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The Maxwell-Boltzmann distribution and temperatue

  1. Sep 16, 2014 #1
    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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  3. Sep 16, 2014 #2

    vanhees71

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    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].
     
  4. Sep 16, 2014 #3
    Can it also be done more generally from the definition of temperature in terms of entropy?

    $$T = (\partial S/\partial U)^{-1}?$$
     
  5. Sep 17, 2014 #4

    vanhees71

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    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]
     
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