- #1

Kashmir

- 439

- 73

The text is long but it is straightforward. The question is at last about the equation given at end

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

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\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

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\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

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\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**,**

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\lim _{\varepsilon^{*} \rightarrow \infty} \frac{\Sigma_{1\left(\varepsilon^{*}\right)}}{(\pi / 6) \varepsilon^{* 3 / 2}}=1.

##

* Why is the above equation true?