Mean inter-particle distance

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 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 ${\displaystyle N}$ ways 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

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

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

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,

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

• Wigner–Seitz radius