I am looking at the start of the consider how to calculate the entropy of a monatomic ideal gas.

We need to determine the number of microstates in $E \leq \mathcal{H}(\Gamma) \leq E+\Delta$. The volume of the this region is $\omega(E,V,N)$. The Hamiltonian has the form

$$\mathcal{H}(\Gamma)=\sum_{i=1}^{3N}\frac{p_i^2}{2m} $$

where the position coordinates restricted to box with volume $V$. The equation $\mathcal{H}(\Gamma)=E$ describes the hypersphere with radius $\sqrt{2mE}$. Thus

$$\omega(E,V,N)=V^NS_{3N}(\sqrt{2mE})\Delta $$

How has this last formula been found?


1 Answer 1


The number $\Omega$ of microstates accessible to the particle system is defined as : $$ \Omega=\omega(E)\Delta $$ where $\omega(E)$ is the density of state. To determine this guy, you need first to calculate the number $\chi(E)$ of µ-states corresponding to an energy inferior or equal to $E$.

Let us call the associated phase space region $\Sigma_{[0,E]}\equiv\lbrace\Gamma,0\leqslant\mathcal{H}(\Gamma)\leqslant E\rbrace$.

Then, it basically reads : $$ \chi(E)=\frac{1}{h^{3N}}\int_{\Sigma_{[0,E]}}\mathrm{d}\Gamma \quad \text{with} \quad \mathrm{d}\Gamma=\prod^N_{i=1}\mathrm{d}\textbf{q}_i\,\mathrm{d}\textbf{p}_i $$ $h^{3N}$ is here the element phase space volume corresponding to one µ-state. It is quite straighforward to compute : $$ \chi(E)=\frac{1}{h^{3N}}\left[\prod^N_{i=1}\int_{\Sigma_{[E,E+\Delta]}}\mathrm{d}\textbf{q}_i\right]\left[\prod^N_{i=1}\int_{\Sigma_{[0,E]}}\mathrm{d}\textbf{p}_i\right]=\frac{1}{h^{3N}}\,V^N\times V_{3N}(\sqrt{2mE}) $$ where $V_{3N}(\text{r})$ is the volume of a $3N$ dimension hyperball with a $\text{r}$ radius, and $\Sigma_{[E,E+\Delta]}$ is the phase space region $\lbrace\Gamma,E\leqslant\mathcal{H}(\Gamma)\leqslant E+\Delta \rbrace$.

For a 3D gas, we have for a given particle $i$ : $$ \int_{\Sigma_{[E,E+\Delta]}}\mathrm{d}q_{i,x}\mathrm{d}q_{i,y}\mathrm{d}q_{i,z}=V\quad\text{volume of the gas} $$ and for $N$ particles : $$ \int_{\Sigma_{[0,E]}}\prod_{i=1}^N\mathrm{d}p_{i,x}\mathrm{d}p_{i,y}\mathrm{d}p_{i,z}=V_{3N}(\sqrt{2mE}) $$ by integrating over all possible direction of the total momentum $\sum_i\textbf{p}_i$ and all magnitudes below the energy shell $\lbrace\Gamma,\mathcal{H}(\Gamma)=E \rbrace$ so that : $$ \left|\sum_i\textbf{p}_i\right|=\sqrt{\sum_i\textbf{p}_i^2}=\sqrt{2mE}\quad\text{with}\quad E=\frac{1}{2m}\sum_i\textbf{p}_i^2 $$ Since all direction are possible, the integration is performed over a $3N$-ball.

Then, it follows : $$ \omega(E)=\frac{\mathrm{d}\chi}{\mathrm{d}E}(E)=\frac{1}{h^{3N}}\,V^N\times S_{3N}(\sqrt{2mE}) $$

  • $\begingroup$ "$\chi(E)=\frac{1}{h^{3N}}\left[\prod^N_{i=1}\int_{\Sigma_{[E,E+\Delta]}}\mathrm{d}\textbf{q}_i\right]\left[\prod^N_{i=1}\int_{\Sigma_{[0,E]}}\mathrm{d}\textbf{p}_i\right]=\frac{1}{h^{3N}}\,V^N\times V_{3N}(\sqrt{2mE})$". I dont see how this works. Why is it $\sum_{[E,E+\Delta]}?$ and how have you evaluated the things in the square brackets? $\endgroup$
    – Trajan
    Commented Mar 9, 2015 at 22:20
  • $\begingroup$ See edit $\uparrow$. $\endgroup$
    – dolun
    Commented Mar 12, 2015 at 12:47

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