# What does it mean that a normal is inside a solid angle?

I'm reading through some stuff. And there are stuff like this mentioned. Assume we have some surface $\mathcal{S}$ we are focusing on some small element $d\mathcal{S}$ let's define $D(n)$ as the distribution of the normals over $d\mathcal{S}$, for some some solid angle $d\omega$ we have

$$D(n)d\omega$$

is the probability that the normal $n$ belongs to the solid angle $d\omega$, where $n$ is the normal built upon a point $p$ inside $d\mathcal{S}$. The question is, what is the actual (formal) meaning that $n$ belong to the solid angle $d\omega$?

I can have some intuition in case where the solid angle is actually built as a cone, but in the general case we have

$$d\omega = \sin \theta d\theta d\phi$$

It's a bit hard to imagine what happens here. My guess could be that this has something to do with the Gauss map $\mathcal{G}_p$ where $p\in d\mathcal{S}$, but I'm not entirely sure.

Your solid angle here is essentially a rectangular pyramid with the tip at the origin. The base is $\sin \theta d\theta \times d\phi$ and you want the normal to be within that pyramid. It is no different from the cone case except for the shape of the region.
• I was looking for a more "rigorous" definition of the statement $n \in d \omega$, and what actually puzzles me is that the solid angle is not a set, it's a scalar value, saying that "a vector belong to a scalar" doesn't make much sense to me. – user8469759 Jan 22 '18 at 15:40
• Yes, that is correct. The solid angle corresponds to a portion of the sphere. In this case it is a small bit in $\phi$ by a small bit in $\theta$ but the bit in $\theta$ is scaled down by $\sin \theta$ which is used to make each bit have the same solid angle. – Ross Millikan Jan 23 '18 at 0:37