134 Twonewideas
2.Internalenergy. AlthoughTisfixedbythe ensemble, theinternalenergyofany
one assembly is not. The energy of an assembly can have any valueE,with a
probability given by (12.6). The one thing we can be sure about is that theaverage
value ofEisU,for that was one of the startingpoints. However, (12. 6 ) contains
thefullstatistical informationaboutE,and that we shall now examine in greater
detail.
3 .Thedistributionfunction P(E). For assemblies ofanyreasonablesize,it turns
out thatP(E)is a very strongly peaked function indeed around its maximum value
(whichthus effectivelyequalsthe mean valueU).This comes aboutbecause the
assembly‘densityof microstates’ functionG(E)is a rapidlyrisingfunction ofE,
whereas the exponential factor in (12.6) is a rapidly falling function. Consider
(without workingthe problem outindetail) a specific case, where the answer
is alreadywell known from Chapter 6. This is where each assemblyconsists of
Ndilute gas particles, to which MB statistics apply. Suppose thatG(E)∝En.
Then (equation (12. 6 ))P(E)∝Enexp(−E/kkkBT).The maximumisgivenby
dP/dE =0, i.e.(check it!)whenE =nkkkBT.In our example we know that
U= 3 / 2 NkkkBT,so theindexnmustbe equalto 3 N/2. Thisisafantasticallyhigh
powerfor therisingfactor. Correspondinglythefallingexponentialfactorhas an
exponent− 3 N/2, quite enough to frighten any normal calculator.
4.Fluctuations. Continuing the previous point,P(E)also enables us toworkout
theprobablefluctuations ofEaroundthe meanvalueU.Thesharpness ofthe
maximum is readily characterized from the second derivatived^2 P/dE^2 at the
maximum. A Taylor expansion ofP(E)about the maximum showsthat a useful
measure ofthewidthEofthe curveisgivenby d^2 P/dE^2 =P(U)/(E)^2.
For the above example, this givesE=n^1 /^2 kkkBT,or more usefully the fractional
widthE/U= 1 /
√
n√√
≈ 1 /
√
N
√√
.TherelativefluctuationsinEarelargefor a small
assemblyat afixedtemperature,but reduce as about 1/√
N
√√
for alarge assembly
at a givenT.
5 .The same answers for the same problems. Naturallysince no new assumptions
have been made, the problems we have alreadytackled will still have the same
solutions attacked by the new method. For instance let us consider an assembly
ofNdistinguishable particles asinChapter 2. We can calculateZZAbyone oftwo
methods.
(i) We can use the ideas of the factorization of a partition function elaborated in
ourdiscussion ofdiatomicgases (section 7.1). Since the particles are weakly
interacting, the microstate simplyspecifies the state of each individual particle
in turn, and the various one-particle energies (allNof them) simply add to give
E,the assemblyenergy. Therefore the assemblypartition function (12.5a) is
given simply by
ZZA=ZN
each particle contributing a factorZ(equation (2.24)). We can immediately
checkthat thisis correct, sinceF=−kkkBTlnZZA(equation (12.7))=−NkkkBT
lnZ(identicalto (2.28)).