Given a large number N of distinguishable particles distributed among M boxes, we know that the total number of possible microstates is $M^N$ and that the number of microstates with a distribution among the boxes given by the configuration $[n_1, n_2, ..., n_M]$ is given by $$\Omega= \frac{N!}{\Pi_{j=1}^M n_j!}$$ I need to show that the most likely distribution sees the particles equally distributed among the M boxes. I know I need to use the Lagrangian variation of parameters, as the number of particles $N$ is constant. By doing so I got that $\ln(n_j)+\alpha =0 $ where $\alpha$ is my Lagrangian parameter, for all $j$. I just do not know how to get $n_j=\frac{N}{M}$ from here... Is my reasoning wrong?


You're essentially done. You know that $n_j$ is constant in $j$; from the constraint that $\sum_j n_j=N$ you get $n_j=N/M$. That's the beauty of Lagrange multipliers!

  • $\begingroup$ Oh that's easier than I thought! Just to confirm, I get the condition that $n_j$ is constant in $j$ through the $ln(n_j)+\alpha =0$ equation, right? $\endgroup$
    – Agnese
    Apr 2 at 20:39
  • 1
    $\begingroup$ Yes, exactly $\log n_j=-\alpha$ for all j $\endgroup$ Apr 2 at 21:24

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