A calculation of microstates

Question

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?

Answer 1

The condition: \begin{equation} \left(n_x^2 + n_y^2+n_z^2\right) \leq \epsilon^*, \end{equation} 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 $\epsilon^*$.