Understanding the Differences Between Bose and Classical Gases

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SUMMARY

This discussion focuses on the distinctions between Bose-Einstein (BE) gases and classical Maxwell-Boltzmann (MB) gases, particularly in the context of entropy calculations. The Sackur-Tetrode equation is derived from classical statistics, leading to the conclusion that classical particles are treated as distinguishable, which differs from the indistinguishable nature of Bose gases. The participants clarify that while the entropy formula for classical gases appears similar to that of Bose gases, the underlying statistical mechanics differ due to the treatment of particle indistinguishability and the distribution functions involved.

PREREQUISITES
  • Understanding of statistical mechanics principles
  • Familiarity with the Sackur-Tetrode equation
  • Knowledge of Bose-Einstein statistics
  • Concept of partition functions in thermodynamics
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  • Study the derivation of the Sackur-Tetrode equation in detail
  • Explore the implications of indistinguishability in quantum statistics
  • Learn about the conditions under which Bose-Einstein condensation occurs
  • Investigate the differences in entropy calculations between classical and quantum gases
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Physicists, students of statistical mechanics, and anyone interested in the fundamental differences between classical and quantum gas behaviors.

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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