0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ "$\chi(E)=\frac{1}{h^{3N}}\left[\prod^N_{i=1}\int_{\Sigma_{[E,E+\Delta]}}\mathrm{d}\textbf{q}_i\right]\left[\prod^N_{i=1}\int_{\Sigma_{[0,E]}}\mathrm{d}\textbf{p}_i\right]=\frac{1}{h^{3N}}\,V^N\times V_{3N}(\sqrt{2mE})$". I dont see how this works. Why is it $\sum_{[E,E+\Delta]}?$ and how have you evaluated the things in the square brackets? $\endgroup$ – Permian Mar 9 '15 at 22:20
  • $\begingroup$ See edit $\uparrow$. $\endgroup$ – dolun Mar 12 '15 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.