Binary Star Inclination Angle probability distribution

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Miviato
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Can someone explain why the probability of the inclination angle of a binary system being less than i_0 is 1-cos(i_0)
i.e. why the fractional distribution of binary stars is df = sini * di, where i is the inclination angle?

Where does the sin i come from? Why is not not uniformly distributed across angles?

Thanks.
 
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You might find George Gamow's One, Two, Three, Infinity an interesting read. It covers this and a number of other interesting probability and statistical problems and other tidbits.
 
I found the book but can't find the section with this explanation -- could you please point you where it is?
Thanks!
 
I read and understood it, but I don't quite see how it answers my question -- for the binary the distance doesn't really matter, you can change the distance without changing the inclination angle...
 
Miviato said:
I don't quite see how it answers my question --
Clearly, I'm going to have to do a little reading and correct my intuitive interpretation of
Miviato said:
probability of the inclination angle of a binary system being less than i_0 is 1-cos(i_0)
"inclination angle." Didn't mean to drag your question off on a "tangent," with a hasty leap --- thought I'd found another application of some of the neat little tricks Gamow collected.
 
Look at it this way: what is the probability on an observer seeing Earth at a certain latitude?

1 degree of latitude from equator to 1 degree latitude is a band around Earth 1 degree wide but 360 degrees long - all around the length of equator. 1 degree of latitude from latitude 89 to pole at latitude 90 is a small circle, with radius 1 degree, but accordingly only a tiny circumference.

I understand that the values of inclination are counted so that inclination 0 is the viewer looking at one pole (latitude 90, the pole where the orbit is counterclockwise), so very improbable, inclination 90 is the viewer looking at the plane/equator of orbit (latitude 0, most probable) and inclination 180 is the viewer looking at the other pole (latitude also 90, the pole where the orbit is clockwise), so also very improbable.