Say I have a 3-ball with radius $R$. If I randomly pick 2 points from the inside of the ball, the probability that the euclidean distance between the points (labeled 1 and 2) takes on a particular value $r = r_{12} = r_{21}$ is given by the probability density function (PDF)
\begin{equation} P_3 (r) = \frac{3 r^{2}}{R^3} - \frac{9 r^{3}}{4 R^4} + \frac{3 r^{5}}{16 R^6} \end{equation}
as described in https://arxiv.org/pdf/math-ph/0201046.pdf, equation 15.
If I were to pick $N$ points from the inside of this ball simultaneously, there would be $N(N-1)/2$ distances between pairs of different points. Is it possible to express the PDF $P(r_1, r_2, \dots{}, r_{N(N-1)/2}$), where $r_1, r_2, \dots{}, r_{N(N-1)/2}$ are distances between pairs of different points, using the pair-wise PDF $P_3(r)$ ? Does there exist some other closed-form solution for such distribution, or a solution for some shape other than a ball ?
In this case, it is obviously not $P_3(r) \times P_3(r) \times \dots{} \times P_3(r)$, because, for example, when 2 points are at a distance $2R$, a third point cannot be at a distance $2R$ from both of them, but such PDF would allow it.
There are $3N$ degrees of freedom for the locations of the $N$ points. Once $N\geq 8$, we have $3N < N(N-1)/2$, so most $N(N-1)/2$-tuples of distances cannot be realized (this is true even if the $N$ points are no longer constrained to a ball). The set of possible $N(N-1)/2$-tuples of distances has positive codimension in $\mathbb{R}^{N(N-1)/2}$, hence has Lebesgue measure $0$. Thus the probability measure on the set of distances is absolutely continuous with respect to Lebesgue measure, so there cannot exist a probability density function.