Chapter 6 Reptation - Laboratory for Aero & HydroDynamics

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Chapter 6
Reptation
6.1 Introduction
Reptation theory describes the rheological behaviour of linear polymers in highly mu-
tually entangled conformations. Entanglements are created only when the density of
polymers is high enough. This means that the polymeric liquid we will be consider-
ing here is a melt or a highly concentrated solution. A second important condition for
entanglements to arise is a large molecular weight.
The notion of entanglements (cid:12)rst arose in the molecular theory of rubber elasticity.
The linear response to deformations of an elastic solid is characterised by the elastic
modulus G , (cid:28) = G γ. This modulus is proportional to kT times the number of
0 xy 0
degrees of freedom per unit volume that do not relax subsequent to the deformation.
Early molecular-based theories of rubber elasticity describe the rubber as composed
of ghost-chains. A strand between crosslinks is modelled to have the same amount of
con(cid:12)guration space available, as if no neighbouring chains are present. Because ghost-
chains are not restricted when sampling phase space, the number of non-relaxed degrees
of freedom equals the number of crosslinks per unit volume. The relaxation modulus is
thus proportional to the number of crosslinks.
For systems where the polymer strands between crosslinks are long, the elastic mod-
ulus calculated in this way is much smaller than experimentally observed values. The
reason for this is the entanglement of chains with neighbouring chains. This entangled
state makes it impossible for a strand (between crosslinks) to sample as large a piece of
phase space as a polymer without neighbours. The entanglements give rise to topolog-
ical constraints. The neighbours do not only hinder the movement of the strand, they
actually make it impossible for the chain to sample as large a space as a ghost-chain.
This results in a reduction of the entropy and consequently the elastic modulus is in-
fluenced. For a highly entangled state the number of degrees of freedom that can not
relax to equilibrium, after a deformation is applied, is proportional to the number of
entanglements. Thus
c
G = kT ; (6.1)
0
N
e
175
176 CHAPTER 6. REPTATION
Figure 6.1: The black polymer is constrained in its sidewise motion because of entanglements with
neighbouring chains. A tube is used to model the space the polymer can sample without crossing the
(semi) topological constraints. The picture is somewhat misleading because the tube cross-sectionwill
be much wider than the one depicted here.
where c is the number of monomers per unit volume and N can be interpreted as the
e
number of monomers in a strand between two entanglement points (in fact Eq. (6.1) is
the experimental de(cid:12)nition of N ).
e
In this setting of rubber elasticity Edwards [1] (cid:12)rst introduced the tube-picture. A
strand between crosslinks is modelled as if it is surrounded by a tube. The tube denotes
the fraction of con(cid:12)guration space the strand can sample. It is characterised by one typ-
ical length scale a, often called the tube-diameter (for the reason for this see appendix
6.B). It is the typical distance between entanglements. The length scale not only charac-
terises the diameter of the tube but also a step size. Because entanglements are thought
to be distributed randomly the tube is taken to have a random walk con(cid:12)guration with
step size a.
The idea of the tube was carried over to the (cid:12)eld of polymer melts by De Gennes
[2]. Now polymers are not connected by crosslinks anymore (at least not permanently).
Entanglements no longer constitute permanent constraints. Therefore the tube is no
longer permanent. Sidewise motion is constrained but motion along the contour of the
tube is not constrained. Subsequent to a deformation the stretch of polymer chains will
relax quickly. Changes in orientation are much slower. The relaxation of orientation can
onlyproceedviadi(cid:11)usionofthechainoutofthetube. Thismotionwastermedreptation
by De Gennes. When the chain reptates out at one side a new tube segment is created
at this location (illustrated by Fig. 6.3). At the other side a part of the tube disappears.
Based on this picture Doi and Edwards developed their constitutive equation [3, 4, 5]
(discussed in x6.2).
