- Track 2[[Student version, January 17, 2003]] 339
Let us writeA(x)for theautocorrelation function〈ˆt(s)·ˆt(s+x)〉;for a long chain this quantity
does not depend on the starting pointschosen. Then the above reasoning implies that Equation 9.29
can be rewritten as
A(sAB+sBC)=A(sAB)×A(sBC). (9.30)
The only function with this property is the exponential,A(x)=eqxfor some constantq.
Tofinish the proof of Equation 9.28, then, we only need to show that the constantqequals
− 1 /A. But for very small ∆s A,thermal fluctuations can hardly bend the rod at all (recall
Equation 9.4). Consider a circular arc in whichˆtbends by a small angleθin theξζplane ass
increases by ∆s.That is, supposeˆtchanges fromˆt(s)=ζˆtoˆt(s+∆s)=(ζˆ+θˆξ)/
√
1+θ^2 ,which
is again a unit vector. Adapting Equation 9.4 on page 304 for this situation yields the elastic energy
cost as (^12 kBTA)×(∆s)×(∆s/θ)−^2 ,or(AkBT/(2∆s))(θ)^2 .This expression is a quadratic function
ofθ.The equipartition of energy (Your Turn 6f on page 194) then tells us that the thermal average
of this quantity will be^12 kBT,orthat
A
∆s
〈θ^2 〉=1.
Repeating the argument for bends in theξηplane, and remembering thatθis small, gives that
A(∆s)=〈ˆt(s)·ˆt(s+∆s)〉 =〈
ζˆ·
ζˆ+θξζˆξ+θηζηˆ
√
1+(θξζ)^2 +(θηζ)^2〉
≈ 1 −^12 〈(θξζ)^2 〉−^12 〈(θηζ)^2 〉
=1−∆s/A.Comparing toA(x)=eqs≈1+qs+···,indeed gives thatq=− 1 /A. This finally establishes
Equation 9.28, and with it the interpretation ofAas a persistence length.
- To make the connection between an elastic rod model and its corresponding FJC model, we now
consider the mean-squared end-to-end length〈r^2 〉of an elastic rod. Since the rod segment ats
points in the directionˆt(s), wehaver=
∫Ltot
0 dsˆt(s), so〈r^2 〉 =〈(∫Ltot
0 ds^1 ˆt(s^1 ))
·
(∫Ltot
0 ds^2 ˆt(s^2 ))〉
=
∫Ltot0ds 1∫Ltot0ds 2 〈ˆt(s 1 )·ˆt(s 2 )〉=∫Ltot0ds 1∫Ltot0ds 2 e−|s^1 −s^2 |/A=2
∫Ltot0ds 1∫Ltots 1ds 2 e−(s^2 −s^1 )/A=2∫Ltot0ds 1∫Ltot−s 10dxe−x/A,wherex≡s 2 −s 1 .For a long rod the first integral is dominated by values ofs 1 far from the end,
so we may replace the upper limit of the second integral by infinity:
〈r^2 〉=2ALtot. (long, unstretched elastic rod) (9.31)This is a reassuring result: Exactly as in our simple discussion of random walks (Equation 4.4 on
page 104),the mean-square end-to-end separation of a semiflexible polymer is proportional to its
contour length.
Wenow compute〈r^2 〉for a freely jointed chain consisting of segments of lengthLsegand compare
it to Equation 9.31. In this case we have a discrete sum over theN=Ltot/Lsegsegments:
〈r^2 〉=∑N
i,j=1〈(Lsegˆti)·(Lsegˆtj)〉=(Lseg)
2[∑N
i=1〈(ˆti)(^2) 〉+2∑N
i<j〈ˆti·ˆtj〉