# Relation between grand potential and expected number of particles in an energy state?

During the review of my lecture notes I stumbled upon an equation that gives me some trouble understanding. The big task that motivates the following is to express the entropy $$S$$ with the expected number of particles in an energy state $$\langle n_i \rangle$$.

Since the entropy also relates with the grand potential we are looking for an expression that gives a relation between the grand Potential $$\Omega$$ and $$\langle n_i \rangle$$ first.

An expression for $$\langle n_i \rangle$$ is for example $$\langle n_i \rangle = \frac{1}{e^{\beta(E_i-\mu)}+\gamma} \quad \text{with} \quad \gamma= \begin{cases} +1,\,& \text{Fermi-Dirac}\\ -1,\,& \text{Bose-Einstein}\\ 0^+,\,& \text{Maxwell-Boltzmann} \end{cases}.$$

Now my notes make the equation, where I can't understand the second equality $$\Omega = -\frac{1}{\beta}\ln \mathcal{Z}_G \stackrel{?}{=} \sum_i (E_i-\mu)\langle n_i\rangle.$$

I've seen an expression for $$\ln \mathcal{Z}_G$$ that looks like $$\ln \mathcal{Z}_G = \frac{1}{\gamma}\sum_i \ln\left[ 1+ \gamma e^{-\beta(E_i-\mu)}\right],$$

but I don't know if this can help me in any way. I tried to find the relation by doing some algebra, but I never seem to get to the equality $$-\frac{1}{\beta}\ln \mathcal{Z}_G = \sum_i (E_i-\mu)\langle n_i\rangle$$. I had the idea that maybe one needs to do some kind of approximation, but then again I am clueless what and how.

It would be great if someone could show how I get from the LHS to the RHS

## 2 Answers

The trick is to use the definition of a derivative of a logarithm. Lets do the calculus for the FD case:

$$-\frac{1}{\beta}\ln \mathcal{Z}_G \stackrel{?}{=} \sum_i (E_i-\mu)\langle n_i\rangle$$

If you see in the right side of the equation, you have a case $$\frac{f'(x)}{f(x)}=\frac{d(\ln f(x))}{dx}$$. Remember the definition of $$\langle n_i \rangle$$

$$\langle n_i \rangle=-\frac{1}{\beta}\frac{\partial \ln Z_G}{\partial E_i}$$

and you will get the left side.

• Oh this is neat. Thank you very much! – MLK Feb 8 '20 at 13:09

There is no way to understand how to get to the equality $$\Omega = \sum_i (E_i - \mu)\langle n_i\rangle$$ because it is wrong for $$T>0$$. From thermodynamics, we know $$S = -\frac1T\left(\Omega - {\cal E} +\mu N\right).$$ And for ideal gases $${\cal E} - \mu N = \sum_i(E_i-\mu)\langle n_i \rangle.$$ Hence we have $$\Omega -\sum_i (E_i-\mu)\langle n_i \rangle = -TS.$$ It is the last equality that gives a possibility to express entropy $$S$$ in terms of expected numbers of particles.