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Brownian dynamics of polymersDumbbell and Rouse models
G. Marrucci
Universit di Napoli Federico II
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( )
++=
20
222
23
exp,,R
zyxC zyxP
RF 20
3RkT =
C k R
k C k R
zyxk Pk S ln
2
3ln
2
3ln 2
0
2
20
222
+=+++== R
x,y,z
R
RRF
=== ST E TSU E ;
Elastic force in a Gaussian chain
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The stress tensor TAssume temporarily that all chains have the same end-to-end
vector R
n
R
If is number of chain per unit volume,chains being cut per unit area are (Rn )
If F is the force in each spring, the totalforce across unit area is FRn . Hence:
T = FR
In reality, because of distribution, it is: T =
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Linear force: F =(3 kT / R o)R
( ) Rd 3
= RRRRR
RRT 20
3RkT =
where (R ) is the distribution function of end-to-end vector
At equilibrium, (R ) is isotropic and Gaussian.
Dynamic problem = find (R , t), from which < RR > as a function of time t, hence T (t)
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The simplest dynamics:Brownian particles with 1 degree of freedom
x
Distribution function of interest = particle concentration = c
Balance equation(number of particles is conserved)
x
N
t
c x
=
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Free particles (force free)
xc
DN x =
Flux is due to Brownian motion (diffusion) only. Hence:
=
xc
Dxt
c
where D = [m 2/s] is the diffusion coefficient.
Putting N x in the balance equation, the well known diffusion
equation is obtained:
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Particles in a force field
xxx
F ccN == v
Suppose now that the particles are in a field of force that movesthem, e.g., towards the right side
x
If v x is the particle velocity due to the force field, to the diffusionflux we must add the term:
where is the friction coefficient
of a particle
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Using Einstein equation linking to one another the diffusion
and friction coefficients, the c equation becomes:
+
=
kT
E
xc
x
c
xD
t
c
(where we have assumed that D does not vary with x).
Finaly assume that the Brownian particles are immersed in asolvent moving with velocity v x. Obviously, particles are alsoconvected, and the complete c equation is obtained:
( )xcxkT E
xc
xc
xD
t c
v
+
=
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Generalization to more degrees of freedom is straightforward.Balance becomes:
N=
t c
Accounting for all 3 possible contributions to the flux N , thegeneral equation becomes:
diffusion force convection
The 3D equation
( )vckT E
ccDt c
+=
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The simplest molecular model of
polymer viscoelasticity.The dumbbell model
R
Two friction beadsconnected by a spring
( )RkRRRR
+
=
kT E
Dt
diffusionD=kT /
spring convection due tovelocity gradient k
End-to-end vectordistribution function: (R , t )
Smoluchowski equation or Fokker-Planck equation
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RRT 20
3R
kT =
( ) Rd 3 = RRRRR
( )RkRRRR
+
=
kT E
Dt
Recall that for a linear force:
where:
Multiplying RR to all terms, and integrating over R space
T
RDD
dt d kRRRRkRRIRR ++= 2
0
62
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Constitutive Equation
IRR3
20R
eq=
220
6M
D
R=
T Rdt d
kRRRRkRRIRR ++= 3
20
Call relaxation time
Previous equation is rewritten as
showing exponential relaxation to equilibrium value
Equivalently, in terms of stress T
T kT dt d
kTTkTIT
++=
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Nonlinear terms can be linearized for slow flows ( k small)Since T eq = kT I , linearized equation becomes
( )T kT kT
dt
d kkTI
T ++=
At a steady state, this reduces to:
( ) termsisotropickT T ++= kkT
which identifies the viscosity as the quantity:
kT =
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Shear flows11
22
33
0
0
0 0
T
T
T
=
T
0 0
0 0 0
0 0 0
=
k22 0
0 0 0
0 0 0
T =
k T
i22
0 0
0 0
0 0 0
T T
=
T ki
22 33T T equilibrium value kT = = =
22d T kT dt = + = +
( )( ) 1 t t kT e = Start up of flow at a fixed shear rate:
( )t kT t kT G = = = t 0
t kT = =
Tangential component behaves like in linear viscoelasticity
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Conclusions on dumbbell modelpositive
Normal stresses in shearflows Elongational strain
hardening With a nonlinear force
law, predicts shearthinning
negative
Shear thinning should bepredicted also forGaussian chains
No prediction of maximain shear start up
In any event, polymersshow more than one
relaxation time More complex models are
needed
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Rouse model
Rouse model predicts a set of relaxation times, given by:
..3,2,1;2 == ppR
p where 2M RouseR =
Each p is the relaxation time of subchains of mass M/p
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Relaxation modulus G (t )
( ) ( )21
expRouse chain Rousep
G t kT p t =
=
Notice that all exponentials have the same coefficient (weight)
For t 0, G (t ) grows indefinitely (approaching the glassy response)
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Interpretation of the Rouse form forG(t)
Assume there are chains per unit volume. At time t = p, theunrelaxed subchains are p, and each contributes to the modulusa term of order kT. Hence:
( )21
21
=
== RRp t
kT kT kT pt G
This power law holds true as long as t remains smaller than R.For t > R, the modulus must decay exponentially. Hence:
( ) ( )RR
t t
kT t G
=
exp21
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Frequency response of Rousechains
logGlogG
logR-1
kT
G G
slope 1/2
slope 1slope 2
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Limits of Rouse model Rouse model is OK for unentangledpolymers
In dilute solutions, it requires Zimmscorrection to account for hydrodynamicinteractions
Nonlinear response is similar todumbbell
In melts or concentrated solutions, it isOK up to M=Mc Rouse model predicts that the viscosity
is proportional to M
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Zero-shear
viscosityof linearpolymers
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