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.

Flat And Newtonian Bridges — Examples

1 · Prerequisites

2 · Summary

This page collects the worked computations of gr-flat-and-newtonian-bridges. On the flat model, the perturbation $h_{00} = - 2 φ$ , $h_{i j} = - 2 φ δ_{i j}$ with the mathematical quadratic potential $φ = ∣ x ∣^{2} / 2$ gives $Γ^{(1) i}_{00} = x^{i}$ , ${Ric}_{00}^{(1)} = 3$ , ${Ric}_{i j}^{(1)} = 3 δ_{i j}$ , $S^{(1)} = 6$ , $G_{00}^{(1)} = 6 = 2 Δ φ$ and $G_{0 i}^{(1)} = G_{i j}^{(1)} = 0$ (First-order curvature of the diagonal perturbation h_00 = -2 phi, h_ij = -2 phi delta_ij for the explicit potential phi = |x|^2/2). The free-fall and clock identities are checked at a timelike vector of coordinate speed $c / 2$ , and the low-velocity expansion bounds are verified at relative speed $s = c / 2$ by exact rational comparisons (Straight timelike geodesics of the flat model: constant coordinate velocity, four-velocity and the clock-reading identity, An exact observer-relative energy and momentum computation at relative speed c/2, with the expansion bounds of the low-velocity lemma checked by exact rational comparison).

For the Newtonian bridge, the spatially factored weak-field example with $φ (x) = ∣ x ∣^{2} / (2 L^{2})$ has exact potential $Δ Φ = 3 ε c^{2} / L^{2}$ and hence exact Newtonian source $\overset{ˉ}{ρ} = 3 ε c^{2} / (4 π G_{N} L^{2})$ . The Einstein-family coordinate density $ρ_{ε} = T_{00} / c^{2}$ is within the stated general component error of this source; the example supplies no independent dust density or dust model (Explicit stationary weak-field data with a dimensionless quadratic potential: exact Poisson source and the first-order Einstein coefficient). For the same scaled quadratic profile, the Newtonian comparison solution is a harmonic oscillator of frequency one. If (H1) holds and $∣ X_{0} ∣ + ∣ W_{0} ∣ < 2$ , a compact convex $K^{'} \subset B_{2} (0)$ contains both the Newtonian path and the scaled geodesic range; then (H2)-(H3) hold with $K_{2} = 1$ (The scaled Newtonian comparison system for the quadratic potential: explicit harmonic solutions, a compact convex containing ball and the hypotheses of the scaled trajectory theorem).

Two counterexamples delimit the bridges. A first-order vacuum family ${\overset{ˉ}{h}}_{a b} = - 4 x^{1} δ_{a 0} δ_{b 0}$ has $G^{(1)} = 0$ while its exact metric has nonzero ${Ric}_{00}$ of order $λ^{2}$ , so linearized vacuum does not imply exact vacuum (A first-order vacuum family whose exact metric has nonzero Ricci tensor at order epsilon^2: a harmonic diagonal perturbation with phi = x^1). Quotienting the flat model by a timelike translation gives a flat, connected, time-oriented Lorentzian four-manifold with a closed future timelike curve and no global chart with constant Minkowski components; vanishing curvature does not imply global Minkowski geometry (A flat Lorentzian four-manifold that is not isometric to Minkowski space: the rolled-up flat model R^4 / Z by a timelike translation).

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passOpen item page →

First-order curvature of the diagonal perturbation h_00 = -2 phi, h_ij = -2 phi delta_ij for the explicit potential phi = |x|^2/2

Example

Work on the flat model of The flat model spacetime: the Minkowski metric on R^4 and its inertial chart with its inertial chart $x$ and take for $Ω$ the open unit ball of $R^{4}$ (identified with its image under $x$ ). Take the flat linearized family of Flat linearized perturbations: families, trace reversal, gauge transformation and harmonic gauge with normalized remainder $q \equiv 0$ and first-order perturbation

$h_{00} = - 2 φ, h_{i j} = - 2 φ δ_{i j}, h_{0 i} = 0, φ (x) : = \frac{1}{2} ((x^{1})^{2} + (x^{2})^{2} + (x^{3})^{2}),$

so that $g (λ)_{a b} = η_{a b} + λ h_{a b}$ . Then $(ε_{1}, h, q)$ with $ε_{1} : = 1 / 8$ and $q \equiv 0$ is a flat linearized family over $Ω$ , and with all indices raised and lowered by $η$ :

$Γ^{(1) i}_{00} = x^{i}$ , $Γ^{(1) i}_{0 j} = Γ^{(1) i}_{j 0} = 0$ and $Γ^{(1) i}_{j k} = - δ_{k}^{i} x_{j} - δ_{j}^{i} x_{k} + δ_{j k} x^{i}$ , while $Γ^{(1) 0}_{00} = Γ^{(1) 0}_{i j} = 0$ and $Γ^{(1) 0}_{0 i} = Γ^{(1) 0}_{i 0} = x_{i}$ ; here $x_{j} : = δ_{j m} x^{m}$ , and $i, j, k$ range over $1, 2, 3$ .
${Ric}_{00}^{(1)} = 3$ , ${Ric}_{0 i}^{(1)} = 0$ , ${Ric}_{i j}^{(1)} = 3 δ_{i j}$ and $S^{(1)} = 6$ .
The trace-reversed perturbation is ${\overset{ˉ}{h}}_{00} = - 4 φ$ , ${\overset{ˉ}{h}}_{0 i} = 0$ , ${\overset{ˉ}{h}}_{i j} = 0$ , and the closed formula of First-order curvature of a flat linearized family, gauge transformations and the quadratic remainder clause (c) gives $G_{00}^{(1)} = 2 Δ φ = 6$ and $G_{0 i}^{(1)} = G_{i j}^{(1)} = 0$ ; these values satisfy $G_{a b}^{(1)} = {Ric}_{a b}^{(1)} - \frac{1}{2} η_{a b} S^{(1)}$ with the tensors of clause 2, as the first-order expansion of $G = Ric - \frac{1}{2} S g$ (The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors) requires.

Facts & Assumptions

Given: The flat model $(M, g, N)$ with inertial chart $x$ , the open unit ball $Ω \subseteq M$ , the potential $φ$ above, and the flat linearized family $(ε_{1}, h, q)$ with $ε_{1} = 1 / 8$ , $q \equiv 0$ and the component fields $h_{00} = - 2 φ$ , $h_{0 i} = 0$ , $h_{i j} = - 2 φ δ_{i j}$ .

[L1]

[L2]

Indices on first-order quantities are raised and lowered with $η$ : $\partial^{a} = η^{a b} \partial_{b}$ , $h^{a b} = η^{a c} η^{b d} h_{c d}$ , $□ = η^{a b} \partial_{a} \partial_{b}$ , $h = η^{a b} h_{a b}$ , ${\overset{ˉ}{h}}_{a b} = h_{a b} - \frac{1}{2} η_{a b} h$ (Flat linearized perturbations: families, trace reversal, gauge transformation and harmonic gauge clause (c), The inverse metric, musical isomorphisms and index raising).

[L3]

First-order formulas (First-order curvature of a flat linearized family, gauge transformations and the quadratic remainder clauses (a)-(c)): $Γ^{(1) a}_{b c} = \frac{1}{2} η^{a d} (\partial_{b} h_{c d} + \partial_{c} h_{b d} - \partial_{d} h_{b c})$ ; ${Ric}_{a b}^{(1)} = \frac{1}{2} (\partial_{c} \partial_{a} h^{c}_{b} + \partial_{c} \partial_{b} h^{c}_{a} - □ h_{a b} - \partial_{a} \partial_{b} h)$ ; $S^{(1)} = \partial_{a} \partial_{b} h^{a b} - □ h$ ; $G_{a b}^{(1)} = - \frac{1}{2} □ {\overset{ˉ}{h}}_{a b} + \partial^{c} \partial_{(a} {\overset{ˉ}{h}}_{b) c} - \frac{1}{2} η_{a b} \partial^{c} \partial^{d} {\overset{ˉ}{h}}_{c d}$ ; here $h^{c}_{b} = η^{c a} h_{a b}$ .

[L4]

For $φ = \frac{1}{2} ((x^{1})^{2} + (x^{2})^{2} + (x^{3})^{2})$ on $Ω$ : $\partial_{i} φ = x_{i}$ , $\partial_{i} \partial_{j} φ = δ_{i j}$ and $Δ φ = 3$ (The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field). The field $φ$ has $x^{0}$ -independent components, so for every component field $f$ that is stationary (independent of $x^{0}$ ) the d'Alembertian reduces to the Laplacian, $□ f = Δ f$ ; in particular $□ φ = 3$ and $□ x^{i} = 0$ . The components of $h$ are stationary.

