0
$\begingroup$

The following text is from the book Introdução à Física Estatística (Introduction to Statistical Physics), Silvio Salinas. There's something in it that i didn't understand.

In order to highlight the role of probabilistic knowledge in defining entropy, Boltzmann proposed a model in which the particles of a gas could be found in a discrete set of energy values, {$ \epsilon_j ; j = 1,2,3... $}. The microscopic state of the Boltzmann gas is specified by supplying the energy of each of the particles. So, let's define the set of "occupation numbers", {$N_j$}, Where $N_1$ designates the number of particles with energy $\epsilon_1$, $N_2$ designates the number of particles with energy $\epsilon_2$, and so on. Therefore, given the set of occupation numbers {$N_j$}, the total energy E and the total number of particles N, we can write in the total number of microstates accessible to the system in the form $$\Omega(\{N_j\},E,N) = \frac{N!}{N_1!N_2!N_3!...} \tag1$$ with the constraints $$ N = \sum_{j} N_j \quad \text{and} \quad E=\sum_{j}\epsilon_j N_j \tag2 $$ the probability of finding the system in these conditions will be proportional to $\Omega (\{ N_j \},E,N)$, with the constraints imposed by the equations (2). So in order to find the occupation numbers in equilibrium, we can maximize $\Omega$ with respect to the set {$N_1$}. Using Lagrange multipliers method, we have the function $$ f(\{ N_j \}, \lambda_1, \lambda_2) = \log \Omega (\{ N_j \}, E, N)) + \lambda_1 (N - \sum_{j} N_j) + \lambda_2 (E - \sum_{j} \epsilon_j N_J). \tag3$$

Using Stirling's Approximation, we can also write in $$ \frac{\partial f}{\partial N_k} = - \log N_k - \lambda_1 - \lambda_2 \epsilon_k = 0 \tag4$$

Eliminating the Lagrange's multiplier $\lambda_1$, we have the equilibrium distribution $$ \frac{N_k}{N} = \frac{\exp(- \lambda_2 \epsilon_k)}{Z_1} \tag5 $$ where the normalization factor $Z_1$ is defined as $$ Z_1 \equiv \sum_{j} \exp(- \lambda_2 \epsilon_j) \tag6 $$

I didn't understand how the equation (5) was obtained, that is, how that $\lambda_1$ was eliminated in order to obtain equation (5).

$\endgroup$
1
$\begingroup$

We use the Lagrange multiplier $\lambda_1$ to enforce the constraint $N - \sum_j N_j = 0$, that is, total number of particles sums to $N$. Solving (4) yields

$$N_k = e^{-\lambda_1} e^{-\lambda_2 \epsilon_k}$$

$$N = \sum_k N_k = e^{-\lambda_1} \sum_k e^{-\lambda_2 \epsilon_k}$$

so that means that $e^{-\lambda_1} = \frac{N}{Z_1}$, using the normalization factor defined in (6).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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