(from E. B. Vinberg, *Volumes of non-Euclidean Polyhedra*,
Russian Math. Surveys **48**:2 (1993) 15-45.)

**Let x_{0} , x_{1} , x_{2} , x_{3}
be the four vertices of a spherical tetrahedron
in the unit S^{3} in R^{4};
The length of the edge (x_{i} , x_{j})
is the angle r_{ij} = arccos(x_{i} .
x_{j}). Consider the
four faces obtained by excluding x_{0} , x_{1} , x_{2} , x_{3}
in turn.
Each of these faces is the
intersection of S^{3} with a unique 3-plane through the origin;
let e_{0} be the inward-pointing
normal to the 0-th plane, etc., scaled so that
**

(*) e_{i} . x_{i} =1 (the other e_{i} . x_{j} are 0).

Then the dihedral angle d_{ij} between the i-th and j-th face is given by cos(d_{ij}) = - e_{i} . e_{j} / sqrt((e_{i} . e_{i}) (e_{j} . e_{j})).

**
For a non-degenerate simplex, the sets ( x_{1} ... x_{4}) and
(e_{1} ... e_{4}) are
both bases of R^{4}. Writing x_{i} =
\sum a_{ij} e_{j} and
e_{i} = \sum b_{ij} x_{j},
it is clear that the matrices (a_{ij}) and
(b_{ij}) are inverse to each
other. Dotting the first equation with x_{k} and the second
with e_{k}
and invoking (*) yields**

x_{i} . x_{k} = a_{ik}

e_{i} . e_{k} = b_{ik}.

**
So the calculation goes
{ r_{ij}} --> {x_{i} . x_{j}}
--> {e_{i} . e_{j}} --> {d_{ij}}.
**

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