# Distances between randomly distributed points in a ball

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.

• +1 for interesting read. However I'm not really sure what you are asking? You are asking the probability of getting $N(N-1)/2$ exact distances – marshal craft Apr 29 '17 at 12:53
• So $r$ is the range of Euclidean distance? Sorry I need to read the paper. – marshal craft Apr 29 '17 at 13:05
• Also shouldn't the original PDF already account for that? – marshal craft Apr 29 '17 at 13:14
• I am looking for a PDF with $N(N-1)/2$ arguments which correspond to individual pair-wise distances between the $N$ points. The distances themselves are coupled in some way due to the requirement that all points fit inside the ball, as described in the example of why the joint PDF is not the product of pair-wise PDFs described in the paper. I simply switched $s$ from eq. 15 from the paper for $r$. – matlab-oh-no Apr 29 '17 at 13:41

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.
• I don't fully understand the arguments and terms used -- why most configurations suddenly cannot be realised when $N$ exceeds some amount, why the PDF does not exist -- but then again I'm not a mathematician. Thank you very much for the answer. – matlab-oh-no May 1 '17 at 14:34
• Possibly this analogy will help. Suppose I pick a random number by flipping a coin, choosing $0$ if the coin lands heads, and $1$ if the coin lands tails. There cannot be a PDF $P(x)$ describing this situation, because $P(x)$ would need to be $0$ everywhere except $x=0$ and $x=1$, and would have to be infinite at $x=0,1$. PDFs really only work for distributions that are spread out, rather than concentrated on only a few values. The same thing is true in your situation: already for $N=5$, it should be true that knowing $9$ of the pairwise distances leaves only a few possibilities for the tenth. – Julian Rosen May 1 '17 at 15:20