0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user8469759 Jan 22 '18 at 15:40
  • $\begingroup$ Just came up in my mind that maybe the test of a generic solid angle could be done in terms of testing if a point in a unit sphere belongs to a connected region on the surface of the sphere. $\endgroup$ – user8469759 Jan 22 '18 at 16:40
  • $\begingroup$ 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. $\endgroup$ – Ross Millikan Jan 23 '18 at 0:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.