Can One Prescribe a Solid Angle to the Vertices of the Platonic Solid

Question

Can One Prescribe a solid angle to the vertices of the platonic solid?

The canonical example given to demonstrate the concept of a solid angle gives one a conical spherical cap whose base encloses some area. The solid angle is the higher-dimensional analogue of the plane angle and is prescribed to the vertex of this cone-like cap. The solid angle, Ω, is then defined in relation to the area, A, enclosed by the base and r, the spherical radius: Ω=𝐴/𝑟^2 .

Can this same concept be used to ascribe a solid angle to the vertices with regular solid such as the platonic solid? It is an almost trivial fact that the sum of the angles of a quadrilateral in regular flat space that is 360 degrees or 2π radians. Is there, for example, a higher-dimensional analog for a sum of solid angles at the vertices within a regular cube?

Answer 1

Yes. In fact, it is possible to assign solid angles to vertices of all polyhedron.

A common way to define the solid angle at a vertex $v$ is put a small sphere of radius $\epsilon$ centered at $v$, look at area of those portion of sphere inside polyhedron and divide it by $\epsilon^2$. Up to a factor of $2$, you can define the dihedral angle between two adjacent faces in same manner.

For 3-d polyhedron, the solid angle $\Omega_v$ subtended at each vertex $v$, the dihedral angle $\theta_e$ substend at each edge $e$ and the number of faces $F$ satisfy an interesting relation $$\frac1{4\pi}\sum_v \Omega_v - \frac1{2\pi}\sum_e \theta_e + \frac12 F - 1 = 0\tag{*1}$$

As an example, a cube has $8$ vertices with solid angle $\Omega = \frac{\pi}{2}$, $12$ edges with dihedral angle $\theta = \frac{\pi}{2}$ and $F = 6$ faces, we do have

$$\frac{8}{4\pi}\frac{\pi}{2} - \frac{12}{2\pi}\frac{\pi}{2} + \frac{6}{2} - 1 = 1 - 3 + 3 - 1 = 0$$

As an application, we know a regular tetrahedron has $4$ vertices, $6$ edges and $4$ faces. Instead of computing the solid angle $\Omega$ directly, it is much simpler to figure out the dihedral angle $\theta$ equals to $\cos^{-1}\frac13$. This leads to $$\frac{4}{4\pi}\Omega - \frac{6}{2\pi}\cos^{-1}\left(\frac13\right) + \frac{4}{2} - 1 = 0$$ and hence $$\Omega = 3\cos^{-1}\left(\frac13\right) - \pi = \cos^{-1}\left(\frac{3}{3} - \frac{4}{3^3}\right) = \cos^{-1}\left(\frac{23}{27}\right)$$

Matching the number you can find on wiki entry of regular tetrahedron.

Look at wiki entry of Gram-Euler theorem for generalization of the idea of solid angle and relation $(*1)$ in higher dimension.