Kinetic Theory for Rigid Dumbbells

Kinetic Theory for Rigid Dumbbells Hector D. Ceniceros hdc@math.ucsb.edu University of California @ Santa Barbara Mathematics Department University of California, Santa Barbara Universidade de São Paulo, November 25 2010 H. D. Ceniceros (UCSB) Complex Fluids 10/2010 1 / 12

Outline 1 Discussion: Elastic Dumbbell 2 Rigid Dumbells 3 Concentrated Solutions H. D. Ceniceros (UCSB) Complex Fluids 10/2010 2 / 12

Dilute Solution of Elastic Dumbbells!" #" #"!" $!" $ #" $" ψ t ( = Q Q ψ ) H. D. Ceniceros (UCSB) Complex Fluids 10/2010 3 / 12

Dilute Solution of Elastic Dumbbells!" #" #"!" $!" $ #" ψ t $" ( = Q Q ψ ) Substituting Q we get the Smoluchowski or Fokker-Planck Equation ψ t = Q [{ v Q 2ζ Fc (Q) + 1ζ } ] (Fe2 Fe1 ) ψ + 2kT ζ 2 Q ψ H. D. Ceniceros (UCSB) Complex Fluids 10/2010 3 / 12

Dilute Solution of Elastic Dumbbells!" #" #"!" $!" $ #" ψ t $" ( = Q Q ψ ) Substituting Q we get the Smoluchowski or Fokker-Planck Equation ψ t = Q [{ v Q 2ζ Fc (Q) + 1ζ } ] (Fe2 Fe1 ) ψ + 2kT ζ 2 Q ψ When the flow is no longer uniform ψ t has to be substitute for Dψ Dt H. D. Ceniceros (UCSB) Complex Fluids 10/2010 3 / 12

Dilute Solution of Elastic Dumbbells!" #" #"!" $!" $ #" ψ t $" ( = Q Q ψ ) Substituting Q we get the Smoluchowski or Fokker-Planck Equation Dψ Dt = Q [{ v Q 2ζ Fc (Q) + 1ζ } ] (Fe2 Fe1 ) ψ + 2kT ζ 2 Q ψ When the flow is no longer uniform ψ t has to be substitute for Dψ Dt H. D. Ceniceros (UCSB) Complex Fluids 10/2010 3 / 12

Flow-Structure, Coupled System The Flow pi + µ 2 v + τ p = 0 v = 0 H. D. Ceniceros (UCSB) Complex Fluids 10/2010 4 / 12

Flow-Structure, Coupled System The Flow The microstructure Dψ Dt = Q pi + µ 2 v + τ p = 0 v = 0 [{ v Q 2ζ Fc (Q) + 1ζ (Fe2 Fe1 ) } ψ τ p = n QF c + 1 2 n Q(F e 1 Fe 2 ) nkt I ] + 2kT ζ 2 Q ψ H. D. Ceniceros (UCSB) Complex Fluids 10/2010 4 / 12

Flow-Structure, Coupled System The Flow The microstructure Dψ Dt = Q pi + µ 2 v + τ p = 0 v = 0 [{ v Q 2ζ Fc (Q) + 1ζ (Fe2 Fe1 ) } ψ τ p = n QF c + 1 2 n Q(F e 1 Fe 2 ) nkt I ] + 2kT ζ 2 Q ψ Recall QF c = QF c ψ(q, t)dq, etc H. D. Ceniceros (UCSB) Complex Fluids 10/2010 4 / 12

Oldroyd B: Hookean Springs Hookean springs, F c = HQ (and no external forces) τ p = nh A nkt I, A = QQ H. D. Ceniceros (UCSB) Complex Fluids 10/2010 5 / 12

Oldroyd B: Hookean Springs Hookean springs, F c = HQ (and no external forces) Note that because τ p = nh A nkt I, A = 4kT ζ I 4 ζ HA A = QQ we have a closed system and we don t need ψ to evaluate A and τ p. H. D. Ceniceros (UCSB) Complex Fluids 10/2010 5 / 12

Oldroyd B: Hookean Springs Hookean springs, F c = HQ (and no external forces) Note that because τ p = nh A nkt I, A = 4kT ζ I 4 ζ HA A = QQ we have a closed system and we don t need ψ to evaluate A and τ p. Oldroyd B Model pi + µ 2 v + τ p = 0 v = 0 τ p + λ H τ p = 2nkT λ H D H. D. Ceniceros (UCSB) Complex Fluids 10/2010 5 / 12

Oldroyd B: Hookean Springs Hookean springs, F c = HQ (and no external forces) Note that because τ p = nh A nkt I, A = 4kT ζ I 4 ζ HA A = QQ we have a closed system and we don t need ψ to evaluate A and τ p. Oldroyd B Model pi + µ 2 v + τ p = 0 v = 0 τ p + λ H τ p = 2nkT λ H D Note τ p = Dτ p Dt v τ p τ p v T H. D. Ceniceros (UCSB) Complex Fluids 10/2010 5 / 12

