What is the significance of Δx and Δp in the ideal gas entropy equation?

[etotheipi]

Up to an undetermined constant $a$ the entropy of an ideal gas goes like$$S = k_B N\ln \left[ a^\frac{3}{2} \left(\frac{V}{N}\right) \left(\frac{E}{N}\right)^{\frac{3}{2}} \right]$$In some notes is written:

We can consider that E/N is the average energy of each molecule and V/N is the average volume occupied by each molecule (both are therefore intensive while S is extensive). How do we relate this equation to the Boltzmann’s law of Eq. (1) and how do we determine the constant a? Clearly, if we let$$\frac{(\Delta p)^2}{2m} = \frac{E}{N},\quad (\Delta x)^3 = \frac{V}{N}$$then, roughly speaking, $\Delta p$ is the magnitude of the change of momentum of the molecules and $\Delta x$ is that of the change of position

And then they identify $\Omega = \left(\frac{\Delta x \Delta p}{w}\right)^{3N}$ from Boltzmann's formula, with $w=\sqrt{2m/a}$.

I am confused about what $\Delta x$ and $\Delta p$ represent here. At first glance it looks like the product $\Delta x \Delta p$ represents some area element in the $x$/$p_x$ phase space, but if that is true, I don't know how we obtain the relation $\frac{(\Delta p)^2}{2m} = \frac{E}{N}$... if anything I would have thought it be something like$$\frac{1}{N} \sum_i^N \frac{p_i^2}{2m} = \frac{E}{N}$$I hoped someone could clarify. Thank you!

 

[vanhees71]

$\Omega$ is the phase-space measure. The problem in classical statistics is the lack of a physical determination of the phase-space measure. This is of course immediately cured by quantum theory. As to be expected it turns out that $w=h=2 \pi \hbar$, counting the number of free-particle states per phase-space-volume element. Just put free particles in a finite-volume box like a cube and assume periodic boundary conditions (it should be periodic rather than rigid boundary conditions, because otherwise you don't have a cleanly defined self-adjoint momentum operator). Then you can calculate everything and finally take the "thermodynamic limit" $V \rightarrow \infty$ for intensive quantities like particle-number density, energy density etc. The limit is not always trivial, as already the example of Bose-Einstein condensation of a free gas shows.

 

[Livio Arshavin Leiva]

I think $\Delta p$ is considered as the variance of the momentum of the particle population. It should then be calculated as
$$(\Delta p)^2=\left( \frac{1}{N}\sum_{i=1}^{N}p_i^2\right)-\bar{p}^2$$
but $\bar{p}=0$ since it's an ideal gas.