[L5]

If a symmetric bilinear form on a four-dimensional real vector space has, in some basis, the matrix $diag (- 1, 1, 1, 1) + E$ with $∣ E_{i j} ∣ < 1 / 8$ for all $i, j$ , then it has inertia $(3, 1, 0)$ (Lorentzian metrics are stable under small compactly supported symmetric perturbations clause (a)).

Verification

technique · direct; verify that the family is admissible, insert the explicit components into the first-order formulas of [L3], and cross-check the Einstein tensor through the trace-reversed closed formula of [L3] and through the first-order expansion of $G=\operatorname{Ric}-\tfrac12Sg$

1.1L1L5algebra

(Admissibility.) In the coordinate frame the components of $λ h$ are $λ h_{00} = - 2 λ φ$ , $λ h_{i j} = - 2 λ φ δ_{i j}$ and $0$ otherwise, so the matrix of $g (λ) = η + λ h$ in the coordinate basis is $diag (- 1, 1, 1, 1) + E (λ)$ with $E (λ) = - 2 λ φ I_{4}$ , the entries of which satisfy $∣ E (λ)_{i j} ∣ = 2 ∣ λ ∣ φ < 2 \cdot \frac{1}{8} \cdot \frac{1}{2} = \frac{1}{8}$ on $Ω$ , because $φ < \frac{1}{2}$ there; all components are stationary and smooth, so $g (λ)$ is a smooth symmetric component field. By [L5] $g (λ)$ has inertia $(3, 1, 0)$ for every $0 < ∣ λ ∣ < ε_{1}$ , and $q \equiv 0$ is smooth; hence $(ε_{1}, h, q)$ satisfies clause (a) of [L1].

1.2L2L3L4algebra

(First-order connection.) By [L2] and [L3], $Γ^{(1) i}_{00} = \frac{1}{2} η^{i a} (2 \partial_{0} h_{a 0} - \partial_{a} h_{00})$ . The time derivatives vanish and $\partial_{a} h_{00} = - 2 \partial_{a} φ$ , so $Γ^{(1) i}_{00} = η^{i a} \partial_{a} φ = \partial^{i} φ = x^{i}$ . Similarly $Γ^{(1) i}_{0 j} = \frac{1}{2} η^{i a} (\partial_{0} h_{a j} + \partial_{j} h_{a 0} - \partial_{a} h_{0 j}) = 0$ and $Γ^{(1) i}_{j 0} = 0$ because the spatial components $h_{k 0} = h_{0 k}$ vanish and $η^{i 0} = 0$ . For the spatial pairs $Γ^{(1) i}_{j k} = \frac{1}{2} η^{i a} (\partial_{j} h_{a k} + \partial_{k} h_{a j} - \partial_{a} h_{j k}) = \frac{1}{2} δ^{i a} (- 2 \partial_{j} φ δ_{a k} - 2 \partial_{k} φ δ_{a j} + 2 \partial_{a} φ δ_{j k}) = - δ_{k}^{i} x_{j} - δ_{j}^{i} x_{k} + δ_{j k} x^{i}$ . For the components with upper index $0$ , $Γ^{(1) 0}_{b c} = \frac{1}{2} η^{0 a} (\partial_{b} h_{c a} + \partial_{c} h_{b a} - \partial_{a} h_{b c}) = - \frac{1}{2} (\partial_{b} h_{c 0} + \partial_{c} h_{b 0} - \partial_{0} h_{b c})$ , since $η^{0 a}$ is nonzero only at $a = 0$ : for $(b, c) = (0, 0)$ and for $b, c$ both spatial this vanishes ( $h_{i 0} = 0$ for spatial $i$ and all component time derivatives vanish), while $Γ^{(1) 0}_{0 i} = Γ^{(1) 0}_{i 0} = - \frac{1}{2} \partial_{i} h_{00} = \partial_{i} φ = x_{i}$ .

1.3L2algebra

(Trace data.) By [L2], $h = η^{a b} h_{a b} = η^{00} h_{00} + η^{i j} h_{i j} = 2 φ - 6 φ = - 4 φ$ , and the mixed components $h^{c}_{b} = η^{c a} h_{a b}$ are $h^{0}_{0} = η^{00} h_{00} = 2 φ$ , $h^{i}_{j} = η^{i k} h_{k j} = - 2 φ δ_{j}^{i}$ , $h^{0}_{i} = η^{0 a} h_{a i} = 0$ and $h^{i}_{0} = η^{i k} h_{k 0} = 0$ .

2.1step 1.3L2L3L4algebra

(First-order Ricci tensor.) By [L3] and step 1.3: for the $(00)$ component, $\partial_{c} \partial_{0} h^{c}_{0} = 0$ because all components of $h^{c}_{0}$ are stationary, and $- □ h_{00} - \partial_{0} \partial_{0} h = 2 □ φ = 2 Δ φ = 6$ (the field $h$ is stationary), so ${Ric}_{00}^{(1)} = \frac{1}{2} (0 + 0 + 6 + 0) = 3$ . For the $(0 i)$ component, $\partial_{c} \partial_{0} h^{c}_{i} = 0$ and $\partial_{c} \partial_{i} h^{c}_{0} = 0$ because the fields $h^{c}_{i}$ are stationary and $h^{c}_{0}$ vanishes for $c \neq 0$ and is stationary for $c = 0$ ; and $□ h_{0 i} = \partial_{0} \partial_{i} h = 0$ because $h_{0 i} = 0$ and $h$ is stationary; so ${Ric}_{0 i}^{(1)} = 0$ . For the $(i j)$ component, $\partial_{c} \partial_{i} h^{c}_{j} = \partial_{k} \partial_{i} (- 2 φ δ_{j}^{k}) = - 2 δ_{i j}$ , likewise $\partial_{c} \partial_{j} h^{c}_{i} = - 2 δ_{i j}$ , $- □ h_{i j} = 2 δ_{i j} Δ φ = 6 δ_{i j}$ and $- \partial_{i} \partial_{j} h = 4 \partial_{i} \partial_{j} φ = 4 δ_{i j}$ , so ${Ric}_{i j}^{(1)} = \frac{1}{2} (- 2 - 2 + 6 + 4) δ_{i j} = 3 δ_{i j}$ .

3.1step 2.1L2L3L4algebra

(Scalar curvature.) By [L3] and [L2], $\partial_{a} \partial_{b} h^{a b} = \partial_{i} \partial_{j} h^{i j} = \partial_{i} \partial_{j} (- 2 φ δ^{i j}) = - 2 Δ φ = - 6$ , since $h^{00} = - 2 φ$ , $h^{0 i} = 0$ , $h^{i j} = - 2 φ δ^{i j}$ and all time derivatives vanish; and $□ h = □ (- 4 φ) = - 4 Δ φ = - 12$ . Hence $S^{(1)} = - 6 + 12 = 6$ . As an arithmetic check on the defining relation $S = g^{a b} {Ric}_{a b}$ of The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors, whose first-order expansion at $λ = 0$ is $S^{(1)} = η^{a b} {Ric}_{a b}^{(1)}$ because $Ric (0) = 0$ : the values of step 2.1 give $- {Ric}_{00}^{(1)} + δ^{i j} {Ric}_{i j}^{(1)} = - 3 + 9 = 6$ , in agreement.

4.1step 1.3step 2.1step 3.1L2L3L4algebra∎

(Einstein tensor and cross-check.) By [L2] the trace reversal of step 1.3 is ${\overset{ˉ}{h}}_{00} = h_{00} - \frac{1}{2} η_{00} h = - 2 φ - \frac{1}{2} (- 1) (- 4 φ) = - 4 φ$ , ${\overset{ˉ}{h}}_{i j} = - 2 φ δ_{i j} - \frac{1}{2} δ_{i j} (- 4 φ) = 0$ and ${\overset{ˉ}{h}}_{0 i} = 0$ ; that is, ${\overset{ˉ}{h}}_{a b} = - 4 φ δ_{a 0} δ_{b 0}$ . The closed formula of [L3] then gives: for $(00)$ , $\partial^{c} \partial^{d} {\overset{ˉ}{h}}_{c d} = η^{00} η^{00} \partial_{0} \partial_{0} {\overset{ˉ}{h}}_{00} = 0$ and $\partial^{c} \partial_{(0} {\overset{ˉ}{h}}_{0) c} = 0$ by stationarity, so $G_{00}^{(1)} = - \frac{1}{2} □ (- 4 φ) = 2 Δ φ = 6$ ; for $(0 i)$ and $(i j)$ every term of the closed formula involves a component ${\overset{ˉ}{h}}_{0 \cdot}$ or a time derivative, hence vanishes, so $G_{0 i}^{(1)} = G_{i j}^{(1)} = 0$ . Finally, the values satisfy $G_{a b}^{(1)} = {Ric}_{a b}^{(1)} - \frac{1}{2} η_{a b} S^{(1)}$ : $3 - \frac{1}{2} (- 1) (6) = 6$ for $(00)$ , $3 δ_{i j} - \frac{1}{2} δ_{i j} \cdot 6 = 0$ for $(i j)$ and $0 - 0 = 0$ for $(0 i)$ , which is the first-order expansion at $λ = 0$ of the defining relation $G = Ric - \frac{1}{2} S g$ of The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors, using $S (0) = 0$ and $g (0) = η$ .

