ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-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 rest frame of the future timelike displacement (5,3,0,0)

Example

Worked computation in an inertial frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ with $η_{0} = diag (- 1, 1, 1, 1)$ . Let $d : = (5, 3, 0, 0)$ , a future timelike coordinate vector with $- η (d, d) = 25 - 9 = 16$ , so $T : = \sqrt{- η (d, d)} = 4$ and $n : = d / T = (5 / 4, 3 / 4, 0, 0)$ .

(a) $n$ is future unit timelike: $η (n, n) = - 25 / 16 + 9 / 16 = - 1$ and $n^{0} = 5 / 4 > 3 / 4 = ∣ n_{s} ∣$ .

(b) The boost parameter $β : = n_{s} / n^{0} = (3 / 5, 0, 0)$ has $∣ β ∣ = 3 / 5 < 1$ and $γ = (1 - ∣ β ∣^{2})^{- 1 / 2} = 5 / 4$ , and the passive boost $B (β)$ of The passive Lorentz boost with parameter beta sends $n$ to $e_{0}$ , as clause (iv) of The passive boost is a proper orthochronous Lorentz map with inverse B(-beta) requires, since $γ (1, 3 / 5, 0, 0) = (5 / 4, 3 / 4, 0, 0) = n$ .

(c) The basis $(e_{0}^{'}, e_{1}^{'}, e_{2}^{'}, e_{3}^{'}) : = (n, B (- β) e_{1}, B (- β) e_{2}, B (- β) e_{3})$ has coordinate columns $e_{0}^{'} = (5 / 4, 3 / 4, 0, 0)$ , $e_{1}^{'} = (3 / 4, 5 / 4, 0, 0)$ , $e_{2}^{'} = (0, 0, 1, 0)$ , $e_{3}^{'} = (0, 0, 0, 1)$ ; the checks $η (e_{0}^{'}, e_{0}^{'}) = - 1$ , $η (e_{1}^{'}, e_{1}^{'}) = - 9 / 16 + 25 / 16 = 1$ , $η (e_{0}^{'}, e_{1}^{'}) = - (5 / 4) (3 / 4) + (3 / 4) (5 / 4) = 0$ , $η (e_{2}^{'}, e_{2}^{'}) = η (e_{3}^{'}, e_{3}^{'}) = 1$ and $η (e_{i}^{'}, e_{j}^{'}) = 0$ for the remaining pairs show that $(o; e_{0}^{'}, e_{1}^{'}, e_{2}^{'}, e_{3}^{'})$ is an inertial frame with first vector $n$ .

(d) Its chart $x^{'}$ assigns to $d$ the coordinates $(4, 0, 0, 0)$ : indeed $d = 4 n = 4 e_{0}^{'}$ , so $E^{' - 1} (d) = 4 E^{' - 1} (e_{0}^{'}) = (4, 0, 0, 0)$ ; in particular the two endpoints of any displacement $4 n$ have equal spatial coordinates in $x^{'}$ and time coordinates differing by $4$ .

Facts & Assumptions

Given: An inertial frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ with chart $x$ , $η_{0} = diag (- 1, 1, 1, 1)$ , the coordinate vector $d = (5, 3, 0, 0)$ , $T = \sqrt{- η (d, d)}$ , $n = d / T$ , and the boost parameter $β = n_{s} / n^{0}$ of The passive Lorentz boost with parameter beta (Inertial frames and inertial charts of the affine Minkowski model).

[L1]

Pairing in the frame: $η (r, r^{'}) = - r^{0} r^{' 0} + r^{1} r^{' 1} + r^{2} r^{' 2} + r^{3} r^{' 3}$ for coordinate vectors $r, r^{'}$ , and $η$ is symmetric and bilinear (Inertial frames and inertial charts of the affine Minkowski model, Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space).

[L2]

Causal classification and cone: $r$ is timelike when $η (r, r) < 0$ and future timelike exactly when $r^{0} > ∣ r_{s} ∣$ in a frame with future first vector (Timelike, null, spacelike and causal vectors of a Lorentz vector space, The timelike set has exactly two convex components, described in orthonormal coordinates).

