Suppose, I have a system of $N$ particles, and $k$-degenerate energy states, such that the total energy of the system is $\epsilon$. Suppose, $n_i$ be the number of particles in the $i$-th energy state, with energy $E_i$. Then I have the constraints :
$$\sum_{i}n_i=N$$
$$\sum_i n_iE_i=\epsilon$$
Now any set of occupation numbers {$n_i$} that fulfill the above criteria, would be possible mode of distributing the total energy among the $N$ particles, that satisfies the constraints. Moreover, let us consider $W_i$ to be the number of ways of distributing the particles to obtain {$n_i$}. In case of distinguishable particles,
$$W_i=N!\prod_i\frac{g_i^{n_i}}{n_i!}$$
Suppose I want to now find out the total number of microstates of this system. I have two ways of thinking about this, and I don't see which one is correct.
Case 1 :
All sets of {$n_i$} that satisfy the constraints would appear as the system. Moreover, $W_i$ is the number of ways {$n_i$} would occur. So the total number of microstates would be the total number of ways in which all the sets of {$n_i$} occur.
Hence, $$\Omega = \sum_{ n_i} W_i$$
Case 2 :
While all sets of {$n_i$} are equally likely to occur, in reality, it can be shown that the set that maximizes $W_i$, is the most probable distribution, that is going to appear.
Since, the only possible mode of distribution, is now this particular mode, the total number of microstates would just be the number of ways of realizing this particular mode.
So, $$\Omega = W_i$$
I can't understand which one is the correct way of thinking.
For example, if there are huge number of particles, then the number of particles in a particular level $i$ would be nothing but $P(i)N$, where $P(i)$ is the probability of a particle being in the $i$-th level.
In this case, we would know the number of particles in each level ( average occupancy ), and the total number of microstates would be the number of ways of attaining this occupancy {$n_i$}.
So, this system behaves like case 2
However, now consider $2$ distinguishable particles, in $4$ energy levels, with energies $0,E,2E,3E$, and the total energy of the system must be $3E$.
In this case the occupation set {$n_0,n_1,n_2,n_3$}, can take two values {$1,0,0,1$} and {$0,1,1,0$}. For each of these, $W_i=2$, and the total number of microstates is $4$.
This system behaves like case 1.
Can anyone clear this out for me, and tell me the correct way of thinking about it.
Best Answer
Note: it would be better to indicate the sum of the total microstates as $\Omega=\sum_{\{n_i\}}^* W{\{n_i\}}$, where $^*$ indicates that the sum is performed over the sets $\{n_i\}$ that satisfy the constraint on total energy end total number of particles.
That said, case 2 is simply the thermodynamic limit of case 1. When the number of microstates $\Omega$ is very high, it is convenient to consider the configuration $\{\overline{n}_i\}$ that maximies $W{\{n_i\}}$ and treat the other contributions to the sum as fluctuations (of course these fluctuations are more important for small systems). This is a kind of saddle point approximation.