The Maxwell-Boltzmann distribution and temperatue

[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?

 

[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 $k_{\text{B}} T=1/\beta$.

 

[center o bass]

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 $k_{\text{B}} T=1/\beta$.

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

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

 

[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,
$$\mathrm{d} U = T \mathrm{d} S + \mu \mathrm{d} \overline{N} - p \mathrm{d} V,$$
leads to the identification
$$T=\frac{1}{\beta}, \quad \mu=\frac{\alpha}{\beta} = T \alpha, \quad p = \left (\frac{\partial \ln Z}{\partial V} \right )_{T,\mu}.$$

 

[sardar]

I can certainly understand your curiosity about the derivation of the relationship between the Maxwell-Boltzmann distribution and temperature. This relationship is a fundamental concept in statistical mechanics and is crucial in understanding the behavior of particles in a system at the microscopic level.

To answer your question, the derivation of the Maxwell-Boltzmann distribution involves using statistical mechanics and the principles of thermodynamics. The key to understanding this relationship lies in the concept of entropy, which is a measure of the disorder or randomness in a system.

In statistical mechanics, the entropy of a system is related to the number of microstates that can correspond to a particular macrostate. A macrostate is a set of conditions that describe the overall properties of a system, such as temperature, pressure, and volume. On the other hand, a microstate is a specific arrangement of the particles within the system.

The Maxwell-Boltzmann distribution is derived by maximizing the number of microstates that can correspond to a particular macrostate. This is done by considering the energy levels of the particles in the system and how they are distributed among these levels. The more ways the particles can be distributed among the energy levels, the higher the entropy and the more likely that particular macrostate will occur.

Using this approach, it can be shown that the probability of finding a particle in a particular energy level is proportional to the Boltzmann factor, which is given by $e^{-\beta\epsilon_i}$. This is where the constant $\beta$ comes into play, which is related to the temperature of the system. The higher the temperature, the larger the value of $\beta$ and the more likely it is for particles to occupy higher energy levels.

Furthermore, the Boltzmann factor is related to the Boltzmann constant, $k$, which is a fundamental constant in thermodynamics and is related to the gas constant and Avogadro's number. This is where the relationship between $\beta$ and temperature, $T$, arises, as $\beta = \frac{1}{kT}$.

In summary, the derivation of the Maxwell-Boltzmann distribution involves considering the entropy of a system and maximizing the number of microstates that correspond to a particular macrostate. This leads to the Boltzmann factor, which is related to the temperature of the system through the Boltzmann constant. Therefore, we can derive the fact that $\beta = kT$ from