# A calculation of microstates

Pathria, Statistical mechanics pg 11,4ed

In order to find the number of microstates $$\Omega(N,V,E$$) author writes

" In other words, we have to determine the total number of (independent) ways of satisfying the equation $$\sum_{r=1}^{3 N} \varepsilon_{r}=E, "$$ Where $$E$$ is the total energy of system and $$\varepsilon_{r}$$ is the energy of $$r$$th degree of freedom.

" Now, the energy eigenvalues for a free, nonrelativistic particle confined to a cubical box of side $$L\left(V=L^{3}\right)$$, under the condition that the wave function $$\psi(\boldsymbol{r})$$ vanishes everywhere on the boundary, are given by $$\varepsilon\left(n_{x}, n_{y}, n_{z}\right)=\frac{h^{2}}{8 m L^{2}}\left(n_{x}^{2}+n_{y}^{2}+n_{z}^{2}\right) ; \quad n_{x}, n_{y}, n_{z}=1,2,3, \ldots,$$ where $$h$$ is Planck's constant and $$m$$ the mass of the particle. The number of distinct eigenfunctions (or microstates) for a particle of energy $$\varepsilon$$ would, therefore, be equal to the number of independent, positive-integral solutions of the equation $$\left(n_{x}^{2}+n_{y}^{2}+n_{z}^{2}\right)=\frac{8 m V^{2 / 3} \varepsilon}{h^{2}}=\varepsilon^{*} .$$ We may denote this number by $$\Omega(1, \varepsilon, V)$$. Extending the argument, it follows that the desired number $$\Omega(N, E, V)$$ would be equal to the number of independent, positiveintegral solutions of the equation $$\sum_{r=1}^{3 N} n_{r}^{2}=\frac{8 m V^{2 / 3} E}{h^{2}}=E^{*}$$"

"... the number $$\Omega(N, V, E)$$, or better $$\Omega_{N}\left(E^{*}\right)$$ is equal to the number of positiveintegral lattice points lying on the surface of a $$3 N$$-dimensional sphere of radius $$\sqrt{E}^{*}$$ . The number $$\Sigma_{N}\left(E^{*}\right)$$, which denotes the number of positive-integral lattice points lying on or within the surface of a $$3 N$$-dimensional sphere of radius $$\sqrt{E}^{*}$$. In terms of our physical problem, this would correspond to the number, $$\Sigma(N, V, E)$$, of microstates of the given system consistent with all macrostates characterized by the specified values of the parameters $$N$$ and $$V$$ but having energy less than or equal to $$E$$.

$$\Sigma(N, V, E)=\sum_{E^{\prime} \leq E} \Omega\left(N, V, E^{\prime}\right)$$ or $$\Sigma_{N}\left(E^{*}\right)=\sum_{E^{*} \leq E^{*}} \Omega_{N}\left(E^{{*\prime}}\right) .$$

"...let us examine the behavior of the numbers $$\Omega_{1}\left(\varepsilon^{*}\right)$$ and $$\Sigma_{1}\left(\varepsilon^{*}\right)$$, which correspond to the case of a single particle confined to the given volume $$V$$. The number $$\Sigma_{1}\left(\varepsilon^{*}\right)$$, on the other hand, exhibits a much smoother asymptotic behavior. From the geometry of the problem, we note that, asymptotically, $$\Sigma_{1}\left(\varepsilon^{*}\right)$$ should be equal to the volume of an octant of a three-dimensional sphere of radius $$\sqrt{\varepsilon}^{*}$$, that is, $$\lim _{\varepsilon^{*} \rightarrow \infty} \frac{\Sigma_{1\left(\varepsilon^{*}\right)}}{(\pi / 6) \varepsilon^{* 3 / 2}}=1.$$

• Why is the above equation true,I can't understand the highlighted text?

The condition: $$$$\left(n_x^2 + n_y^2+n_z^2\right) \leq \epsilon^*,$$$$ together with $$n_i \in \mathbb{Z}_+$$ indicates one octant of a sphere in $$n$$-space in the limit $$\epsilon^* \rightarrow \infty$$. This condition comes from the similar condition for $$\Omega$$ but with equality replaced with inequality because now we admit also the states with energy lower than $$\epsilon^*$$.
• Why should the number of those integral points in the first octant which satisfy $$$\left(n_x^2 + n_y^2+n_z^2\right) \leq \epsilon^*,$$$ be equal to the volume of the octant? Commented Mar 31, 2022 at 16:14
• Each point corresponds to a cube of volume 1, so counting the points is the same as calculating the volume. The only differences are related to the boundary, which is somewhat rough in the finite $\epsilon^*$, but all those problems are washed away ni the limit $\epsilon^* \rightarrow \infty$. Commented Mar 31, 2022 at 19:42
• The difference between the volume of the octant of the ball of radius ${\epsilon^*}^{1/2}$ and the number of cubes fitting inside said octant is smaller than the number of the cubes cut through by on1 octant of the sphere of the same radius. This number of cubes is of the same order as the area of said sphere. The area of the sphere will be negligible in comparison with the volume in the limit of infinite radius. Commented Apr 1, 2022 at 10:08