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During my studying of physics, I've been introduced to a concept of a solid angle. I think that I do understand it pretty good, however, I'm stuck with one certain problem.

We know that a solid angle is $S/r^2$ where $S$ is the area subtended by a cone with the vertex in the center of a sphere with radius $r$.

Suppose we have some arbitrary surface that encloses some volume. And suppose I want to insert a cone into that surface, such that the cone will cover/cut some tiny area $\Delta S$ on that surface (that will be its base). I want to know the solid angle that subtends this surface.

Illustration

I do not understand, why I should take the projection of $\Delta A_2$ here (which is a vector for area $\Delta S_2$), in order to calculate the solid angle $\Delta \Omega$. Why they claim $\Delta A_2 \cos \theta$ is "the radial projection of $\Delta A_2$ onto a sphere $S_2$ or radius $r_2$"? What if the area is not necessarily sphere-like? Is there any mathematical proof of this?

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    $\begingroup$ $\displaystyle{\large{\rm d}\Omega_{\,\,\vec{r}}\stackrel{\rm def.}{=} {\hat{r}\cdot{\rm d}\vec{S} \over r^{2}}}$ where $\displaystyle{\large\hat{r}\stackrel{\rm def.}{=}{\vec{r} \over r}}$ $\endgroup$ – Felix Marin Jun 7 '14 at 3:10
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When a plane shape of area $A$ is tilted by the angle $\theta>0$ versus the horizontal plane and then projected orthogonally onto this plane, then the area of the "shadow" is exactly $\cos\theta\>A$.

Now your "infinitesimal" surface element $\Delta A$ centered at the point $p\in S$ can be considered as such a plane shape. This shape is then projected orthogonally on a sphere of radius $r$ passing through $p$ and having normal $n$ there. The shadow $\Delta A_n$ covers such a tiny part of the sphere that it can as well be considered as plane. Thereforet we can apply the formula found above for the "linear model" of this situation, and we obtain $|\Delta A_n|=\cos\theta\>|\Delta A|$.

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I suppose that your “tiny” arrea is in fact assumed to be infinitesimally small. Otherwise you'd only have approximations. For exact results you'd have to integrate infinitesimal angles computed from infinitesimal areas. In either case, you can understand the following by assuming that your surface area is flat enough that you won't have to distinguish between it and a planar area.

Consider first the case of a sphere. That is the one used in the definition of the solid angle, so if everything is a sphere, all is well. Now imagine yourself sitting at the center of the sphere and looking outward in the direction of the angle you're measuring. Notice how a sufficiently small piece of the surface of the sphere will be almost within a plane perpendicular to your line of sight. That perpendicular plane is the tangent plane of the sphere. Now suppose you take that plane, and rotate it so it is no longer in tangential position. Then the same angle-measuring cone as before will cut out a bigger area from the rotated plane. Therefore simply measuring area that way will not be enough to describe the solid angle. In order to describe the solid angle, you have to get back to the sphere-like situation where the area you measure is perpendicular to your line of sight. That's what this “radial projection” of $\Delta A_2$ onto $\Delta A_{2n}=\Delta A_2\cos\theta$ is doing.

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The solid angle $(\omega)$ subtended at the apex point by any section of a right cone with apex angle $2\alpha$, is given as $$\omega=2\pi(1-\cos\alpha)$$ Above, expression shows that the solid angle is dependent only on the apex angle $2\alpha$ which doesn't depend on the radius and normal height of cone. Hence, the area of interception of a right cone with the sphere, having a radius $R$, is $$=2\pi R^2(1-\cos\alpha)$$ Thus the solid angle, subtended by the surface intercepted by a cone having its vertex at the center of any sphere, is $$=\frac{2\pi R^2(1-\cos\alpha)}{R^2}=2\pi (1-\cos\alpha)$$ Above expression shows that the solid angle does not depend on the radius $R$ of the sphere.

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