Bhattacharjee J.K., Bhattacharyya S. Non-Linear Dynamics Near and Far from Equilibrium
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9.1 Introduction 283
which is the time required for the centre of mass to diffuse through a distance
comparable to the size of the polymer.
To study the internal motion of the polymer chain, we consider the mean square
displacement of the n
th
segment, i.e.
d
2
=
[
R(n, t) −
R(n, 0)]
2
Using Eq.(9.1.23) we can write this as
d
2
=
X
0α
(t) −X
0α
(0) +2
∞
p=1
cos
np π
N
{X
pα
(t) −X
pα
(0)}
×
X
0α
(t) −X
0α
(0) +2
∞
q=1
cos
nqπ
N
{X
qα
(t) −X
qα
(0)}
=
(X
0α
(t) −X
0α
(0))(X
0α
(t) −X
0α
(0))
+4
∞
p=1
cos
2
nπp
N
(X
pα
(t) −X
pα
(0))(X
pα
(t) −X
pα
(0))
= 6D
CM
t +
4N
2
b
2
π
2
∞
p=1
cos
2
πpn
N
(1 −e
−p
2
t/τ
1
)
1
p
2
(9.1.37)
where we have written τ
p
as τ
1
/p
2
.Ift τ
1
(the relaxation time for rotational
motion), the centre of mass motion dominates. For t τ
1
, the internal modes dom-
inate and the second term in Eq.(9.1.37) can be significant. This can be evaluated by
replacing the sum by an integral and substituting the average value for cos
2
nπpN .
Since
cos
2
θ=1/2,
we have
d
2
=6D
CM
t +
2N
2
b
2
π
2
∞
p=1
cos
2
πpn
N
(1 −e
−p
2
t/τ
1
)
p
2
= 6D
CM
t +
2N
2
b
2
π
2
t
τ
1
∞
0
dy
1 −e
−y
2
y
2
(9.1.38)
This expression clearly shows the crossover from a linear in t behaviour to a
square root t behaviour on the time scale τ
1
. The above picture of the dynamics of
a polymer chain is called the Rouse Model. Its predictions are
• i) D
CM
∝N
−1
∝M
−1
• ii) τ ∝M
2
284 9 Polymers
where M is the mass of the polymer. The experimental measurements yield
D
CM
∝ M
−ν
τ ∝ M
3ν
(9.1.39)
In an ideal state ν =1/2 and in a good solvent ν 3/5. The discrepancy could
come from three different sources
• a) The response of a monomer to an applied force has been taken to be local.
However, the monomer will distort the velocity field all over the fluid and this
can affect a distant monomer. This is called the backflow effect
.
• b) Ideal chain elasticity, as expressed through the energy expression fails for a
good solvent.
• c) Rouse model ignores the fact that real chains do not cross. The repulsion
expressed in the second term of Eq.(9.1.38) is absent in the model just discussed.
We will include the correlation due to the backflow effect by looking at the velocity
V
n
of the bead at n due to a force
F
m
at m. We will write this response in the form
of a mobility matrix µ
mn
defined as
V
n
=
m
µ
nm
F
m
(9.1.40)
We picture the beads as a set of spheres in a fluid of viscosity η. The fluid velocity
v satisfies
η∇
2
v =
∇P (9.1.41)
where we have assumed a steady state situation and the velocity is low enough for
the nonlinear term in the Navier-Stokes equation to be dropped. It is an incom-
pressible flow with
∇.v =0
Since µ
nm
depends on the positions of all the spheres, calculating it is very dif-
ficult. If we make the simplifying assumption that the average distance between
neighbouring spheres is much larger than the radius of the sphere, we can make
a few simplifying approximations. Focussing on the n
th
sphere whose velocity is
V
n
, we can write the viscous drag on it as
−6πηa(
V
n
−
v
(
R
n
))
The velocity field
v
is created by all spheres except the n
th
one. Consequently,
this is a flow field under the action of forces
F
m
at all locations ‘m
other than ‘n
and we have
η∇
2
v
(r
n
) =
∇P −
m=n
δ(r −
R
m
)
F
m
(9.1.42)
with
∇.
v
=0 (9.1.43)
9.1 Introduction 285
In Fourier space
−ηk
2
v
α
(
k) =ik
α
P −
m=n
e
i
k.
