Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have found an answer on this site to the question of determining the mean straight-line distance between 2 randomly chosen points in a disc of radius r. I'm now trying to find an answer to the same question except involving a sphere of radius r rather than a disc. Any guidance on this question would be appreciated.

share|improve this question
    
Most of the times (i am pretty sure in this case too) these proofs generalize straight forward to higher dimensions. Can you point out your solution for the disk? Do you understand it and did you already try on your own to generalize it? –  Listing Jul 7 '12 at 17:56
1  
It should be much easier on a sphere: Due to symmetry, you may as well keep one point fixed, say the south pole of the sphere. Rescaling so the radius is 1 (just multiply the answer by $r$ afterwards) and assuming the sphere is centered at the origin, the distance from a given point on the sphere to the south pole will be $\sqrt{x^2+y^2+(z+1)^2}=\sqrt{2+2z}$. This should be straightforward to integrate over the sphere. (I just now noted the word “within” in the question title. Are you looking at points in the ball rather than the sphere?) –  Harald Hanche-Olsen Jul 7 '12 at 17:57
    
@HaraldHanche-Olsen. I've found solutions to similar questions involving points on a circle and points on the surface of a sphere. But now my question involves points within a sphere. Yes, sorry, the ball might be the better word. –  Brian Jul 7 '12 at 18:01
1  
@Listing, if the random simplex volume problem is an indicator, there's no easy way to generalize to higher dimensions. For example already the expression for the area of a random triangle (ie, with vertices independently and uniformly distributed in the unit square) is significantly more complex than the length of a random line segment in [0,1], to say nothing about the volume of a random tetrahedron. –  alancalvitti Jul 7 '12 at 18:18
    
Using the two points and the center of the ball, you could form a plane on which is a disc with the two points in question and thus reduces to the same problem. Though there's the issue if the 3 points are co-linear... kind of an edge case to iron out EDIT: actually just choose any 3rd point in ball such that its not co-linear with the 2 random points of interest, from that you can get your plane, don't have to use the center... –  Russ Jul 7 '12 at 18:22

3 Answers 3

up vote 6 down vote accepted

(a) Two random points on the unit sphere $S^2$:

Assume the first point at the north pole ($\theta=0$) of $S^2$. Then the distance to a point at latitude $\theta$ is $2\sin{\theta\over2}$. Therefore the mean distance between the north pole and the second point is given by $${1\over 4\pi}\int_0^\pi 2\sin{\theta\over2}\cdot 2\pi\sin\theta\ d\theta={4\over3}\ .$$ $$ $$

(b) Two random points in the unit ball of ${\mathbb R}^3\ $:

Let ${\bf X}$ and ${\bf Y}$ be the two random points. Then $R:=|{\bf X}|$, $\ S:=|{\bf Y}|$, and $\Theta:=\angle({\bf X},{\bf Y})$ are independent random variables with densities $$f_R(r)=3r^2\quad (0\leq r\leq 1)\ ,\qquad f_S(s)=3s^2\quad(0\leq s\leq 1)\ ,$$ and $$f_\Theta(\theta)={1\over2}\sin\theta\quad(0\leq\theta\leq\pi)\ .$$ (Concerning $f_R$ and $f_S$ note that the volume included between $r$ and $r+dr$ is proportional to $r^2$. For $f_\Theta$ you may assume ${\bf X}$ pointing due north. The abstract surface area between $\theta$ and $\theta+d\theta$ is then proportional to $\sin\theta$, as in (a).)

It follows that the mean distance $\delta$ between ${\bf X}$ and ${\bf Y}$ is given by $$\delta=\int_0^1\int_0^1\int_0^\pi \sqrt{r^2+s^2-2rs\cos\theta}\ f_R(r) f_S(s)f_\Theta(\theta) d\theta\ ds\ dr\ .$$ The innermost integral computes to $$\eqalign{{1\over2}\int_0^\pi \sqrt{r^2+s^2-2rs\cos\theta}\ \sin\theta\ d\theta&={1\over 6rs}\bigl(r^2+s^2-2r s\cos\theta\bigr)^{3/2}\Biggr|_0^\pi \cr &={1\over 6rs}\bigl((r+s)^3-|r-s|^3\bigr)\ . \cr}$$ In the sequel we assume $s\leq r$ and compensate this by a factor of $2$. We are then left with $$\delta=\int_0^1\int_0^r 3r s(6r^2 s +2s^3)\ ds\ dr={36\over35}\ .$$ It should not be too difficult to set a similar computation up that is valid for a ball in any ${\mathbb R}^n$, $\ n\geq 2$.

share|improve this answer
    
Yes, @ChristianBatter, that was the value I got. Your solution is much more concise than mine, so thank you for providing it. For the similar problem of two points on a circle of radius 1 I found that the average distance is 4/pi, which makes me wonder if this upward trend continues as we continue to go up in dimensions, and whether or not there is some limit as n -> infinity. –  Brian Jul 7 '12 at 22:29
    
Perfect! Thank you so much. –  Brian Jul 8 '12 at 17:24

The average distance can be calculated as a triple integral in polar coordinates:

enter image description here

We ran into it some years back when doing research on proteins (http://www.ncbi.nlm.nih.gov/pubmed/9514112).

share|improve this answer

This isn't a complete answer, but a start, a rough one at that.

Given two points $p_0$ and $p_1$ find a third point $p_2$ such that $N=(p_0-p_2)\times(p_1-p_2) \neq \varnothing$ Where $\varnothing$ is the null vector. Define $n = \frac{N}{|N|}$ that is normalize the vector $N$, without normalization the distance won't be correct below.

Now we have a plane where $p$ is any point on it defined as: $$ n\cdot(p-p_2)=0 \\ n\cdot p-n\cdot p_2=0 \\ n\cdot p = n\cdot p_2\\ $$ Recall the general equation of a plane is $ax+by+cz=d$ where $d$ is the distance from the origin to the plane. Expanding out the above equation gives us $$a=n_x,b=n_y,c=n_z \\ d=n_xp_{2x}+n_yp_{2y}+n_zp_{2z} $$ Radius of the resulting circle in the plane with a sphere of radius $r$ is given as $R = \sqrt{r^2-d^2} $ You may want to negate $n$ if $d$ is negative as that will allow you to have the center of the circle at $(da,db,dc)$

We haven't defined an axes for our plane so its difficult to map from our 3D points onto it. This is where I'm stopping for now. (I might come back and finish this up) I have a feeling this approach is more of a hassle than just computing $p_0-p_1$ and calculating its length. Namely because we have to normalize a vector ($N$) just to get a distance and thus the radius.

Wiki pages I used for this: Plane and Plane-sphere intersection

share|improve this answer
1  
It's not clear that restricting to this plane is going to help any. If the points are chosen uniformly in the original ball, they won't be uniformly distributed when we look only at the plane they define. (Intuitively, the problem is that the possible planes are "closer together" near the center at the ball, so an inifinitesimal area of each plane will be represent less of an infinitesimal volume than a similar area closer to the periphery). –  Henning Makholm Jul 7 '12 at 21:30
    
Yeah when I set out to write this answer I thought you just needed to use the same method for a disc but in the ball. I should have noticed the probability distributions tag and that you are going to have a large number of points –  Russ Jul 7 '12 at 22:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.