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passOpen item page →

Straight timelike geodesics of the flat model: constant coordinate velocity, four-velocity and the clock-reading identity

Example

Let $(M, g, N)$ be the flat model of The flat model spacetime: the Minkowski metric on R^4 and its inertial chart with inertial chart $x$ , and let $c > 0$ . Take the initial event $p$ with $x (p) = 0$ and the tangent vector

$U_{0} : = c v, v : = (\frac{2}{\sqrt{3}}, \frac{1}{\sqrt{3}}, 0, 0), η (v, v) = - 1, v^{0} > 0,$

so that $g_{p} (U_{0}, U_{0}) = - c^{2}$ and $U_{0}$ is future timelike. Then:

The maximal geodesic of the metric-selected connection with $x (γ (0)) = 0$ and velocity $v$ at $0$ is $x (γ (s)) = s v$ on $R$ ; it is a future timelike worldline, and $τ (s) = s / c$ is its proper time up to an additive constant, with $d γ / d τ = c v = U_{0}$ and $g (d γ / d τ, d γ / d τ) = - c^{2}$ .
Its coordinate velocity in the parametrization $t = x^{0} / c$ is the constant $w = (c / 2, 0, 0)$ and $∣ w ∣ = c / 2 < c$ .
Between coordinate times $t_{0} < t_{1}$ the proper time elapsed on the worldline is $Δ τ = \sqrt{1 - ∣ w ∣^{2} / c^{2}} (t_{1} - t_{0}) = \frac{\sqrt{3}}{2} (t_{1} - t_{0})$ , and this agrees with the direct parametrization $t = v^{0} τ$ .

Facts & Assumptions

Given: The flat model $(M, g, N)$ with inertial chart $x$ , the event $p$ with $x (p) = 0$ , and the vector $v = (2 / \sqrt{3}, 1 / \sqrt{3}, 0, 0)$ .

[L1]

In the inertial chart, $g_{a b} = η_{a b} = diag (- 1, 1, 1, 1)$ and $g_{p} (c \partial_{0}, c \partial_{0}) = - c^{2}$ ; timelike, null and causal classifications are those of $η$ (The flat model spacetime: the Minkowski metric on R^4 and its inertial chart).

[L2]

A smooth curve $γ : I \to M$ is a geodesic of the metric-selected connection if and only if there are $x_{0}, v \in R^{4}$ with $x (γ (s)) = x_{0} + s v$ ; the speed $g (γ^{'}, γ^{'}) = η (v, v)$ is constant; the maximal geodesic with $x (γ (0)) = x_{0}$ and velocity $v$ at $0$ has domain $R$ ; if $η (v, v) < 0$ and $v^{0} > 0$ it is a future timelike worldline and $τ (s) = \sqrt{- η (v, v)} (s - s_{0}) / c$ is proper time up to an additive constant, with $d γ / d τ$ the physical four-velocity $U$ and $g (U, U) = - c^{2}$ . Its coordinate velocity in the parametrization $x (γ (t)) = (c t, q (t))$ is $w = (c / v^{0}) (v^{1}, v^{2}, v^{3})$ and $∣ w ∣ < c$ (The flat model: vanishing connection and curvature, translation Killing fields and straight geodesics clauses (d)-(e), Timelike worldlines, proper time, length and four-velocity on a Lorentzian four-manifold).

[L3]

For a $C^{1}$ future timelike worldline $γ$ on an interval with nonempty interior with proper-time function $τ$ , one has $τ^{'} (t) = \frac{1}{c} \sqrt{- g_{γ (t)} (γ^{'} (t), γ^{'} (t))} > 0$ , and $τ$ is strictly increasing; reparametrizing by $τ$ gives tangent $U$ with $g (U, U) = - c^{2}$ (Proper time is well defined and reparametrization invariant; the four-velocity is its proper-time tangent clauses 1 and 3, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

For $v = (2 / \sqrt{3}, 1 / \sqrt{3}, 0, 0)$ : $η (v, v) = - (4 / 3) + (1 / 3) = - 1$ , $v^{0} = 2 / \sqrt{3} > 0$ , $1 / v^{0} = \sqrt{3} / 2$ , and $1 - 1 / (v^{0})^{2} = 1 - 3 / 4 = 1 / 4$ ; also $c^{2} η (v, v) = - c^{2}$ , and $\sqrt{3 / 4} = \sqrt{3} / 2$ (Rational powers $a^{r}$ of a positive base).

Verification

technique · direct; evaluate the flat-model geodesic, velocity and proper-time statements of [L2] and [L3] on the explicit vector $v$, and compare the clock reading with the direct affine parametrization

1.1L1L2L4algebra

(The worldline.) By [L4], $η (v, v) = - 1$ and $v^{0} > 0$ , so $U_{0} = c v$ satisfies $g_{p} (U_{0}, U_{0}) = c^{2} η (v, v) = - c^{2}$ by [L1] and is future timelike. By [L2] the maximal geodesic with $x (γ (0)) = 0$ and velocity $v$ has $x (γ (s)) = s v$ on $R$ and is a future timelike worldline; its speed is the constant $η (v, v) = - 1$ ; and $τ (s) = \sqrt{- η (v, v)} (s - 0) / c = s / c$ is proper time up to an additive constant, with $d γ / d τ = c v = U_{0}$ and $g (d γ / d τ, d γ / d τ) = c^{2} η (v, v) = - c^{2}$ .

1.2L2L4algebra

(Coordinate velocity.) In the parametrization $x (γ (t)) = (c t, q (t))$ of [L2] clause (e), the coordinate velocity is $w = (c / v^{0}) (v^{1}, v^{2}, v^{3}) = (\sqrt{3} / 2) c (1 / \sqrt{3}, 0, 0) = (c / 2, 0, 0)$ using [L4]; hence $∣ w ∣ = c / 2 < c$ and $1 - ∣ w ∣^{2} / c^{2} = 1 - 1 / 4 = 3 / 4$ , so $\sqrt{1 - ∣ w ∣^{2} / c^{2}} = \sqrt{3} / 2$ .

2.1step 1.2L1L3algebra

(Clock reading via the proper-time density.) Parametrize the worldline by $t = x^{0} / c$ ; its tangent is $γ_{t}^{'} = (c, w)$ with components $(c, c / 2, 0, 0)$ , so by [L1] $g (γ_{t}^{'}, γ_{t}^{'}) = c^{2} (- 1 + 1 / 4) = - 3 c^{2} / 4$ and [L3] gives $τ^{'} (t) = \frac{1}{c} \sqrt{3 c^{2} / 4} = \sqrt{3} / 2 > 0$ . Integrating over $[t_{0}, t_{1}]$ gives $Δ τ = (t_{1} - t_{0}) \sqrt{3} / 2 = \sqrt{1 - ∣ w ∣^{2} / c^{2}} (t_{1} - t_{0})$ by step 1.2.

3.1step 1.1step 1.2step 2.1L4algebra∎

(Cross-check with the affine parametrization.) On the parametrization $γ (s) = s v$ one has $x^{0} (γ (s)) = s v^{0}$ , that is $t = v^{0} s / c$ , and step 1.1 identifies $τ = s / c$ , so $τ = t / v^{0} = (\sqrt{3} / 2) t$ by [L4]; hence between $t_{0}$ and $t_{1}$ the elapsed proper time is $(t_{1} - t_{0}) / v^{0} = (t_{1} - t_{0}) \sqrt{3} / 2$ , in agreement with step 2.1. The identity $1 / v^{0} = \sqrt{1 - ∣ w ∣^{2} / c^{2}}$ is exactly $η (v, v) = - 1$ : $1 - ∣ w ∣^{2} / c^{2} = 1 - ∣ v^{1} ∣^{2} / (v^{0})^{2} = (1 + ∣ v^{1} ∣^{2} - ∣ v^{1} ∣^{2}) / (v^{0})^{2} = 1 / (v^{0})^{2}$ .