Non-Linear Dumbbells and a Closure Approximation The (Warner) FENE spring: F c = HQ 1 Q 2 /Q 2 0 Q Q 0 QF c f (A) So we cannot longer obtain a closed system. H. D. Ceniceros (UCSB) Complex Fluids 10/2010 6 / 12

Non-Linear Dumbbells and a Closure Approximation The (Warner) FENE spring: F c = HQ 1 Q 2 /Q 2 0 Q Q 0 QF c f (A) So we cannot longer obtain a closed system. Peterlin Approximation F c = HQ 1 Q 2 /Q 2 0 and QF c = H 1 tr(a)/q 2 0 A = f (tra)ha H. D. Ceniceros (UCSB) Complex Fluids 10/2010 6 / 12

Non-Linear Dumbbells and a Closure Approximation The (Warner) FENE spring: F c = HQ 1 Q 2 /Q 2 0 Q Q 0 QF c f (A) So we cannot longer obtain a closed system. Peterlin Approximation F c = HQ 1 Q 2 /Q 2 0 and QF c = H 1 tr(a)/q 2 0 A = f (tra)ha FENE-P Model pi + µ 2 v + τ p = 0; v = 0 τ p = nf (tra)ha nkt I A = 4kT ζ I 4 ζ f (tra)ha H. D. Ceniceros (UCSB) Complex Fluids 10/2010 6 / 12

Rigid Dumbbells %!"! #! "# $#!"# "# $!"# $ "# &# H. D. Ceniceros (UCSB) Complex Fluids 10/2010 7 / 12

Rigid Dumbbells %!"! #! "# $#!"# "# $!"# $ "# &# This is a system with a constraint. We don t know the force of the constraint acting on each bead. H. D. Ceniceros (UCSB) Complex Fluids 10/2010 7 / 12

Rigid Dumbbells %!"! #! "# $#!"# "# $!"# $ "# &# This is a system with a constraint. We don t know the force of the constraint acting on each bead. But at r c 1 ( ) F h 1 2 + Fh 1 + 1 ( ) F b 1 2 + Fn 1 + 1 ( F e 2 1 + F e ) 1 = 0 (1) H. D. Ceniceros (UCSB) Complex Fluids 10/2010 7 / 12

Equations of Motion Also the forces along the connector Lu = r 1 r 1 do not generate any motion perpendicular to it: [( ) ( ) (I uu) F h 1 Fh 1 + F b 1 Fb 1 + ( F e 1 1) ] Fe = 0 (2) H. D. Ceniceros (UCSB) Complex Fluids 10/2010 8 / 12

Equations of Motion Also the forces along the connector Lu = r 1 r 1 do not generate any motion perpendicular to it: [( ) ( ) (I uu) F h 1 Fh 1 + F b 1 Fb 1 + ( F e 1 1) ] Fe = 0 (2) We need expressions for the forces: F h j = ζ ( ṙ j v j ) = ζ ( ṙ c + 1 2 jl u v 0 v r c 1 2 jl v u ) F b j = j kt R ln ψ L where R = u u is the gradient on the surface of the sphere: R = e θ θ + e φ 1 sin θ φ H. D. Ceniceros (UCSB) Complex Fluids 10/2010 8 / 12

Equations of Motion II Substituting these forces in (1) and (2) we get: ṙ c = v 0 + v r c + 1 ( F e 2ζ 1 + F e 1) u = v u v : uuu 1 6λ R ln ψ + 1 [ ( (I uu) F e ζl 1 F e 1)] where λ = ζl 2 /12kT is a time constant. H. D. Ceniceros (UCSB) Complex Fluids 10/2010 9 / 12

Equations of Motion II Substituting these forces in (1) and (2) we get: ṙ c = v 0 + v r c + 1 ( F e 2ζ 1 + F e 1) u = v u v : uuu 1 6λ R ln ψ + 1 [ ( (I uu) F e ζl 1 F e 1)] where λ = ζl 2 /12kT is a time constant. We write the last term as (2/L)RΦ, Φ=potential independent of r c. Also observing that v u v : uuu = u v u and setting D r = 1/(6ζ) we get ( u = u v u D r R ln ψ + 1 ) kt RΦ H. D. Ceniceros (UCSB) Complex Fluids 10/2010 9 / 12

Equations of Motion II Substituting these forces in (1) and (2) we get: ṙ c = v 0 + v r c + 1 ( F e 2ζ 1 + F e 1) u = v u v : uuu 1 6λ R ln ψ + 1 [ ( (I uu) F e ζl 1 F e 1)] where λ = ζl 2 /12kT is a time constant. We write the last term as (2/L)RΦ, Φ=potential independent of r c. Also observing that v u v : uuu = u v u and setting D r = 1/(6ζ) we get ( u = u v u D r R ln ψ + 1 ) kt RΦ We can now substitute into the continuity equation ψ t = R ( u ψ) H. D. Ceniceros (UCSB) Complex Fluids 10/2010 9 / 12

