LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck pass

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.

The displacement of a future timelike worldline is future timelike

Statement

Let $c > 0$ and let $x : [a, b] \to A$ be a future timelike worldline with $a < b$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) that is piecewise $C^{1}$ with partition $a = s_{0} < s_{1} < \dots < s_{m} = b$ in the weak sense of Curves, regular causal curves and future timelike worldlines in affine Minkowski space, with future timelike interior tangents and finite one-sided difference-quotient tangents. Put $p : = x (a)$ , $q : = x (b)$ and $d : = q - p$ . Then $d$ is future timelike. More precisely, writing $ξ^{a} : = x_{A}^{a} \circ x$ for the coordinate path of the chart $x_{A}$ of any inertial frame, one has $ξ^{0} (b) - ξ^{0} (a) > 0 and ∣ ξ_{v e c} (b) - ξ_{v e c} (a) ∣ < ξ^{0} (b) - ξ^{0} (a),$ where $∣ \cdot ∣$ is the Euclidean spatial norm of The Euclidean spatial structure and cross product of an inertial frame. Consequently $p \neq q$ , the number $\sqrt{- η (d, d)}$ is a positive real number, and in every inertial chart the time component of $d$ is positive and dominates the Euclidean norm of its spatial part, which is exactly the coordinate description of future timelikeness in that chart. Completeness of the real numbers is inherited from the improper-integrability supplier; no compactness assumption beyond the given compact interval or choice principle is used.

Facts & Assumptions

Given: A real $c > 0$ ; an inertial frame with chart $x_{A}$ and coordinate path $y : = ξ = (y^{0}, z)$ of a piecewise $C^{1}$ future timelike worldline $x : [a, b] \to A$ , $a < b$ , with partition $a = s_{0} < \dots < s_{m} = b$ , future timelike interior tangents and finite one-sided difference-quotient tangents (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Oriented time-oriented affine Minkowski space (the spacetime primitives)).

[F1]

The weak piecewise class (Curves, regular causal curves and future timelike worldlines in affine Minkowski space): $x$ is $C^{1}$ on every open piece $(s_{i - 1}, s_{i})$ with future timelike tangent $x^{'}$ there, and the finite one-sided difference-quotient derivatives at both ends of every piece exist; no continuous extension of the open-piece derivative to the closed piece is assumed.

[F2]

Improper integrability and endpoint increments (Improper integrability and endpoint increments of weak piecewise future causal curves): for the piece $(α, β) = (s_{i - 1}, s_{i})$ with the notation $\int_{(α, β)} g = \lim_{u ↓ α} \int_{u}^{r} g + \lim_{v ↑ β} \int_{r}^{v} g$ of that lemma, one has $y^{0}^{'} \geq ∣ z^{'} ∣ \geq 0$ and $y^{0}^{'} > 0$ on $(α, β)$ ; every coordinate derivative is absolutely improperly Riemann integrable with signed improper integral its endpoint increment, so $\int_{(α, β)} y^{j}^{'} = y^{j} (β) - y^{j} (α)$ in the oriented convention of The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ; the improper values are additive under subdivision; and $D = x (b) - x (a)$ is nonzero future causal, with $D^{0} > 0$ and $∣ D_{s p a t i a l} ∣ \leq D^{0}$ in every inertial chart.

[F3]

Chart metric and cone coordinates (Inertial frames and inertial charts of the affine Minkowski model, The timelike set has exactly two convex components, described in orthonormal coordinates, The Euclidean spatial structure and cross product of an inertial frame, Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space): in the chart of an inertial frame (first basis vector future) a vector $v$ with coordinates $(v^{0}, v_{s})$ is future timelike exactly when $v^{0} > ∣ v_{s} ∣$ , and $- η (v, v) = (v^{0})^{2} - ∣ v_{s} ∣^{2}$ ; $η$ is bilinear.

[L4]

Norm inequalities (The inner-product norm is definite, homogeneous, and satisfies the triangle inequality, For $a \leq b$ and $f : [a, b] \to R^{m}$ integrable when $a < b$ , $∥ \int_{a}^{b} f ∥_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ ; for $a < b$ , $∥ f ∥_{2}$ is integrable, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral): the spatial Euclidean norm satisfies the triangle inequality, finite sums included; and for vector-valued functions on an interval, $∥ \int_{c}^{d} f ∥ \leq \int_{c}^{d} ∥ f ∥$ whenever $f$ is Riemann integrable.

[L5]

Monotonicity and positivity of the Riemann integral (If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ , A continuous $f \geq 0$ on $[a, b]$ with $\int_{a}^{b} f = 0$ is identically $0$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ): if $g_{1} \leq g_{2}$ pointwise on $[c, d]$ then $\int_{c}^{d} g_{1} \leq \int_{c}^{d} g_{2}$ ; and if a nonnegative continuous real function on $[c, d]$ has positive value at some point, then its integral over $[c, d]$ is positive.

