The Maxwell-Boltzmann distribution and temperatue

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In summary, the derivation of the Maxwell Boltzmann distribution involves maximizing the number of ways to obtain a particular macrostate with respect to the distribution of particles in their energy states. This leads to the equation ##\frac{n_i}{n} = \frac{1}{Z} e^{-\beta \epsilon_i}##, where ##n_i## is the number of particles in energy level ##\epsilon_i##, ##n## is the total number of particles, and ##\beta## is a constant derived from the temperature and Boltzmann constant. This can also be derived from the definition of temperature in terms of entropy. The first law of thermodynamics can then be used to identify ##T## as ##1/\beta##
  • #1
center o bass
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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?
 
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  • #2
Take an ideal gas. The partion sum of a non-relativistic ideal gas is
[tex]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 ).[/tex]
Then the total mean energy, i.e., thermodynamically spoken the internal energy, is given by
[tex]U=-\frac{1}{Z} \partial_{\beta} Z=-\partial_{\beta} \ln Z.[/tex]
The Gauß integral can be done exactly, and the derivative gives
[tex]U=\frac{3}{2 \beta}=\frac{3}{2} k_{\text{B}} T,[/tex]
and this identifies [itex]k_{\text{B}} T=1/\beta[/itex].
 
  • #3
vanhees71 said:
Take an ideal gas. The partion sum of a non-relativistic ideal gas is
[tex]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 ).[/tex]
Then the total mean energy, i.e., thermodynamically spoken the internal energy, is given by
[tex]U=-\frac{1}{Z} \partial_{\beta} Z=-\partial_{\beta} \ln Z.[/tex]
The Gauß integral can be done exactly, and the derivative gives
[tex]U=\frac{3}{2 \beta}=\frac{3}{2} k_{\text{B}} T,[/tex]
and this identifies [itex]k_{\text{B}} T=1/\beta[/itex].

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

$$T = (\partial S/\partial U)^{-1}?$$
 
  • #4
Ok, let's see. We start from the grand-canonical partition sum
[tex]Z[\beta,\alpha]=\mathrm{Tr} \exp(-\beta \hat{H} + \alpha \hat{N}).[/tex]
The Statistical Operator then is
[tex]\hat{R}= \frac{1}{Z} \exp(-\beta \hat{H} + \alpha \hat{N}),[/tex]
and thus the entropy (setting the Boltzmann constant to 1, so that we measure temperatures in units of energy at the end):
[tex]S=\mathrm{Tr} [\hat{R} \ln \hat{R}] = -(-\beta \langle H \rangle + \alpha \overline{N}- \ln Z)=\beta U - \alpha \overline{N} + \ln Z.[/tex]
As independent variables we take [itex]\alpha[/itex] and [/itex]\beta[/itex] as well as the box volume of the gas [itex]V[/itex] as external parameter.

Then you find
[tex]\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}.[/tex]
Now we have
[tex]\frac{\partial \ln Z}{\partial \beta}=-U, \quad \frac{\partial \ln Z}{\partial \alpha}=\overline{N},[/tex]
and thus
[tex]\mathrm{d} S=\beta \mathrm{d} U - \alpha \mathrm{d} \overline{N} + \mathrm{d} V \frac{\partial \ln Z}{\partial V}.[/tex]
To comare with the 1st Law of Thermodynamics we resolve this to [itex]\mathrm{d} U[/itex]. This gives
[tex]\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}.[/tex]
comparing this to the First Law,
[tex]\mathrm{d} U = T \mathrm{d} S + \mu \mathrm{d} \overline{N} - p \mathrm{d} V,[/tex]
leads to the identification
[tex]T=\frac{1}{\beta}, \quad \mu=\frac{\alpha}{\beta} = T \alpha, \quad p = \left (\frac{\partial \ln Z}{\partial V} \right )_{T,\mu}.[/tex]
 
  • #5


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
 

Related to The Maxwell-Boltzmann distribution and temperatue

1. What is the Maxwell-Boltzmann distribution?

The Maxwell-Boltzmann distribution is a mathematical description of the distribution of speeds for particles in a gas. It describes the relationship between temperature and the velocity of particles in a gas.

2. How does temperature affect the Maxwell-Boltzmann distribution?

The temperature of a gas affects the shape and spread of the Maxwell-Boltzmann distribution curve. As temperature increases, the curve shifts to the right and becomes broader, indicating a higher average velocity and a wider range of velocities for the particles.

3. What is the significance of the Maxwell-Boltzmann distribution?

The Maxwell-Boltzmann distribution is significant because it helps us understand the behavior of particles in a gas and how temperature affects that behavior. It also provides a basis for understanding thermodynamic properties such as pressure, volume, and energy.

4. How is the Maxwell-Boltzmann distribution related to the kinetic theory of gases?

The Maxwell-Boltzmann distribution is a fundamental component of the kinetic theory of gases. It explains the relationship between the average kinetic energy of particles in a gas and the temperature of the gas. It also helps to explain the behavior of gases at different temperatures and pressures.

5. Can the Maxwell-Boltzmann distribution be applied to all gases?

Yes, the Maxwell-Boltzmann distribution can be applied to all gases, as long as they are in thermal equilibrium. It is a fundamental concept in gas kinetic theory and is used to describe the behavior of both ideal and real gases.

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