# Ideal gas in ensemble

I want to calculate the phase space density for a single ideal gas particle in a microcanonical ensemble.

I know that the partition function is given by the well-known expression that you find for example here link.

Now the phase space density for the n-particle system is $$\rho(q,p) = \frac{\delta(E-H(q,p))}{\Omega(N,V,E)}$$

I guess the phase space density for the i-th particle can be derived by calculating $$\rho_i(q_i,p_i)=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \rho(q,p) dp_1..dp_{i-1} dp_{i+1}..dp_{n} dq_1..dq_{i-1} dq_{i+1}..dq_{n},$$

but I have no idea how to integrate this. My first idea was that this would be $$\rho_i(q_i,p_i) = \frac{\Omega(N-1,V,E-\frac{p_i^2}{2m})}{\Omega(N,V,E)},$$ since this calculation is somehow similar to the one for the partition function.

If anything is unclear, please let me know.

• Perhaps you are allowed to take the large N limit? So you should obtain the canonical ensemble for the single particle I'd say Nov 8 '14 at 16:12
• nope, no $N-$ limit. I mean my general question is: If you knwo the phase space density for N particles, how do you get the one for a single particle? Nov 8 '14 at 16:18

There are two issues in this question, one is how to compute the "reduced" phase space density (i.e. the marginal distribution on the reduced phase space), and the other is how to deal with continuum phase spaces and distributions on it.

Let's face the first problem, in a simplified situation. Suppose you have two sets

$A = \{a_{1},a_{2}\}$ and $B=\{b_{(1,1)},b_{(1,2)},b_{(2,1)},b_{(2,2)},b_{(2,3)}\}$

on the set $A \times B$ you define a probability distribution in the following way

$P(a_{i},b_{(m,n)}) = \delta(i,m)/N$,

where $N$ is a normalization factor. You can easily see that its 5. Now if you want to compute the marginal probability for the set A only, you have to compute

$p(a_i) = \sum_{b \in B} P(a_i,b)$

You can see that you obtain $p(a_i) = (i+1)/5$. This can be read as the ratio between the "volume" of the subset $\{b_{(m,n)}\}$ of $B$ compatible to the request $m=i$ and the volume of the subset of $A \times B$ where $P$ does not vanish.

If you get to understand how it works in this case, then you can understand how does it work for particles: now $A = \mathbb{R^3}, B = \mathbb{R}^{3(N-1)}$ the formula for the reduced density it's exactly the integral you wrote

$\rho_i(p_i) = \int_{R^{3(N-1)}} \delta(1/2m(p_i^2 + \sum_{j\neq i} pj^2) - E) \ \Pi_{j \neq i} dp_j = 1/\Omega(N,E) \int_S \Pi_{j \neq i} dp_j \\= \Omega(N-1,E-p_i^2/2m)/\Omega(N,E)$

where $S$ is the subset of $\mathbb{R^{3(N-1)}}$ where the delta does not vanish.

• (sorry I forgot the q variables but it does not change anything) Nov 8 '14 at 17:02
• thanks, but could you explain a little bit more how this integral gets you your final result? (also there should be a typo in the delta distribution ( = should be - I think)... Nov 9 '14 at 10:20
• Yes there was typo, sorry. The integral is computed in the same way you compute the microcanonical distribution, delta(\sum_i p_i^2 - E), normalization factor. The intuitive way is that the delta vanish everywhere but on a certain region, the integral then becomes the integral of 1 on this region that is its volume. You can be more formal of course: the region will probably have a zero measure so the integral should be zero strictly speaking for example if you have a single particle the region is a sphere with radius E. This has a zero volume but non zero area... Nov 9 '14 at 11:16
• ...is this area we mean by "volume" in the normalization factor. Nov 9 '14 at 11:17
• I still don't trust this result. I mean look, your first integral depends on $p_i$, but then all other integrals don't, so somehow this is strange... Nov 9 '14 at 12:03