Mean inter-particle distance

Ideal gas

Nearest neighbor distribution

We want to calculate probability distribution function of distance to the nearest neighbor (NN) particle. (The problem was first considered by Paul Hertz; for a modern derivation see, e.g.,.) Let us assume ${\displaystyle N}$ particles inside a sphere having volume ${\displaystyle V}$, so that ${\displaystyle n=N/V}$. Note that since the particles in the ideal gas are non-interacting, the probability of finding a particle at a certain distance from another particle is the same as the probability of finding a particle at the same distance from any other point; we shall use the center of the sphere.

An NN particle at a distance ${\displaystyle r}$ means exactly one of the ${\displaystyle N}$ particles resides at that distance while the rest ${\displaystyle N-1}$ particles are at larger distances, i.e., they are somewhere outside the sphere with radius ${\displaystyle r}$.

The probability to find a particle at the distance from the origin between ${\displaystyle r}$ and ${\displaystyle r+dr}$ is ${\displaystyle (4\pi r^2/V)dr}$, plus we have ${\displaystyle N}$ kinds of way to choose which particle, while the probability to find a particle outside that sphere is ${\displaystyle 1-4\pi r^3/3V}$. The sought-for expression is then

${\displaystyle P_N(r)dr=4\pi r^{2}dr\frac{N}{V}{\left(1-\frac{4\pi }{3}{r}^3/V\right)}^{N-1}=\frac{3}{a}{\left(\frac{r}{a}\right)}^{2}dr{\left(1-{\left(\frac{r}{a}\right)}^3\frac{1}{N}\right)}^{N-1}}$

where we substituted

${\displaystyle \frac{1}{V}=\frac{3}{4\pi N{a}^3}.}$

Note that ${\displaystyle a}$ is the Wigner-Seitz radius. Finally, taking the ${\displaystyle N\to \mathrm{\infty }}$ limit and using ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\left(1+\frac{1}{x}\right)}^x=e}$, we obtain

${\displaystyle P(r)=\frac{3}{a}{\left(\frac{r}{a}\right)}^2{e}^{-(r/a{)}^3}.}$

One can immediately check that

${\displaystyle {\int }_0^{\mathrm{\infty }}P(r)dr=1.}$

The distribution peaks at

${\displaystyle r_{\text{peak}}={\left(2/3\right)}^{1/3}a\approx 0.874a.}$

Mean distance and higher moments

${\displaystyle ⟨r^k⟩={\int }_0^{\mathrm{\infty }}P(r)r^{k}dr=3a^k{\int }_0^{\mathrm{\infty }}{x}^{k+2}{e}^{-x^3}dx,}$

or, using the ${\displaystyle t=x^3}$ substitution,

${\displaystyle ⟨r^k⟩=a^k{\int }_0^{\mathrm{\infty }}{t}^{k/3}{e}^{-t}dt=a^k\mathrm{\Gamma }(1+\frac{k}{3}),}$

where ${\displaystyle \mathrm{\Gamma }}$ is the gamma function. Thus,

${\displaystyle ⟨r^k⟩=a^k\mathrm{\Gamma }(1+\frac{k}{3}).}$

In particular,

${\displaystyle ⟨r⟩=a\mathrm{\Gamma }(\frac{4}{3})=\frac{a}{3}\mathrm{\Gamma }(\frac{1}{3})\approx 0.893a.}$