Statistical interpretation of Entropy

I'm preparing my statistical physics course, and while writing the lecture notes it says that a system with non distinguishable particles has much less microstates asociated with a particular macrostate. Hence, a system with distinguishable particles has much more microstates asociated with a particular microstate. Entropy $S$ is related with the number of microstates $\Omega$ via:

$$S=k \ln (\Omega)$$

where $k$ is Boltzmann's constant. Then my question is

Given the same macrostate for both systems, the entropy of the first is much lower than the entropy of the second?

Thanks for your time.

In the microcanonical ensemble, the probability of each microstate is assumed to be the same. In this case, the Gibbs entropy formula reduces to Boltzmann’s equation, which you gave above. Assuming there are two systems in the same macrostate, since the number of microstates is less in a system with indistinguishable particles than in a system with distinguishable particles (reason: combinatorially, there are more ways to arrange distinguishable objects), the entropy—which increases when the probability that the system will be in more microstates increases—will be lower for the system with indistinguishable particles in the microcanonical ensemble.

For the canonical or grand canonical ensembles, the probability of each microstate is different, and so the entropy may or may not be lower.

I am not sure what you mean by “much” lower, but note that the logarithm function may reduce large differences.

• Thank you for your anser. Why is not the same probability of each microstate the same in the canonical and grand canonical ensembles? I see it should be the same, by the ergodic hypothesis. – Jorge Lavín Feb 1 '12 at 19:35
• The ergodic hypothesis states that microstates with the same energy have equal probabilities. Outside the microcanonical ensemble, energy of the system is allowed to fluctuate. – user2963 Feb 1 '12 at 19:45
• +1 for mentioning the combinatorial reasoning. – onurcanbektas Apr 4 at 5:46

Let's examine a really simple "lattice gas" example. We have a lattice consisting of three cells, each of which can either be occupied with a particle or not. Let's say that the macrostate is $n=2$, i.e. there are 2 particles to be placed in each of the three cells. If we consider the case of indistinguishable particles, there are three possible microstates compatible with this macrostate:

x x .
x . x
. x x

(where x represents a particle and . is an empty cell.) So we have $S=k\log 3$ for this case. But if we consider the two particles to be different from one another, we have six microstates:

1 2 .
2 1 .
1 . 2
2 . 1
. 1 2
. 2 1

and thus $S=k\log 6$, which is greater than the other entropy by $k\log 2$. In this case the entropy was only slightly higher, but in general the number of microstates for distinguishable particles is greater than that for indistinguishable ones by a factor of $N!$, so the entropy will be greater by an amount $k\log N!$. With $N$ equal to Avogadro's number, $k\log N!$ is of the order 500 $\mathrm{JK^{-1}}$.