[L6]

Order arithmetic of $R$ (Ordered field): the order is transitive, and a finite sum of positive reals is positive.

[L7]

Integral linearity (Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ ): for integrable real functions on a compact truncation, $\int (f - g) = \int f - \int g$ . If the two improper integrals are finite, passing to their one-sided limits proves the same identity for their difference.

Proof

technique · direct

1.1givenF1F3

Fix the inertial frame and write $y = ξ = x_{A} \circ x = (y^{0}, z)$ with $z = (y^{1}, y^{2}, y^{3})$ ; the partition has $m \geq 1$ pieces, and on each open piece $(s_{i - 1}, s_{i})$ the tangent $x^{'}$ is future timelike, so by [F3] $y^{0}^{'} (u) > ∣ z^{'} (u) ∣ \geq 0$ at every $u \in (s_{i - 1}, s_{i})$ .

2.1step 1.1F1F2

The hypotheses of [F2] are satisfied: $x$ is $C^{1}$ on the open pieces, the one-sided difference-quotient derivatives are finite, and the interior tangents are future timelike, hence in particular nonzero future causal; therefore clauses (a)-(d) of [F2] apply to each piece $(α, β) = (s_{i - 1}, s_{i})$ .

3.1step 2.1F2L5L7

Per piece: the difference $g_{i} : = y^{0}^{'} - ∣ z^{'} ∣$ is continuous and positive on $(α, β)$ by step 1.1, so for any $α < u < v < β$ its integral over $[u, v]$ is positive by [L5]; the improper integral $\int_{(α, β)} g_{i}$ dominates that truncated integral, and the additive decomposition of [F2] gives $\int_{(α, β)} g_{i} = \int_{(α, β)} y^{0}^{'} - \int_{(α, β)} ∣ z^{'} ∣$ , so $\int_{(α, β)} ∣ z^{'} ∣ < \int_{(α, β)} y^{0}^{'} = y^{0} (β) - y^{0} (α)$ with the last equality from [F2]; in particular $y^{0} (β) - y^{0} (α) > 0$ .

3.2step 2.1F2L4

Per piece the spatial increment satisfies $∣ z (β) - z (α) ∣ = ∥ \int_{(α, β)} z^{'} ∥ \leq \int_{(α, β)} ∣ z^{'} ∣$ : for $α < u < v < β$ the vector norm inequality of [L4] gives $∥ \int_{u}^{v} z^{'} ∥ \leq \int_{u}^{v} ∣ z^{'} ∣$ , and letting $u ↓ α$ , $v ↑ β$ with the endpoint-increment limits of [F2] and continuity of the norm yields the claim.

4.1step 3.2F2L4L6

Summing the finitely many pieces, with $Δ_{i} z : = z (s_{i}) - z (s_{i - 1})$ : $∣ z (b) - z (a) ∣ \leq \sum_{i = 1}^{m} ∣ Δ_{i} z ∣ \leq \sum_{i = 1}^{m} \int_{(s_{i - 1}, s_{i})} ∣ z^{'} ∣ < \sum_{i = 1}^{m} \int_{(s_{i - 1}, s_{i})} y^{0}^{'} = y^{0} (b) - y^{0} (a)$ , where the first inequality is the finite triangle inequality [L4], the second is step 3.2, the strict inequality is step 3.1 together with positivity of the finite sum of the positive differences $\int_{(s_{i - 1}, s_{i})} (y^{0}^{'} - ∣ z^{'} ∣)$ [L6], and the final equality is additivity over the partition from [F2]; in particular $y^{0} (b) - y^{0} (a) > 0$ and $∣ z (b) - z (a) ∣ < y^{0} (b) - y^{0} (a)$ .

5.1step 4.1F3L4

Consequently, in this chart the displacement $d = q - p$ has coordinates $(y^{0} (b) - y^{0} (a), z (b) - z (a))$ with positive time component strictly dominating the spatial norm, so $d$ is future timelike by the coordinate criterion [F3]; since the inertial chart was arbitrary, the same conclusion holds in every inertial chart. Hence $p \neq q$ ( $d \neq 0$ because it is timelike), and by bilinearity [F3] one has $- η (d, d) = (y^{0} (b) - y^{0} (a))^{2} - ∣ z (b) - z (a) ∣^{2} > 0$ , so $\sqrt{- η (d, d)}$ is a positive real number, and the time component of $d$ dominates in every inertial chart — the coordinate description of future timelikeness.

6.1step 5.1∎

The argument uses only the weak piecewise class on the compact interval $[a, b]$ , truncated Riemann integrals and their one-sided limits, and the finite sum over the $m$ pieces: completeness of $R$ is used by [F2] to obtain finite improper limits from monotone bounded truncation integrals, no compactness assumption beyond the interval $[a, b]$ is made, and no choice principle is used.

The displacement of a future timelike worldline is future timelike

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources