Understanding the Differences Between Bose and Classical Gases

In summary: No. This equation says that out of N particles, we have arranged n1 in g1 states with energy E1. But the order in which we do this doesn't matter.
  • #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?
 
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  • #2
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.
 
  • #3
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...?
 

What is the difference between Bose gas and Classical gas?

Bose gas and Classical gas are two different types of gases that behave differently based on the principles of quantum mechanics. Classical gas follows the classical laws of physics, while Bose gas follows the laws of quantum mechanics.

What is Bose gas?

Bose gas is a type of gas composed of particles that follow the Bose-Einstein statistics, meaning they can occupy the same quantum state at the same time. This leads to unique behavior, such as Bose-Einstein condensation.

What is Classical gas?

Classical gas is a type of gas composed of particles that follow the Maxwell-Boltzmann statistics, meaning they cannot occupy the same quantum state at the same time. This results in a more random and chaotic behavior compared to Bose gas.

What is Bose-Einstein condensation?

Bose-Einstein condensation is a phenomenon that occurs in Bose gases when the temperature is lowered to a critical point. At this point, a large number of particles occupy the same quantum state, resulting in a superfluid state with zero viscosity and unique properties.

What are the practical applications of Bose gas and Classical gas?

Bose gas has been used in various fields, such as creating superfluids for precision measurements and developing Bose-Einstein condensates for quantum computing. Classical gas has practical applications in fields such as thermodynamics, fluid dynamics, and atmospheric science.

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