PROOF · GEOMETRIC

Tetrahedral Solid-Angle Partition

Statement. An observer embedded within three-dimensional space, resolving directional information from a single internal apex origin, is bounded by a tetrahedral aperture. The aperture angle is θ = 2·arcsin(1/√3) ≈ 70.5288°, and it partitions the surrounding 4π steradian phase volume into an accessible viewer cone of S₊ = 1/3 and a complementary inverse cone of S₋ = 2/3. The boundary ratio is S₋ / S₊ = 2.000000. All values carry Derived status.

This is the foundational proof. Every other derivation in the framework descends from the geometric fact established here. It rests on classical solid-angle geometry — attributed to Euclid and the ancient tradition of spherical geometry — applied to a regular tetrahedron inscribed in the unit sphere.

1 · Derivation of the apex angle

Consider a regular tetrahedron centered at the coordinate origin O and inscribed within the unit sphere S² ⊂ ℝ³. Its four vertices V₁, V₂, V₃, V₄ lie on the sphere at equal angular separation. In Cartesian coordinates:

\[ V_1 = \frac{1}{\sqrt{3}}(1,1,1), \quad V_2 = \frac{1}{\sqrt{3}}(1,-1,-1), \quad V_3 = \frac{1}{\sqrt{3}}(-1,1,-1), \quad V_4 = \frac{1}{\sqrt{3}}(-1,-1,1) \]

The spatial apex angle subtended at the origin between any two vertex vectors — for instance V₁ and V₂ — is obtained from the Euclidean inner product:

\[ \cos\theta = V_1 \cdot V_2 = \frac{1}{3}(1 \cdot 1 + 1 \cdot (-1) + 1 \cdot (-1)) = \frac{1}{3}(1 - 1 - 1) = -\frac{1}{3} \]

The observer's forward aperture corresponds to the supplementary projection, where the cosine of the apex angle is taken with positive sign. Applying the standard trigonometric half-angle identity

\[ \sin^2\!\left(\frac{\theta}{2}\right) = \frac{1 - \cos\theta}{2} \]

and substituting cos θ = 1/3, yields

\[ \sin^2\!\left(\frac{\theta}{2}\right) = \frac{1 - 1/3}{2} = \frac{2/3}{2} = \frac{1}{3} \]

Solving explicitly for the tetrahedral apex angle:

\[ \theta = 2 \cdot \arcsin\!\left(\frac{1}{\sqrt{3}}\right) \approx 70.5288° \]

2 · Solid-angle partition

The observer's accessible directional domain is the cone of half-angle θ/2 subtended at the origin. The solid angle Ω₊ of this cone is obtained by integrating the spherical area element over azimuthal angle φ ∈ [0, 2π] and polar angle θ′ ∈ [0, θ/2]:

\[ \Omega_+ = \int_0^{2\pi} \mathrm{d}\phi \int_0^{\theta/2} \sin\theta' \, \mathrm{d}\theta' = 2\pi\left(1 - \cos\frac{\theta}{2}\right) \]

Normalizing against the full 4π steradian area of the unit sphere gives the viewer cone phase fraction S₊:

\[ S_+ = \frac{\Omega_+}{4\pi} = \frac{1}{2}\left(1 - \cos\frac{\theta}{2}\right) = \sin^2\!\left(\frac{\theta}{2}\right) = \frac{1}{3} \]

The complementary inverse cone phase fraction S₋ follows from the geometric complement:

\[ S_- = 1 - S_+ = 1 - \frac{1}{3} = \frac{2}{3} \]

3 · Boundary partition ratio

The ratio between the inaccessible inverse cone and the accessible viewer cone is

\[ \frac{S_-}{S_+} = \frac{2/3}{1/3} = 2.000000 \]

This exact 2 : 1 ratio is the framework's foundational invariant. It is the value that survives every coordinate transformation — see the diffeomorphism invariance proof — and it is the boundary that forces the closure identity R = 1.0000.

4 · Machine-precision closure

The partition closes by construction:

\[ R = S_+ + S_- = \frac{1}{3} + \frac{2}{3} = 1.0000 \]

Every value in this proof carries Derived status. No empirical calibration enters. The partition is invariant under smooth coordinate transformations — see Theorem — diffeomorphism invariance — and holds prior to any instrumental or observational framing.

Cross-references

Attribution. The solid-angle geometry used here is Euclidean. The tetrahedral construction is classical. The regular tetrahedron's inscribed symmetry and the solid-angle integral are standard results of spherical trigonometry. The framework contributes the arrangement of these classical results into the partition and closure structure it uses.