2
How would we distinguish between states with energy
()
12
21
and
ab
εε
⎫
+
⎪
⎬
⎪
+
⎭
, if both particles are of the same species?
You cannot.
So, in the partition function, you must divide by two to not over count the
number of system states!
For
N
particles, with energy
(
)
( )
( )
123
abc
E
α
εεε
=+++
…
,
divide by
!
N
, the number of arrangements of permuting 1,2,3,… among a,b,c,… .
But, what to do about the system state
(
)
( )
( )
112
E
…
.
where you do not have
N
numbers to permute among the
N
particles?
Then,
!
N
division
wouldn’t be correct.
But, if number of available configurations
Ω
is much greater than
N
(
N
Ω>>
) then the chance of obtaining two molecules in same state, i. e., a configuration
where
(
)
( )
11
ε
+
, is vanishingly small.
Thus,
N
! division is correct for
N
.
identical
!
indistinguishable
L.BOLTZMANN
N
Zz
N
=
Ω
>> N
When is
Ω
>>
N
?
For a particle in box,
3
2
2
2
83
where
62
B
mL
kT
h
πε
⎛⎞
Ω=
=
⎜⎟
⎝⎠
for the average particle at temperature
T
.
Combining this with the condition
N
Ω >>
leads to
3
2
2
12
6
B
mk T
N
hV
π
>>
Therefore, Boltzmann behavior is favored by high
T,
large mass,
m
, and low density,
NV
ρ
=
.