Biological Physics: Energy, Information, Life

6. Track 2[[Student version, January 17, 2003]] 209

wiped out when we normalized our probability distribution for system “a.” So the second term is
actually the leading one.
Since this is such an important point, let’s make it yet again in a slightly different way. Since
system “B” is macroscopic, we can subdivide it equally intoM=1000 little systems, each one
itself macroscopic and weakly interacting with the others and with “a.” We know that each one
is overwhelmingly likely to shareEBequally, since they’re all identical and have come to the same
temperature, so each has the same number of allowed states

Ωi=eSi/kB=Ω 0 ,iexp

(

δE

M

dSi

dEi+···

)

i=1,...,M.

HereδE=Etot−Ea,Ω 0 ,iis the number of allowed states when subsystemihas energyEtot/M,
and the dots denote terms with higher powers of 1/M.The derivative dSi/dEiis just 1/Tand is
independent ofM,sowejust takeMlarge enough to justify truncating the Taylor series expansion
after the first term. The total number of states available to system “B” is then (Ωi)M,which is
indeed a constant times e−Ea/kBT,and we recover Equation 6.23.

3. The equipartition formula is not valid in quantum statistical physics: Modes whose excitation
   energy exceeds the thermal energy get “frozen out” of equipartition. Historically, this observation
   held the key to understanding black-body radiation, and hence to the creation of quantum theory
   itself.

6.6.3′

1. The entropySadefined above can’t simply be added toSBto get the total system entropy,
   because we can’t assume that “a” is weakly interacting with “B” (“a” may not be macroscopic,
   hence surface energies needn’t be smaller than interior energies).
   But suppose that “a”isitself macroscopic (but still smaller than “B”). Then the fluctuations in
   Eabecome negligible, so we can omit the averaging symbol in〈Ea〉.Inaddition, all microstates with
   this energy are equally probable, so that−

∑

i=1PilnPi=(

∑

i=1P)(lnP

− (^1) )=1×ln((Ωa)− (^1) )− (^1) =
ln Ωa,andSareduces to the usual form, Equation 6.5. Thus the formula Equation 6.32 derived in
Section 6.6.3 reduces to our prior formula (6.14) for the special case of a macroscopic subsystem.

2. The Gibbs free energy (Equation 6.16) has a similar generalization to microscopic subsystems,
   namely
   Ga=〈Ea〉+p〈Va〉−TSa. (6.37)


3. Equation 6.32 gives an important formula for the free energy of an ideal gas of uniform tempera-
   ture, but nonuniform densityc(r). (For example, an external force like gravity may act on the gas.)
   Suppose a container hasNmolecules, each moving independently of the others but with specified
   potential energyU(r). Divide space into many small volume elements ∆v.Then the probability for
   any one molecule to be in the element centered atrisP(r)=c(r)∆v/N. Equation 6.32 on page
   197 then gives the free energy per molecule asF 1 =

∑

r(c(r)∆v/N)(U(r)−kBTln(c(r)∆v/N)).
(We can omit the kinetic energy, which just adds a constant toF 1 .) Multiplying byNgives the
total free energy.

Your Turn 6k

a. Show that

F=

∫

d^3 rc(r)(U(r)+kBTln(c(r)/c∗)), (6.38)

for some constantc∗.Whydon’t we care about the value of this constant?