Kinetic Theory for Elastic Dumbbells

Kinetic Theory for Elastic Dumbbells

Hector D. [email protected]

University of California @ Santa BarbaraMathematics Department

University of California, Santa Barbara

Universidade de Sao Paulo, November 11 2010

H. D. Ceniceros (UCSB) Complex Fluids 10/2010 1 / 28

Outline

1 Discussion: Equilibrium Configurations

2 Elastic Dumbbells


Bead-Spring Chain

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Convenient coordinates, relative to the center of mass rc :

Rj = rj − rc , rc =1

mT

N+1∑j=1

mjrj

Kinetic energy can be easily written as

K =12

mT r2c +

12

N+1∑j=1

mj R2j


Generalized Coordinates

The Hamiltonian (H = K + φ) needs to be expressed in terms ofgeneralized coordinates q1, . . . ,qn and generalized momenta:

pα =∂K∂qα

α = 1, . . .n.

The Hamiltonian is appropriately written as

H =1

2mTp2

c +12

Gαβpαpβ + Φ(rc ,q1, . . . ,qM).

(Recall qα = Gαβpβ , pα = gαβqβ, and Gg = I ) and the chain evolvesaccording to Hamilton’s canonical equations:

∂H∂rc

= −pc ,∂H∂pc

= rc

∂H∂qα

= −pα,∂H∂pα

= qα


Distribution functions at Equilibrium

Solution in a thermal bath at temperature T (canonical ensemble), nmacromolecules per unit volume.

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Each macromolecule can have many different configurations andmomenta for a given energy (range).

feq(rc ,q,pc ,p)drcdqdpcdp

is the number of macromolecules in the solution having a configurationin the range drcdq about rc ,q and with momenta in the range dpcdpabout pc ,p. feq(rc ,q,pc ,p) = nV × pdf


Configurational Distribution Function

From Statistical Mechanics we know that the pdf is proportional to theBoltzmann factor e−H/kT .Configurational distribution function:

Ψeq(q) = n∫

e−H/kT dpcdp∫ ∫ ∫e−H/kT dqdpcdp

:= nψeq(q).

Substituting H and canceling out the integral wrt pc we get

ψeq(q) =

√|g(q)|e−

1kT Φ(q)∫ √

|g(q)|e−1

kT Φ(q)dq

All we need is Φ describing the ”springs” and the chain’s geometricquantity |g|. For the Rouse Model (Hookean springs) Φ = 1

2H∑

k Q2k .

ψeq(Q1, . . . ,QN) =N∏

j=1

(H

2πKT

)3/2

e−(H/2KT )Q2j


Average Values

Average of B(rc ,Q,pc ,p) in phase-space of a single macromolecule:

< B >eq=

∫ ∫ ∫ ∫BfeqdrcdQdpcdp∫ ∫ ∫ ∫feqdrcdQdpcdp

=1

nV

∫ ∫ ∫ ∫BfeqdrcdQdpcdp

and the momentum-space average is defined as

JBKeq =

∫ ∫Bfeqdpcdp∫ ∫feqdpcdp

=1

Ψeq

∫ ∫Bfeqdpcdp

Note that formally

< B >eq=1

nV

∫ ∫JBKeqΨeqdrcdQ

If B = B(Q) and Φ = Φ(Q) then

< B >eq=

∫BψeqdQ


Dilute Solution of Elastic Dumbbells

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Newtonian solvent of viscosity ηs. Solution is dilute enough so thatinteractions of macromolecules with themselves are neglected.

rc =12

(r1 + r2), Q = r2 − r1

Let f (r1, r2,p1,p2, t) be the distribution function in phase-space andlet F (r1, r2, r1, r2, t) be the corresponding distribution inconfiguration-velocity space.


Assumptions

1 The flow is homogeneous, i.e. ∇v = constant.v = mass averaged velocity.

2 F (r1, r2, r1, r2, t) = Ψ(r1, r2, t)Θ(r1, r2, r1, r2, t)where Θ is the (normalized) velocity-space distribution.

3 Ψ(r1, r2, t) = nψ(Q, t).4 Θ is Maxwellian about the solution’s velocity v at rc :

Θeq(r1, r2) = C exp{− 1

KT

[12

m(r1 − v)2 +12

m(r2 − v)2]}

Velocity distribution in the flow system is the same as that of thesolution at equilibrium=”equilibration in momentum-space”.

