What is the solid angle $d\Omega$ in radiative transfer?

Question

The Wikipedia article for radiative transfer gives the following definition:

In terms of the spectral radiance, $I_{\nu }$, the energy flowing across an area element of area $da$, located at $\mathbf{r}$ in time $dt$, in the solid angle $d\Omega$ about the direction ${\hat {\mathbf {n} }}$ in the frequency interval $\nu +d\nu$, is $$dE_{\nu }=I_{\nu }(\mathbf {r} ,{\hat {\mathbf {n} }},t)\cos \theta \ d\nu \,da\,d\Omega \,dt,$$ where $\theta$ is the angle that the unit direction vector ${\hat {\mathbf {n} }}$ makes with a normal to the area element.

It isn't clear to me what the solid angle $d\Omega$ is supposed to be. I would greatly appreciate it if people would please take the time to explain this.

Related: https://en.wikipedia.org/wiki/Linear_transport_theory

Answer 1

A solid angle is a 3 dimensional angle. See this link and this image with red box:

Answer 2

The concept of solid angle is not special to your use-case, but used much more widely.

The solid angle $\Omega$ is simply defined as the ratio of a surface area element $A$ of a sphere and its squared radius $R^2$: $$\Omega=\frac{A}{R^2}$$

_{(image from this math question)}

The same goes when you have a small infinitesimal solid angle $d\Omega$ and surface area element $dA$: $$d\Omega=\frac{dA}{R^2}$$