When increasing molecular weight (i.e. increasing chain length) a polymer melt will
make the transition from an unentangled state to an entangled state. After this tran-
6.1. INTRODUCTION 177
sition there will be a split in time scales. The time scale corresponding to stretch
relaxation will be comparable to the relaxation of an unentangled polymer. The time
scale corresponding to relaxation of orientation, the reptation or disengagement time (cid:28) ,
d
is much larger. According to the (pure) reptation picture it scales asM3. For time scales
somewhere between the stretch time and the reptation time not much happens. In this
range of time scales the response is characterised by a constant value of the relaxation
modulus, the rubber plateau modulus, which scales as Eq. (6.1). This modulus does not
depend on the molecular weight (G / M0) only on entanglement density.
0
One of the most famous graphs supporting reptation theory is Fig. 6.2. It clearly
illustrates the transition in the dynamics when increasing the molecular weight. In
the unentangled state the viscosity is proportional to M and in the entangled regime
it is proportional to M3:4. Reptation predicts M3 (/ G (cid:28) ), which is as close to the
0 d
experimental value as one might expect of such a simple theory. On the other hand the
deviation indicates that there is more to melt rheology than reptation only. Doi (cid:12)rst
identi(cid:12)ed the 0:4 deviation in the exponent as a possible end-e(cid:11)ect [7]. Fluctuations of
the chain length near the ends of the tube cause extra rapid relaxation of the orientation
there. Doi’s treatment of tube length fluctuations gave however a too narrow range of
molecular weights where the 3:4 exponent is valid. In the description of Milner and
McLeish [8] this shortcoming is no longer present.
The Doi-Edwards equation makes a number of predictions that are at least quali-
tatively correct. These results have convinced the majority of researchers in the (cid:12)eld
that the tube-picture is the most promising road toward a quantitative description of
rheology of entangled polymer systems. However, the Doi-Edwards model also makes
predictions that are completely o(cid:11). One important example is the excessive shear thin-
ning that is predicted. The model predicts that for high shear rates the stress decreases
with increasing shear rates. This is in fact not found in experiments. Although the
viscosity of melts is very shear thinning the stress itself always increases with increasing
shear rates. These, and other mispredictions, are now thought to be caused by the lack
of self-consistency of the Doi-Edwards model. The surrounding chains also relax causing
a certain amount of relaxation of the tube constraint.
Doi and Edwards assumed that the lack of self-consistency only gives minor errors.
This turned out not to be true. In branched polymers the e(cid:11)ect even gives rise to the
dominant relaxation mechanism: dynamic dilution. For polydisperse systems relaxation
of neighbouring polymers gives rise to the double reptation mechanism [9, 10]. But
also for monodisperse linear polymers the relaxation of neighbouring polymers is im-
portant, especially at high shear rates. The dominant mechanism by which constraints
are released at high shear rates was named convective constraint release by Marrucci
[11]. It counteracts the strong alignment which gives rise to the excessive shear thinning
predicted by the Doi-Edwards equation.
The last few years there has been signi(cid:12)cant progress in the development of con-
stitutive relations for entangled polymeric systems. The (cid:12)rst major step forward is the
identi(cid:12)cation of (probably) all important relaxation mechanisms. The last of these is the
convective constraint mechanism proposed by Marrucci. The second step forward is the
178 CHAPTER 6. REPTATION
3.4
M
1.0
M
0 polydi-methyl
h siloxane
g
o
l

+
polyiso-buthylene

t
n
a
t
s
n
o
c polyethylene
polybutadiene
polytetra-methyl
p-silphenylene
siloxane
polymethyl
methacrylate
polyethylene
glycol
polyvinyl acetate
polystyrene
-1 0 1 2 3 4 5
constant + log M
Figure6.2: Afamouspictureinreptationtheory(from[6]). Foralargerangeofdi(cid:11)erentlinearpolymer
melts the dependence of viscosity onmolecular weightshowssimilar scaling. For smallM the viscosity
increaseslinearlywith M. For largeM the viscosityscales asM3:4. This changein dynamics is caused
by the occurrence of entanglements when M becomes large enough. Reptation theory tries to describe
the dynamics of highly entangled polymer melts.
