Edited by Chris H.
Rycroft
∗
December 5, 2006
1 Restoring force for highly stretched polymers
1.1
A globally valid asymptotic approximation
Using the substitution t = ua the expression can be rewritten as
P
N
( r ) =
=
2 π
2 π
1
2
2
1 r a 2
� r
∞
0
� t a sin( rt/a )
� sin t t �
N
0
∞ t sin( rt/a )
� sin t �
N t dt a dt
The integrand is even, and can be rewritten as
P
N
( r ) =
=
=
=
4 π 2
1 a 2
1 r
8 iπ 2 a 2 r
�
∞
−∞ t sin( rt/a )
� sin( t ) t
�
N dt
�
∞
� u e irt/a
− e
− irt/a
�
� sin( t ) �
N t
4 iπ
1
2
1 a 2 r
�
−∞
∞ te irt/a
� sin( t )
−∞ t
�
∞
�
N dt
4 iπ 2 a 2 r
−∞ t exp [ − N f ( t )] dt dt where f ( t ) = − irt
N a
− log
� sin t �
.
t
We define ξ = r/N a and keep it fixed.
The first and second derivatives are f f
��
�
( t ) = − iξ − cot t +
( t ) = 1 + cot
2 t −
1 t
1 t 2
.
∗
Solution to problem 1 based on sections of Random Walks and Random Environments by Barry Hughes.
Solutions
2 and 3 based on previous years.
1

M.
Z.
Bazant – 18.366
Random Walks and Diffusion – Problem Set 5 Solutions 2
By putting f
�
( t ) = 0, we find that there is a saddle point at t
0
= iL
− 1
( ξ ) where L is the Langevin function L ( ξ ) = coth ξ − 1 /ξ .
Over the range 0 < ξ < 1 the Langevin function is monotonic and thus has a well-defined inverse.
We see that the second derivative of f can be written as f
��
( t ) = 1 − coth
2
1
( − it ) +
( − it ) 2
� 1
= 1 − coth( − it ) −
( − it )
�
2
−
2 coth( − it )
( − it )
= 1 − ξ
2 −
2 ξ
L
− 1 ( ξ )
.
2
+
( − it ) 2
This function is positive, so we deform the contour of integration to the line Im( t ) = L
− 1
( ξ ).
Our main contribution comes from the above saddle point and we obtain
�
P
N
( r ) ∼
L
− 1
( ξ )
4 π 2 ra 2 N (1 − ξ 2
2 π
− 2 ξ/L − 1 ( ξ )) e
− N ξL
− 1
( ξ )
� sinh( L
− 1
L − 1 ( ξ )
( ξ )) �
N
.
∼
∼
ξ
�
L
− 1
( ξ )
8 π 3 N 3 a 6 (1 − ξ 2 − 2 ξ/L − 1 ( ξ )) e
− N ξL
− 1
( ξ )
� sinh( L
− 1
( ξ )) �
N
L − 1 ( ξ )
.
r
�
8 π 3 N a 4
L
− 1 ( r/aN )
(1 − ( r/aN ) 2 − 2 r/aN L − 1 ( r/aN )) e
− rL
− 1
( r/aN ) /a
� sinh( L
− 1 ( r/aN )) �
N
L
− 1 ( r/aN )
.
1.2
Free energy
The free energy is given by
F = T S
= − T log P
N
( r )
∼
T
2
− T log � 8 π log L
3
− 1
N a
(
4
(1 r/aN
− (
) + r/aN
T log
) r
2 − 2 r/aN L
− 1
( r/aN )) � + rT L
− 1
( r/N a ) a
− T N log
� sinh( L
L − 1
− 1
( r/N a )) �
.
( r/N a )
The restoring force is given by f = − dF dr
= −
T ( − 2 r/a
2
N
2 − 2 /aN L
− 1
( r/N a ) + 2 rM ( r/N a ) / ( aN L
− 1
( r/N a ))
2
)
2(1 − ( r/aN ) 2 − 2 r/aN L − 1 ( r/aN ))
−
T L
− 1
( r/N a ) T rM ( r/N a )
− a N a 2
M ( r/aN )
+ T aN L − 1 ( r/aN )
− T /r
=
+
T N coth( L
− 1
( r/N a ) M ( r/N a ) T N M ( r/N a )
−
N a N aL − 1 ( r/N a )
T ( − 2 r/a
2
N
2 − 2 /aN L
− 1
( r/N A ) + 2 rM ( r/N a ) / ( aN L
− 1
( r/N a ))
2
)
2(1 − ( r/aN ) 2 − 2 r/aN L − 1 ( r/aN ))
−
T L
− 1
( r/N a ) a
M ( r/aN )
+ T aN L − 1 ( r/aN )
− T /r

M.
Z.
Bazant – 18.366
Random Walks and Diffusion – Problem Set 5 Solutions
22
20
18
16
14
12
10
8
6
4
-8 -6 -4 -2 0 r
2 4
Central
Saddle Point
6 8
3
Figure 1: Comparison between the two approximations for free energy for the case when a = T = 1 and N = 10.
where M ( z ) is the first derivative of L
− 1
( z ).
By the inverse function theorem, we know that
M ( z ) =
1
L � ( L − 1 ( z ))
.
We compare this to the central region approximation, shown on problem set 1 to be
P c
N
( r ) ∼
� 3
2 πa 2 N
�
3 / 2
� 3 r
2 exp −
2 a 2 N
�
.
The free energy for this expression is
F c
= −
3
2 log
� 3
2 πa 2 N
�
+
3 r
2
2 a 2 N
.
and the restoring force is f c
∼ − a
3 r
2 N
.
A comparison between the two expressions for free energy and restoring force are shown in figures
1 and 2 respectively.
We see a good match in the central region between the two approximations.
As r → a , we see that the restoring force begins to get very large, as would be expected for a highly stretched polymer.

