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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
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 2 / 28
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
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 3 / 28
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
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 4 / 28
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
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 5 / 28
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
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 6 / 28
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
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 7 / 28
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.
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 8 / 28
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.
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 9 / 28
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.
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 10 / 28
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 .
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 11 / 28
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)
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 12 / 28
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
)
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 13 / 28
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
⟩H. D. Ceniceros (UCSB) Complex Fluids 10/2010 15 / 28
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
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 16 / 28
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.
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 17 / 28
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
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 18 / 28
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)⟩
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 20 / 28
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
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 21 / 28
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
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 22 / 28
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)
⟩H. D. Ceniceros (UCSB) Complex Fluids 10/2010 23 / 28
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
⟩H. D. Ceniceros (UCSB) Complex Fluids 10/2010 24 / 28
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
⟩
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 25 / 28
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)⟩
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 26 / 28
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
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 27 / 28
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
H. D. Ceniceros (UCSB) Complex Fluids 10/2010 28 / 28