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I Deriving the thermodynamic beta from Lagrange Multipliers

  1. Aug 22, 2016 #1
    I'm nearly at the end of this derivation but totally stuck so I'd appreciate a nudge in the right direction

    Consider a set of N identical but distinguishable particles in a system of energy E. These particles are to be placed in energy levels ##E_i## for ##i = 1, 2 .. r##. Assume that we have ##n_i## particles in each energy level. The two constraints we impose are that ##\sum_{i}^{r}n_i = N## and ##\sum_{i}^{r}E_i n_i = E##.

    The number of microstates in a given macrostate is given by
    \begin{equation}
    \Omega = \frac{N!}{\sum_{i}^r n_{i}!}
    \end{equation}

    We want to maximize this and for ease of notation, we work with ##\ln\Omega## and we use Stirling's approximation (##\ln x! = x\ln x - x##) to obtain
    \begin{equation}
    \ln\Omega = N\ln N - N - \sum_{i}^{r}n_i\ln n_i - n_i
    \end{equation}

    Maximizing this function subject to the constraints ##\sum_{i}^{r}n_i = N## and ##\sum_{i}^{r}E_i n_i = E## is a classic Lagrange multiplier problem. We represent the undetermined multipliers to be ##\alpha## and ##\beta## for the two constraints and obtain
    \begin{align}
    \frac{\partial\ln\Omega}{\partial n_i} &= \alpha\frac{\partial n_i}{\partial n_i} + \beta\frac{\partial E_i n_i}{\partial n_i} \\ \nonumber
    \ln n_i &= \alpha + \beta E_i \\ \nonumber
    \therefore n_i &= e^{\alpha}e^{\beta E_i}
    \end{align}

    Now, we use the first constraint equation to determine ##\alpha##. We get
    \begin{align}
    \sum_i^r n_i &= N \\ \nonumber
    \sum_i^r e^{\alpha}e^{\beta E_i} &= N \\ \nonumber
    e^\alpha &= \frac{N}{\sum_i^re^{\beta E_i}} \\ \nonumber
    e^\alpha &= \frac{N}{Z}
    \end{align}

    We have introduced the partition function, ##Z=\sum_i^re^{\beta E_i}## in the last line. Next, we have the second constraint equation that determines ##\beta##
    \begin{align}
    \sum_i^r E_i n_i &= E \\ \nonumber
    \frac{\sum_{i}^{r} E_i e^{\beta E_i}}{\sum_i^r e^{\beta E_i}} &= E \\ \nonumber
    \end{align}

    I'm assuming I should somehow connect ##E## with ##T## so let's say ##E=Nk_B T##. Then we have
    \begin{align}
    \frac{N}{Z}\frac{\partial Z}{\partial\beta} &= E \\
    \frac{\partial\ln Z}{\partial\beta} &= k_B T
    \end{align}

    How do I get to ##\beta = -\frac{1}{k_B T}## here? Notice that this derivation requires an extra minus sign compared to the usual definition of ##\beta## and this should come out naturally too, shouldn't it?
     
  2. jcsd
  3. Aug 23, 2016 #2

    vanhees71

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    Science Advisor
    2016 Award

    Of course, it's easier to flip the sign of ##\beta## from the very beginning, because then you have ##\beta>0## for the usual case where the Hamiltonian of the system is bounded from below but can become infinite (e.g., for an ideal gas you have ##H=\vec{p}^2/(2m) \geq 0##. Then you have
    $$Z(\beta,\alpha)=\sum_i \exp(-\beta E_i + \alpha n_i).$$
    Now to get the interpretation of ##\beta## and ##\alpha## in terms of the thermodynamical quantities note that
    $$U=\langle E \rangle=-\partial_{\beta} \ln Z, \quad \mathcal{N}=\langle N \rangle=\partial_{\alpha} \ln Z.$$
    Further the probability distribution is
    $$P_i=\frac{1}{Z} \exp(-\beta E_i+\alpha \mathcal{N}).$$
    This implies that the entropy is given by
    $$S=-k_B \sum_i P_i \ln P_i \rangle=k_B (\ln Z+\beta U-\alpha \mathcal{N}).$$
    With the above relations you get
    $$\frac{1}{k_B} \mathrm{d} S = \beta \mathrm{d} U - \alpha \mathrm{d} \mathcal{N}.$$
    Then comparing this to the 1st Law
    $$\mathrm{d} U=T \mathrm{d} S +\mu \mathrm{d} \mathcal{N}$$
    you'll get
    $$\beta=\frac{1}{k_B T}, \quad \alpha=\frac{\mu}{k_B T}.$$
     
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