R
m
F
mα
(9.1.44)
The divergence-free condition k
al
v
al
(
k) =0 leads to
k
2
P =−i
m=n
e
i
k.
R
m
F
mα
k
α
(9.1.45)
Inserting in Eq.(9.1.44)
ηk
2
v
α
(
k) =
β,m=n
e
i
k.
R
m
[δ
αβ
−
k
α
k
β
k
2
]F
mβ
(9.1.46)
From Eq.(9.1.41)
V
nα
−v
α
(
R
n
) =
F
nα
6πηa
or
V
nα
=
F
nα
6πηa
+
1
η
β,m=n
d
3
k
e
i
k.(
R
m
−
R
n
k
2
δ
αβ
−
k
α
k
β
k
2
=
F
nα
6πηa
+
β,m=n
T
αβ
(
R
n
−
R
m
)F
mβ
(9.1.47)
where Fourier- transforming yields
T
αβ
(r)=
1
8πηr
&
r
α
r
β
r
2
+δ
αβ
'
(9.1.48)
The mobility matrix µ
mn
is
µ
mn
=
1
6πηa
for m =n
= T(
R
m
−
R
n
) for m =n (9.1.49)
The Langevin equation of motion now becomes
d
dt
R
n,α
=−
m
µ
mn
∂E
∂R
mα
+g
nα
(9.1.50)
(There would have been an additional term but it does not show up because
m
∂µ
mn
∂R
m
=0).
286 9 Polymers
The correlation of g
n
is specified as
g
nα
=0
g
nα
(t
1
)g
mβ
(t
2
)=2(µ
mn
)
αβ
k
B
Tδ(t
1
−t
2
) (9.1.51)
Substituting for E, the equation of motion is
˙
R
nα
=k
m,β
(µ
mn
)
αβ
(R
m+1,β
+R
m−1,β
−2R
m,β
) +g
nα
(9.1.52)
Since µ
mn
depends on R
nα
, this is nonlinear and hence extremely complicated.
The simplification that is done is to use (µ
mn
)
αβ
which is defined as
(µ
mn
)
αβ
=
d
3
r
3
2π|n −m|b
2
3/2
exp
−
3
2
r
2
|n −m|b
2
×
1
8πη
1
r
r
α
r
β
r
2
+δ
αβ
=¯µ
mn
δ
αβ
(9.1.53)
where
¯µ
mn
=
1
ηb
1
6π
3
|m −n|
An analysis by decomposition into normal modes now leads to the more realistic
result
D
CM
∝M
−1/2
,τ∝M
3/2
(9.1.54)
References
1. P. De Gennes, Scaling Concepts in Polymer Physics, Cornell university Press (1979);
2nd ed. (1985).
2. P. De Gennes, Introduction to Polymer Dynamics C.U.P. (1990).
3. M. Doi and S. F. Edwards, The Theory of Polymer Dynamics ‘The International Series
of Monographs on Physics’ O.U.P. (1988).
Appendix
Functional Integration
The ultimate mathematical step that one has to perform in the problems that we
have discussed so far is an average. This is expressed in terms of the N-point
correlation function
G
N
(x
1
,x
2
....x
N
) =φ(x
1
)φ(x
2
)......φ(x
N
)
=
1
Z
D[φ]φ(x
1
)φ(x
2
)......φ(x
N
)e
−
Fd
D
x
. (A.0.1)
Here we have used a scalar field in defining G
N
for simplicity but extension to
vector fields is quite straightforward. In the above expression, the partition function
Z and the free energy F are
Z(h(x)) =
D[φ]e
−
F (h(x))d
D
x
(A.0.2)
F (h(x) =
1
2
(
∇φ)
2
+
1
2
m
2
φ
2
+
λ
4!
(φ
2
)
2
−h(x)φ (A.0.3)
the h →0 limits of the above expressions. It follows that
G
N
(x
1
,x
2
....x
N
) =
1
Z
δ
N
Z(h(x))
δh(x
1
)δh(x
2
).....δh(x
N
)
h=0
(A.0.4)
The integral (A.0.1) can be solved exactly only if λ =0. For λ =0 one can make
use of perturbation technique to achieve the desired accuracy for small λ, otherwise
one has to fall back on computers for numerical evaluation.
