Volume of momentum space of an ideal gas

[LightPhoton]

In *An Introduction to Thermal Physics* by Schroeder, while deriving the multiplicity of an ideal gas makes the following statements (image below):

Even in quantum mechanics, the number of allowed wavefunctions is infinite.
But the number of independent wavefunctions (in a technical sense that’s defined
in Appendix A) is finite, if the total available position space and momentum space
are limited. I like to picture it as in Figure 2.9. In this one-dimensional example,
the number of distinct position states is L/Δ x, while the number of distinct momentum states is L_p /Δp_x The total number of distinct states is the product,


$$\frac{L\,L_p}{\Delta x\,\Delta p_x}=\frac{L\,L_p}{h}$$
according to the uncertainty principle. In three dimensions, the lengths become
volumes and there are three factors of h:
$$\Omega=\frac{V\,V_p}{h^3}$$


Here we have an ideal monoatomic gas's single atom in a cuboidal box, whose momentum is constrained in the following manner

$$p^2_x+p^2_y+p^2_z=2mU$$

Where m is its mass and U is its kinetic energy.

Since this is the equation of a sphere, we can multiply by a small thickness Δp_r to make it a volume.




Now the calculation for the momentum volume in Schroeder seems off to me, in that the volume here is a sphere rather than a cubical box, thus instead of using $\Delta p_x,\Delta p_y,\Delta p_y$, shouldn't $\Delta p_r,\Delta p_\theta,\Delta p_\phi$ (uncertainties in spherical coordinates) be used? this would alter the uncertainty products as well, making the overall constants a bit different.

 

[anuttarasammyak]

Phase space cell of volme h is like a piece of Lego block any shapes of which are allowed keeping volume h. I do not share your concern because I think any figure in phase space can be constructed by the cells , like sphere is made of square Lego blocks.