[L3]

Rest-frame lemma: for a future timelike displacement $d$ with $T = \sqrt{- η (d, d)}$ and $n = d / T$ , the vector $n$ is future unit timelike, and for every event $o$ there is an inertial frame with first vector $n$ ; in any such frame the chart assigns to $d$ the coordinates $(T, 0, 0, 0)$ (Existence of an inertial frame whose time axis is a prescribed future timelike direction).

[L4]

Boosts: for $β \in R^{3}$ with $0 < ∣ β ∣ < 1$ the passive boost $B (β)$ acts on coordinates by $x^{' 0} = γ (x^{0} - β \cdot x)$ , $x^{'} = x + \frac{γ - 1}{∣ β ∣^{2}} (β \cdot x) β - γ β x^{0}$ , with $γ = (1 - ∣ β ∣^{2})^{- 1 / 2}$ ; for $β = 0$ set $B (0) = {id}_{V}$ ; for every $∣ β ∣ < 1$ , $B (β)$ is a proper orthochronous Lorentz map with inverse $B (- β)$ , and if $u$ is future unit timelike with coordinates $(u^{0}, u_{s})$ then $B (u_{s} / u^{0}) u = e_{0}$ ; for $β \cdot x = 0$ and $x^{0} = 0$ the formula returns $x$ (The passive Lorentz boost with parameter beta, The passive boost is a proper orthochronous Lorentz map with inverse B(-beta), Linear and affine Lorentz maps, and the Lorentz group).

[L5]

Square roots and monotonicity: every real $c \geq 0$ has a unique $\sqrt{c} \geq 0$ with $(\sqrt{c})^{2} = c$ , so $\sqrt{16} = 4$ and $\sqrt{9} = 3$ ; for $x, y \geq 0$ , $x < y ⟺ x^{2} < y^{2}$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Squaring is monotone on the nonnegatives).

[L6]

Order arithmetic: the order is trichotomous and transitive, and multiplication by a positive real preserves inequalities (Ordered field).

Verification

technique · direct

1.1givenL1L5algebra

By [L1], $η (d, d) = - 25 + 9 = - 16$ , so $- η (d, d) = 16 > 0$ and $T = \sqrt{16} = 4$ by [L5]; hence $n = d / T = (5 / 4, 3 / 4, 0, 0)$ and $n_{s} = (3 / 4, 0, 0)$ .

2.1step 1.1L1L2L3algebra

For (a): $η (n, n) = - (5 / 4)^{2} + (3 / 4)^{2} = - 25 / 16 + 9 / 16 = - 1$ by [L1], so $n$ is timelike, and $n^{0} = 5 / 4 > 3 / 4 = ∣ n_{s} ∣$ , so $n$ is future unit timelike by [L2] and [L3].

3.1step 2.1L4L5L6algebra

For (b): $β = n_{s} / n^{0} = (3 / 5, 0, 0)$ with $∣ β ∣ = 3 / 5$ and $∣ β ∣^{2} = 9 / 25 < 1$ , so $B (β)$ is defined; $γ = (1 - 9 / 25)^{- 1 / 2} = (16 / 25)^{- 1 / 2} = 5 / 4$ , because $(5 / 4)^{2} = 25 / 16$ and $5 / 4 > 0$ by [L5]; and $γ (1, 3 / 5, 0, 0) = (5 / 4) (1, 3 / 5, 0, 0) = (5 / 4, 3 / 4, 0, 0) = n$ , which is the displayed application of clause (iv) of [L4] to the future unit timelike vector $n$ .

