Chemical potential and fugacity

[Kaguro]

I am trying to learn statistical physics. While learning MB statistics, my textbook defined chemical potential as $\mu = (\frac{\partial F}{\partial N})_{V,T}$. That's nice.

Later when I started on Quantum statistics, my textbook described all three distribution functions via:
$n_i = \frac{g_i}{e^{\alpha + \beta E_i} + \kappa}$
We had already found out the value of beta from MB statistics (using MB distr. function. Why would that apply here is another mystery altogether)

Then suddenly book said:
$n_i = \frac{g_i}{e^{\frac{E_i - \mu}{K_B T}} + \kappa}$

Where we define chemical potential via the relation $\alpha = -\mu/kT$ (and its exponential is called fugacity)

How and why did the book define the same thing twice!?

 

[Kaguro]

To all future people who want to know: I found out how can we show equivalence.

We need to define mu only once, but neither in these places. We need to define mu in the fundamental thermodynamic equation:

First consider Grand Canonical Ensemble, that is, allow particle number to change.

Generalise first and second laws into this:

$dU = TdS - PdV + \mu dN$
We define $\mu$ as the rate of change of energy per unit change in particle number.

Then from this relation, $\mu = (\frac{\partial U}{\partial N})_{S,V}$
F = U-TS
So, $dF=-PdV - SdT + \mu dN$
$\Rightarrow \mu = (\frac{\partial F}{\partial N})_{T,V}$ (Yaay!)

Now while deriving the distribution function by maximizing the log of number of microstates and using method of Lagrange multipliers we got:

$d(lnW) = \alpha dN + \beta dE$
So, $\alpha = (\frac{\partial ln(W)}{\partial N})_{E,V}$

But we know $S=k_B ln(W)$ (separate derivation for that. But it is standalone)
$\Rightarrow lnW = \frac{S}{k_B}$
$\Rightarrow \alpha = \frac{1}{k_B} (\frac{\partial S}{\partial N})_{E,V}$

Now, $TdS=dU+PdV-\mu dN$
$(\frac{\partial S}{\partial N})_{U,V} = \frac{-\mu}{T}$

Therefore, $\alpha = \frac{-\mu}{kT}$ (Yaaaay!)

Similarly we can show that $\beta = \frac{1}{kT}$

Hence Proved.