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What is the average of a random hemispherical distribution 
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#1
Jul614, 04:27 PM

P: 2

Hello,
I'm trying to write a monte carlo simulation for an optical analysis. Half the area of a sphere is within 60 degrees of the poles. Hence, I'm assuming half of randomly directed radiation should fall within 60 degrees of the poles, when radiation is generated at the center of the sphere. I have tried 3 algorithms so far for generating a random angle, and I test each one by averaging all of the angles. It seems to me that the average angle of a truly random distribution should be 60 degrees because half the angles will fall below 60 degrees and half above 60 degrees in a uniform distribution. So far all 3 algorithms give me the same value ~57.3 when averaged over hundreds of millions of runs. So I'm thinking, either the pseudorandom number generator on python is making the same mistake in each algorithm, or I'm not correct that the average will be 60 degrees. Could someone please confirm that the average of a hemispherical distribution of angles would be 60? THANKS! 


#2
Jul614, 07:47 PM

P: 327

It's tricky to generate a point uniformly distributed on the surface of a sphere. In particular, generating the latitude and longitude uniformly will not work. You can find a discussion here: http://mathworld.wolfram.com/SpherePointPicking.html
The definition of "uniformly distributed" is that at point's probability of landing in a region is proportional to the area of the region. You are correct that 1/2 of the points should be within 60 degrees of the poles, because the two "spherical caps" (the regions within 60 degrees of a pole) each have area [itex]\pi r^2[/itex], so the total area within 60 degrees of the poles is [itex]2 \pi r^2[/itex]. Since the area of a sphere is [itex]4 \pi r^2[/itex], the probability of a point's landing within the spherical caps is [itex]\frac{2 \pi r^2}{4 \pi r^2} = \frac{1}{2}[/itex]. [Edit] On rereading your post, though, I see you are averaging the angles. That's not the same as counting the points that lie in a specified region. So no, I don't think the average angle should be 60 degrees. I'll look into this more later, but I don't have time to right now.[/edit] 


#3
Jul714, 02:31 PM

Sci Advisor
PF Gold
P: 2,080

As awkward states, the distribution of points on a sphere does not uniform have a uniform distribution in polar angle theta. What's more, the "average angle" of points uniformly distributed on a sphere does not have a 2D answer since a spherical surface has no center. (In 3D, of course, the average is the origin at the center of the sphere.)
There are many ways to compute samples on a spherethere's even a method based on Archimedes' observation in 200 BC that the area of cylinder (neglecting endcaps) is the same as that of an inscribed sphere. In practice, some work better than others when considering finite precision of computational systems. My favorite approach, which is pretty foolproof, is this: Muller, M.E., “A Note on a Method For Generating Points Uniformly on NDimensional Spheres,” Comm. of the ACM, vol. 2, p. 1920 (1959). 


#4
Jul714, 04:43 PM

P: 327

What is the average of a random hemispherical distribution
OK, as I wrote earlier, the most direct way to check that you are generating uniformly distributed points is to check that the number of points in a region on the surface of the sphere is proportional to the area of the region; that's the definition of "uniformly distributed". But since you're looking at the average angle, let's see if we can figure out what that should be.
I'm assuming that [itex]\phi[/itex] is zero at the "north pole" and you are generating points uniformly distributed on the hemisphere where [itex]0 \leq \phi \leq \pi/2[/itex]. If so, then the joint pdf of [itex](\theta, \phi)[/itex] is $$f(\theta, \phi) = \frac{1}{2 \pi} \sin \phi$$ for [itex]0 \leq \theta \leq 2 \pi[/itex] and [itex]0 \leq \phi \leq \pi / 2[/itex], so the expected value of [itex]\phi[/itex] is $$\int_0^{2 \pi} \int_0^{\pi /2} \phi \; \sin \phi \; d\phi \; d\theta = 1 \text{ radian} = 57.3 \text{ degrees}$$ (calculus left as an exercise for the reader). So it appears from this result that you are correctly generating the uniformly distributed points. 


#5
Jul814, 02:08 PM

Mentor
P: 11,815

This just boils down to "the median (here: 60°) is not always the same as the average (here: 57.3°)".
Why don't you check how many angles are below / above 60°? This should give 1/2 as answer. 


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