5 Inertia is neglected. Small microscopic Re.


Forces Acting on the Beads

A hydrodynamic drag force Fhj of the beads as they move through

the viscous fluid.A brownian force Fb

j . Due to the many collisions of the solventmolecules with the dumbbell beads.An intramolecular force Fφj resulting from the spring, −∇rjφ.

An external force Fej . e.g. gravitational, electromagnetic, etc.

Hydrodynamic forces tend to orient and distort the macromoleculeswhereas intramolecular forces tend to restore to equilibriumconfigurations. Brownian forces tend to randomize the orientation ofthe macromolecules.

Because inertia is neglected mrj = 0 for j = 1,2 and hence

Fhj + Fb

j + Fφj + Fej = 0, j = 1,2.


Forces II

The hydrodynamics drag:

Fhj = −ζ ·

[JrjK− (vj + v′j)

]vj = impose flow at bead j and v′j is the perturbation of the flowfrom the other bead ( hydrodynamic interaction).The (statistically averaged) Brownian force:

Fbj = − 1

Ψ∇rj ·

[Jm(rj − v)(rj − v)KΨ

]≈ −kT ∇rj ln Ψ when Maxwellian velocity distribution is used(Reading: Doi and Edwards, Chap. 3).The intramolecular:

Fφj = −∇rjφ

Connector force Fc = Fφ1 = −Fφ2 .


Equations for rc and Q

Neglecting v′j , setting vj = v0 +∇v · rj and using the Maxwellian for Fbj :

−ζ(JrjK− v0 −∇v · rj

)− kT ∇rj ln Ψ + Fφj + Fe

j = 0, j = 1,2.

Now, using the chain rule

∇rj ln Ψ = ∇rj ln nψ =∂rc

∂rj· ∂ ln nψ

∂rc+∂Q∂rj· ∂ ln nψ

∂Q=∂Q∂rj· ∂ ln nψ

∂Q

as nψ is a function of Q only.Now add/subtract to get evolution eqs for rc and Q:

JrcK = v0 +∇v · rc +12ζ(Fe

1 + Fe2)

JQK = ∇v ·Q− 2kTζ∇Q lnψ − 2

ζFc +

1ζ

(Fe

2 − Fe1)


Continuity Equation for Ψ

Continuity equation for Ψ(rc ,Q):

∂Ψ

∂t= −

[∇rc · (JrcKΨ) +∇Q ·

(JQKΨ

)]Rate of change =flux

Recall Ψ = nψ(Q) and

JrcK = v0 +∇v · rc +12ζ(Fe

1 + Fe2)

If the external forces are independent of rc then∇rc · (JrcKΨ) = nψtr(∇v) = 0 and

∂Ψ

∂t= −∇Q ·

(JQKΨ

)


Fokker-Planck (Smoluchowski)Equation

Substituting JQK we get the Smoluchowski or Fokker-Planck Equation

∂ψ

∂t= −∇Q ·

[{∇v ·Q− 2

ζFc(Q) +

1ζ

(Fe2 − Fe

1)

}ψ

]+

2kTζ∇2

Qψ

One can derive an evolution equation for the average of B(Q):∫B∂ψ

∂tdQ = −

∫B∇Q ·

(JQKψ

)dQ

=

∫JQKψ · ∇QB dQ

where integration by parts was used. That is

d 〈B〉dt

=⟨JQK · ∇QB

⟩H. D. Ceniceros (UCSB) Complex Fluids 10/2010 14 / 28

Evolution Equation for Observables

Recall JQK = ∇v ·Q− 2kTζ ∇Q lnψ − 2

ζFc + 1ζ

(Fe

2 − Fe1

)Now,

(∇v ·Q) · ∇QB = (∇v)ijQj∂B∂Qi

= ∇v : (Q · ∇QB)

