What is the solid angle in higher dimensions?

Question

I'm struggling with understanding the generalization of a solid angle to higher dimensions. What is the intuition behind a solid angle in general and how does it generalize in higher dimensions?

Edit: One thing that is not very clear to me:

For 3D space $d\Omega = \dfrac{dS}{r^2}$ so the solid angle is the ratio of the area subtended to the square of the radius, how can I write the n dimensional infinitesimal solid angle?

Answer 1

You generalize by generalizing the notation. Let $dS$ be the surface "area" $(n-1)$-form on $\Bbb R^n-\{0\}$ which gives the "area" on the unit sphere. And then you set $d\Omega = \dfrac{dS}{r^{n-1}}$ as an $(n-1)$-form on $\Bbb R^n-\{0\}$. Integrating this over a chunk of hypersurface in $\Bbb R^n$ gives the solid angle it subtends. (And yes, one can write down explicit formulas.)

Answer 2

This definition from wikipedia generalizes:

In geometry, a solid angle (symbol: Ω) is a measure of the amount of the field of view from some particular point that a given object covers. That is, it is a measure of how large the object appears to an observer looking from that point. The point from which the object is viewed is called the apex of the solid angle, and the object is said to subtend its solid angle from that point.

Center a unit sphere at the vertex of the angle and calculate the fraction of the $n-1$-dimensional surface area of the sphere your angle cuts off.

Edit in response to comment.

In three dimensions:

To find the solid angle subtended by a rectangular plate you need the coordinates of the vertices and the coordinates of the point $P$ you're looking from. Project the vertices from $P$ onto the surface of a unit sphere centered at $P$. Joint those vertices to form a quadrilateral on the sphere with edges sections of great circles. Then find the area of that quadrilateral (as a fraction of the area of the sphere). I think this will call for a fair amount of spherical trigonometry. Also see Area of Spherical Polygon

For a weird surface you'll have to find a way to find it's weird projection onto the sphere, and then the area of the projection.