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 N particles inside a sphere having volume V, so that 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 r means exactly one of the N particles resides at that distance while the rest N - 1 particles are at larger distances, i.e., they are somewhere outside the sphere with radius r. The probability to find a particle at the distance from the origin between r and r + dr is (4 \pi r^2/V) dr, plus we have N ways to choose which particle, while the probability to find a particle outside that sphere is 1 - 4\pi r^3/3V. The sought-for expression is then :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 : \frac{1}{V} = \frac{3}{4 \pi N a^{3}}. Note that a is the
Wigner-Seitz radius. Finally, taking the N \rightarrow \infty limit and using \lim_{x \rightarrow \infty}\left(1 + \frac{1}{x}\right)^x = e, we obtain :P(r) = \frac{3}{a}\left(\frac{r}{a}\right)^2 e^{-(r/a)^3}\,. One can immediately check that :\int_{0}^{\infty}P(r)dr = 1\,. The distribution peaks at :r_{\text{peak}} = \left(2/3\right)^{1/3} a \approx 0.874 a\,.
Mean distance and higher moments : \langle r^k \rangle = \int_{0}^{\infty}P(r) r^k dr = 3 a^k\int_{0}^{\infty}x^{k+2}e^{-x^3}dx\,, or, using the t = x^3 substitution, : \langle r^k \rangle = a^k \int_{0}^{\infty}t^{k/3}e^{-t}dt = a^k \Gamma(1 + \frac{k}{3})\,, where \Gamma is the
gamma function. Thus, : \langle r^k \rangle = a^k \Gamma(1 + \frac{k}{3})\,. In particular, : \langle r \rangle = a \Gamma(\frac{4}{3}) = \frac{a}{3} \Gamma(\frac{1}{3}) \approx 0.893 a\,. ==References==