4.1step 3.1L4L6algebra

For (c): the coordinate action of [L4] with parameter $- β$ gives $B (- β) e_{1}$ the column $(3 / 4, 5 / 4, 0, 0)$ : from $- β = (- 3 / 5, 0, 0)$ and $x = e_{1}$ , with spatial part $x_{v e c} = (1, 0, 0)$ , one gets $x^{' 0} = γ (x^{0} - (- β) \cdot x_{v e c}) = γ (0 + \frac{3}{5}) = \frac{5}{4} \cdot \frac{3}{5} = \frac{3}{4}$ and $x_{v e c}^{'} = x_{v e c} + \frac{γ - 1}{∣ β ∣^{2}} ((- β) \cdot x_{v e c}) (- β) - γ (- β) x^{0} = (1, 0, 0) + \frac{1 / 4}{9 / 25} \cdot (- \frac{3}{5}) \cdot (- \frac{3}{5}, 0, 0) = (1, 0, 0) + (\frac{1}{4}, 0, 0) = (\frac{5}{4}, 0, 0)$ , that is $B (- β) e_{1} = (3 / 4, 5 / 4, 0, 0)$ ; for $x = e_{2}$ and $x = e_{3}$ one has $β \cdot x = 0$ and $x^{0} = 0$ , so the formula returns $x$ , giving $B (- β) e_{2} = e_{2}$ and $B (- β) e_{3} = e_{3}$ ; and for $x = e_{0} = (1, 0, 0, 0)$ it gives $x^{' 0} = γ = 5 / 4$ and $x_{v e c}^{'} = - γ (- β) x^{0} = γ β = (3 / 4, 0, 0)$ , so $B (- β) e_{0} = (5 / 4, 3 / 4, 0, 0) = n = e_{0}^{'}$ ; these are the coordinate columns of the basis of the example, and $e_{a}^{'} = B (- β) e_{a}$ for every $a = 0, 1, 2, 3$ .

5.1step 4.1L1L2L3L4algebra

For the orthonormality checks of (c): $η (e_{0}^{'}, e_{0}^{'}) = - 25 / 16 + 9 / 16 = - 1$ , $η (e_{1}^{'}, e_{1}^{'}) = - 9 / 16 + 25 / 16 = 1$ , $η (e_{0}^{'}, e_{1}^{'}) = - (5 / 4) (3 / 4) + (3 / 4) (5 / 4) = 0$ , $η (e_{2}^{'}, e_{2}^{'}) = 1$ , $η (e_{3}^{'}, e_{3}^{'}) = 1$ , and the remaining cross pairings vanish by the diagonal form [L1]; since by step 4.1 the basis $(e_{0}^{'}, \dots, e_{3}^{'})$ is the image of the inertial basis $(e_{0}, \dots, e_{3})$ under the proper orthochronous Lorentz map $B (- β)$ of [L4], orthonormality also follows from metric preservation, the first vector $e_{0}^{'} = n$ is future timelike by step 2.1, and the basis is positively oriented because $B (- β)$ is proper and preserves the oriented volume form and the orientation class; therefore $(o; e_{0}^{'}, e_{1}^{'}, e_{2}^{'}, e_{3}^{'})$ is an inertial frame with first vector $n$ , as [L3] predicts.

6.1step 5.1L3algebra

For (d): $d = T n = 4 e_{0}^{'}$ by step 1.1 and step 4.1, so for the chart $x^{'} = E^{' - 1} (\cdot - o)$ of the frame of step 5.1 linearity gives $x^{'} (o + λ d) = E^{' - 1} (λ d) = λ E^{' - 1} (d) = λ (4, 0, 0, 0)$ because $E^{' - 1} (e_{0}^{'}) = (1, 0, 0, 0)$ ; in particular $x^{'}$ assigns to $d$ the coordinates $(4, 0, 0, 0)$ , and the two endpoints of any displacement $4 n$ have equal spatial coordinates and time coordinates differing by $4$ .

7.1step 2.1step 3.1step 4.1step 5.1step 6.1L3∎

Steps 2.1, 3.1, 5.1 and 6.1 verify (a), (b), (c) and (d) respectively: the example is the concrete instance $d = (5, 3, 0, 0)$ , $T = 4$ , $n = (5 / 4, 3 / 4, 0, 0)$ , $β = (3 / 5, 0, 0)$ of Existence of an inertial frame whose time axis is a prescribed future timelike direction, and the displayed basis is exactly the basis $e_{a}^{'} = B (- β) e_{a}$ constructed in step 4.1 and certified inertial in step 5.1.

Worked rest frame of the future timelike displacement (5,3,0,0)

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources