# The Maxwell-Boltzmann distribution and temperatue

1. Sep 16, 2014

### center o bass

The derivation of the maxwell boltzmann distribution involves maximizing the number of ways to obtain a particular macrostate with respect to how the particles are distributed in their respective energy states. One then arrives at
$$\frac{n_i}{n} = \frac{1}{Z} e^{- \beta \epsilon_i},$$
where $n_i, n, \epsilon_i$ respectively denotes the number of particles in the energy level $\epsilon_i$, the total number of particles, and the energy level $\epsilon_i$.
Now, it is often just taken out of thin air that $\beta = k T$ where $T$ is temperature and $k$ is the Boltzmann constant -- but this surely can be derived.

My question is how can we derive this fact?

2. Sep 16, 2014

### vanhees71

Take an ideal gas. The partion sum of a non-relativistic ideal gas is
$$Z=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{V}{(2 \pi)^3 \hbar^3} \exp \left (-\beta \frac{\vec{p}^2}{2m} \right ).$$
Then the total mean energy, i.e., thermodynamically spoken the internal energy, is given by
$$U=-\frac{1}{Z} \partial_{\beta} Z=-\partial_{\beta} \ln Z.$$
The Gauß integral can be done exactly, and the derivative gives
$$U=\frac{3}{2 \beta}=\frac{3}{2} k_{\text{B}} T,$$
and this identifies $k_{\text{B}} T=1/\beta$.

3. Sep 16, 2014

### center o bass

Can it also be done more generally from the definition of temperature in terms of entropy?

$$T = (\partial S/\partial U)^{-1}?$$

4. Sep 17, 2014

### vanhees71

Ok, let's see. We start from the grand-canonical partition sum
$$Z[\beta,\alpha]=\mathrm{Tr} \exp(-\beta \hat{H} + \alpha \hat{N}).$$
The Statistical Operator then is
$$\hat{R}= \frac{1}{Z} \exp(-\beta \hat{H} + \alpha \hat{N}),$$
and thus the entropy (setting the Boltzmann constant to 1, so that we measure temperatures in units of energy at the end):
$$S=\mathrm{Tr} [\hat{R} \ln \hat{R}] = -(-\beta \langle H \rangle + \alpha \overline{N}- \ln Z)=\beta U - \alpha \overline{N} + \ln Z.$$
As independent variables we take $\alpha$ and [/itex]\beta[/itex] as well as the box volume of the gas $V$ as external parameter.

Then you find
$$\mathrm{d} S=\beta \mathrm{d} U + U \mathrm{d} \beta - \mathrm{d} \alpha \overline{N} - \alpha \mathrm{d} \overline{N} + \mathrm{d} \beta \frac{\partial \ln Z}{\partial \beta} + \mathrm{d} \alpha \frac{\partial \ln Z}{\partial \alpha} + \mathrm{d} V \frac{\partial \ln Z}{\partial V}.$$
Now we have
$$\frac{\partial \ln Z}{\partial \beta}=-U, \quad \frac{\partial \ln Z}{\partial \alpha}=\overline{N},$$
and thus
$$\mathrm{d} S=\beta \mathrm{d} U - \alpha \mathrm{d} \overline{N} + \mathrm{d} V \frac{\partial \ln Z}{\partial V}.$$
To comare with the 1st Law of Thermodynamics we resolve this to $\mathrm{d} U$. This gives
$$\mathrm{d} U = \frac{1}{\beta} \mathrm{d} S + \frac{\alpha}{\beta} \mathrm{d} \overline{N} -\mathrm{d} V \frac{\partial \ln Z}{\partial V}.$$
comparing this to the First Law,
$$\mathrm{d} U = T \mathrm{d} S + \mu \mathrm{d} \overline{N} - p \mathrm{d} V,$$
$$T=\frac{1}{\beta}, \quad \mu=\frac{\alpha}{\beta} = T \alpha, \quad p = \left (\frac{\partial \ln Z}{\partial V} \right )_{T,\mu}.$$