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I have some points that sit on the hemisphere in spherical coordinates: $\theta \in [0,\pi/2]$, $\phi \in [0, 2\pi]$ (ie so a hemisphere around the vector (1,0,0) (spherical coordinates).

I should note the convention I'm using: $\theta$ is "elevation" measured from +y-axis.

I want to rotate each vector in spherical coordinates in that hemisphere without having to convert to Cartesian coordinates first.

The subproblem is I am trying create a couple thousand random vectors in a hemisphere centered around some normal.

The "cheap" way to do this is to generate uniformly distributed points on the unit sphere, and discard those that don't have a dot product with the normal > 0.

But I want to avoid creating vectors that I'll have to discard anyway, so I'd rather generate them correctly the first time around. So generate spherical vectors with $\theta \in [0,\pi/2]$, $\phi \in [0, 2\pi]$, then rotate by $\theta, \phi$ of the normal vector these random points are supposed to surround.

To get evenly distributed points on a hemisphere around the spherical vector (1,0,0):

$ \text{Spherical Vector} (r=1, \theta=\cos^{-1}( \sqrt{ \zeta_1 } ), \phi=2 \pi \zeta_2 ) $

Where $\zeta_1, \zeta_2$ are a random variables with values between 0 and 1.

What this does is distribute the elevation component with GREATER emphasis on the equator (LOTS of samples with $\theta=\pi/2$, less values at poles)

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    @anon: Look at Royden's *[Real Analysis, 3rd ed.](http://www.amazon.com/dp/0024041513)* for example. I always thoguht $\varepsilon$ stood for element, but according to [Earliest Uses of Symbols of Set Theory and Logic](http://jeff560.tripod.com/set.html) "Giuseppe Peano (1858-1932) used an epsilon for membership in *Arithmetices prinicipia nova methodo exposita*, Turin 1889 (page vi, x). He stated that the symbol was an abbreviation for est; the entire work is in Latin."2011-08-23

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Why not just reverse the ones with dot product <0 instead of discarding them? You will still have a uniform distribution in the hemisphere.

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    I do$n$'t know of a matrix approach. You could chase through the trigonometry, but I don't have the equations available.2011-08-23