In lecture notes, Atkin's Physical Chemistry, etc. the Boltzmann Distribution is derived by first claiming that the system takes on properties of its weightiest configuration. The weight is therefore maximised, utilising Lagrange multipliers.

\begin{align} \nabla \ln{W} - \alpha \ \nabla \sum_i n_i - \beta \ \nabla \sum_i n_i\epsilon_i &= 0 \\ \to \frac{\partial{\ln{W}}}{\partial{n_i}} - \alpha - \beta \epsilon_i &= 0 \end{align}

The Stirling Approximation transforms the derivative to something we can work with, and the Boltzmann distribution is eventually derived.

\begin{align} \frac{\partial \ln{W}}{\partial n_i} &= -\ln{\frac{n_i}{N}} \\ &\vdots \\ \frac{n_i}{N} &= \frac{e^{-\beta\epsilon_i}}{\displaystyle \sum_i e^{-\beta\epsilon_i}} \hfill \end{align}

The derivation is left there with no mention of the value of $\beta$, though it is later claimed $\beta = \frac{1}{kT}$. How is the value of $\beta$ ascertained?

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    $\begingroup$ Isn't that the definition of temperature? $\endgroup$ Apr 1, 2018 at 12:02
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    $\begingroup$ Hardly an expert on this, but afaik you can derive $\beta = 1/kT$ if you define $S = k\ln W$, but I've also seen some cases where one defines $\beta = 1/kT$ and derives $S = k\ln W$ from it. I guess either way it is a postulate. $\endgroup$ Apr 1, 2018 at 13:31

2 Answers 2


The relationship between $\beta$ and $T$ comes from experimental observations like $E=\frac{3}{2}RT$, i.e. if you add $E$ amount of energy to a mole of monatomic gas, then you raise the temperature by $\frac{3}{2}R$ degrees.

$E = N_A\langle \varepsilon \rangle = N_A \sum_i p_i \varepsilon_i = \frac{N_A}{q} \sum_i \varepsilon_i e^{-\beta\varepsilon_i}$

$\varepsilon_i e^{-\beta \varepsilon_i} = - \frac{d }{d \beta} e^{-\beta \varepsilon_i} \to \langle \varepsilon \rangle = - \frac{1}{q} \frac{\partial q}{\partial\beta}$

$q = \left( \frac{2\pi m}{h^2\beta} \right)^\frac{3}{2}V \to E = \frac{3N_A}{2\beta} \to \beta = \frac{N_A}{RT} $

  • $\begingroup$ What OP cited is a derivation of the Boltzmann distribution starting from a micro-canonical ensemble (which has a fixed energy). Here, $\beta$ arises as a Lagrange multiplier, and one needs to relate it to the thermodynamic tenperature $T = 1/(\partial S / \partial E)$. On the other hand, you started from a partition function, which means you are working with a canonical ensemble that has a fixed temperature. In this case, you have $\beta=1/kT$ from the very beginning, and you don't need to compare against the internal energy of the monatomic ideal gas to derive what is already there. $\endgroup$
    – higgsss
    Apr 3, 2018 at 1:59

What OP cited is the derivation of the Boltzmann distribution starting from the micro-canonical ensemble, characterized by a fixed energy and a fixed number of constituent particles. In mathematical terms, the problem statement is as follows:

Maximize the function $\ln W(\{n_i\})$ under the constraints $\sum_i n_i \epsilon_i = E$ and $\sum_i n_i = N$.

The solution is a function of $E$ and $N$, so let's define

$\ln \Omega(E, N)$ := (constrained maximum of $\ln W$ as a function of $E$ and $N$),

$\tilde{n}_i (E, N)$ := (value of $n_i$ at the constrained maximum).

Notice that $\ln\Omega = \ln W(\{\tilde{n}_i\})$. Also, this quantity is what appears in Boltzmann's entropy formula $S=k\ln\Omega$ because $\Omega$ is the number of available states for given values of $E$ and $N$. This observation enables us to relate $\ln\Omega$ to the thermodynamic temperature: \begin{equation} \frac{\partial \ln \Omega}{\partial E} = \frac{1}{k}\frac{\partial S}{\partial E} = \frac{1}{kT}. \end{equation} Now, OP's question can be rephrased as follows:

When deriving the Boltzmann distribution starting from the micro-canonical ensemble, $\beta$ is introduced as a Lagrange multiplier associated with the constraint $\sum_i n_i \epsilon_i = E$. How can we relate $\beta$ to the thermodynamic temperature, i.e., establish that $\beta = \frac{\partial \ln \Omega}{\partial E}$?

The underlying principle is really the chain rule in multivariate calculus: \begin{equation} \begin{split} \frac{\partial \ln \Omega}{\partial E} &= \sum_i \frac{\partial \ln W}{\partial n_i}\bigg|_{n_i = \tilde{n}_i} \frac{\partial \tilde{n}_i}{\partial E}\\ &= \sum_i (\alpha + \beta\epsilon_i)\frac{\partial \tilde{n}_i}{\partial E}\\ &= \alpha\frac{\partial}{\partial E}\Big(\sum_i \tilde{n}_{i}\Big) + \beta\frac{\partial}{\partial E}\Big(\sum_i \tilde{n}_{i}\epsilon_i\Big)\\ &= \alpha \frac{\partial N}{\partial E} + \beta \frac{\partial E}{\partial E}\\ &= \beta. \end{split} \end{equation}

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    $\begingroup$ I'm not sure I follow where $\beta = \frac{\partial \ln{\Omega}}{\partial E}$ is coming from. Could you clarify that? $\endgroup$
    – Jacob
    Apr 1, 2018 at 17:40
  • $\begingroup$ If you are talking about where I rephrased your question, the reason is that $\Omega (E, N)$ is what enters Boltzmann's relation $S = k \ln \Omega$, where $S$ is a function of $E$ and $N$. Then, one should have $\partial \ln\Omega /\partial E = k^{-1}\partial S / \partial E = 1/kT$. $\endgroup$
    – higgsss
    Apr 1, 2018 at 17:51

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