Wikipedia gives an excellent treatise about solid angles in 1-2-3-Dimensions. But how about n-D? I read once some notes from a seminar held during WWII in Switzerland, and one result concerned spatial angles in even dimensions (I have forgotten the reference), but I would like to have a similar general definition, likely computationally nice.

So my question is: "What is a general expression for spatial angles in $\mathbb{R}^d$, given that origo-based vectors $x_1..x_{d+1}$ are known". It can be recursive of course.

And yes, I have an application on mind.

here is my reference: http://en.wikipedia.org/wiki/Solid_angle

Thank you for any help :)

  • 1
    $\begingroup$ Hello jjepsuomi: isn't what you need contained in "Solid angles in arbitrary dimensions" in the reference above? $\endgroup$ – Avitus Aug 15 '13 at 9:23
  • $\begingroup$ +1 Hi @Avitus The global angle factor is just the sum of all possible angles on a unit sphere $S_d(0,1)$ (area of its surface). What I need is the way to compute the area given the $d+1$ corner points. $\endgroup$ – jjepsuomi Aug 15 '13 at 9:32

I prefer to use fractions of all-space (ie the sphere = 1), rather than solid radians. This means that the solid angle occupied by a point on a polyhedron edge is the same as its dihedral angle, and the point on the meeting of faces in general are also likewise such.

I am not exactly sure if there is a means when one is given general vectors. Even the specific cases (like a regular simplex in N dimensions) are hard to find without fancy calculus. To this end, one tries to fill the inside of a sphere with a solid, and fill the exterior (just outside the unit sphere) with a least-enclosing solid.

Currently, of the simplex, it is known that the volume $S_n$, lies $1 < S_{n-1} < S_{n} < \frac 12\sqrt{n}$, of a volume of an orthotope whose diagonals are the radius of the sphere, or $1/n!$ of the solid radian. The packing of spheres in $N=8$ and $N=24$ are dense enough to fall in this range $S_8 > 1.33, \;\; S_{24} > 1.6 $. The volume of the interior solid, is $\sqrt{2^n/n}$ of the simplex inscribed in the sphere. At eight dimensions, this means $5 \frac 13$ of simplex, but we can get as many as $7 \frac 19$ times this volume in there, the extra lieing within 1/30 of the surface.

The solid angle at the orthotope converges on $2^n/2\sqrt{n}$ of the simplex, but only E8 has an integer number of simplices (17280), and orthotopes (2160) at a vertex.

The main problem is that even for the current estimate for an interior figure (which lies within 3% of the surface), raising something like $1.03^n$ can get quite large for even small values of $n$, eg $1.03^{24}=2.032$. This means that there is nearly as much room in the orange-peel (the outer 30/31 part), as there in the orange! The errors come rediably large.

There are of course, vectors that form the corner groups of symmetries, which allow one to experiment with the processes. For example, the vectors $(1,0,0,..), (1,1,0,..), ... (1,1..1)$, form a simplex whose volume is $1/(2^n. n!)$, this is the symmetry of a measure polytope.

  • $\begingroup$ +1 Good comment @wendy.krieger. I can change my question to get the answer of a space angle as a fraction of all-space then. ;) The question itself still stays (e.g. in 3D: one has $d$ vectors, and $d+1$th is at the origo. What is the spatial fraction cut out from the unit sphere by those $d$ vectors. $\endgroup$ – jjepsuomi Aug 15 '13 at 9:42

Let me cite a theorem by Ribando (Measuring Solid Angles Beyond Dimension Three, Discrete Comput Geom 36:479–487 (2006)), which is a rediscovery of a result of Aomoto (Analytic structure of Schlafli function, Nagoya Math. J. 68:1-16 (1977)) as described in the paper by Beck, Robins and Sam (Positivity Theorems for Solid-Angle Polynomials, Beitrage zur Algebra und Geometrie, Vol. 51, No. 2, 493-507 (2010)). I cite Ribando's result as its statement more to my taste:

Let $\Omega \subseteq \Bbb{R}^n$ be a solid-angle spanned by unit vectors $\lbrace v_1 , \dots , v_n \rbrace$, let $V$ be the matrix whose ith column is $v_i$ , and let $\alpha _{ij} = v_i \cdot v_j$ as above. Let $T_{\alpha}$ be the following infinite multivariable Taylor series: $$T_{\alpha} = \dfrac{det \ V}{(4 \pi )^{n/2}} \sum _{a \in \Bbb{N}^{{n \choose 2}}} \left[ \dfrac{(-2)^{\sum _{i < j} a_{ij}}}{ \Pi _{i<j} a_{ij}!} \Pi _{i} \Gamma \left( \dfrac{1 + \sum _{m \neq i} a_{im}}{2} \right) \right] \alpha^{a}$$ The series $T_{\alpha}$ agrees with the normalized measure of solid-angle $\Omega$ whenever $T_{\alpha}$ converges.

  • $\begingroup$ Thank you, much appreciated :) $\endgroup$ – jjepsuomi Mar 9 '15 at 7:31
  • $\begingroup$ Is there any non-infinite sum exact formula for some dimension greater than 3? $\endgroup$ – guillefix Mar 2 at 14:33
  • $\begingroup$ What do you mean non-infinite sum? The summation in Ribando's formula for $T_\alpha$ is a finite sum $\endgroup$ – AB Balbuena Apr 16 at 3:14

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