηb 3π 3 pn) analogous to Rouse but: τ p = α 1 η( N/b) 3 k B T p 3/2 and k B T D cm = α 2 η N/b (1) with α and α 2 =.196 for theta solutions (Gaussian chain) high Q approximation: S intern (Q, t) = 12 du exp[ u ( Γ (Qb) 2 q t) 2/3 h(u/[ Γq t] 2/3 )] with Γ q = k BT 6πη Q3 h(u) = 2 π dx cos(xu)(1 e x3/2 / 2 )/x 2

27 Zimm Dynamics: Single Chain Scattering in Solution I Scaling: Γ Q 3 S(Q, t) = f[(k B T/6πη) 2/3 (Q 3 t) 2/3 ], solely depends on the solvent viscosity η! S(Q,t)/S(Q) [kt/(6πη) Q 3 t] (2/3)