2
$\begingroup$

I'm stuck halfway through a derivation of the Boltzmann distribution using the principle of maximum entropy.

Let us consider a particle that may occupy any discrete energy level $\mathcal{E}_i$. The probability of occupying this level is $p_i$. I wish to now maximize the entropy $H = -\sum_i p_i\log (p_i)$, subject to constraints $\sum_i p_i = 1$ and $\sum_i p_i\mathcal{E}_i = \mu$. That is, the average energy is known.

I write the Lagrangian $\mathcal{L} = \sum_i p_i\log (p_i) + \eta(\sum_i p_i - 1) + \lambda(\sum_i p_i\mathcal{E}_i - \mu)$. With the method of Lagrange multipliers, I can set $\frac{\partial\mathcal{L}}{\partial p_j} = 0$, $\frac{\partial\mathcal{L}}{\partial\eta} = 0$ and $\frac{\partial\mathcal{L}}{\partial\lambda} = 0$.

From the derivatives, I get

$p_i = e^{\eta -1}e^{-\lambda\mathcal{E}_i}$

$\sum_i e^{\eta -1}e^{-\lambda\mathcal{E}_i}\mathcal{E}_i = \mu$

$\sum_i e^{\eta -1}e^{-\lambda\mathcal{E}_i} = 1$

1) I'm not sure how to complete the argument and find $\lambda$ and $\eta$. I seem to get a sign error or something else is wrong. How do I get to $\lambda = \frac{1}{\mu}$?

2) Can someone show how to extend this argument for continuous energies?

$\endgroup$
2
$\begingroup$

Your target formula $\lambda=1/\mu$, where $\mu$ is the average energy, is wrong.

(NB throughout this answer, I'm following your usage of $\mu$ as the specified "mean value" of the energy; any casual readers should note that it is not the chemical potential).

This derivation goes back to two papers of Jaynes: Phys Rev, 106, 620 (1957) and Phys Rev, 108, 171 (1957). I've also seen copies of these papers made available online, perhaps you can search for them. It is also fairly standard in textbooks.

As with many Lagrange multiplier questions, direct solution of the equations to get values of the multipliers themselves is avoided. For the "normalization" multiplier, $\eta$ in your notation, it is more convenient to replace it by the partition function $Z$, writing $$ p_i = \frac{\exp(-\lambda \mathcal{E}_i)}{Z}, \quad\text{where}\quad Z = \sum_i \exp(-\lambda \mathcal{E}_i) . $$ The remaining multiplier $\lambda$ is eventually related to the derivative of the maximized entropy with respect to the average energy $\mu$. In the language of those two papers, given constraints on the average values of several functions $\langle f_1(x)\rangle$, $\langle f_2(x)\rangle$, etc, where $x$ takes discrete values $x_i$ characterizing each state $i$, and $\langle f(x)\rangle$ is short for $\sum_i p_i \, f(x_i)$, the probability distribution which maximizes the entropy can be written $$ p_i = \exp[-(\lambda_0+\lambda_1f_1(x_i)+\lambda_2f_2(x_i)+\ldots)] $$ where $\lambda_0=\ln Z$ and the other $\lambda_i$ are the remaining Lagrange multipliers. Noting from this that $$ -\ln p_i = \lambda_0+\lambda_1f_1(x_i)+\lambda_2f_2(x_i)+\ldots $$ and substituting into $S=-\sum_i p_i \ln p_i$ (your function $H$), it follows that the maximum value of the entropy can be expressed $$ S_{\text{max}} = \lambda_0 + \lambda_1 \langle f_1(x)\rangle + \lambda_2 \langle f_2(x)\rangle + \ldots . $$ From this, $$ \lambda_i = \frac{\partial S_{\text{max}}}{\partial \langle f_i(x)\rangle} . $$ In your case, $$ S_{\text{max}} = \ln Z + \lambda \langle E\rangle = \ln Z + \lambda \mu $$ this eventually leads to $$ \lambda =\frac{\partial S_{\text{max}}}{\partial \mu} =\frac{\partial S_{\text{max}}}{\partial \langle E\rangle} = \frac{1}{T} $$ thereby defining the temperature, as it is used in standard thermodynamics (all in units of $k_B=1$).

Generalizing to the continuum case is discussed on the relevant Wikipedia page, and involves calculus of variations, with summations replaced by integrals, and Lagrange multipliers playing their usual role. It is necessary, though, to introduce a measure, for the integration. You might formulate the problem as an integral over coordinates $x$, or coordinates-and-momenta, $(x,p)$. Or you might formulate it as an integral over energies. In any case, you need to consider the density of states (per unit whatever the integration variable is).

$\endgroup$
  • $\begingroup$ Can I ask why $S_{max} = \lambda_0 +\lambda_1\langle f_1(x)\rangle + ..$? $\endgroup$ – user1936752 Jan 12 at 15:04
  • $\begingroup$ I've edited my answer to make this (hopefully) more clear. $\endgroup$ – LonelyProf Jan 12 at 17:02
  • $\begingroup$ Yep, that line makes it clear, thank you for a great answer! $\endgroup$ – user1936752 Jan 12 at 18:54

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.