# For ideal classical gasses, in terms of the energy levels why do we ignore whether the particles are fermions or bosons?

I am confused as to why when dealing with ideal classical gasses, the dependency of the particles being either fermions or bosons is ignored. How does this relate to the energy levels within the system?

I thought it had something to do with the fact that ideal classical gasses are at temperatures where the thermal energy kT is much greater than the spacing between the energy levels but I am not entirely certain whether this is the correct way I should be thinking.

Any knowledge to make me a little less ignorant on the matter will be greatly appreciated.

• Also in the electron gas, $kT$ is much larger than the level spacing. – Pieter Dec 15 '17 at 23:55

Because both the Fermi-Dirac distribution and the Bose-Einstein distribution are well approximated by the Maxwell-Boltzmann distribution in the limit of low density. Basically, the assumption behind the ideal gas is that the gas density is low enough that collisions are not a significant factor in describing the dynamics of the gas, allowing us to go from the thermodynamic distribution of 1 particle to the gas. When density starts becoming high, the first correction is usually described by the Van der Waals equation. As the temperature drops or density goes up further, you have to start worrying about the Boson/Fermion distinction.

More precisely, it's not about $kT$ compared to some energy level spacing, it's compared to the chemical potential. In detail, it's having $kT$ high enough that \begin{align} \mathrm{BE}(E) & = \frac{1}{\operatorname{e}^{(E-\mu)/kT}-1} \mathrm{\ and} \\ \mathrm{FD}(E) & = \frac{1}{\operatorname{e}^{(E-\mu)/kT}+1} \end{align} are adequately approximated by \begin{align} \mathrm{MB}(E) & = \operatorname{e}^{-E/kT}. \end{align}

Note that the approximation is only "good" at the low end ($E<kT$) when $\mu<0$.

For Fermions, the chemical potential is the Fermi energy or greater, which is controlled by the particle density. I'm having trouble finding a reference for how to find the chemical potential for the Bose-Einstein distribution. ResearchGate hosts a plot of the chemical potential of Heliums 3 and 4 at low temperature ($\mu/k$ is around $-2$ and $-7$ Kelvin for them, respectively).

• It's NOT "having $kT$ high enough"! The requirement for the approximation of BE and FD distribution by the Botzmann distribution is $kT<< E$. – freecharly Dec 16 '17 at 1:42
• @freecharly You sure that it's in the low temperature limit that BD andFD are well approximated by MB? Please show your work. – Sean E. Lake Dec 16 '17 at 1:48
• The condition $E>> kT$ appears to be the mathematical requirement that the BE and DE distribution functions are both approximated by $e−E/kT$. – freecharly Dec 16 '17 at 2:20
• Thanks, @freecharly, for helping me catch my error. For what it's worth, $E$ cannot be used to compare distributions as a whole since it is the variable in the distribution. The parameters $kT$ and $\mu$ control the shape of the distribution, overall, so they're the only valid pieces of information when discussing the shape of the distribution. – Sean E. Lake Dec 16 '17 at 3:11
• “the assumption behind the ideal gas is that the gas density is low enough that collisions are not a significant factor in describing the dynamics of the gas” – actually, collisions are important: to ensure equipartition. Really, what you need for an ideal gas is a free path that's much longer than the particle length scales, but still significantly shorter than the system size. The latter is generally given at any near-atmospheric pressure, but notably often not in outer space, where you always need to consider whether ideal MHD is an approriate model or you need a kinetic description. – leftaroundabout Dec 16 '17 at 12:53

Another way of seeing it is that the distance between atoms is large compared to their de-Broglie wavelength. Then it does bot matter that one should use statistics of indistinguishable particles - it would in principle still be possible to follow a particle most of the time.

There are already good answers; I'll just add another way to see this. Let $n$ be the number of particles in a particular quantum state. Using Maxwell-Boltzmann statistics, you can calculate a probability distribution $p_{\text{MB}}(n)$.

Fermions modify this distribution by forbidding more than one particle in the same state, $$p_{\text{Fermi}}(2) = p _{\text{Fermi}}(3) = \ldots = 0.$$ Bosons modify this distribution by preferring to 'clump up', i.e. one tends to see groups of bosons in the same state, $$p_{\text{Bose}}(2) \gtrsim p_{\text{MB}}(2), \quad p_{\text{Bose}}(3) \gtrsim p_{\text{MB}}(3), \ldots$$ In all cases, the average $\langle n \rangle$ is the same since there are the same number of total particles.

The limit where all of these distributions are the same is the low-density limit $\langle n \rangle \ll 1$. In this case, the overwhelming majority of the probability is concentrated in $p(0)$ with a tiny bit in $p(1)$. The modifications that the Fermi and Bose distributions make to $p(2)$ and higher are negligible.

Since there's one quantum state for every Planck's constant of phase space area, $\langle n \rangle \ll 1$ is equivalent to $$(\text{typical momentum})(\text{typical distance between neighboring particles}) \gg h.$$ As already mentioned, this is equivalent to saying that the particles are separated by a distance much greater than their de Broglie wavelength.

• Thank you very much. I wish there was more than one green button. All the answers were quite helpful but this really made things make more sense to me! – C.Pic Dec 17 '17 at 21:10

Classical ideal gas is approximation such that number of energy states is very large. (g>>n) So the particles don't compete to occupy same energy state. Actually, we ignore because this case(particles with same state) is very unlikely.

And if number of states is not large enough, then we must consider quantum statistics.