Equidistribution of points on a sphere

1. Jul 25, 2012

iironiic

I am struggling to find a way to count the number of irrational points defined recursively which satisfies specific conditions. This is the only hurdle I need to overcome, but the problem is proving itself extremely difficult.

Here is my construction of a recursion. My goal is to essentially prove that this recursion will guarantee a set of equidistributed points.

Let $α$ and $β$ be any positive irrational value. Define $u_n = \{n α\} = n α \mod 1$ and $v_n = \{n β\} = n β \mod 1$.

Consider the spherical coordinates, defined as follows:

$θ_n = 2 π u_n$

$ψ_n = \arccos{2 v_n - 1}$

This construction guarentees uniform distribution of points on the surface of the sphere. To show equidistribution, consider a spherical cap $S$ with height $h$ on the unit sphere (sphere with radius one). Given a particular $n \in \mathbb{N}$, my goal is to find the number of points that lie in $S$. Inevitably, what I want to show is as follows:

$\displaystyle \lim_{n \rightarrow \infty} \frac{\# p_n \in S}{n} = \frac{2 π h}{4π} = \frac{h}{2}$.

This is equivalent to finding the number of $n \in \mathbb{N}$ (we'll call this number $k$) satisfying,

$\displaystyle u_n \in \left(0, \frac{\arccos{\frac{1-h}{2\sqrt{v_n(1-v_n)}}}}{2 \pi} \right) \cup \left(1- \frac{\arccos{\frac{1-h}{2\sqrt{v_n(1-v_n)}}}}{2 \pi}, 1 \right)$

$\displaystyle v_n \in \left( \frac{1-\sqrt{1-(1-h)^2}}{2} , \frac{1+\sqrt{1-(1-h)^2}}{2}\right)$

For all $0 \leq h \leq 1$. I used Mathematica to find $k$ and it seems that $\frac{k}{n} \rightarrow \frac{h}{2}$ where $n$ denotes the number of points on the surface of the entire sphere.

My question is, how do you use the definition of $u_n$ and $v_n$ to determine at least, the bounds for $k$? Thank you!

2. Jul 26, 2012

cosmik debris

I'll follow this with interest. We had to equally distribute points on a sphere (to do with cell towers). The only way of doing it that we could find was by simulated annealing.

3. Jul 26, 2012

haruspex

You mean this, right?
$ψ_n = \arccos(2 v_n - 1)/itex] I think you mean, for a given [itex]m \in \mathbb{N}$ you want the number of points n <= m in S.
Your expressions beyond that look much too complicated. Have you oriented the cap in the most convenient way? I get cos(2πun) = cos(θn) > 1 - h as the only constraint.