M.
Z.
Bazant – 18.366
Random Walks and Diffusion – Problem Set 5 Solutions
10
8
6
4
2
0
-2
-4
-6
-8
-10
-8 -6 -4 -2 0 r
2 4
Central
Saddle Point
6 8
4
Figure 2: Comparison between the two approximations for the restoring force for the case when a = T = 1 and N = 10.
2 Linear Polymer Structure
2.1
Mean Total Energy
We define η = ασ 2 /kT so that p ( θ ) ∝ e η cos θ .
The normalizing constant can be found as
A ( η ) =
�
2 π
�
π e
η cos θ sin θdθdφ
0
= 2 π e
η
0
− e
− η
η
= 4 π sinh
η
η
.
Thus we have the normalized p ( θ ) as
1 p ( θ ) =
A ( η ) e
η cos θ
=
η
4 π sinh η e
η cos θ
.
To get � E
N
� we first calculate the correlation coefficient ρ ( T ), which is
ρ ( T ) =
� Δ
�x n
· Δ x� n +1
�
= � cos θ � .
a 2
We can get this easily using the derivative of A ( η ):
1
� cos θ � =
A ( η )
�
0
2 π
�
0
π
= cos θe
η cos θ sin
1 dA ( η )
= coth η −
A ( η ) dη
1
.
η
θdθdφ

M.
Z.
Bazant – 18.366
Random Walks and Diffusion – Problem Set 5 Solutions 5
Therefore
� E
N
� = − ( N − 1) αa
2
ρ ( T ) = − ( N − 1) αa
2
� 1 � coth η − .
η
2.2
Asymptotic scaling
Since two adjacent steps have correlation ρ ( T ), the correlation between n -th and n + m -th steps is generally given by
� Δ
�x n
· Δ x� n + m
�
= ρ ( T ) m
.
a 2
From the lecture we know
� R
2
N
� ∼
1 + ρ ( T ) a
2
1 − ρ ( T ) and a eff
( T ) ∼
�
1 + ρ ( T )
1 − ρ ( T ) a as N → ∞ .
Thus the asymptotic behaviors of ρ ( T ) and a eff
( T ) are
ρ ( T ) = e
η
+ e
− η e η − e − η
1
−
η
=
⎧
⎪
⎨
⎪
⎩
η 3 ασ
3
1
=
−
1 k
B
= 1 −
η
2
T k
B
T
ασ 2 a eff
( T )
∼ a
⎧
⎪
⎩
⎪
⎨
1 +
�
3 ασ
2 k
2 ασ
B
2
T k
B
T
(
(
T
T
→ ∞
→
(
(
0)
η
η
.
)
→ 0
→ ∞ or or
T
T
→ ∞
→
)
0) ,
3 A persistent L´ flight
First we take the sum of our non-independent steps, each expressed in terms of the independent steps, denoted with primes.
X
N
N
�
= Δ x n
= (1 + ρ )Δ x
�
1
+
N − 1
�
Δ x
� n
+ (1 − ρ )Δ x
�
N
.
n =1 n =2
The structure function, P k ), of the PDF of � Δ x n is given by the product of the structure functions associated with each step.
The length scales of the first and last steps are renormalized by 1 + ρ and 1 − ρ respectively.
Thus the structure function for the entire walk is given by
P
ˆ
N
( k ) = exp {− a (1 + ρ ) | k | − a ( N − 2) | k | − a (1 − ρ ) | k |} = e
− aN | k | and the corresponding PDF as found in Homework 1 is given by
P
N
( x ) = aN
π ( x 2 + a 2 N 2 )
.
Thus the half-width scales as Δ x
1 / 2
∼ aN and it doesn’t depend on ρ at all.

M.
Z.
Bazant – 18.366
Random Walks and Diffusion – Problem Set 5 Solutions 6
4 A continuous-time random walk
As shown in the lecture notes (2005, lecture 23), if the waiting time distribution is ψ ( t ) = e
− t
, then the number of steps N that have taken place by time t follows a Poisson distribution
P ( N, t ) = t
N e − t
N !
.
We also know that the probability of a Bernoulli walker being at a location x after N steps is
P ( x, N ) =
�
2
− N �
N
N
+ x
�
2
0 for for x x
+
+
N
N even odd.
By summing over all possible numbers of steps taken, we find that the probability disribution of the walker being at a location x after time t is
∞
�
P ( x, t ) = P ( N, t ) P ( N, t )
N =0
Only the terms in this sum of the form N = x + 2 m where m = 0 , 1 , 2 , . . .
will have a non-zero contribution, so
P ( x, t ) =
∞
�
P ( x + 2 m, t ) P ( x + 2 m, t ) m =0
∞
�
2
− x − 2 �
+2 m m
� t x +2 m e
− t
=
( x + 2 m )!
m =0
∞
�
( t/ 2) x +2 m e
− t
=
( x + m )!
m !
m =0
= I x
( t ) e
− t where I x
( t ) is the modified Bessel function.