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Statistical test for spherical uniformity?

  1. Jan 18, 2008 #1
    I have a large sample of vectors in 3-space, and I would like to how uniformly they are distributed. Ultimately I'd like to know how uniform the angular distribution as a function of magnitude. What I mean by that is that if I divide the space into spherical shells, will the vectors whose magnitudes fall into each shell be distributed uniformly throughout the shell?

    Thanks for any suggestions.
  2. jcsd
  3. Jan 18, 2008 #2


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    Can you provide a little more detail?

    Is each shell also divided into discrete subregions? If so, that makes it a discrete problem and you can use the chi-square test: http://en.wikipedia.org/wiki/Chi-square_test

    If you are looking at it as a continuous uniform distribution, then you cannot use categorical tests such as chi-squared, but may be able to use other tests.
  4. Jan 18, 2008 #3
    The problem I'm looking at is fairly straight-forward. I have a large number of objects all moving from one point in space with some distribution of velocity vectors. I'd like to track the spatial density of objects over time, and it's easier to do this if I can make the assumption that the objects are uniformly distributed spherically. That way I can just divide my space into spherical shells of uniform thickness and count the number of objects in each shell to get the density at that radius. My simplification amounts to saying that the density will be a function of radial distance, but not of direction.

    I'd like to know how valid this assumption is, however. Since I have a discrete set of vectors, then of necessity I would likely divide each shell into cells of equal volume in order to test the assumption. In that case, then I think I'm just asking how much variation there is in the number of objects per cell for a given radius.

    In any case, since I posted my original question, I've gotten my hands on a book entitled, "Statistical Analysis of Spherical Data" by N. I. Fisher et. al. Hm. If the answer isn't in there, I'll eat my hat.
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