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Equidistribution of points on a sphere

  1. Jul 25, 2012 #1
    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 [itex]α[/itex] and [itex]β[/itex] be any positive irrational value. Define [itex]u_n = \{n α\} = n α \mod 1[/itex] and [itex]v_n = \{n β\} = n β \mod 1[/itex].

    Consider the spherical coordinates, defined as follows:

    [itex]θ_n = 2 π u_n[/itex]

    [itex]ψ_n = \arccos{2 v_n - 1}[/itex]


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

    [itex]\displaystyle \lim_{n \rightarrow \infty} \frac{\# p_n \in S}{n} = \frac{2 π h}{4π} = \frac{h}{2}[/itex].

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

    [itex]\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)[/itex]

    [itex] \displaystyle v_n \in \left( \frac{1-\sqrt{1-(1-h)^2}}{2} , \frac{1+\sqrt{1-(1-h)^2}}{2}\right)[/itex]

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

    My question is, how do you use the definition of [itex]u_n[/itex] and [itex]v_n[/itex] to determine at least, the bounds for [itex]k[/itex]? Thank you!
     
  2. jcsd
  3. Jul 26, 2012 #2
    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.
     
  4. Jul 26, 2012 #3

    haruspex

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    You mean this, right?
    [itex]ψ_n = \arccos(2 v_n - 1)/itex]
    I think you mean, for a given [itex]m \in \mathbb{N}[/itex] 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.
     
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