6.6. Microscopic systems[[Student version, January 17, 2003]] 197
proceed spontaneously at a negligible rate, due to their high activation barriers. Chapter 10 will
discuss how cells use molecular devices—enzymes—to facilitate such reactions when and where they
are desired.
Another aspect of exponential relaxation will prove useful later. Suppose we follow a single
molecule, which we observe to be in state #1 initially. Eventually the molecule will jump to state
#2. Now we ask, how long does itstayin #2 before jumping back? There is no single answer to
this question—sometimes thisdwell timewill be short, other times long. But we can say something
definite about theprobability distributionof dwell times,P 2 → 1 (t).^5
Example Find this distribution.
Solution: First imagine a collection ofN 0 molecules, all starting in state #2. The
numberN(t)surviving in this state after an elapsed timetobeys the equationN(t+dt)=(1−k+dt)N(t),withN(0) =N 0.The solution to this equation isN(t)=N 0 e−k+t.Nowuse the multiplication rule
(Equation 3.15 on page 70): The probability for a molecule to survive till timet,
andthen hop in the next interval dt,isthe product (N(t)/N 0 )×(k+dt). Calling
this probabilityP 2 → 1 (t)dtgivesP 2 → 1 (t)=k+e−k+t. (6.31)Notice that this distribution is properly normalized:∫∞
0 P^2 →^1 (t)dt=1.Similarly, the distribution of dwell times to hop in the other direction equalsk-e−k-t.
6.6.3 The minimum free energy principle also applies to microscopic subsystems
Section 6.6.1 found the Boltzmann distribution by requiring that every microstate of a combined
system a+B be equally probable, or in other words that the entropy of the total system be maximum.
Just as in our discussion of macroscopic subsystems, though (Section 6.5), it’s better to characterize
our result in a way that directs all our attention onto the subsystem of interest and away from the
reservoir. For the case of a macroscopic system “a,” Idea 6.14 did this by introducing the free
energyFa=Ea−TSa,which depended on system “B” only through its temperature. This section
will extend the result to a corresponding formula for the case of a microscopic subsystem “a.”
In the microscopic case the energy of “a” can have large relative fluctuations, so it has no definite
energyEa.Accordingly, we must first replaceEabyits average over all allowed states of “a,” or
〈Ea〉=
∑
jPjEj.Todefine the entropySa,note that the Boltzmann distribution gives different
probabilitiesPjto different statesjof system “a.” Accordingly we must define the entropy of “a”
using Shannon’s formula (Equation 6.3), which givesSa=−kB
∑
jPjlnPj.Areasonable extension
of the macroscopic free energy formula is then
Fa=〈Ea〉−TSa. free energy of a molecular-scale subsystem (6.32)(^5) Some authors use the synonym “waiting time” for dwell time.