The weight, $W$ (the same $W$ as in Boltzmann formula $S=k\text{ln}W$) of a configuration {$N_0, N_1, N_2$, ...} is given by

$$W=\frac{N!}{N_A!N_B!N_C!...} \tag{1}$$

where $N$ is the total number of molecules in the system and $N_j$ is number of molecules with state $j$.

The denominator is to account for different permutations of indistinguishable molecules. For example, molecule labelled 1 and 2 below have exactly the same energy and are of a same element. If the sample is a perfect gas, molecule 1 and 2 will be indistinguishable. The simple N factorial overcounts by counting such many different permutations which are indistinguishable multiple times, and the denominator in (1) is to correct this. Molecule 1 and 3 are distinguishable; they have different energies.

enter image description here

If we proceed to operate on (1) using Stirling's approximation and through some simple algebraic manipulations, the equation for entropy below can be deduced:

$$S(T) = \frac{U(T)-U(0)}{T}+Nk\text{ln}q \tag{2}$$

Because (2) is derived from (1), (2) has already taken into account of different permutations of indistinguishable molecules, just like (1).

Now here is the question, the entropy of perfect gas is actually given by

$$S(T) = \frac{U(T)-U(0)}{T}+Nk\text{ln}q -Nk(\text{lnN}-1)\tag{3}$$

(3) is derived using canonical partition function and

$$Q = \frac{q^N}{N!} \tag{4}$$

which is true for independent, indistinguishable molecules.

(2) can be reached similarly, but using

$$Q = q^N \tag{5}$$

which is true for independent and distinguishable molecules.

What is wrong with the argument above? I thought (2) has already accounted for the fact that the molecules are indistinguishable, but proven otherwise by canonical partition function, or did I understand the indistinguishability wrongly.

  • 4
    $\begingroup$ The $W$ you give is the number of ways of placing $N$ distinguishable objects into $r$ distinguishable boxes so that they contain $N_A,N_B\cdots $ each. Is that what you wanted? $\endgroup$
    – porphyrin
    Commented Aug 17, 2021 at 6:28
  • 2
    $\begingroup$ You should also have a look at the Gibbs Paradox. The partition function is divided by $N!$ because the $N!$ permutations of the molecules among themselves do not lead to physically different situations so the number of states summed in the partition function is too large by this amount. $\endgroup$
    – porphyrin
    Commented Aug 17, 2021 at 17:33

1 Answer 1


Inspired by user porphyrin.

A common, potentially confusing way of approaching configuration is to think of each energy state as a distinguishable box, and each molecule as a marble.

The number of ways of arranging N marbles in a line is N!. But the exact arrangement of marbles in one box is irrelevant. If molecule 1 and 2 are in state A, they both have energy $\epsilon_A$, and that's all. There is no such thing as 'arrangement of molecules in a state'. State is not a physical entity; if two molecules are in a same state they have the same energy; it is not like the two molecules are really enclosed by a confinement.

For illustration, the two cases shown below are counted separately by N! but they are just exactly the same physically because molecule 1 and 2 have energy $\epsilon_A$, molecule 3, 4, 5 have energy $\epsilon_B$, and so on.

enter image description here

When making the diagram above, the molecules have already been assumed to be distinguishable; they can be labelled with different numbers, and you can tell which one is 1, which one is 2...

$\therefore$ The equation


is for distinguishable molecules.

For indistinguishable molecule, the labelling should not be done:

enter image description here

But that gives rise to another question...This will mean that for indistinguishable molecules the W of any configuration is 1, so $S=k\text{ln}W = 0$?


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