Biological Physics: Energy, Information, Life

10.2. Purely mechanical machines[[Student version, January 17, 2003]] 367

• Thus the net number of ratchets in the distribution crossingx=a+^12 ∆xfrom left
  to right is^12 M[P(a)−P(a+∆x)]∆x≈−(∆x)
  2
  2 M
  d
  dx

∣∣

x=aP(x). (Here and below we

drop terms of cubic and higher order in the small quantity ∆x.)

• Wecan compactly restate the last result as−MDddPx∆t,whereDis the diffusion
  constant for the movement of the ratchet along its axis in the surrounding viscous
  medium. (RecallD=(∆x)^2 /(2∆t)from Equation 4.5b on page 104).

Now we add the effect of an external force:

• Each ratchet also drifts under the influence of the force−dUdxtot,whereUtot(x)isthe
  potential energy function sketched in Figure 10.11c.


• The average drift velocity of those ratchets located at x = a is vdrift =
  −kDBT ddx

∣∣

x=aUtot.(Toget this expression, write the force as−dVtot/dxand use

the Einstein relation, Equation 4.15 on page 108, to express the viscous friction

coefficient in terms ofD.)

• The net number of ratchets crossingx=ain time ∆tfrom the left thus gets a second
  contribution,M×P(a)vdrift∆t,or−MDkBTPdUdtotx ∆t.

The arguments just given yielded two contributions to the number of systems crossing a given point
in time ∆t.Adding these contributions and dividing by ∆tgives:

j(1d)≡net number crossing per time =−MD

(dP

dx

+

1

kBT

P

dUtot

dx

)

. (10.3)

(In this one-dimensional problem the appropriate dimensions for a flux areT−^1 .) In order for the
probability distributionP(x)tobetime-independent, we now require that probability is not piling
up anywhere. This requirement means that the expression in Equation 10.3 must be independent
ofx.(Asimilar argument led us to the diffusion equation, Equation 4.19 on page 118.) In this
context the resulting formula is called theSmoluchowski equation:

0=

d

dx

(

dP

dx+

1

kBTP

dUtot

dx

)

. (10.4)

The equilibrium case Wewantto find some periodic solutions to Equation 10.4 and interpret
them. First suppose that the potentialUtot(x)isitself periodic: Utot(x+L)=Utot(x). This
situation corresponds to the unloaded G-ratchet (Figure 10.11a), or to the S-ratchet (Figure 10.11c)
withf=/L.

Example

Show that in this case the Boltzmann distribution is a solution of Equation 10.4,

find the net probability per time to crossx,and explain why your result makes

physical sense.

Solution: Weexpect that the system will just come to equilibrium, where it makes

no net progress at all. Indeed, takingP(x)=Ae−Utot(x)/kBTgives a periodic, time-

independent probability distribution. Equation 10.3 then gives thatj(1d)(x)=0

everywhere. Hence thisP(x)isindeed a solution to the Smoluchowski equation

withnonet motion.

Sincej(1d)=0,Sullivan’s first claim was right (see page 365): The unloaded G-ratchet makes no
net progress in either direction. We can also confirm Sullivan’s physical reasoning for this claim:
Indeed the function e−Utot(x)/kBTpeaks at the lowest-energy points, so each ratchet spends a lot of
its time poised to hopbackwardwhenever a chance thermal fluctuation permits this.