Molecular modelling of entangled polymer fluids under flow The ...

46 CHAPTER 2. INTRODUCTION TO MOLECULAR RHEOLOGY
Note that in the limit of a large potential F ≫ 1 the particle will always accept the
opportunity to hop to the right. Similarly, a hop to the left is evaluated by normalising
the Boltzmann distribution between −a...a and finding the probability of the particle
being between −a...0. This gives
P − (a) =
1
. (2.57)
1 + eαa From these probabilities the first and second moments of the particle displacement
distribution can be found.
〈x(t)〉 = νt [P + (a) − P − (a)]
[ e αa ]
− 1
= νta
e αa + 1
Taking αa to be small and expanding the exponentials to first order.
≈ νta
[ αa
]
2
= νta2
2k B T F
Using F = 3k BT
a 2
∂ 2 R x(s)
∂s 2
from equation 2.50 gives
〈x(t)〉 = 3ν 2
∂ 2 R x (s)
∂s 2 t (2.58)
The second moment follows similarly
〈
x 2 (t) 〉 = νta 2 [P + (a) + P − (a)]
= νta 2 .
(2.59)
These moments can be mapped to an effective Langevin equation
where
dx
dt = 1
ζ eff
F eff + g(t). (2.60)
〈
g(t)g(t ′ ) 〉 = 2k BT
ζ eff
δ(t − t ′ ). (2.61)
The unknown quantities ζ eff and F eff are chosen so that equation 2.60 has the same
distribution of x(t) as the obstructed system solved above. Taking a direct average of
equation 2.60 gives
〈x(t)〉 = F eff
ζ eff
t. (2.62)