Why/how does integral of solid angle = pi?

[ck99]

Hi folks, can someone help explain this in words of one syllable or less? I am looking at a text that compares flux and intensity of a distant source, and it states that

∫∫dΩ = ∏

I know that

dΩ = sinθ dθ d∅

but I don't understand where the given result comes from. What are the limits of integration here, and how does it all work? I assume that we treat the distant source as a circle projected onto our sphere of view, but in that case surely the size of that circle must come into the answer somehow? (EG a large source like the sun would subtend a larger area than Jupiter, they can't both have an area of ∏!)

 

[HallsofIvy]

Since you know that d\Omega= sin(\theta)d\theta d\phi you know that \int\int d\Omega= \int\int sin(\theta) d\theta d\phi. And that, assuming for the moment that the limits in one variable do not involve the other, is -cos(\theta)\phi. Since we are free to orient a coordinate system any way we want, we can assume that the lower limits of integration are \theta= 0, \phi= 0. And we then want upper limits so that -cos(\theta)\phi= \pi. There are several different combinations of \theta and \phi that will do that. Taking \phi= \pi, the largest possible, we would need -cos(\theta)+ cos(0)= -cos(\theta)+ 1= 1, cos(\theta)= 0 which gives \theta= \pi/2. But taking \phi= \pi/2 we would need -cos(\theta)+ cos(0)= 1- cos(\theta)= 2, cos(\theta)= -1 so \theta= \pi.

\theta from 0 to \pi/2 and \phi from 0 to \pi or \theta from 0 to \pi and \phi from 0 to \pi/2. Which do you prefer? Since you are free to orient you coordinate system however you like are they really different?

 

[ck99]

Thank you for your response; I understand now how that result is reached but now I don't see how it relates to my problem. In this text, it says that flux F = L/(4∏D²) where D is the distance from the source and L is its luminosity. Then it says that intensity I = F/Ω, which implies that I = L/4D² if Ω = ∏, but in fact it states that I = L/4R² where R is the radius of the source!

This last result would be true if Ω = R² / ∏D² so I did some research, and the closest I can find is that the solid angle subtended by a distant source is given by area/D² or ∏R²/D². (that last result comes from page 5 of this link http://www.drdrbill.com/downloads/optics/photometry/Solid_Angle.pdf )

So I have three different answers, and I can't see how they link together!