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# Bose Einstein distribution

Let's take our example of 3 particles and 2 levels. The four macrostates for indistinguishable particles could be represented as follows:
```OOO|
|OOO
O|OO
OO|O
```

The number of macrostates that we have is the number of combinations; of 3 particles within 3 + (2 - 1) places C(3, 4) = 4!/3!(4 - 3)! = 4.

Generally,

With N particles and n levels,the number of macrostates is: Ω = C(N,N+n-1) = (N+n-1)!/N!(n-1)!

For the Maxwell-Boltzman function, we have seen that the macrostate had the expression Ω = N! Π(gini/ni!); the particles were distibguishable. Here, the particles are indistinguishable (bosons) , the factor N! Π(1/ni) is equal to 1, because all the combinations C(n,N) that lead to it are equal to 1.

Now, if the level "i" contains gi degeneracies; the number of ways to palce ni particles ( bosons) is exatly C(ni, ni + gi -1) = (ni + gi -1)!/ni!(gi - 1)!

The total number of macrostates is then:
Ω = Π (ni + gi -1)!/ni!(gi - 1)!
Which are subject to the folowing constrainsts: Σ ni = N [i: 1 → n], and
Σ niEi = E
ln(Ω)= Σ ln(ni + gi -1)! - ln(ni!) - ln((gi - 1)!)
Using Stirling's approximation, we get:
ln(Ω)= Σ (ni + gi -1)ln(ni + gi -1) - (ni + gi - 1) - niln ni + ni - (gi - 1)ln((gi - 1)) + (gi - 1)
ln(Ω)= Σ (ni + gi -1)ln(ni + gi -1) - niln ni - (gi - 1)ln(gi - 1)
We assume that:
gi and ni are >> 1.
Then:
ln(Ω)= Σ (ni + gi)ln(ni + gi) - ni ln ni - gi ln gi
Differentiating to maximaze, we get:
d(ln(Ω))= Σ dniln(ni + gi) - dni ln ni = Σ dni ln[(ni + gi)/ni]

The combination with the following constraints
α Σ dni = 0
-β Σ dni Ei = 0
gives: Σ [ln[(ni + gi)/ni] + α - βEi]dni = 0
We have then: (ni + gi)/ni = exp[ - α] exp[βEi]
Therefore:

The Bose-Einstein population numbers is :
ni = gi/(exp[ - α] exp[βEi] - 1) Web ScientificSentence