Good luck with this one --- a very long time ago I briefly became interested in these matters, but quickly despaired. It's a great shame that the natural way to measure angles in higher dimensional space is so impractical! WFL On 7/1/20, Dan Asimov <dasimov@earthlink.net> wrote:
I recently became curious to know how to calculate the volume of a regular spherical tetrahedron — that is, a subset of the unit 3-sphere S^3 bounded by 4 spherical triangles, such that all 6 dihedral angles are equal.
The 2D case is just a "regular spherical triangle" with its 3 angles equal. It's well known that if the common angle is t, then its area is
A(t) = 3t - π.
I've perused a few papers on the volume of a spherical tetrahedron, and this is astonishingly more complicated than the 2D case.
We can, however use the limiting cases and the regular 4-polytopes to assert a few of its values. Let V(t) denote the volume for dihedral angle = t, and let
a = arccos(1/3).
Note that 2π^2 is the volume of the entire 3-sphere S^3.
Then:
V(a) = 0
V(2π/5) = π^2/300
V(π/2) = π^2/8
V(2π/3) = 2π^2
V(π) = π^2
V(2π-a) = 2π^2
If we only knew the *form* of V(t), this might be (more than) enough to determine it for all t. But I don't.
—Dan
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