Classical Thermodynamics

Statistical Thermodynamics

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Statistical Mechanics

### Thermodynamics

 Statistical Thermodynamics

Bosons & Fermions

1. Bose Einstein distribution

Let's take our example of 3 particles and 2 levels. The four macrostates for indistinguishable particles case that we had could be represented as follows:

```OOO|
|OOO
O|OO
OO|O
```
The number of macrostates that we have is the number of arrangments; that 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)!

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 contrainsts: 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) - ni)ln(ni)+ ni) - ((gi - 1))ln((gi - 1)) + ((gi - 1))
ln(ω)= Σ (ni + gi -1)ln(ni + gi -1) - ni)ln(ni) - ((gi - 1))ln((gi - 1))
We assume 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] = 0
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)

5. Fermi-Dirac distribution

The case of fermions is a little bit similar, except that each sublevel within a degeneracy can take just ONE or ZERO particle.
O| |O| | | | | |O| ...
Here the number of ways to place ni in gi degeneracies is:
C(nii, gi) = gi!/ni!(gi - ni)!
Then : Total Ω = Π gi!/ni!(gi - ni)!
Using Stirling's approximatio, we get:
ln(Ω) = Σ giln gi - gi - niln i + ni - (gi - ni)ln(gi - ni) + (gi - ni)
= Σ giln gi - niln ni - (gi - ni)ln(gi - ni)
Differentiating to maximaze, we get:
dln(Ω) = Σ - dniln ni - dni + dniln(gi - ni) + d ni = Σ dniln[(gi - ni)/ ni] = 0
It follows that by adding the same above constraints:
Σ dni {ln[(gi - ni)/ ni] + α - βEi]} = 0    (5.)
Which gives: (gi - ni)/ ni = exp[- α] exp[βEi]
Therefore:
The Fermi-Dirac population numbers is :

ni = gi/(exp[- α] exp[βEi] + 1)

2. Finale expressions for the distribution of fermions and bosons

### First law:

dQ = dU + dW
U is the internal energy of the system E.
E = Σ ni Ei. Therefore:
dU = dE = Σ dni Ei + Σ ni dEi.
For a change in the volume then in the number (mass) of the gas, we have:
dW = - PdV , then:
dW = PdV = Σ dniEni
dQ = Σ ni dEi

### Second law:

Now for a system that is a Grand canonical ensemble, where nothing is fixed, we add Q (heat) and N (particles), then:
dQ + dN = dU
Therefore: dQ + μdN = dU + dW , where μ is called the Chemical potential. Then:
dQ = TdS = dU + dW - μdN

We have seen from the equation (5.) that :
Σ dni {ln[(gi - ni)/ ni] + α - βEi]} = 0
or
Σ dni {ln[(gi - ni)/ ni]} = Σ dni {- α + βEi]}
With:
Σ dni [- α + βEi] = - αN + βΣ dni Ei]

Since : dni Ei = d(ni Ei) - ni dEi, we can write:
Σ dni [- α + βEi] = - α dN + βΣ {d(ni Ei) - ni dEi}

From S = k ln(Ω) and then dS = k dln(Ω, we have:
dS = k Σ dni {ln[(gi - ni)/ ni]}
= k Σ dni {- α + βEi]}
Then dQ = TdS = kT Σ dni {ln[(gi - ni)/ ni]} = kT Σ dni {- α + βEi]}
= - α kT dN + kT βΣ {d(ni Ei) - ni dEi}
= dU + dW - μdN
Since kT = 1/β then:
dU = kT βΣ d(ni Ei)= Σ d(ni Ei) = dE = dU
dW = - kT βΣ nidEi = - Σ nidEi
μdN = α kT dN
Thus:
α = μ/kT

Finaly,

The Fermi-Dirac population numbers is :
ni = gi/(exp [- (μ - Ei)/kT] + 1)

The Bose-Einstein population numbers is :
ni = gi/(exp [- (μ - Ei)/kT] - 1)

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