ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Worked causal classification of explicit displacements

Example

Worked classification in an inertial frame of the model. Let $(A, V, η, C^{+})$ be an oriented time-oriented affine Minkowski space and let $x$ be an inertial chart with frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ and $E (u) = \sum_{a} u^{a} e_{a}$ . Vector coordinates mean $E^{- 1} (v) \in R^{4}$ , so $E^{- 1} (e_{a}) = (δ_{a}^{0}, δ_{a}^{1}, δ_{a}^{2}, δ_{a}^{3})$ . The chart $x$ itself acts on events in $A$ , not on vectors in $V$ . Computing with the components $η_{a b} = diag (- 1, 1, 1, 1)_{a b}$ :

$v = (1, 1, 0, 0)$ has $η (v, v) = 0$ and $v^{0} = 1 = ∣ v_{v e c} ∣$ , so $v$ is future null;
$w = (2, 1, 0, 0)$ has $η (w, w) = - 4 + 1 = - 3 < 0$ and $w^{0} = 2 > 1 = ∣ w_{v e c} ∣$ , so $w$ is future timelike;
$z = (1, 2, 0, 0)$ has $η (z, z) = - 1 + 4 = 3 > 0$ , so $z$ is spacelike;
$u = (- 1, 1, 0, 0)$ has $η (u, u) = 0$ and $u^{0} = - 1 < 0$ , so $u$ is past null;
$e_{0} = (1, 0, 0, 0)$ is future timelike with $η (e_{0}, e_{0}) = - 1$ , and $- e_{0}$ is past timelike.

In particular no future causal displacement is spacelike, and the coordinate descriptions $v \neq 0$ , $v^{0} \geq ∣ v_{v e c} ∣$ (future causal) and $v \neq 0$ , $v^{0} \leq - ∣ v_{v e c} ∣$ (past causal) reproduce the sign classification. The nonzero condition excludes the cone vertex, which belongs to both closed cones but is not causal.

Facts & Assumptions

Given: An inertial chart $x$ of $(A, V, η, C^{+})$ with frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ and the coordinate vectors listed in the Example.

[L1]

In an $η$ -orthonormal basis $η (v, w) = - v^{0} w^{0} + v^{1} w^{1} + v^{2} w^{2} + v^{3} w^{3}$ , so $η (v, v) = - (v^{0})^{2} + ∣ v_{v e c} ∣^{2}$ (The timelike set has exactly two convex components, described in orthonormal coordinates, statement).

[L2]

A vector is timelike when $η (v, v) < 0$ , null when $v \neq 0$ and $η (v, v) = 0$ , spacelike when $η (v, v) > 0$ (Timelike, null, spacelike and causal vectors of a Lorentz vector space).

[L3]

For a fixed time orientation, the closed cones are $cl (C^{+}) = {v^{0} \geq ∣ v_{v e c} ∣}$ and $cl (C^{-}) = {v^{0} \leq - ∣ v_{v e c} ∣}$ , and a nonzero null vector is future null exactly when it lies in $cl (C^{+})$ and past null exactly when it lies in $cl (C^{-})$ (Time orientation, future and past timelike/null/causal vectors, and the closed cones, The timelike set has exactly two convex components, described in orthonormal coordinates).

[L4]

An inertial frame's basis is $η$ -orthonormal and its first vector is future, and the frame's coordinates of $v$ are its coordinates in the ordered basis (Inertial frames and inertial charts of the affine Minkowski model).

Verification

technique · direct computation with the diagonal form

1.1L1L2L3algebra

The displacement $v = (1, 1, 0, 0)$ satisfies $η (v, v) = - 1 + 1 = 0$ and $v \neq 0$ , so $v$ is null by [L2]; since $v^{0} = 1 = ∣ v_{v e c} ∣$ , the vector lies in $cl (C^{+})$ and is future null by [L3].

1.2L1L2L3algebra

The displacement $w = (2, 1, 0, 0)$ satisfies $η (w, w) = - 4 + 1 = - 3 < 0$ , so $w$ is timelike by [L2]; since $w^{0} = 2 > 1 = ∣ w_{v e c} ∣$ , it lies in $C^{+}$ and is future timelike by [L3].

1.3L1L2algebra

The displacement $z = (1, 2, 0, 0)$ satisfies $η (z, z) = - 1 + 4 = 3 > 0$ , so $z$ is spacelike by [L2].

1.4L1L2L3algebra

The displacement $u = (- 1, 1, 0, 0)$ satisfies $η (u, u) = - 1 + 1 = 0$ and $u \neq 0$ , so $u$ is null by [L2]; since $u^{0} = - 1 = - ∣ u_{v e c} ∣$ , it lies in $cl (C^{-})$ and is past null by [L3].

1.5L1L2L3L4algebra

The frame vector $e_{0}$ has coordinates $(1, 0, 0, 0)$ by [L4], so $η (e_{0}, e_{0}) = - 1 < 0$ and $e_{0}^{0} = 1 > 0 = ∣ (e_{0})_{v e c} ∣$ : $e_{0}$ is future timelike; negating, $- e_{0}$ has $η (- e_{0}, - e_{0}) = - 1 < 0$ and $(- e_{0})^{0} = - 1 < 0$ , so $- e_{0}$ is past timelike.

2.1step 1.1step 1.2step 1.3step 1.4step 1.5L1L3∎

For any displacement $y$ , membership in $cl (C^{+})$ is exactly the condition $y^{0} \geq ∣ y_{v e c} ∣$ by [L3], which forces $η (y, y) = - (y^{0})^{2} + ∣ y_{v e c} ∣^{2} \leq 0$ by [L1]; hence no future causal displacement is spacelike, and the coordinate inequalities of [L3] reproduce the classifications computed in steps 1.1 through 1.5 for the five displayed displacements.

Depends on

Dependency tree · 2 levels

Sources

B. Crowell, Special Relativity, rev. 2020-12-10, sections 1.3-1.4 pp. 23-31 and Ch. 2 pp. 43-51 (signature (+---) converted to (-+++))