Remarks

Reading. The worldline, velocity and clock identities verified here are the mathematical content of the free-fall and clock clauses of The Minkowski vacuum model with Lambda = 0: field law, free-fall kinematics and clock readings; that item additionally adopts the postulates that assign a freely falling test body and an ideal clock to this worldline. No dynamics, no matter model and no measurement occur in this example. [scope]

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passOpen item page →

An exact observer-relative energy and momentum computation at relative speed c/2, with the expansion bounds of the low-velocity lemma checked by exact rational comparison

Example

Work on the flat model of The flat model spacetime: the Minkowski metric on R^4 and its inertial chart with inertial chart $x$ and let $c > 0$ and $m > 0$ . At $p$ with $x (p) = 0$ take the observer $n : = \partial_{0} ∣_{p}$ and the tangent vector

$P : = m c u, u : = \frac{2}{\sqrt{3}} \partial_{0} ∣_{p} + \frac{1}{\sqrt{3}} \partial_{1} ∣_{p},$

so that $g_{p} (u, u) = - 1$ and $P$ is future timelike with rest mass $m$ (Observer-relative energy, spatial momentum and rest mass of a tangent vector). Then the observer data of the low-velocity lemma are

$h_{n} (u) = \frac{1}{\sqrt{3}} \partial_{1} ∣_{p}, γ = \frac{2}{\sqrt{3}}, w = \frac{c}{2} \partial_{1} ∣_{p}, s = \frac{c}{2},$

and the exact energy and momentum are

$E_{n} (P) = \frac{2 m c^{2}}{\sqrt{3}}, q_{n} (P) = \frac{m c}{\sqrt{3}} \partial_{1} ∣_{p}, ∣ q_{n} (P) ∣ = \frac{m c}{\sqrt{3}} .$

With $v_{0} : = c / 2$ , $z_{0} = 1 / 4$ this realizes $s = v_{0} < c$ and the exact identities

$E_{n} (P) = m c^{2} (1 - \frac{s^{2}}{c^{2}})^{- 1 / 2}, ∣ q_{n} (P) ∣ = m s (1 - \frac{s^{2}}{c^{2}})^{- 1 / 2}$

of Low-velocity expansions of observer energy and momentum with uniform error bounds, and the two uniform error bounds of that lemma read

$∣ E_{n} (P) - m c^{2} - \frac{1}{2} m s^{2} ∣ = \frac{m c^{2} (16 - 9 \sqrt{3})}{8 \sqrt{3}} \leq \frac{m c^{2}}{12 \sqrt{3}}, ∣ ∣ q_{n} (P) ∣ - m s ∣ = \frac{m c (2 - \sqrt{3})}{2 \sqrt{3}} \leq \frac{m c}{6 \sqrt{3}},$

which are strict, finite checks of the bounds at the single relative speed $s = c / 2$ .

Facts & Assumptions

Given: The flat model with inertial chart $x$ , the point $p$ with $x (p) = 0$ , the observer $n = \partial_{0} ∣_{p}$ , the vector $u = \frac{2}{\sqrt{3}} \partial_{0} ∣_{p} + \frac{1}{\sqrt{3}} \partial_{1} ∣_{p}$ , and $P = m c u$ with $c, m > 0$ .

[L1]

An observer at $p$ is a future-directed vector $n$ with $g_{p} (n, n) = - 1$ ; the spatial projection is $h_{n} (v) = v + g_{p} (n, v) n$ (Observers, rest spaces and the spatial projection at an event clauses (a) and (c)). The projections satisfy $h_{n} (n) = 0$ , $h_{n} (v) = v$ for $g_{p} (v, n) = 0$ , and $g_{p} (h_{n} (v), h_{n} (v)) = g_{p} (v, v) + g_{p} (v, n)^{2}$ (The observer rest-space decomposition and the spatial projection clauses 1-2).

[L2]

For $P$ future timelike with rest mass $m > 0$ , $U = P / m$ , $u = U / c$ , $γ = - g_{p} (n, u)$ : $γ \geq 1$ , $E_{n} (P) = m c^{2} γ$ , $q_{n} (P) = m γ w$ , $g_{p} (q_{n}, q_{n}) = m^{2} c^{2} (γ^{2} - 1)$ with $w = c h_{n} (u) / γ$ , and $E_{n} (P) = - c g_{p} (P, n)$ , $q_{n} (P) = h_{n} (P)$ (Observer-relative energy, spatial momentum and rest mass of a tangent vector clauses (a)-(b), Positivity, relative-velocity identities and the mass shell of observer-relative energy and momentum clause 3).

[L3]

With $s : = \sqrt{g_{p} (w, w)}$ : $s < c$ , $γ = (1 - s^{2} / c^{2})^{- 1 / 2}$ , $E_{n} (P) = m c^{2} (1 - s^{2} / c^{2})^{- 1 / 2}$ , $∣ q_{n} (P) ∣ = m s (1 - s^{2} / c^{2})^{- 1 / 2}$ , and for every $0 \leq v_{0} < c$ with $z_{0} = v_{0}^{2} / c^{2} < 1$ and all $s \leq v_{0}$ the two error bounds of the statement of Low-velocity expansions of observer energy and momentum with uniform error bounds hold.

[L4]

In the inertial chart $g_{a b} = η_{a b} = diag (- 1, 1, 1, 1)$ ; $η (\partial_{0}, \partial_{0}) = - 1$ and $g_{p} (\partial_{1}, \partial_{1}) = 1$ (The flat model spacetime: the Minkowski metric on R^4 and its inertial chart, Timelike, null, spacelike and causal vectors of a Lorentz vector space).

[L5]

Arithmetic: $\sqrt{3}$ is the positive root of $3$ (Rational powers $a^{r}$ of a positive base) and $\frac{5}{3} < \sqrt{3} < 2$ , since $\frac{25}{9} < 3 < 4$ . Hence $\frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$ , $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ , $\frac{1}{\sqrt{3}} - \frac{1}{2} = \frac{2 - \sqrt{3}}{2 \sqrt{3}} > 0$ , $9 \sqrt{3} < 16$ since $(9 \sqrt{3})^{2} = 243 < 256 = 16^{2}$ , and $27 \sqrt{3} > 46$ since $(27 \sqrt{3})^{2} = 2187 > 2116 = 46^{2}$ , both by squaring positive quantities; hence $\frac{16 - 9 \sqrt{3}}{8 \sqrt{3}} > 0$ .

Verification

technique · direct; evaluate the observer data and the mass-shell identities on the explicit vector $u$, then verify the two expansion bounds by exact algebraic comparison

1.1L1L2L4algebra

(Observer data.) By [L4], $g_{p} (n, n) = - 1$ , so $n$ is a unit future-directed timelike vector, i.e. an observer. Set $u$ as given; then $g_{p} (u, u) = \frac{4}{3} g_{p} (\partial_{0}, \partial_{0}) + \frac{1}{3} g_{p} (\partial_{1}, \partial_{1}) = - \frac{4}{3} + \frac{1}{3} = - 1$ , so $u$ is future timelike with $g_{p} (u, u) = - 1$ , and $P = m c u$ is future timelike with rest mass $m (P) = m$ by [L2]. Further $g_{p} (n, u) = - \frac{2}{\sqrt{3}}$ , so by [L1] $h_{n} (u) = u + g_{p} (n, u) n = \frac{2}{\sqrt{3}} \partial_{0} + \frac{1}{\sqrt{3}} \partial_{1} - \frac{2}{\sqrt{3}} \partial_{0} = \frac{1}{\sqrt{3}} \partial_{1}$ . Hence $γ = - g_{p} (n, u) = \frac{2}{\sqrt{3}} \geq 1$ , $w = c h_{n} (u) / γ = \frac{c}{2} \partial_{1}$ , and $s = \sqrt{g_{p} (w, w)} = \frac{c}{2}$ by [L4].

2.1step 1.1L1L2L4algebra

(Exact energy and momentum.) By [L2] and [L4]: $E_{n} (P) = - c g_{p} (P, n) = - m c^{2} g_{p} (u, \partial_{0}) = m c^{2} \cdot \frac{2}{\sqrt{3}}$ and $q_{n} (P) = h_{n} (P) = m c h_{n} (u) = \frac{m c}{\sqrt{3}} \partial_{1}$ , so $∣ q_{n} (P) ∣ = \sqrt{g_{p} (q_{n}, q_{n})} = \frac{m c}{\sqrt{3}}$ since $g_{p} (\partial_{1}, \partial_{1}) = 1$ .

3.1step 1.1step 2.1L3L5algebra

(Exact identities at $s = c / 2$ .) By step 1.1, $s = c / 2 < c$ and $1 - s^{2} / c^{2} = 3 / 4$ , so $(1 - s^{2} / c^{2})^{- 1 / 2} = 2 / \sqrt{3}$ ; step 2.1 gives $E_{n} (P) = m c^{2} \cdot \frac{2}{\sqrt{3}} = m c^{2} (1 - s^{2} / c^{2})^{- 1 / 2}$ and $∣ q_{n} (P) ∣ = \frac{m c}{\sqrt{3}} = m \cdot \frac{c}{2} \cdot \frac{2}{\sqrt{3}} = m s (1 - s^{2} / c^{2})^{- 1 / 2}$ , which are the exact identities of [L3].

4.1step 3.1L3L5algebra

(Energy bound.) By step 3.1, $∣ E_{n} (P) - m c^{2} - \frac{1}{2} m s^{2} ∣ = m c^{2} ∣ \frac{2}{\sqrt{3}} - 1 - \frac{1}{8} ∣ = m c^{2} (\frac{2}{\sqrt{3}} - \frac{9}{8}) = m c^{2} \frac{16 - 9 \sqrt{3}}{8 \sqrt{3}}$ , the sign being positive by [L5]. The bound of [L3] at $v_{0} = c / 2$ is $\frac{3}{8} (\frac{3}{4})^{- 5 / 2} m \frac{(c / 2)^{4}}{c^{2}} = \frac{3}{8} \cdot \frac{32}{9 \sqrt{3}} \cdot \frac{m c^{2}}{16} = \frac{m c^{2}}{12 \sqrt{3}}$ using $(\frac{4}{3})^{5 / 2} = \frac{16}{9} \cdot \frac{2}{\sqrt{3}}$ and $\sqrt{3} > 0$ . The inequality $\frac{16 - 9 \sqrt{3}}{8 \sqrt{3}} \leq \frac{1}{12 \sqrt{3}}$ is equivalent, upon multiplying by the positive $24 \sqrt{3}$ , to $3 (16 - 9 \sqrt{3}) \leq 2$ , i.e. $27 \sqrt{3} \geq 46$ , which holds by [L5].

5.1step 3.1step 4.1L3L5algebra∎

(Momentum bound.) By step 3.1, $∣ ∣ q_{n} (P) ∣ - m s ∣ = m c ∣ \frac{1}{\sqrt{3}} - \frac{1}{2} ∣ = m c (\frac{1}{\sqrt{3}} - \frac{1}{2}) = m c \frac{2 - \sqrt{3}}{2 \sqrt{3}}$ by [L5]. The bound of [L3] at $v_{0} = c / 2$ is $\frac{1}{2} (\frac{3}{4})^{- 3 / 2} m \frac{(c / 2)^{3}}{c^{2}} = \frac{1}{2} \cdot \frac{8}{3 \sqrt{3}} \cdot \frac{m c}{8} = \frac{m c}{6 \sqrt{3}}$ using $(\frac{4}{3})^{3 / 2} = \frac{4}{3} \cdot \frac{2}{\sqrt{3}}$ . The inequality $\frac{2 - \sqrt{3}}{2 \sqrt{3}} \leq \frac{1}{6 \sqrt{3}}$ is equivalent, upon multiplying by the positive $6 \sqrt{3}$ , to $3 (2 - \sqrt{3}) \leq 1$ , i.e. $3 \sqrt{3} \geq 5$ , which holds since $27 > 25$ and both sides are positive. Both bounds of the lemma are therefore satisfied strictly at the relative speed $s = c / 2$ , and the exact identities of step 3.1 verify its statements $s < c$ , $γ = (1 - s^{2} / c^{2})^{- 1 / 2}$ and the two displayed formulas.

Scope. The example is a finite algebraic check of the kinematic identities and error bounds at one relative speed; it involves no dynamics, no conservation law and no measurement. The choices $s = c / 2$ and the direction $\partial_{1}$ are arbitrary normalizations; the points checked are the exact identities and the two strict inequalities.

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passOpen item page →

Explicit stationary weak-field data with a dimensionless quadratic potential: exact Poisson source and the first-order Einstein coefficient

Example

Work on the flat model of The flat model spacetime: the Minkowski metric on R^4 and its inertial chart with inertial chart $x$ , let $L > 0$ be a length, take spacetime region $Ω = R \times B_{L} (0) \subseteq R^{4}$ with spatial base $U = B_{L} (0) \subseteq R^{3}$ , and let $K = {\overline{B}}_{L / 2} (0) \subset U$ . Set $c > 0$ , $G_{N} > 0$ , $κ : = 8 π G_{N} / c^{4}$ , $C : = 1$ , $ε_{0} : = 1 / 2$ , choose the stationary-data remainder interval parameter $ε_{1}^{d a t a} : = 1$ , set $r \equiv 0$ on that interval, and

$φ (x) : = \frac{(x^{1})^{2} + (x^{2})^{2} + (x^{3})^{2}}{2 L^{2}} .$

Then:

$(Ω, K, L, C, ε_{0}, φ, r)$ is stationary weak-field data in the sense of Stationary weak-field data: the diagonal ansatz, its bounds, stationarity and the Einstein-family condition clause (a), with $g_{a b} (ε) = η_{a b} + ε h_{a b}$ , $h_{00} = - 2 φ$ , $h_{i j} = - 2 φ δ_{i j}$ , $h_{0 i} = 0$ ; the family $(δ_{1}, h, q)$ with $δ_{1} : = 1 / 2$ and $q \equiv 0$ is a flat linearized family and $g (ε) - η = ε h$ .
Defining $T_{a b} (ε) : = G_{a b} (g (ε)) / κ$ and $ρ_{ε} : = T_{00} (ε) / c^{2}$ (The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors) makes the data an Einstein family with coupling $κ$ (Stationary weak-field data: the diagonal ansatz, its bounds, stationarity and the Einstein-family condition clause (c)). This smooth symmetric source is defined by the metric; no dust model or independent rest-mass density is asserted.
For $Φ = ε c^{2} φ$ , $Δ Φ = 3 ε c^{2} / L^{2}$ exactly. The constant Newtonian source

$\overset{ˉ}{ρ} : = \frac{3 ε c^{2}}{4 π G_{N} L^{2}}$

satisfies $Δ Φ = 4 π G_{N} \overset{ˉ}{ρ}$ exactly on $U$ . The distinct coordinate density read from the Einstein-family tensor satisfies $∣ ρ_{ε} - \overset{ˉ}{ρ} ∣ \leq C_{2} ε^{2} c^{2} / (8 π G_{N} L^{2})$ on $K$ for $0 < ε \leq ε_{*}^{'}$ , by the general $00$ -component hierarchy; this does not use the hierarchy theorem's independent-dust conclusion.
The first-order Einstein coefficient of $A (ε) : = ε h$ is

$G_{00}^{(1)} (A (ε)) = \frac{6 ε}{L^{2}} = \frac{2 Δ Φ}{c^{2}}, G_{0 i}^{(1)} (A (ε)) = G_{i j}^{(1)} (A (ε)) = 0,$

so the leading term of the hierarchy is exactly the first-order coefficient of the family.

Facts & Assumptions

Given: The flat model with inertial chart $x$ , spacetime region $Ω = R \times B_{L} (0)$ , spatial base $U = B_{L} (0)$ , compact spatial set $K = {\overline{B}}_{L / 2} (0)$ , the function $φ$ and $r \equiv 0$ above, $C = 1$ , $ε_{0} = 1 / 2$ , $ε_{1}^{d a t a} = 1$ , $δ_{1} = 1 / 2$ , $c > 0$ , $G_{N} > 0$ and $κ = 8 π G_{N} / c^{4}$ .

[L1]

Stationary weak-field data have metric components $g_{00} = - (1 + 2 ε φ) + r_{00}$ , $g_{0 i} = r_{0 i}$ and $g_{i j} = (1 - 2 ε φ) δ_{i j} + r_{i j}$ pulled back from $U$ ; the metric must be Lorentzian on $Ω$ and the spatial derivatives through order two obey the stated bounds on $K$ . An Einstein family has a smooth symmetric field $T$ with $G = κ T$ and $ρ_{ε} = T_{00} / c^{2}$ (Stationary weak-field data: the diagonal ansatz, its bounds, stationarity and the Einstein-family condition clauses (a)-(c), Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model, The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors).

[L2]

For every stationary weak-field Einstein family, the general component hierarchy gives $∣ κ T_{00} - 2 Δ Φ / c^{2} ∣ \leq C_{2} ε^{2} / L^{2}$ on $K$ for $0 < ε \leq ε_{*}^{'}$ (The weak-field Einstein hierarchy: linearized Einstein tensor, stress components and the Poisson equation clauses (b)-(c)).

[L3]

For $φ (x) = ∣ x ∣^{2} / (2 L^{2})$ , $\partial_{i} φ = x_{i} / L^{2}$ , $\partial_{i} \partial_{j} φ = δ_{i j} / L^{2}$ and $Δ φ = 3 / L^{2}$ ; the field is stationary (The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field).

[L4]

The first-order formulas are linear in $h$ and give ${\overset{ˉ}{h}}_{00} = - 4 φ$ , ${\overset{ˉ}{h}}_{0 i} = {\overset{ˉ}{h}}_{i j} = 0$ and $G_{a b}^{(1)} = - \frac{1}{2} □ {\overset{ˉ}{h}}_{a b} + \partial^{c} \partial_{(a} {\overset{ˉ}{h}}_{b) c} - \frac{1}{2} η_{a b} \partial^{c} \partial^{d} {\overset{ˉ}{h}}_{c d}$ ; for stationary fields $□ = Δ$ (First-order curvature of a flat linearized family, gauge transformations and the quadratic remainder clause (c), Flat linearized perturbations: families, trace reversal, gauge transformation and harmonic gauge clauses (a) and (c)).

[L5]

On $U = B_{L} (0)$ , $0 \leq φ < 1 / 2$ ; for $0 < ε \leq 1 / 2$ the diagonal metric entries $- (1 + 2 ε φ)$ and $1 - 2 ε φ$ have signs $-, +, +, +$ . Also $κ c^{2} = 8 π G_{N} / c^{2}$ .

Verification

technique · check the spatially factored data, compute the exact Newtonian Poisson source and the first-order Einstein tensor, then use the general coordinate-density estimate without imposing dust assumptions

1.1L1L3L5algebra

(Admissible data.) The fields are smooth and factor through the spatial projection. For $0 < ε \leq 1 / 2$ and $x \in U$ , $0 \leq 2 ε φ < 1 / 2$ , so [L5] gives a diagonal Lorentzian metric of signature $(-, +, +, +)$ on all of $Ω$ . On $K$ , $∣ φ ∣ \leq 1 / 8$ , $∣ \partial_{i} φ ∣ \leq 1 / (2 L)$ and $∣ \partial_{i} \partial_{j} φ ∣ \leq 1 / L^{2}$ , so the bounds hold with $C = 1$ ; $r = 0$ satisfies its bounds on $(- ε_{1}^{d a t a}, ε_{1}^{d a t a})$ , with $ε_{1}^{d a t a} = 1 > ε_{0} = 1 / 2$ . Taking $δ_{1} = 1 / 2$ and $q = 0$ gives the stated flat linearized family.

1.2L3L4algebra

(First-order coefficient.) By [L3], $Δ φ = 3 / L^{2}$ . The trace reversal has ${\overset{ˉ}{h}}_{00} = - 4 φ$ and all other components zero. Formula [L4], stationarity and $□ = Δ$ give $G_{00}^{(1)} (h) = 2 Δ φ = 6 / L^{2}$ and $G_{0 i}^{(1)} (h) = G_{i j}^{(1)} (h) = 0$ . Linearity gives $G_{00}^{(1)} (A (ε)) = 6 ε / L^{2}$ and the remaining displayed components vanish.

2.1step 1.1L1L5algebra

(Einstein family.) Define $T_{a b} (ε) = G_{a b} (g (ε)) / κ$ on $Ω$ . The metric-selected connection and curvature depend smoothly on the metric and its derivatives; hence $T$ is a smooth symmetric component field and $G = κ T$ identically. Thus the data are an Einstein family by [L1], with coordinate density $ρ_{ε} = T_{00} / c^{2}$ . This construction is a mathematical source prescription, not a dust realization.

3.1step 2.1L2L3L5algebra

(Exact Poisson source and coordinate-density estimate.) By [L3], $Δ Φ = ε c^{2} Δ φ = 3 ε c^{2} / L^{2}$ . Defining $\overset{ˉ}{ρ} = 3 ε c^{2} / (4 π G_{N} L^{2})$ gives $Δ Φ = 4 π G_{N} \overset{ˉ}{ρ}$ exactly. By [L2], $∣ κ c^{2} ρ_{ε} - 6 ε / L^{2} ∣ \leq C_{2} ε^{2} / L^{2} .$ Since $κ c^{2} \overset{ˉ}{ρ} = 6 ε / L^{2}$ and $κ c^{2} = 8 π G_{N} / c^{2}$ , division by $κ c^{2} > 0$ gives $∣ ρ_{ε} - \overset{ˉ}{ρ} ∣ \leq \frac{C_{2} ε^{2} c^{2}}{8 π G_{N} L^{2}} .$ This compares the exact Newtonian source with the relativistic coordinate density to the order established by the component hierarchy; it does not assert equality or an independent dust source.

4.1step 1.2step 3.1L4algebra∎

(Leading term.) By steps 1.2 and 3.1, $G_{00}^{(1)} (A (ε)) = 6 ε / L^{2} = 2 Δ Φ / c^{2}$ , and the other first-order Einstein components vanish as stated. This completes the verification.

Remarks

Non-vacuity and scope. The example exhibits a data tuple satisfying the general Einstein-component hypotheses of the hierarchy theorem with $C = 1$ and arbitrary $L > 0$ ; it asserts no independent dust realization, and computes the dictionary quantities $Φ$ , $\overset{ˉ}{ρ}$ and $G^{(1)}$ for it. It is a conditional statement about the model family: it asserts that the quadratic potential solves the Poisson equation with constant density, not that any physical body realizes the density, and it involves no observation.

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passOpen item page →

The scaled Newtonian comparison system for the quadratic potential: explicit harmonic solutions, a compact convex containing ball and the hypotheses of the scaled trajectory theorem

Example

Work on the flat model with inertial chart $x$ , let $L > 0$ be a length, $Ω = R \times B_{2 L} (0)$ with spatial base $U = B_{2 L} (0)$ , and $K = {\overline{B}}_{L / 2} (0) \subset U$ . Set $C : = 1$ , $ε_{0} : = 1 / 16$ , $r \equiv 0$ and

$φ (x) : = \frac{∣ x ∣^{2}}{2 L^{2}}, Φ = ε c^{2} φ .$

Every unmarked norm $∣ \cdot ∣$ of a vector in $R^{3}$ below means $∥ \cdot ∥_{2}$ from the Euclidean inner product (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ). The gradient is the coordinate vector of partial derivatives (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case), and the Hessian bound below uses the induced operator norm $∥ D^{2} ψ ∥_{o p}$ of Scaled trajectory limit: weak-field geodesics converge to the Newtonian system, with rate O(epsilon) for identical initial data.

These are stationary weak-field data (Stationary weak-field data: the diagonal ansatz, its bounds, stationarity and the Einstein-family condition). The scaled spatial base is $U_{L} = B_{2} (0)$ and the scaled potential is

$ψ (X) = φ (L X) = \frac{1}{2} ∣ X ∣^{2}, \nabla ψ (X) = X, D^{2} ψ = I_{3} .$

For any $S_{*} > 0$ and initial vectors $X_{0}, W_{0} \in R^{3}$ satisfying $R_{0} : = ∣ X_{0} ∣ + ∣ W_{0} ∣ < 2$ , the Newtonian comparison system of Scaled trajectory limit: weak-field geodesics converge to the Newtonian system, with rate O(epsilon) for identical initial data

$X_{*}^{'} (S) = W_{*} (S), W_{*}^{'} (S) = - \nabla ψ (X_{*} (S)) = - X_{*} (S)$

has the solution

$X_{*} (S) = \cos S X_{0} + \sin S W_{0}, W_{*} (S) = - \sin S X_{0} + \cos S W_{0} .$

Its spatial range is contained in ${\overline{B}}_{R_{0}} (0) : = {X : ∥ X ∥_{2} \leq R_{0}}$ , with ${\overline{B}}_{0} (0) = {0}$ . Set $R^{'} : = \max {1 / 2, R_{0}} < 2$ and $K^{'} : = {\overline{B}}_{R^{'}} (0)$ . This is a compact convex subset of $U_{L}$ containing the comparison trajectory. For the trajectory estimate take $0 < ε \leq ε_{*}$ from Scaled trajectory limit: weak-field geodesics converge to the Newtonian system, with rate O(epsilon) for identical initial data. If (H1) holds, then $X_{ε} (S) \in {\overline{B}}_{1 / 2} (0) \subseteq K^{'}$ as well, so (H2) and (H3) hold with $K_{2} = 1$ . With matched scaled initial data, the conclusion (ii) of the scaled trajectory theorem gives $\sup_{[0, S_{*}]} D (S) \leq C_{3} ε S_{*} e^{2 S_{*}} = O (ε)$ .

Facts & Assumptions

Given: The flat model with inertial chart $x$ , $L > 0$ , $Ω = R \times B_{2 L} (0)$ , $U = B_{2 L} (0)$ , $K = {\overline{B}}_{L / 2} (0)$ , $C = 1$ , $ε_{0} = 1 / 16$ , the functions $φ$ and $r \equiv 0$ above, and $c > 0$ .

