4 ms·
FTA: > One solution is to pick λ ∈ [-180°, 180°) as before and then set φ = cos^(-1)(2x - 1), where x is uniformly distributed and x ∈ [0, 1). There is a fasc
by ggchappell 11y ago
FTA:
> One solution is to pick λ ∈ [-180°, 180°) as before and then set φ = cos^(-1)(2x - 1), where x is uniformly distributed and x ∈ [0, 1).
There is a fascinating fact hidden in the above sentence: if a point (x,y,z) is uniformly distributed on the surface of a sphere -- say, of radius 1, centered at (0,0,0) -- then each of x, y, and z is uniformly distributed over the interval [-1, 1].
Honestly, that's amazing.
- fsloth 11y agoAre you sure? If z is distributed over the surface of the sphere there is far less surface at the extremums -1 and 1. Doesn't z being uniform in the range (-1,1) lead to points being actually more densly concentrated near the poles? Or did I misunderstand what you meant?
- kasdkashdkas 11y agoNear the poles the surface becomes very flat and therefore there is as much surface area in a slice of constant height than anywhere else. Archimedes was a clever fella :-)
- fsloth 11y agoOh, I was totally confused, of course the numbers in every dimension are evenly distributed, silly me - I was thinking of great circles. Thanks :)
- xtacy 11y agoSomething doesn't add up in your observation. (x,y,z) can't all be uniformly distributed in [-1,1] because of the constraint x^2+y^2+z^2=1.
- kasdkashdkas 11y agoThey are only individually distributed uniformly in [-1,1], not jointly.
- semaphoreP 11y agoI think that's a consequence of spherical symmetry. I'm not sure I have the entire logic worked out, but because it's isotropic, then the distributions need to be uniform I think.
- kasdkashdkas 11y agoIt's a bit more subtle (the corresponding statement for circles is not correct).
- fizbin 11y agoNor for 3-spheres (i.e. the 3-dimensional surface defined as all the points in 4-d space at distance 1 from the origin). Nor for higher dimensions than that. It's peculiar to the 2-sphere ("ordinary" spheres).
- ggchappell 11y agoAh, I didn't know that. I imagine there is a trend where, in higher dimensions, coordinates have a greater tendency to be near zero? In a 0-sphere, x is either -1 or +1. In a 1-sphere, each of x, y, is (informally speaking) more likely to be near +/- 1 than near 0. A 2-sphere gives us uniform distribution for each coordinate. So I suppose that the coordinates of a 3-sphere are more likely to be near 0 than near +/- 1, and this tendency is more pronounced, the higher the dimension gets. (?) n-spheres are funny things. Intuition about them is often misleading. (See, for example, the comments expressing skepticism about my original uniform-distribution observation.) ---- EDIT. Ooo, some interesting questions here. What is the limiting behavior of the distribution of x on the n-sphere, as n -> +infinity? I imagine the graph looks like a narrower & narrower spike at x = 0. Now, suppose we scale the graph of the distribution horizontally by some appropriate function of n. That is, let f_n be the probability distribution of x on the n-sphere. Is there some function g:Z -> R so that the function x -> f_n(x / g(n)) has a limit as n -> +infinity? Does it approach (wild guess) a normal distribution? If so, is this fact (even wilder guess) a special case of some kind of central-limit-theorem-ish statement that holds for geometrical objects?
- fizbin 11y agoYes, it is a really weird property of 2-spheres that that holds. One consequence is that yet another way to pick a point uniformly on the sphere is to choose z ∈ [-1, 1] uniformly, then choose λ ∈ [-180°, 180°) uniformly and use the point (cos(λ)*sqrt(1-z*z), sin(λ)*sqrt(1-z*z), z)
- ampqoi 11y agoI believe that this is due to a theorem of Archimedes: http://en.wikipedia.org/wiki/On_the_Sphere_and_Cylinder http://en.wikipedia.org/wiki/On_the_Sphere_and_Cylinder