198 Chapter 6. Entropy, temperature, and free energy[[Student version, January 17, 2003]]
Your Turn 6g
Following the steps in Section 6.1 on page 174, show that the Boltzmann probability distribution
is the one that minimizes the quantityFadefined in Equation 6.32.Thus if initially the subsystem has a probability distribution differing from the Boltzmann formula
(for example, just after releasing a constraint), it is out of equilibrium and can in principle be
harnessed to do useful work.
Example Whatisthe minimal value ofFa? Show that it’s just−kBTlnZ,where theZis
defined as the normalization constant for thePj:Z=
∑
je−Ej/kBT. partition function (6.33)Solution: Use the probability distribution you found in Your Turn 6g to evaluate
the minimum free energy:Fa = 〈Ea〉−TSa
=∑
jZ−^1 e−Ej/kBTEj+kBT∑
jZ−^1 e−Ej/kBTln(
Z−^1 e−Ej/kBT)
=
∑
jZ−^1 e−Ej/kBTEj+kBT∑
jZ−^1 e−Ej/kBT(
ln (e−Ej/kBT)−lnZ)
The second term equals−(∑
jZ− (^1) e−Ej/kBTEj)−kBTlnZ,soFa=−kBTlnZ.
In the above formulas the summation extends over all allowed states; ifMallowed states all have the
same value ofEj,then they contributeMe−Ej/kBTto the sum. (We say thatMis thedegeneracy
of that energy level.) The trick of evaluating the free energy by finding the partition function will
prove useful when we work out entropic forces in Chapters 7 and 9.
T 2 Section 6.6.3′on page 209 makes some additional comments about free energy.
6.6.4 The free energy determines the populations of complex two-state systems
The discussion of simple two-state systems in Sections 6.6.1–6.6.2 may seem too oversimplified to
beuseful for any real system. Surely the complex macromolecules of interest to biologists never
literally have just two relevant states!
Suppose as before that subsystem “a” is itself a complex system with many states, but that
the states may usefully be divided into two classes (or “ensembles of substates”). For example, the
system may be a macromolecule; states #1,...,NImay be conformations with an overall “open”
shape, while states #NI+1,...,NI+NIIconstitute the “closed” shape. We callNIandNIIthe
“multiplicities” of the open and closed conformations.
Consider first the special situation in which all the states in each class have roughly the same
energy. ThenPI/PII=(NIe−EI/kBT)/(NIIe−EII/kBT): in this special case we just weight the Boltz-
mann probabilities of the two classes by their respective multiplicities.
More generally, the probability to be in classiisPI=Z−^1
∑NI
j=1e−Ej/kBT,and similarly for
classii,whereZis the full partition function (Equation 6.33). Then the ratio of probabilities is
PI/PII=ZI/ZII,whereZIis the part of the partition function from classiand so on.