I have a system consisting of N distinguishable particles. Each particle has two states, one with energy E and the other with energy 0. The number of particles in the state with energy 0 is $n_{0}$ and the number of particles in the second state is $n_{1}$. The number of particles in the different states satisfy $N=n_{0}+n_{1}$. Now I want to compute the number of available microstates in the microcanonical ensemble that are in agreement with the energy constraint $U=n_{1}*E$.

Since the particles are distinguishable I can arrange them in $N!$ ways. And since I have two states the total number of possibilities to arrange this system is $2*N!=2*(n_{0}+n_{1})$. Hence the partition function is:

$$ \Omega(U,N)=2*(n_{0}+n_{1})! $$

Now I want to compute the entropy in the microcanonical ensemble and maximize it with respect to U. I guess the maximum will be at $n_{0}=n_{1}=N/2$ But I don't know how to show this. My feeling somehow tells me I should use Lagrange multipliers and maximize $\Omega$.


2 Answers 2


Since the particles are distinguishable, the answer is simply the number of possible combinations of $n_1$ elements drawn from a set of $N$ elements:

$$ \Omega (U,N) = \binom{N}{n_1} $$

Where $U = n_1 E$.

You can easily show that the entropy is maximized at $n_1 = N/2$ via the properties of the number of combinations - it is maximized at that number.

  • $\begingroup$ Thank your for your comment. So if I understand correctly I to obtain the total number of realizations I have to make a sum over the binomial coefficient $\sum_{n_{1}=1}^{N}{N \choose n_{1}}$. And hence the number of states constrained by the energy is the equation you give. I think this I understand. But how do I show the maximum entropy condition I mean for even N I would just follow the N/2 approach but what would beif N is odd? $\endgroup$
    – zodiac
    Sep 28, 2018 at 6:16
  • $\begingroup$ I don't understand where the summation comes from. At any rate, that sum evaluates to just $\sum_{n_{1}=1}^{N}{N \choose n_{1}} = 2^N$. $\endgroup$
    – Al Nejati
    Sep 28, 2018 at 19:25
  • $\begingroup$ Yes correct but this is exactly the point I dont't get. For example if I have 3 particles A,B,C I and I arrange them in 2 boxes I can find much more then $2^N$ states since the particles are distinguishable. This I can proof by sketching the boxes and putting the particles in. So how do I count those states. I am sorry but I really don't understand this $\endgroup$
    – zodiac
    Oct 1, 2018 at 6:58

I figured my problem now out maybe it is also of interest for others. I have in total N labeled particles and hence I can arrange them in $N!$ ways. Now I distribute the particles in two boxes. The boxes dont't have any drawers or anything and hence in the end I am not able to discriminate between the different orders in which the particles where put into the boxes. In other words my result has to be independent of the drawing process. And hence I have to divide the total number of arrangements $N!$ by the number of particle permutations for every box. This is then just:

$$ W=\frac{N!}{n_{1}!n_{0}!} = { N \choose n_{1}} $$ like already pointed out by Al Nejati. Next the entropy in the microcannonical ensemble is given by (k_{b}=1):

$$ S=ln\left ( \frac{N!}{n_{1}!n_{0}!} \right )=ln(N!)-ln(n_{1}!)-ln(n_{0}!)\approx \\ Nln(N)-N -n_{1}ln(n_{1}) + n_{1} - n_{0}ln(n_{0}) + n_{0}=\\ Nln(N)-n_{1}ln(n_{1})-n_{0}ln(n_{0}) $$

Now I want to maximize the entropy under the constraint U=n_{1}\epsilon. Therefore the function to maximize is:

$$ f(n_{1},N)=Nln(N)-n_{1}ln(n_{1})-n_{0}ln(n_{0}) - \lambda( U - n_{1}\epsilon ) $$ Differentiating with respect to $n_{1}$:

$$ \frac{\partial f}{\partial n_{1}} = ln(n_{1}) + 1 - \lambda\epsilon $$

setting this zero gives the maximum $n_{1}$: $$ n_{1}=e^{-1+\lambda\epsilon} $$


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