Nearest neighbor distribution

PDF of the NN distances in an ideal gas.

We want to calculate probability distribution function of distance to the nearest neighbor (NN) particle. (The problem was first considered by Paul Hertz;^[1] for a modern derivation see, e.g.,.^[2]) 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 to find a particle at a certain distance from another particle is the same as probability to find a particle at the same distance from any other point; we shall use the center of the sphere.

An NN particle at 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

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

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

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

One can immediately check that

${\displaystyle \int _{0}^{\infty }P(r)dr=1\,.}$

The distribution peaks at

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

Mean distance and higher moments

${\displaystyle \langle r^{k}\rangle =\int _{0}^{\infty }P(r)r^{k}dr=3a^{k}\int _{0}^{\infty }x^{k+2}e^{-x^{3}}dx\,,}$

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

${\displaystyle \langle r^{k}\rangle =a^{k}\int _{0}^{\infty }t^{k/3}e^{-t}dt=a^{k}\Gamma (1+{\frac {k}{3}})\,,}$

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

${\displaystyle \langle r^{k}\rangle =a^{k}\Gamma (1+{\frac {k}{3}})\,.}$

In particular,

${\displaystyle \langle r\rangle =a\Gamma ({\frac {4}{3}})={\frac {a}{3}}\Gamma ({\frac {1}{3}})\approx 0.893a\,.}$