6.2. THE DOI-EDWARDS MODEL 179
increasing quantitative accuracy of the predictions. This is especially the case for linear
rheological behaviour. Milner and McLeish have made impressively accurate predictions
for the linear rheology of star-polymer systems [12, 13], and also for linear polymers [8].
A third development has to do with the applicability of constitutive equations. Only
recently the equations can be incorporated into macroscopic flow simulations. The con-
(cid:12)guration (cid:12)elds method [14] has made it possible to incorporate kinetic theory based
models into macroscopic simulations.
For most models of entangled systems the deformation (cid:12)elds method, which will be
treated in chapter 7, is more appropriate. The deformation (cid:12)elds method is a method to
incorporate so called integral constitutive equations into flow calculations. These models
construct the stress in a fluid element by performing an integral over its deformation
history. The Doi-Edwards equation is such a model.
In this chapter we will start with discussing the ingredients needed to describe a
entangled polymeric system. The (cid:12)nal goal is to develop a constitutive equation that
incorporates all important e(cid:11)ects and that can also be implemented into a deformation
(cid:12)elds code. This last point is essential for numerical simulations of complex flow (cid:12)elds.
We believe that segment connectivity is a very important property that has to be
modelled correctly. It is well known that in the double step strain experiment the
Doi-Edwards equation is not valid. This is caused by the lack of connectivity in this
model (i.e. the independent alignment approximation). In real flows the deformation
history of a fluid element is so complicated that e(cid:11)ects such as double step strain will be
important. Furthermore important relaxation e(cid:11)ects that occur at endpoints, such as
the so-called contour length fluctuations, are currently only modelled well in the linear
rheological regime, see [8]. In our description these e(cid:11)ects are also modelled accurately
in the nonlinear regime.
The demand of incorporating chain connectivity forces a special form upon the for-
malism. Especiallythedescriptionofthereptativemotioniscastinasomewhatunfamil-
iar form (but is still highly accurate). Although we will gradually develop a constitutive
equation when the chapter progresses, we do not consider this as the (cid:12)nal equation. We
note that, even in the current state of development of reptation theory, there are still
fundamental problems with the description of the so-called constraint release mecha-
nism in the nonlinear regime. We will discuss these problems in appendix 6.A. Not the
equation itself is important, but the methodology. The developed methodology can be
used for a larger class of systems, not only monodisperse linear polymers but also e.g.
polydisperse branched polymers.
6.2 The Doi-Edwards model
The Doi-Edwards constitutive equation is based on the tube-picture, making the most
simple assumptions.
In this section we will give a detailed derivation of the Doi-Edwards. The reason we
do this is because our formalism is aimed at repairing a flaw that is present in the Doi-
Edwards equation (which has to do with chain retraction). Besides repairing the flaw,
180 CHAPTER 6. REPTATION
we will add some relaxation mechanisms which are believed to be important but are not
included in the Doi-Edwards equation. During the derivation it will become clear what
approximations are made and which e(cid:11)ects are not included. In x6.3 the basic formalism
will be presented. In the next section x6.4 it is shown how this formalism can be used
to repair the flaw present in the Doi-Edwards equation.
If one makes a snapshot of a polymer in an entangled melt in equilibrium one sees
no evidence for a characteristic length scale a (the tube diameter). Because a chain is
surrounded by chains with identical chemical properties (i.e. in a melt), the equilibrium
statistics are the same as for a chain submerged in a theta solvent. This means that,
when considering long enough length scales, a chain can be modelled as a random walk,
characterised by a Kuhn step b. The root mean square of the end-to-end distance is
hR2i = b2N; (6.2)
where N is the number of Kuhn steps.
