- Track 2[[Student version, January 17, 2003]] 343
particular azimuthal directionφ.With these considerations in mind, Marko and Siggia considered
afamily of smooth trial functions, azimuthally symmetrical and peaked in the forward direction:
Vw(ˆt)=exp[wˆt·ˆz].Thus, for each value of the applied forcef,wemust evaluate Equation 9.39 using Equations 9.37–
9.38, then find the valuew∗of the parameterwthat maximizesλmax,est,and substitute to get
λ∗.
Letν=ˆt·ˆz.Wefirst need to evaluate
(TVw)(ˆt)=efν/(2kBT)e−A/∫
d^2 nˆexp[
A
ˆn·ˆt+
f
2 kBT
ˆn·ˆz+wnˆ·ˆz]
.
Todo the integral, abbreviateζ=w+ 2 kfBT andA ̃=A/ .The integrand can then be written as
exp[Qmˆ·ˆn], wheremˆis the unit vectormˆ=(A ̃ˆt+ζzˆ)/Qand
Q=‖A ̃ˆt+ζˆz‖=√
A ̃^2 +ζ^2 +2Aζν. ̃Wecan write the integral using spherical polar coordinatesθandφ,choosingmˆ as the polar axis.
Then
∫
d^2 nˆ=∫ 2 π
0 dφ∫ 1
− 1 dμ,whereμ=cosθ,and the integral becomes simply
2 π
Q(e
Q−e−Q).
Toevaluate the numerator of Equation 9.39 we need to do a second integral, overˆt.This time
choose polar coordinatesθ, φwithˆzas the polar axis. Recalling thatν≡ˆt·ˆz=cosθ,wefind
Vw·(TVw)=∫
d^2 ˆtVw(ˆt)(TVw)(ˆt)=e−A ̃ 2 π∫+1
− 1dνeζν
2 π
Q
(eQ−e−Q)=e−A ̃(2π)^2∫A ̃+ζ|A ̃−ζ|dQ
Aζ ̃ e(Q^2 −A ̃^2 −ζ^2 )/(2A ̃)(eQ−e−Q). (9.40)The last step changed variables fromνtoQ.The final integral above is not an elementary function,
but you may recognize it as related to the error function (see Section 4.6.5′on page 134).
Your Turn 9q
Next evaluate the denominator of Equation 9.39 for our trial functionVw.Having evaluated the estimated eigenvalue on the family of trial functions, it is now straightforward
to maximize the result over the parameterwusing mathematical software, obtainingλ∗as a function
ofA, ,andf.For ordinary (double stranded) DNA, it turns out that the answer is nearly
independent of the link lenth ,aslong as < 2 nm.Wecan then finish the calculation by following
the analysis leading to Equation 9.10: For largeN=Ltot/ ,
〈z/Ltot〉=
kBT
Ltotd
df
ln(
W·(TN−^1 V)
)
≈
kBT
d
df
lnλ∗(f). (9.41)The force–extension curve given by the Ritz approximation (Equation 9.41) turns out to be
practically indistinguishable from the exact solution (see Problem 9.7). The exact result cannot be
written in closed form (see Marko & Siggia, 1995; Bouchiat et al., 1999). For reference, however,
here is a simple expression that is very close to the exact result in the limit →0:
〈z/Ltot〉=h(f ̄)+1. 86 h(f ̄)^2 − 3. 80 h(f ̄)^3 +1. 94 h(f ̄)^4 , where h(f ̄)=1−^12(√
f ̄+^94 − 1