The Smoluchowski Equation Smoluchowski Equation for Rigid-rod like Molecules Dψ Dt ( = D r R Rψ + ψ ) kt RΦ R (u v u ψ) (3) H. D. Ceniceros (UCSB) Complex Fluids 10/2010 10 / 12

The Smoluchowski Equation Smoluchowski Equation for Rigid-rod like Molecules Dψ Dt ( = D r R Rψ + ψ ) kt RΦ R (u v u ψ) (3) We can obtain an equation for the second moment A = uu from (3) : A = 2D r (I 3A) 2 v : Q D r kt where Q = uuuu. (RΦ)u + u(rφ) H. D. Ceniceros (UCSB) Complex Fluids 10/2010 10 / 12

The Smoluchowski Equation Smoluchowski Equation for Rigid-rod like Molecules Dψ Dt ( = D r R Rψ + ψ ) kt RΦ R (u v u ψ) (3) We can obtain an equation for the second moment A = uu from (3) : A = 2D r (I 3A) 2 v : Q D r kt (RΦ)u + u(rφ) where Q = uuuu. Clearly, we don t have a close system; an equation for Q would involve the sixth moment and so on... H. D. Ceniceros (UCSB) Complex Fluids 10/2010 10 / 12

The Smoluchowski Equation Smoluchowski Equation for Rigid-rod like Molecules Dψ Dt ( = D r R Rψ + ψ ) kt RΦ R (u v u ψ) (3) We can obtain an equation for the second moment A = uu from (3) : A = 2D r (I 3A) 2 v : Q D r kt (RΦ)u + u(rφ) where Q = uuuu. Clearly, we don t have a close system; an equation for Q would involve the sixth moment and so on... Mathematical Temptation: Closure Approximation, Q f (A). H. D. Ceniceros (UCSB) Complex Fluids 10/2010 10 / 12

The Polymeric Stress Due to the constraint, we can t apply the procedure we used for an elastic dumbbell. But it turns out that the Kramers-Kirkwood formula is still valid: τ p = n 1 j= 1 R j F h j H. D. Ceniceros (UCSB) Complex Fluids 10/2010 11 / 12

The Polymeric Stress Due to the constraint, we can t apply the procedure we used for an elastic dumbbell. But it turns out that the Kramers-Kirkwood formula is still valid: τ p = n 1 j= 1 R j F h j Substituting R j = 1 2 jlu and Fh j = ζ( ṙ j v 0 v r j ) we get τ p = 1 2 nl2 ζ u( u v u) H. D. Ceniceros (UCSB) Complex Fluids 10/2010 11 / 12

The Polymeric Stress Due to the constraint, we can t apply the procedure we used for an elastic dumbbell. But it turns out that the Kramers-Kirkwood formula is still valid: τ p = n 1 j= 1 R j F h j Substituting R j = 1 2 jlu and Fh j = ζ( ṙ j v 0 v r j ) we get τ p = 1 2 nl2 ζ u( u v u) and substituting the expression for u we get (Kramer s) Stress τ p = n kt D r v : Q + 3nkT A + n urφ nkt I H. D. Ceniceros (UCSB) Complex Fluids 10/2010 11 / 12

Doi s Theory for a Concentrated Solution Φ is replaced by a mean field potential describing intermolecular interactions Maier-Saupe Φ = 3 ktua : uu 2 Models excluded volume effects: finite volume of macromolecules. H. D. Ceniceros (UCSB) Complex Fluids 10/2010 12 / 12

Doi s Theory for a Concentrated Solution Φ is replaced by a mean field potential describing intermolecular interactions Maier-Saupe Φ = 3 ktua : uu 2 Models excluded volume effects: finite volume of macromolecules. Marrucci-Greco (adds spatial elasticity) Φ = 3 ( ) 2 ktu A + l2 24 2 A : uu H. D. Ceniceros (UCSB) Complex Fluids 10/2010 12 / 12

Doi s Theory for a Concentrated Solution Φ is replaced by a mean field potential describing intermolecular interactions Maier-Saupe Φ = 3 ktua : uu 2 Models excluded volume effects: finite volume of macromolecules. Marrucci-Greco (adds spatial elasticity) Φ = 3 ( ) 2 ktu A + l2 24 2 A : uu Doi uses a pre-averaged rotational diffusivity D r = D r (1 S 2 ) 2 S 2 = (S : S)/2 and S = A 1 3 I, D r constant. Translation diffusivity neglected. H. D. Ceniceros (UCSB) Complex Fluids 10/2010 12 / 12