Actually, something that doesn't make sense to me...why is it that my first attempt at the problem (see original post) gives you the same answer even though we're treating the flux as though it were crossing some spherical region and not a plane?
My latest thoughts:
F = \int{I cos^2\theta dA/\rho^2}
dA = 2\pi \rho d\rho
F = 2\pi I \int{cos^2\theta d\rho /\rho} = 2\pi I \int{cos^3\theta d\rho /z} = 2\pi I \int{cos\theta sin\theta d\theta} = -I\pi cos^2\theta
Or, when evaluated from 0 to pi/2:
F = I\pi
And I think this is where you...
Ok, great. So then I think up to here I'm good:
F = \int{I cos^{2}\theta dA / \rho^2}
One thing that just occurred to me from my last post is that dA is the infinitesimal area through which the flux is being calculated, in our case, the plane at z=Z. I was treating it as an infinitesimal area...
It seems like we disagree right off the bat with the equation for flux? Isn't this how it's defined in radiative processes? Where d\Omega = dA/r^2. dA is an "infinitesimal amount of surface area that is located a distance r from the source and oriented perpendicular to the position vector r."...
\rho^2 = r^2 + z^2
2\rho d\rho = 2rdr
since z is constant.
I think I understand most of your reasoning, but I'm still not sure how to go about the problem. A disk as viewed from some point P that is off axis will appear as an ellipse -- ok, that much makes sense. So that means the projected...
Hm, ok. Let's see...
F = \int{I \cos{\theta} d\Omega}
d\Omega = dAcos{\theta}/\rho^2
F = I \int{cos{\theta} dA cos{\theta}/\rho^2}
cos{\theta} = z/\rho
F = I \int{(z/\rho)^2 dA/\rho^2} = I \int{(z/\rho)^2 2\pi r dr/\rho^2} = 2\pi I z^2 \int{\rho d\rho / \rho^4} = -\pi I z^2/\rho^2
Not sure...
Ok...so is what I did above in calculating the flux not the same? The cos term arrises from the projection into the line of sight direction. I am incredibly confused.
Ok, fair enough. That being said, I don't think I understand the drawing. I'm not sure what that length (perpendicular from X1 to QX2 is?
Thanks for the help by the way!
Ok, I think you're essentially asking me to do this (page 32), right? I don't think I fully grasped the concept of solid angle. It's the differential of the projected area (projected onto the direction of the observer) divided by the radius of a sphere squared. So, above, I was regurgitating the...
Ah, you mean expressing the solid angle in cylindrical coordinates and in turn the flux? So d\Omega ={z d\phi d\rho}/\rho^{2}?
The problem also says that by assuming Z>>r, you can use the small angle approximation, so in the expression for flux can I let cos\theta = 1, and then F is entirely in...