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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

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