In most practically relevant situations the Kuhn step b is much smaller than a. This
means that a tube segment of length a contains many Kuhn steps. Strands of chains
inside a tube segment will therefore never be far from equilibrium. For this reason the
chainstrandscanbedescribed aslinear entropicsprings. Becauseofthedense packing of
polymer chains the movement of a segment does not create a flow (cid:12)eld. Hydrodynamic
interaction is screened. The dynamics inside the tube is therefore characterised by a
simple Rouse spectrum (and not a Zimm spectrum).
The Rouse time (cid:28) , i.e. the largest time of the Rouse spectrum, corresponds roughly
R
to the equilibration time of polymers with similar molecular weights but in an unentan-
gled state. In an entangled state nothing much happens for times in between the Rouse
time and a reptation time (cid:28) . The chain inside the tube is relaxed, but a correlation of
d
typical length a in the conformation persists. In this range the tube becomes observable.
The extra tension created by the prevention of the relaxation of this degree of freedom
is
3kT
F = (6.3)
a
(this formula can be taken as the de(cid:12)nition of a). The magnitude of the tension is easily
understood. It takes O(kT) of energy, (cid:25) F(cid:1)a, to keep a degree of freedom from relaxing.
The tension has a thermodynamic/entropic origin. It decreases the number density
of monomers within a tube segment. Therefore the entropy per monomer increases,
counteracting the decrease in entropy caused by the tube constraint. The total entropy
inside the tube will become equal to the entropy of a monomer outside the tube (i.e. at
the chain ends). The exact calculation of the ‘chain-in-a-tube’-statistics is performed in
[15].
The length scale a can be used to de(cid:12)ne a new random walk with step size a. This
random walk is called the primitive chain (or also primitive path). Its contour length
L (measured with a ruler of length a) can be obtained by considering the end-to-end
distance
2 2
b N = a (L=a) = aL (6.4)
6.2. THE DOI-EDWARDS MODEL 181
Figure 6.3: The (pure) reptation picture of a step strain. When applying a rapid deformation both
the primitive chain and the surrounding tube deform a(cid:14)nely. In a Rouse time he chain retracts to its
equilibrium length. On longer time scales the chain escapes the con(cid:12)ning tube by means of reptative
motion. Now the residual stress relaxes.
The number of tube segment of length a per chain is L=a = (b2=a2)N. For realistic melts
this number ranges somewhere between 20 and 200. The number of Kuhn steps within
one tube segment is a2=b2. The equilibrium ‘Kuhn-step-density’ per unit tube length is
thus
2
n = a=b : (6.5)
The main thought experiment in developing the Doi-Edwards equation is that of
the single step strain deformation and subsequent relaxation. The experiment has three
stages. First the deformation, then stretch relaxation on a Rouse time scale followed by
orientational relaxation by reptation (see Fig. 6.3).
For the deformation step one assumes that the step is performed so quickly that
no relaxation occurs on this time scale. The deformation is therefore a(cid:14)ne (for long
enough length scales such as a). Furthermore it is assumed that the deformation is not
that large that (cid:12)nite extensibility e(cid:11)ects of the chain become important (i.e. the Hencky
strain (cid:15) (cid:28) log(N)). In this limit the Rouse description stays valid for the chain inside
the tube.
Not only the chain one is considering, but also the surrounding chains (constituting
the tube) deform a(cid:14)nely. This results in the observation that the tube deforms a(cid:14)nely
as well. The way this is described is by posing that individual tube segments deform
a(cid:14)nely. This last statement contains a rather strong assumption, which is not made
explicitly in the Doi-Edwards paper [3].
In the Doi-Edwards papers a switching between two mental pictures is used. The
(cid:12)rst picture is that of the tube. The other is that of slip links. In the second picture
entanglements are modelled as slip links through which the chain can move without
182 CHAPTER 6. REPTATION
friction. Duringthedeformationstepthesliplinksaredisplaced a(cid:14)nely. These sliplinks
now constitute the new endpoints of the chain segments. The important question that is
not raised in the papers is whether the entanglements before and after the deformation
are caused by the same neighbouring chains. (If this is the case the slip link picture is
not a good analogue).
