Biological Physics: Energy, Information, Life

398 Chapter 10. Enzymes and molecular machines[[Student version, January 17, 2003]]

T 2 Track 2

10.2.3′

1. Strictly speaking,tstepin Equation 10.2 should be computed as the mean time for a random
   walker to arrive at an absorber located atx=L,after being released at a reflecting wall atx=0.
   Luckily, this time is given by the same formula,tstep=L^2 /(2D), as the na ̈ıve formula we used to
   get Equation 10.2! (See Berg, 1993, Equation 3.13.)


2. M. Smoluchowski foresaw many of the points made in this chapter around 1912–1915. The
   application of his equation to the theory of escape processes is due to O. Klein (1922). Some
   authors instead use the term “Fokker–Planck equation” for Equation 10.4 on page 367; others
   reserve that term for a related equation involving both position and momentum.


3. Equation 10.7 on page 368 was applicable only in the perfect-ratchet limit. To study the S-
   ratchet in the general nonequilibrium case, we first need the general solution to Equation 10.4 on
   the interval (0,L)with dUtot/dx=f,namelyP(x)=A(be−xf /kBT−1) for any constantsAand
   b.The corresponding probability flux isj(1d)=MfDA/kBT.
   Tofix the unknown constantb,wenext show quite generally that the functionP(x)eUtot(x)/kBT
   must have the same value just above and just below each jump in the potential, even though we
   have already seen thatP(x)itself will not (see the Example on page 367). Multiply both sides of
   Equation 10.3 by eUtot(x)/kBT,tofind

d

dx

(

PeUtot/kBT

)

=−

j(1d)

MD

eUtot/kBT.

Integrating both sides of this equation fromL−δtoL+δ,whereδis a small distance, gives that

P(L−δ)eUtot(L−δ)/kBT=P(L+δ)eUtot(L+δ)/kBT, (10.25)

plus a correction that vanishes asδ→0. That is,P(x)eUtot(x)/kBTis continuous atL,aswasto be
shown.
Imposing Equation 10.25 on our solution atx=Land using the periodicity assumption,P(L+
δ)=P(δ), gives that
P(L−δ)efL ̄ =P(0 +δ)e(fL−)/kbt
b(e−fL/kBT−e−/kBT)=1−e−δ/kBT

or

b=

e/kBT− 1

e(−fL+)/kBT− 1

.

Proceeding as in deriving Equation 10.5,

∆t=

MA

j(1d)

(

−f/kbBT

(

e−fL/kBT− 1

)

−L

)

.

and hence

v=

L

∆t

=−

D

L

(

fL

kBT

) 2 [

fL

kBT

−

(1−e−/kBT)(1−e−fL/kBT)

e−fL/kBT−e−/kBT

]− 1

. (10.26)

Youcan verify from this formula that all four of Sullivan’s claims listed on page 365 are correct: