Brownian dynamics

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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