# Bose gas versus Classical gas

Kaguro
Homework Statement:
Find the entropy of a Bose gas.
Relevant Equations:
Many.
In classical statistics, we derived the partition function of an ideal gas. Then using the MB statistics and the definition of the partition function, we wrote:

$$S = k_BlnZ_N + \beta k_B E$$, where ##Z_N## is the N-particle partition function. Here ##Z_N=Z^N##

This led to the Gibb's paradox. Then we found that the problem was assuming all particles are distinguishable. So to correct that we divided the ##Z_N## by N! and hence derived the Sackur-Tetrode equation.

$$S = \frac{5}{2} N k_B + N k_B ln[ \frac{N}{V}(\frac{2\pi m k_B T}{h^2})^{3/2} ]$$

My question is, doesn't this make the particles same as that of a Bose gas? Identical and indistinguishable.

The only reason MB gas doesn't form BE condensate is because the distribution function doesn't have a -1 in denominator which would force the chemical potential to have a maximum value of 0.

So shouldn't the S-T equation give the value for entropy of a Bose gas too?

Also, in MB we assumed distinguisble particles. The total number of ways to distribute energy is W .

Correction by W'=W/N! has the same conditions as BE statistics, but it doesn't produce the same distribution function. Why?

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## Answers and Replies

Mentor
Homework Statement:: Find the entropy of a Bose gas.
Relevant Equations:: Many.

My question is, doesn't this make the particles same as that of a Bose gas? Identical and indistinguishable.
No, because the counting is still different. For classical particles, particle 1 in state a and particle 2 in state b is not the same as particle 1 in state b and particle 2 in state a. You have to count both multi-particle states. With quantum particles, this would correspond to a single multi-particle state, with one particle in state a and the other in b.

Kaguro
Earlier,
##W=\frac{N!}{n1! n2! n3!...}g_1^{n1} g_2^{n2}... ##

Now we have divided by N! the get:
##W=\frac{1}{n1! n2! n3!...}g_1^{n1} g_2^{n2}... ##

How can I interpret this one?

Doesn't this just mean, out of N particles, we have arranged n1 in g1 states with energy E1. And their order doesn't matter, because we divide by n1! And so on...?