∫∇Q lnψ · ∇QB ψ dQ =

∫∇Qψ · ∇QB dQ

= −∫ψ∇2

QB dQ = −⟨∇2

QB⟩

Thusd 〈B〉

dt= ∇v : 〈Q · ∇QB〉 − 2

ζ

⟨Fc · ∇QB

⟩+

2kTζ

⟨∇2

QB⟩

+1ζ

⟨(Fe

2 − Fe1) · ∇QB


The Configuration Tensor

The average of the second moment A = 〈QQ〉 plays an important rolein the theory.Using the above technique we get:

dAdt−∇v A− A∇vT =

4kTζ

I− 4ζ

⟨QFc⟩

+1ζ

⟨(Fe

2 − Fe1)Q + Q(Fe

2 − Fe1)⟩

The LHS is the upper convected derivative A∇. Thus

A∇ =4kTζ

I− 4ζ

⟨QFc⟩+

1ζ

⟨(Fe

2 − Fe1)Q + Q(Fe

2 − Fe1)⟩

Note that at equilibrium (∇v = 0, Fej = 0) we get⟨

QFc⟩eq = kT I


Dumbbell’s Contribution to the Stress Tensor

Recall σ = σs + σp = −pI + 2ηsD + τ p.Three main contributions (Kramers):

1 An arbitrary plane in the suspension may be straddled by the twobeads and there will be a tension force transmitted along thespring:

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2 A similar contribution when external forces act on the straddleddumbbell.

3 The beads may cross an arbitrary plane and bring with themmomentum.


Intramolecular Potential Contribution σφp

Consider an arbitrary plane of area S and unit normal n.

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How many dumbbells with connector Q will be straddling the planewith bead 1 on the negative side (inside) and bead 2 on the positiveside? n × n ·QS × ψ(Q, t)dQ

n = number of dumbbells per unit volumen ·QS = volume in which bead 1 must be

ψ(Q, t)dQ = probability of finding the dumbbell within dQ of Q


Intramolecular Potential Contribution II

Bead 1 transmits a force Fφ1 = Fc on the fluid parcel. The contributionto the force per unit area of dumbbells of all orientations with bead 1on the negative side is:

n∫

n·Q>0n ·Qψ(Q, t)FcdQ

Similarly, the contribution to the force of the dumbbells of allconfigurations with bead 1 on the + side and bead 2 on the - side is

n∫

n·Q<0(−n ·Q)ψ(Q, t)(−Fc)dQ

Combining these two we get

n ·[n∫

QFcψ(Q, t)dQ]

= n · n⟨QFc⟩

i.e. the intramolecular contribution to the stress is σφp = n 〈QFc〉H. D. Ceniceros (UCSB) Complex Fluids 10/2010 19 / 28

Stress contribution from External Forces σep

Similarly, the force per unit area exerted on the fluid element due toexternal forces acting on the beads is

n∫

n·Q>0n ·Qψ(Q, t)(−Fe

2)dQ + n∫

n·Q<0(−n ·Q)ψ(Q, t)(−Fe

1)dQ

and this is equal to the negative of the force of the fluid element ontothe surrounding fluid due to the external forces

n∫

n·Q>0n ·Qψ(Q, t)(−Fe

1)dQ + n∫

n·Q<0(−n ·Q)ψ(Q, t)(−Fe

2)dQ

Thus, their difference divided by 2 gives us n · σep and

σep =

12

n⟨Q(Fe

1 − Fe2)⟩


Stress Contribution from Bead Motion σbp

Number of beads ”1” with velocity r1 which will cross surface S in ∆t :

n[S(r1 − v) · n∆t ]

The amount of momentum transported across the plane, into the fluidelement is

−n[S(r1 − v) · n∆t ]m(r1 − v)

Thus, the momentum flux coming from beads ”1” of all possiblevelocities and configurations is

−n · n∫

Jm(r1 − v)(r1 − v)Kψ(Q, t)dQ

Hence

σbp = −n

2∑j=1

∫Jm(rj − v)(rj − v)Kψ(Q, t)dQ


The Polymeric Stress, Kramers Formula

Putting together all the contributions we get:

σp = n⟨QFc⟩+

12

n⟨Q(Fe

1 − Fe2)⟩− nm

2∑j=1

⟨(rj − v)(rj − v)

⟩When a Maxwellian distribution is used σb

p simplifies to σbp = −2nkT I

i.e. each bead has an equal contribution of nkT I as a result of theequilibration in momentum space hypothesis.

