tuomlarsen wrote:Thank you very much for the reply, much appreciated!

One more question, though:

As a counterexample to your formula, omega = A/r*r, what if r = 1 and A = 1e6? The correct solid angle in this case (assuming a planar area perpendicular to r) would be approximately pi, not 1e6.

Solid angle is the area "cut out" from unit sphere surface divided by 4*pi (as the total surface area of unit sphere). So if r = 1 and A = 1e6, the omega should be 1e6 / (4*pi), rather than pi, no?

PS: Now reading Wikipedia, there is the formula: omega = 2 * pi (1 - cos theta). I would assume that since the area is tiny, the angle should also be tiny, because cos(tiny angle) is 1, so omega = 2 * pi (1 - 1)..?

Oops! I should have said that the solid angle is 2*pi, not pi. Also, 1e6 = 1000000, which is not tiny. Perhaps you were thinking of

1e-6 (= 0.0000001)?

Edit: (Gotta try the new MathJax support!)

\[

\Omega = 2 \pi \left( 1-\cos \theta \right)

\]

I'm not comfortable with your term "cut out", probably because my many years of school bias me. But if you can figure out a way to cut out 1000000 square meters of cloth from a piece that is only 4*pi square meters, I think you will become very rich.

Here is a better description of my thinking: The surface area of a unit sphere is 4*pi, as you stated. The 1e6 area would project onto nearly half of the sphere, so the solid angle is half of the area of the sphere, or 2*pi.

One way to picture projection is to imagine an infinite sphere concentric with the unit sphere. The infinite sphere radiates light inward toward the unit sphere, but all of the light rays are normal (perpendicular) to the surface of the sphere. The projection of any area onto the unit sphere corresponds exactly to the shadow that would be generated. In the case of the very large area, roughly half of the unit sphere would be shadowed.

Another way of thinking about this projection is from the inside out. If you trace rays at random angles from the center of the sphere, roughly half of them would hit the 1e6 area, so the solid angle of the area must be half that of the unit sphere.

Using the last formula you give, omega = 2 * pi * (1 - cos(pi/2)) = 2*pi*(1-0) = 2*pi

Another MathJax Edit:

\[

\Omega = 2 \pi \left( 1 - \cos \left( \pi / 2 \right) \right) = 2 \pi \left( 1 - 0 \right) = 2 \pi

\]

Ooh, I like MathJax!