- #1

Kaguro

- 221

- 57

- 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?

$$S = k_BlnZ_N + \beta k_B E$$, 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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