[L1]

The metric coefficients factor through $U$ and are Lorentzian for $0 < ε \leq 1 / 16$ : on $U$ , $0 \leq φ < 2$ , so $1 - 2 ε φ > 3 / 4 > 0$ and $1 + 2 ε φ > 0$ . On $K$ , $∣ φ ∣ \leq 1 / 8$ , $∣ \partial_{i} φ ∣ \leq 1 / (2 L)$ and $∣ \partial_{i} \partial_{j} φ ∣ \leq 1 / L^{2}$ ; $r = 0$ satisfies the remainder bounds (Stationary weak-field data: the diagonal ansatz, its bounds, stationarity and the Einstein-family condition clause (a)).

[L2]

Under $X \mapsto L X$ , the chain rule gives $\partial_{i} ψ (X) = L \partial_{i} φ (L X)$ and $D^{2} ψ (X) = L^{2} D^{2} φ (L X)$ ; for the stated potential, $\nabla ψ (X) = X$ and $D^{2} ψ = I_{3}$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder).

[L3]

Sine and cosine derivatives and bounds: $(\cos S)^{'} = - \sin S$ , $(\sin S)^{'} = \cos S$ , and $∣ \cos S ∣, ∣ \sin S ∣ \leq 1$ (The derivatives of sine and cosine are cosine and minus sine, Signs, monotonicity intervals, and ranges of sine and cosine).

[L4]

Closed Euclidean balls are compact and convex; for $v, w$ in the closed ball of radius $R^{'}$ , $∣ (1 - t) v + t w ∣ \leq (1 - t) ∣ v ∣ + t ∣ w ∣ \leq R^{'}$ for $0 \leq t \leq 1$ (Euclidean spheres and closed balls as subspaces of $R^{n}$ , For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact, A convex subset of $R^{m}$ contains every line segment between two of its points, The inner-product norm is definite, homogeneous, and satisfies the triangle inequality).

[L5]

The scaled trajectory theorem requires $K^{'} \subseteq U_{L}$ to be compact and convex, to contain every $X_{ε} (S)$ and the range of a $C^{1}$ comparison solution, and to satisfy $∥ D^{2} ψ ∥_{o p} \leq K_{2}$ on $K^{'}$ ; under (H1) and matched initial data it yields the stated $O (ε)$ bound (Scaled trajectory limit: weak-field geodesics converge to the Newtonian system, with rate O(epsilon) for identical initial data).

[L6]

The norm $∥ \cdot ∥_{2}$ is induced by the Euclidean inner product, the gradient is the coordinate vector of first partial derivatives, and for a totally differentiable map the total derivative has those partials as its coordinate entries (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, A total derivative computes every directional derivative, and its matrix is the Jacobian).

Verification

technique · verify the metric and derivative bounds on the stated spatial domain, solve the harmonic comparison system explicitly, and construct a compact convex subset of the scaled spatial domain containing both paths when (H1) holds

1.1L1L2L6algebra

(Data and scaled potential.) For every $x \in U$ , the metric is diagonal with entries $- (1 + 2 ε φ)$ and $(1 - 2 ε φ) δ_{i j}$ ; by [L1] it is Lorentzian on $Ω$ for $0 < ε \leq ε_{0}$ . The spatial derivative bounds on $K$ follow from [L1], so the tuple is stationary weak-field data. By [L2], $ψ (X) = ∣ X ∣^{2} / 2$ , $\nabla ψ (X) = X$ and $D^{2} ψ = I_{3}$ on $U_{L} = B_{2} (0)$ .

1.2L2L3algebra

(The comparison solution.) Define $X_{*}, W_{*}$ by the displayed formulas. Differentiating with [L3] gives $X_{*}^{'} = - \sin S X_{0} + \cos S W_{0} = W_{*}$ and $W_{*}^{'} = - \cos S X_{0} - \sin S W_{0} = - X_{*} = - \nabla ψ (X_{*})$ by [L2]. Since $\cos 0 = 1$ and $\sin 0 = 0$ , the initial values are $X_{*} (0) = X_{0}$ and $W_{*} (0) = W_{0}$ . These formulas are smooth on $R$ ; the range check below verifies that the position range stays in $U_{L}$ , where the comparison equation is defined.

1.3L2L3L4L5L6algebra

(Compact convex containing ball.) By [L3], $∣ X_{*} (S) ∣ \leq ∣ X_{0} ∣ + ∣ W_{0} ∣ = R_{0}$ for every $S$ . Thus its range lies in ${\overline{B}}_{R^{'}} (0)$ ; by [L4] this is compact and convex. As $R^{'} < 2$ , $K^{'} \subset U_{L}$ . If (H1) holds, then $∣ X_{ε} (S) ∣ \leq 1 / 2$ because $x_{ε} (t) \in K = {\overline{B}}_{L / 2} (0)$ , so $X_{ε} (S) \in K^{'}$ . Moreover $∥ D^{2} ψ ∥_{o p} = 1$ on $K^{'}$ by [L2]. Hence (H2)-(H3) hold with $K_{2} = 1$ under (H1).

2.1step 1.2step 1.3L5algebra∎

(Conclusion and limitation.) For $0 < ε \leq ε_{*}$ , if (H1) holds for the geodesic family with scaled initial data matching $(X_{0}, W_{0})$ , then $D (0) = 0$ . The scaled theorem's conclusion (ii) gives $\sup_{0 \leq S \leq S_{*}} D (S) \leq C_{3} ε S_{*} e^{2 S_{*}} = O (ε)$ . The example verifies the Newtonian trajectory and the domain hypotheses conditional on (H1); it asserts neither existence of such geodesics nor a physical orbit.

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passOpen item page →

A first-order vacuum family whose exact metric has nonzero Ricci tensor at order epsilon^2: a harmonic diagonal perturbation with phi = x^1

Statement refuted

Statement refuted: "Let $(ε_{1}, h, q)$ be a flat linearized family over an open set $Ω$ of the flat model (Flat linearized perturbations: families, trace reversal, gauge transformation and harmonic gauge) whose first-order Einstein tensor vanishes identically, $G_{a b}^{(1)} (h) = 0$ on $Ω$ . Then every metric $g (λ)$ of the family with $0 < ∣ λ ∣ < ε_{1}$ is an exact vacuum solution of the field equation, $G_{a b} (g (λ)) = 0$ on $Ω$ ."

The counterexample is the flat linearized family over the open unit ball $Ω \subseteq R^{4}$ with

$h_{00} = - 2 φ, h_{i j} = - 2 φ δ_{i j}, h_{0 i} = 0, φ (x) : = x^{1}, q \equiv 0, ε_{1} : = \frac{1}{4},$

so that $g (λ)_{a b} = η_{a b} + λ h_{a b}$ . Then:

$(ε_{1}, h, q)$ is a flat linearized family over $Ω$ , and $φ$ is harmonic, $Δ φ = 0$ ;
$G_{a b}^{(1)} (h) = 0$ on $Ω$ , so the hypothesis of the refuted statement holds, and by linearity $G_{a b}^{(1)} (λ h) = λ G_{a b}^{(1)} (h) = 0$ ;
nevertheless the exact Ricci tensor of $g (λ)$ does not vanish: in closed form, with $p : = 1 + 2 λ φ$ and $q : = 1 - 2 λ φ$ ,

${Ric}_{00} (g (λ)) = - \frac{2 λ^{2}}{p q^{2}} on Ω,$

which equals $- 2 λ^{2} \neq 0$ at the origin $x = 0$ for every $0 < ∣ λ ∣ < ε_{1}$ ; hence $G_{a b} (g (λ)) \neq 0$ on $Ω$ and the conclusion of the refuted statement fails.

Facts & Assumptions

Given: The flat model with inertial chart $x$ , $Ω$ the open unit ball, the potential $φ = x^{1}$ , the functions $p = 1 + 2 λ x^{1}$ , $q = 1 - 2 λ x^{1}$ , and the family $g (λ)_{a b} = η_{a b} + λ h_{a b}$ with $h_{00} = - 2 φ$ , $h_{0 i} = 0$ , $h_{i j} = - 2 φ δ_{i j}$ , $λ \in (- ε_{1}, ε_{1})$ , $ε_{1} = 1 / 4$ .

[L1]

A flat linearized family over $Ω$ consists of $ε_{1} > 0$ and smooth fields $h$ , $q$ such that for $0 < ∣ λ ∣ < ε_{1}$ the field $g_{a b} (λ, p) = η_{a b} + λ h_{a b} (p) + λ^{2} q_{a b} (λ, p)$ is the component field of a smooth Lorentzian metric of signature $(-, +, +, +)$ on $Ω$ ; the first-order coefficients are the $λ$ -derivatives at $0$ and indices are raised and lowered with $η$ , with ${\overset{ˉ}{h}}_{a b} = h_{a b} - \frac{1}{2} η_{a b} h$ and $□ = η^{a b} \partial_{a} \partial_{b}$ (Flat linearized perturbations: families, trace reversal, gauge transformation and harmonic gauge clauses (a)-(c)).

