Polymers in Confined Geometry.pdf

2.3. SEMI-FLEXIBLE POLYMERS 11
to infinity as well, ε → ∞, such that the the energy per segment stays fixed
˜ε = ε
N
The total length L of the polymer must be finite as well:
!
= fixed. (2.14)
L = Nl ! = fixed. (2.15)
Further, of course, derivatives are defined as the continuum limit of a finite difference
∂t(s) ti+1 − ti
= lim
.
∂s l→0 l
(2.16)
Here the arc length s, which replaces the index i, is labeling the position along
the contour of the chain. We use partial derivatives to indicate that, in general,
the tangent vector may also depend on time. But, since we are only concerned
with static properties, we will ignore this in the following.
Let us start off the Hamiltonian in eq. (2.13) which gives after a simple transformation,
where −ti+1 · ti = 1
2 [(ti+1 − ti) 2 − 2l 2 ],
N−1
εl
H[t(s)] = lim
l→0 2
ε,N→∞ i=1
2 ti+1 − ti
l
, (2.17)
l
by dropping irrelevant constants. These would only shift the energy zero in the
continuum limit. The sum finally transforms into an integral N−1 i=1 l → L
0 ds,
such that
H[t(s)] = κ
L 2 ∂t(s)
ds =
2 ∂s
κ
L 2 ∂ r(s)
ds
2 ∂s2 2 . (2.18)
0
The constraint |ti| = l transformed into |t(s)| = 1 2 , which means that the space
curve is locally inextensible 3 . The bending modulus is defined as κ = εl = ˜εNl =
˜εL.
2Arc length parametrization enforces the constraint |t(s)| = 1 since
ds 2 ∂x 2 2
2
∂y ∂z
= + + ds
∂s ∂s ∂s
2 ⇒ |t(s)| 2 = 1
=|t(s)| 2 ŕ ŕ
ŕ
= ŕ
∂r(s) ŕ2
ŕ
ŕ ∂s ŕ
3 When the inextensibility constraint is relaxed, then a potential energy for the stretching
of segments must be introduced, e.g. when the force-extension relation for very large forces
is investigated, the segments will start stretching [33]. Normally the backbone segments are
very rigid in length so that stretching requires considerably more energy than bending, therefore
there is a time-scale separation for the relaxations times, where the stretching modes are relaxing
much faster. That is why normally one can focus on an inextensible chain [2].
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