288 Appendix Functional Integration
Here we evaluate Z for λ =0.
Z =
D[φ]e
−
d
D
x [
1
2
(
∇φ)
2
+
1
2
m
2
φ
2
]
=
D[φ]e
−
d
D
xφ[−
1
2
∇
2
+m
2
]φ
=
D[φ]e
−
d
D
xφ B φ
(A.0.5)
where B represents the operator −
1
2
∇
2
+m
2
. Every integral, in principle can be
thought of as the limit of a sum. Hence to evaluate this integral we replace the
continuum by a lattice and at each point i of the lattice we replace φ(x) by φ
i
where the variable φ
i
can range from −∞ to ∞. Under this transformation
d
D
xφ B φ →
ij
φ
i
B
ij
φ
j
with the operational meaning of D[φ] being
D[φ]=
N
i=1
∞
−∞
dφ
i
Therefore the discretized version of Eq.(A.0.5) appears as
Z =
N
i=1
∞
−∞
dφ
i
e
−
jk
φ
j
B
jk
φ
k
(A.0.6)
where the matrix B
jk
is symmetric. Here, antisymmetry would imply vanishing of
the sum. This allows for diagonalization by an orthogonal transformation. From
now on we will use Einstein’s summation convention to imply that repeated indices
are summed over. We consider an orthogonal transformation S that diagonalizes
B such that
B
jk
=S
jm
D
mn
S
T
nk
(A.0.7)
where D is the diagonal matrix
D
mn
=D
mn
δ
mn
.
Starting from
φ
j
B
jk
φ
k
= φ
j
S
jm
D
mn
S
T
nk
φ
k
= ψ
m
D
mn
φ
n
(A.0.8)
Appendix Functional Integration 289
where
φ
j
S
jm
= ψ
m
and φ
m
= S
mn
ψ
n
(A.0.9)
Using the fact that the Jacobian of the transformation is unity, i.e. Det(S) =1, we
have
k
dφ
k
=
k
dψ
k
. (A.0.10)
Taking into account that S is orthogonal, i.e. SS
T
=1wehave
Z =
N
i=1
∞
−∞
dψ e
−
1
2
D
jj
ψ
j
2
=[
∞
−∞
dψ e
−
1
2
D
jj
ψ
j
2
]
N
=
2π
N/2
1
Det(D)
1/2
=
2π
N/2
1
Det(B)
1/2
(A.0.11)
It is apparent from the given structure of F that
φ(x
1
)φ(x
2
)......φ(x
2m+1
)=0 (A.0.12)
in the limit of h →0.
For the two point function we have
N
i=1
∞
−∞
dφ
i
φ
p
φ
q
e
−
1
2
φ
j
B
jk
φ
k
=
N
i=1
∞
−∞
dφ
i
φ
q
B
−1
pr
δ
δφ
r
e
−
1
2
φ
j
B
jk
φ
k
=
N
i=1
∞
−∞
dφ
i
B
−1
pr
δ
rq
e
−
1
2
φ
j
B
jk
φ
k
=B
−1
pr
2π
N/2
1
Det(B)
1/2
(A.0.13)
where in the penultimate step we have integrated by parts. Hence, we can readily
write (by induction)
N
i=1
∞
−∞
dφ
i
φ
i
1
φ
i
2
......φ
i
2n+1
e
−
1
2
φ
j
B
jk
φ
k
=0 (A.0.14)
290 Appendix Functional Integration
and
N
i=1
∞
−∞
dφ
i
φ
i
1
φ
i
2
......φ
i
2n
e
−
1
2
φ
j
B
jk
φ
k
=
2π
N/2
(DetB)
−1/2
B
−1
i
1
,i
2
B
−1
i
3
,i
4
....B
−1
i
2n−1
,i
2n
+all possible pairings
(A.0.15)
Thus,
φ
i
1
φ
i
2
......φ
i
2n
=B
−1
i
1
,i
2
B
−1
i
3
,i
4
....B
−1
i
2n−1
,i
2n
+all possible pairings (A.0.16)
This is the Wick’s theorem in the parlance of functional integration.