After a(cid:14)ne deformation of both the tube and the chain inside this tube, the chain is
no longer in thermodynamic equilibrium. Doi and Edwards assume that the equilibrium
tension in the deformed tube equals the equilibrium tension. Their hand waving argu-
ment for this is that, for a stretched tube segment, the number of entanglements per
unit length is decreased, but that in the perpendicular direction neighbouring polymers
are squeezed inward. According to them it is reasonable to assume the net e(cid:11)ect is
zero. The entropically generated tension in the deformed state equals the equilibrium
tension, Eq. (6.3). The chain retracts until the entropic spring forces balance the tension
Eq. (6.3). This process occurs in the order of a Rouse time. The net e(cid:11)ect is that the
number density of monomers in a tube segment relaxes back to the equilibrium value.
Within the Doi-Edwards picture a deformation step followed by retraction can be
put into mathematical terms with relative ease. Let us start with an equilibrium chain
segment characterised by its end-to-end vector Q , which connects two ‘entanglement-
0
(cid:22)
points’. This vector is a random, isotropically (presumably Gaussian) distributed vector
with an average length a. After a(cid:14)ne deformation this segment becomes
Q = F (cid:1)Q ; (6.6)
0
(cid:22) (cid:22)(cid:22) (cid:22)
where F is the deformation-gradient tensor.
(cid:22)(cid:22)
After the relaxation of the stretch, which occurs in a few Rouse times, the chain in
the segments have equilibrium tension (and equilibrium monomer density). The stress
contribution of a single segment is readily identi(cid:12)ed as (tension time direction times the
connector vector)
3kT QQ
FQ = (cid:22) (cid:22): (6.7)
(cid:22) (cid:22) a jQj
(cid:22)
The number of Kuhn steps in one segment is (a=b2)jQj, using Eq. (6.5). Let c be the
(cid:22)
densityofKuhnstepsperunitvolumeinthemelt. (Thisisofcoursedirectlyproportional
to the number of monomers per unit volume). Then the mean number of segments per
unit volume becomes
cb2 1
: (6.8)
a hjQji
(cid:22)
The stress tensor is given by (something like nhFQi):
(cid:28) (cid:29) (cid:22) (cid:22)
b2 QQ b2
(cid:28)(cid:22)(cid:22)p = 3 a2 ckT j(cid:22)Q(cid:22)j =hjQ(cid:22)ji = 3 a2 ckTQ(cid:22)(cid:22): (6.9)
(cid:22)
Here the tensor Q is the so called Q-tensor. This tensor denotes the mean orientation.
The average deno(cid:22)(cid:22)tes anaverage over(cid:22)(cid:22)all possible initial values Q , which areisotropically
0
(cid:22)
6.2. THE DOI-EDWARDS MODEL 183
distributed. Because the length of Q drops out of the expression one can use a vector
0
(cid:22)
uniformly distributed on a random sphere. Linearising the stress expression for small
shear strains γ, gives (cid:28) = 4=5(b2=a2)ckTγ, thus
xy
4b2
G = ckT: (6.10)
0 5a2
Comparing this expression with Eq. (6.1) gives N = 5a2=4b2, with N the number of
e e
Kuhn steps instead of monomers.
Immediately after the fast relaxation of the stretch all surviving segments contribute
to the stress. Thereafter newly created segments do not contribute to the stress, because
these do not have a preferred direction. To be able to calculate the stress one has to
know how many of the original tube segments are still present at a later time.