Recall σ = σs + σp and evaluating this at equilibrium gives−pI = −psI− nkT I. We arrive at (σ = −pI + 2ηsD + τ p)

Kramers formula

τ p = n⟨QFc⟩+

12

n⟨Q(Fe

1 − Fe2)⟩− nkT I


Other Expressions for τ p

Note⟨QFc⟩ =

⟨(R2 − R1)Fc⟩ = −

⟨(R1Fφ1 + R2Fφ2)

⟩= −

2∑j=1

⟨RjF

φj

⟩12⟨Q(Fe

1 − Fe2)⟩

=12⟨R2(Fe

1 − Fe2)⟩− 1

2⟨R1(Fe

1 − Fe2)⟩

= −12

2∑j=1

⟨RjFe

j

⟩+

12⟨R1Fe

2⟩

+12⟨R2Fe

1⟩

= −2∑

j=1

⟨RjFe

j

⟩as R1 + R2 = 0. Thus,

σp = −n2∑

j=1

⟨Rj(F

φj + Fe

j )⟩− nm

2∑j=1

⟨(rj − v)(rj − v)


Modified-Kramers Formula

We get

Modified-Kramers formula

τ p = −n2∑

j=1

⟨Rj(F

φj + Fe

j )⟩− nkT I

Now if we use Fφj + Fej = −Fh

j − Fbj :

σp = n2∑

j=1

⟨Rj(Fh

j + Fbj )⟩− nm

2∑j=1

⟨(rj − v)(rj − v)

⟩Further manipulation on the term n

∑⟨RjFb

j

⟩, including integration by

parts and assuming the vel dist is even in rj − v we obtain

n∑⟨

RjFbj

⟩= nm

∑⟨Rj Rj


Kramers-Kirkwood Formula

Using the preceding expression for n∑⟨

RjFbj

⟩:

σp = n2∑

j=1

⟨RjFh

j

⟩− 2nm 〈(rc − v)(rc − v)〉

Assuming a Maxwellian velocity, evaluating at equilibriumσ = σs + σp, and recalling σ = −pI + 2ηsD + τ p, we get

Kramers-Kirkwood formula

τ p = n2∑

j=1

⟨RjFh

j

⟩


Giesekus Formula

Kramers formula reads

τ p = n⟨QFc⟩+

12

n⟨Q(Fe

1 − Fe2)⟩− nkT I (1)

On the other hand we know (A = 〈QQ〉)

A∇ =4kTζ

I− 4ζ

⟨QFc⟩+

1ζ

⟨(Fe

2 − Fe1)Q + Q(Fe

2 − Fe1)⟩

or ⟨QFc⟩ = kT I− 1

4ζA∇ +

14⟨(Fe

2 − Fe1)Q + Q(Fe

2 − Fe1)⟩

Substituting into (1) we obtain

Giesekus Formula

τ p = −14ζnA∇ +

14

n⟨(Fe

2 − Fe1)Q−Q(Fe

2 − Fe1)⟩


Hookean Dumbbells

Let’s neglect external forces for now. For Hookean springs, Fc = HQ:

Kramers: τ p = nH A− nkT I

Giesekus: τ p = −14ζn A∇

Note that becauseA∇ =

4kTζ

I− 4ζ

HA

we have a closed system and we don’t need ψ to evaluate A.

λHτ p + nkTλH I =14ζn A; λH = ζ/4H

−14ζn A∇ = −λHτ

∇p − nkTλH I∇ = −λHτ

∇p + 2nkTλHD

Oldroyd B Model

τ p + λHτ∇p = 2nkTλHD


Non-Linear Dumbbells

The (Warner) FENE spring:

Fc =HQ

1−Q2/Q20

Q ≤ Q0

Recall

A∇ =4kTζ

I− 4ζ

⟨QFc⟩+

1ζ

⟨(Fe

2 − Fe1)Q + Q(Fe

2 − Fe1)⟩

So we cannot longer obtain a closed system. We need to solve the FPequation for ψ and then calculate 〈QFc〉.Peterlin Approximation (FENE-P)

Fc =HQ

1− 〈Q2〉 /Q20

and now, simply⟨QFc⟩ =

H1− 〈Q2〉 /Q2

0〈QQ〉 =

H1− tr(A)/Q2

0A


Kinetic Theory for Elastic Dumbbells

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