Total internal solid angle of a convex polyhedron

Question

The total internal angle of a convex polygon with n sides is $(n-2)\pi$. Is there an analogous formula for the total solid angle of a convex polyhedron?

Answer 1

The total solid angle is not fixed but there is the Gram-Euler theorem, which is similar to the Euler characteristic $$V - E + F = 2$$ but for the internal angles:

$$ \sum_V \Omega_v - 2\sum_E \theta_e + \sum_F 2\pi - 4\pi = 0 $$

where

• $\sum_V \Omega_v$ is the sum of the solid angles
• $\sum_E \theta_e$ is the sum of the dihedral angles (the angles between faces, $\times2$ to make it a solid angle)
• $\sum_F 2\pi$ counts the number of faces

The last two terms combine to $2\pi(F-2)$ for $F$ faces, resembling the polygon formula.

The analogy with the Euler characteristic is especially clear when we count in hemispheres of solid angle:

$$ \sum_V \cfrac{\Omega_v}{2\pi} - \sum_E \frac{\theta_e}{\pi} + F = 2 $$

Also, combining with the Euler characteristic, we have

$$ \sum_V \cfrac{\Omega_v}{2\pi} - \sum_E \frac{\theta_e}{\pi} = V - E $$

or equivalently

$$ V - \sum_V \cfrac{\Omega_v}{2\pi} = E - \sum_E \frac{\theta_e}{\pi} $$

which is a positive quantity (at least for convex polyhedra).

So you might say the vertices and edges are collectively short of being "flat" by the same total amount of solid angle, and we may call $2\pi - \Omega_v$ and $2\pi - 2\theta_e$ the "solid angle defect" of a vertex and edge, respectively.

Note that this is distinct from the angle defect of Descartes theorem which was already mentioned.

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As a simple proof, note that we can rearrange to

\begin{equation} \sum_V \Omega_v = 2\sum_E \theta_e - 2\pi(E - V) = \sum_V(\sum_{e\in V}\theta_e - (n_v - 2)\pi) \end{equation}

in which we can recognize a sum of the spherical excess at each vertex: \begin{equation} \Omega_v = \sum_{e\in V}\theta_e - (n_v - 2)\pi \end{equation}

where $n_v$ is the number of edges adjacent to $v$.

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Reasoning in reverse, this suggests the Gram-Euler theorem should also hold for non-convex polyhedra, as long as we take the different Euler characteristic into account; $4\pi \rightarrow 2\pi\chi$.

Answer 2

There is Descartes' theorem, which states that the sum of the angular defects of a (simply connected) polyhedron is $4\pi$.

The angular defect at a vertex is the difference between $2\pi$ and the sum of the angles around the vertex. For example, in (say) a right equilateral triangular prism, the angular defect at each vertex is \begin{equation} 2\pi-\frac{\pi}{3}-\frac{\pi}{2}-\frac{\pi}{2} = \frac{2\pi}{3} \end{equation} There are $6$ vertices, so the angular defects sum to $4\pi$.

This could be thought of as analogous to the polygon formula in the following way. If we regard the exterior angle at a vertex as the "angular defect" at that vertex, then the sum of these angular defects is $2\pi$. This is equivalent to the $(n-2)\pi$ formula for the sum of the interior angles.

Neither formula requires the object to be convex, incidentally. We just have to allow for the possibility of negative angular defects

There are various interesting generalizations of Descartes' formula for surfaces.

Answer 3

It's not difficult to give examples of polyhedra with the same topology but a different sum $S$ of solid angles. This implies that no formula analogous to that of polygons can exist.

Take for instance a regular pyramid with triangular base and height $h$: in the limit $h\to0$ the solid angle at vertex tends to $2\pi$, while solid angles at the base tend to $0$, hence $S\to2\pi$; but in the limit $h\to\infty$ the solid angle at vertex tends to $0$, while solid angles at the base tend to $\pi/3$, hence $S\to\pi$.