At every moment in time the position of tube segment along the total tube contour
(with length L) can be characterised by a variable s. The left tube-end is denoted by
s = 0, the middle by s = 1. For time steps larger than the Rouse time, the length of the
primitive chain can be taken to be constant. The only relevant motion is centre-of-mass
di(cid:11)usion in the tube. For a speci(cid:12)c tube segment the change of s is therefore described
by the stochastic di(cid:11)erential equation
p
2D
c
dS = dW; (6.11)
L=2
where D is the curvilinear centre-of-mass di(cid:11)usion tensor. We here use the standard
c
convention to indicate stochastic processes with an uppercase, here S. (Note that the
dimensionful position is LS=2.) According to the fluctuation-dissipation theorem, or
more precisely the Einstein relation, the di(cid:11)usion tensor is proportional to kT divided
by a friction coe(cid:14)cient. This friction coe(cid:14)cient has a linear dependence on the number
of monomers in the chain, thus D / 1=L / M−1. For all segments of one tube this
c
equationEq.(6.11) holdswith thesameWiener process, onlytheinitialvalueofS di(cid:11)ers.
If S = 0 (or S = 2 which is the other segment end) the segment is annihilated and a new
one is created. The probability that at a position s a segment is present that already
existed at t = 0 obeys the di(cid:11)usion equation
@ 4D @2
c
p(s;t) = p(s;t): (6.12)
@t L2 @s2
with the initial condition p(s;0) = 1 and the boundary condition for the tube end is
p(0;t) = 0. The problem is clearly symmetric around s = 1. Therefore the problem
only has to be considered on s 2 [0;1], with boundary condition @ p(1;t) = 0. This
@s
di(cid:11)usion propblem can easily be solved by a Fourier decomposition using base functions
of the form 2sin[((cid:25)=2+p(cid:25))s],
(cid:20) (cid:21)
X1 4 (cid:25)2D
p(s;t) = sin[((cid:25)=2+p(cid:25))s]exp − c(1+2p)2t : (6.13)
(cid:25)(1+2p) L2
p=0
184 CHAPTER 6. REPTATION
1100 33
10 0
a]a]
PP
) [) [ 1100 22
gg) G(t,) G(t, GGDDEE((tt)) gh() 10 -1
11gg((
--hh 1100 11
tt tt
RR dd 10 -2
1100 00 1100 11 1100 22 1100 33 10 -1 10 0 10 1
tt [[sseecc]] g
Figure 6.4: Left: relaxation after a shear step, for di(cid:11)erent strains γ. The shear stress is plotted as
function of time (experimental data from [16]). The stress is scaled by an amount h(γ)γ. For times
larger than (cid:28)R a perfect superposition is observed. The damping function h(γ) is plotted in the right
graph(thepointsarethe experiments). ThefunctioncomparesfavourablewiththeDoi-Edwardsresult
(solid line), obtained from Eq. (6.9). The so-called independent alignment(dashed line) compares even
better.
The probability that a segment present at t = 0 is still present at t is given by
Z (cid:20) (cid:21)
1 X1 8 (cid:25)2D
P(t) = p(s;t)ds = exp − c(1+2p)2t (6.14)
(cid:25)2(1+2p)2 L2
0
p=0
The largest relaxation time is given by
L2
(cid:28) = : (6.15)
d (cid:25)2D
c
Taking the scaling of the curvilinear di(cid:11)usion coe(cid:14)cient with the number of monomers
into account, one (cid:12)nds that (cid:28) / L3 / M3.
d
After a step strain and the subsequent stretch relaxation, up to t = O((cid:28) ), the stress
R
relaxes further via
15
(cid:28) (t) = G QP(t): (6.16)
p 0
(cid:22)(cid:22) 4 (cid:22)(cid:22)
This is the stress contributions of the segments times the probability a segment is still
present after a time t. This equation shows a time-strain separability of the response
subsequent to a step strain.
Fig. 6.4 shows experimental results [16] for stress relaxation after a step strain. This
picture (and especially the right one) is one of the major reasons for the popularity of
the tube-picture. The left graph illustrates time strain separability. For times t > (cid:28)
R
after a step shear strain γ is applied, the shear stress can be written as
(cid:28) (t) = G(t)h(γ)γ: (6.17)
xy