Understanding the Differences Between Bose and Classical Gases

AI Thread Summary
The discussion focuses on the differences between Bose and classical gases, particularly in the context of the partition function and entropy calculations. It highlights that while both gases consist of indistinguishable particles, the counting methods differ significantly. In classical statistics, particles are treated as distinguishable, leading to the need for corrections like the Sackur-Tetrode equation, which accounts for indistinguishability by dividing by N!. However, this approach does not yield the same distribution function as Bose-Einstein statistics, which inherently considers the indistinguishable nature of quantum particles. The conversation concludes that while the entropy calculations may seem similar, the underlying statistical mechanics principles differentiate the behaviors of Bose and classical gases.
Kaguro
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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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Kaguro said:
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.
 
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...?
 
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