# Degeneracy of Fermionic gas

1. Dec 3, 2016

### Kara386

1. The problem statement, all variables and given/known data
The question asked me to show that the entropy of a fermionic gas is
$S = -k_B \Sigma_i (1-f_i)\ln(1-f_i) +f_i\ln(f_i)$

Using the Fermi-Dirac distribution so $f_i = \frac{n_i}{g_i}$.

2. Relevant equations

3. The attempt at a solution
The number of microstates $\Omega$ is
$\Omega =$ Π$\frac{g_i!}{n_i!(g_i-n_i)!}$ and using $S = k_B ln(\Omega)$ I've arrived at the expression:
$S = -k_B \Sigma_i g_i (1-f_i)\ln(1-f_i) +f_i\ln(f_i)$
So there's an unwanted factor of $g_i$. I'm told it is meant to be there, and I need to explain why it can be set to 1. Something to do with i being the state index and a quantum state can't be degenerate. Could someone explain that to me? Any help is much appreciated! :)

2. Dec 6, 2016

### Oxvillian

Hi Kara386

First observation here is that you need some big brackets in your equations - for example in your first equation, the $k_B$ should be multiplying everything, not just the first term.

The business with the $g_i$ is just bookkeeping. In their formula they're using the index $i$ to label different states, and in your formula you're using the index $i$ to label different energies. Your formula includes a $g_i$ factor in it to allow for the possibility that there might be lots of states with the same energy.

3. Dec 6, 2016

### Kara386

Thanks! I thought that might be it, because although originally the sum was over multiple states taken to be at the same energy, I think the introduction of the average $f_i$ allows the sum to be over individual states.