Statistical Physics, Second Revised and Enlarged Edition

Counting microstatesfor gases 57

Hence our (slightlyapproximate)finalresultfor the number ofmicrostates to an
allowed distribution for a boson gas is

tttBE({ni})=

∏

i

(ni+gi)!

ni!gi!

(5.5)

where againthe product ranges over all groupsi.

5.3.3 Dilutegases

For gases, there are two, and only two, possibilities. The gaseous particles are either
bosons(+ 1 ,symmetric)orfermions(− 1 , antisymmetric). Thereisnohalf-wayhouse.
However, it is profitable to consider a third form of statistics quite explicitly, namely
that of a dilute gas.
The word‘dilute’herehas a specific meaning.Itis to suppose thatfor all groupsi,
the states are very sparsely occupied, i.e.

nigi for alli

Weshallseelater that this conditionholdsfor realgasesinthelimitoflowdensity
and hightemperature, andthatit correspondstoaclassicallimittoeither theFDor
the BE form ofquantum statistics.
Inthisdilutelimit, we can readily see thatbothforms ofstatistics give almost the
same answerfort({ni}).Itis a resultweshouldexpect, since whennigieven
for bosons we would anticipate that almost all the states are unoccupied, and just a
fewhaveasingle occupation. Theexistence or otherwise ofthe exclusion principle
isirrelevantifthesystemis not even temptedtowardsmultiple occupation!
To consider the fermion case first, each factor in(5.4)can be written as

[gi(gi− 1 )(gi− 2 )...(gi−ni+ 1 )]/ni!

wherebothnumerator anddenominatorhavebeendividedby(gi−ni)!.Note that the
numerator ofthisexpressionistheproduct ofnifactors, allofwhichin ourpresent
limit are almost equal to (but a bit less than)gi.Therefore,we have

tttFD≈

∏

i

gnii

ni!

In the boson case, (5.5) can be treated similarlyin the dilute limit. Dividingtop and
bottom bygi!one obtains for the typical term

[(gi+ 1 )(gi+ 2 )...(gi+ni)]/ni!