Appendix.1 Gaussian Theory
In this theory λ =0. To calculate the correlation function, we go to momentum
space which facilitates calculation. We define Fourier transform pair as
φ(x)=
d
D
p
(2π)
D
˜
φ(p) e
−i p.x
˜
φ(p) =
d
D
xφ(x)e
i p.x
(A.1.1)
which allows us to write
d
D
x φ(x) [−∇
2
+m
2
]φ(x) =
d
D
x
d
D
p
(2π)
D
˜
φ(p) e
−i p.x
[−∇
2
+m
2
]
×
d
D
q
(2π)
D
˜
φ(q) e
−i q.x
=
d
D
p
(2π)
D
˜
φ(p) [p
2
+m
2
]
˜
φ(−p) (A.1.2)
The operator B is thus diagonal in momentum space and hence B
−1
=(p
2
+m
2
)
−1
.
The Green’s function G
0
(p) of the Gaussian theory is thus
G
0
(p) =
1
p
2
+m
2
(A.1.3)
Higher order Green’s functions can also be found from the Eq.(A.0.16).
We now try to find the susceptibility for this model which follows from the
fluctuation-dissipation theorem and is given by
Appendix.1 Gaussian Theory 291
χ =G
0
(p =0) =(m
2
)
−1
∝(T −T
c
)
−1
(A.1.4)
Thus γ =1. At T =T
c
G
0
(p) ∝
1
p
2
and the Fourier transform has a conspicuous scaling behaviour
G
0
(x) ∝
1
|x|
D−2
.
Therefore, η =0. The Fourier transform at non-zero m
2
is
G(x) =
d
D
p
(2π)
D
e
−i p.x
p
2
+m
2
. (A.1.5)
This integral is easily done for D =3 but is nontrivial in other dimensions. But the
correlations decay exponentially with a scale m
−1
irrespective of the dimension-
ality D. Hence one gets as the correlation length
ξ =m
−1
∝(T −T
c
)
−1/2
(A.1.6)
resulting in ν =1/2. The specific heat exponent alpha is found from
C ∝
d
D
xd
D
yφ
2
(x)φ
2
(y)
=
d
D
xd
D
y
d
D
p
(2π)
D
˜
φ(p)e
−i p.x
d
D
q
(2π)
D
˜
φ(q)e
−i q.x
×
d
D
r
(2π)
D
˜
φ(r)e
−ir.y
d
D
s
(2π)
D
˜
φ(s)e
−is.y
c
=
d
D
p
(2π)
D
d
D
q
(2π)
D
˜
φ(p)
˜
φ(−p)
˜
φ(q)
˜
φ(−q)
c
=
d
D
p
(2π)
D
G
0
(p)
2
=
d
D
p
(2π)
D
[p
2
+m
2
]
−2
∝m
D−4
∝(T −T
c
)
−
4−D
2
(A.1.7)
Hence α =(4 −D)/2. The spontaneous magnetization index β cannot be obtained
within the Gaussian framework as the theory becomes unstable in the broken
symmetry state. This is a first step towards resolving the complicated issue of
critical exponents and universality. This theory does not yield correct exponents
as the interaction term is missing in the free energy though it satisfies the scaling
laws.
Appendix
Field Theoretic RG
We briefly sketch the basic ideas of Field Theoretic RG for a scalar field. The Free
energy density can be written as
F =
r
0
2
φ
2
+
1
2
(
∇φ)
2
+
u
0
4!
φ
4
. (B.0.1)
We focus on the two point correlation function
(2)
(k) at the critical point. The
crucial point is that we must have a finite
(2)
(k) as the cut-off on the momentum
space variable goes to infinity (or equivalently the lattice spacing tending to zero
in real space). The behaviour of
(2)
(k) at the critical point is
(2)
(k) ∝k
2−η
for k . Dimensional analysis suggests that
(2)
(k) should have the form
(2)
(k) =k
2
k
−η
1 +B
k
ω
1
+........
(B.0.2)
Na
¨
ıve power counting yields k
2
as the dimension of
(2)
(k) which is independent
of the dimensionality of space. Since we are interested in the long-wavelength
behaviour of
(2)
(k), we would obviously like to take the limit k →0. But that
also involves the limit →∞and therein lies the problem since it renders the
leading term meaningless. To overcome the difficulty we introduce an arbitrary
length scale µ in the problem and write
µ
η
(2)
(k) =k
2
k
µ
−η
1 +B
k
ω
1
+........
(B.0.3)