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Why/how does integral of solid angle = pi?

  1. Jun 14, 2013 #1
    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 ∏!)
  2. jcsd
  3. Jun 14, 2013 #2


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

    [itex]\theta[/itex] from 0 to [itex]\pi/2[/itex] and [itex]\phi[/itex] from 0 to [itex]\pi[/itex] or [itex]\theta[/itex] from 0 to [itex]\pi[/itex] and [itex]\phi[/itex] from 0 to [itex]\pi/2[/itex]. Which do you prefer? Since you are free to orient you coordinate system however you like are they really different?
    Last edited by a moderator: Jun 14, 2013
  4. Jun 14, 2013 #3
    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!
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