[L2]

First-order formulas (First-order curvature of a flat linearized family, gauge transformations and the quadratic remainder clauses (b)-(c)): ${Ric}_{a b}^{(1)} = \frac{1}{2} (\partial_{c} \partial_{a} h^{c}_{b} + \partial_{c} \partial_{b} h^{c}_{a} - □ h_{a b} - \partial_{a} \partial_{b} h)$ , $S^{(1)} = \partial_{a} \partial_{b} h^{a b} - □ h$ and $G_{a b}^{(1)} = - \frac{1}{2} □ {\overset{ˉ}{h}}_{a b} + \partial^{c} \partial_{(a} {\overset{ˉ}{h}}_{b) c} - \frac{1}{2} η_{a b} \partial^{c} \partial^{d} {\overset{ˉ}{h}}_{c d}$ , expressions linear in $h$ .

[L3]

The metric-selected connection is unique with Christoffel symbols $Γ^{a}_{b c} = \frac{1}{2} g^{a d} (\partial_{b} g_{c d} + \partial_{c} g_{b d} - \partial_{d} g_{b c})$ computed from the exact metric components of $g (λ)$ (Metric compatibility, torsion freedom and the metric-selected connection class, Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields, Christoffel symbols of an affine connection). Curvature components in a chart are $R^{a}_{b c d} = \partial_{c} Γ^{a}_{d b} - \partial_{d} Γ^{a}_{c b} + Γ^{a}_{c e} Γ^{e}_{d b} - Γ^{a}_{d e} Γ^{e}_{c b}$ and ${Ric}_{b d} = R^{a}_{b a d}$ with $R_{a b c d} = g_{a e} R^{e}_{b c d}$ , $G_{a b} = {Ric}_{a b} - \frac{1}{2} S g_{a b}$ (Tensoriality of the curvature, its coordinate formula and its point-normal metric formula clause (b), The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors clause (c)).

[L4]

For $φ = x^{1}$ : $\partial_{i} φ = δ_{i 1}$ , $Δ φ = 0$ , and $φ$ is stationary (The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field).

[L5]

A diagonal symmetric form with matrix $diag (- p, q, q, q)$ has inertia $(3, 1, 0)$ whenever $p > 0$ and $q > 0$ ; its coordinate eigenvalues are exactly $- p, q, q, q$ .

[L6]

On the open unit ball, $∣ x^{1} ∣ < 1$ , so for $∣ λ ∣ < ε_{1} = 1 / 4$ one has $∣ 2 λ x^{1} ∣ < 1 / 2$ and hence $p = 1 + 2 λ x^{1} > 1 / 2$ , $q = 1 - 2 λ x^{1} > 1 / 2$ . Also $p + q = 2$ , $\partial_{1} p = 2 λ$ , $\partial_{1} q = - 2 λ$ , and $\partial_{1} (1 / q) = 2 λ / q^{2}$ .

Counterexample

technique · direct; verify the family, compute the first-order Einstein coefficient from the closed formula, compute the exact Christoffel symbols and the $(00)$ Ricci component of the exact metric, and evaluate it at the origin

1.1L1L5L6algebra

(The family.) The fields $h$ and $q \equiv 0$ are smooth and stationary, and for $0 < ∣ λ ∣ < ε_{1} = 1 / 4$ the exact metric matrix is $diag (- p, q, q, q)$ . By [L6], both $p$ and $q$ are positive everywhere on $Ω$ , so [L5] gives the Lorentzian signature $(-, +, +, +)$ directly. Thus every $g (λ)$ is a smooth Lorentzian metric on $Ω$ and $(ε_{1}, h, q)$ is a flat linearized family.

1.2L1L2L4algebra

(First-order vacuum.) By [L1], $h = η^{a b} h_{a b} = 2 φ - 6 φ = - 4 φ$ , so ${\overset{ˉ}{h}}_{00} = - 2 φ - \frac{1}{2} (- 1) (- 4 φ) = - 4 φ$ and ${\overset{ˉ}{h}}_{0 i} = 0$ , ${\overset{ˉ}{h}}_{i j} = - 2 φ δ_{i j} - \frac{1}{2} δ_{i j} (- 4 φ) = 0$ ; that is, ${\overset{ˉ}{h}}_{a b} = - 4 φ δ_{a 0} δ_{b 0}$ . With $□ φ = Δ φ = 0$ and stationarity, the closed formula of [L2] gives $G_{00}^{(1)} = - \frac{1}{2} □ (- 4 φ) = 2 Δ φ = 0$ , while $\partial^{c} \partial^{d} {\overset{ˉ}{h}}_{c d} = η^{00} η^{00} \partial_{0} \partial_{0} {\overset{ˉ}{h}}_{00} = 0$ and every divergence term involves ${\overset{ˉ}{h}}_{0 \cdot}$ components or time derivatives; the $(0 i)$ and $(i j)$ components contain only such terms as well. Hence $G_{a b}^{(1)} (h) = 0$ ; by the linearity of [L2], $G_{a b}^{(1)} (λ h) = λ G_{a b}^{(1)} (h) = 0$ for every $λ$ .

1.3L3L6algebra

(Exact Christoffel symbols.) In the coordinate frame $g (λ) = diag (- p, q, q, q)$ , so $g^{00} = - 1 / p$ and $g^{i i} = 1 / q$ ; the general formula of [L3] and [L6] give, and all other components vanish: $Γ^{0}_{01} = Γ^{0}_{10} = \partial_{1} p / (2 p) = λ / p$ ; $Γ^{1}_{00} = \partial_{1} p / (2 q) = λ / q$ ; $Γ^{1}_{11} = \partial_{1} q / (2 q) = - λ / q$ ; $Γ^{1}_{22} = Γ^{1}_{33} = - \partial_{1} q / (2 q) = λ / q$ ; $Γ^{2}_{12} = Γ^{2}_{21} = Γ^{3}_{13} = Γ^{3}_{31} = \partial_{1} q / (2 q) = - λ / q$ . (For example $Γ^{1}_{22} = \frac{1}{2} g^{11} (\partial_{2} g_{12} + \partial_{2} g_{21} - \partial_{1} g_{22}) = \frac{1}{2} (1 / q) (- \partial_{1} q)$ , since only the $\partial_{1}$ derivative of the diagonal entries is nonzero; all components with a time derivative vanish by stationarity.)

2.1step 1.3L3L6algebra

(Exact Ricci component.) By [L3] and step 1.3, $R^{a}_{0 a 0} = \partial_{a} Γ^{a}_{00} - \partial_{0} Γ^{a}_{a 0} + Γ^{a}_{a e} Γ^{e}_{00} - Γ^{a}_{0 e} Γ^{e}_{a 0}$ and ${Ric}_{00} = R^{a}_{0 a 0}$ . The terms are: $\partial_{a} Γ^{a}_{00} = \partial_{1} (λ / q) = 2 λ^{2} / q^{2}$ by [L6]; $\partial_{0} Γ^{a}_{a 0} = 0$ ; $Γ^{a}_{a 1} = Γ^{0}_{01} + Γ^{1}_{11} + Γ^{2}_{21} + Γ^{3}_{31} = λ / p - 3 λ / q$ , so $Γ^{a}_{a e} Γ^{e}_{00} = Γ^{a}_{a 1} Γ^{1}_{00} = λ^{2} / (p q) - 3 λ^{2} / q^{2}$ ; and $Γ^{a}_{0 e} Γ^{e}_{a 0} = Γ^{0}_{01} Γ^{1}_{00} + Γ^{1}_{00} Γ^{0}_{10} = 2 λ^{2} / (p q)$ . Summing, ${Ric}_{00} = 2 λ^{2} / q^{2} + λ^{2} / (p q) - 3 λ^{2} / q^{2} - 2 λ^{2} / (p q) = - λ^{2} / q^{2} - λ^{2} / (p q) = - λ^{2} (p + q) / (p q^{2}) = - 2 λ^{2} / (p q^{2})$ since $p + q = 2$ by [L6].

3.1step 1.2step 2.1L1L3algebra∎

(Refutation.) At the origin $x = 0$ one has $p = q = 1$ , so step 2.1 gives ${Ric}_{00} (g (λ)) = - 2 λ^{2} \neq 0$ for every $0 < ∣ λ ∣ < ε_{1}$ ; hence the exact Ricci tensor does not vanish on $Ω$ . If the exact Einstein tensor vanished, then contracting $G_{a b} = {Ric}_{a b} - \frac{1}{2} S g_{a b}$ of [L3] with $g^{a b}$ would give $S = - g^{a b} G_{a b} = 0$ and then ${Ric}_{a b} = G_{a b} + \frac{1}{2} S g_{a b} = 0$ , contradicting ${Ric}_{00} \neq 0$ . So $G_{a b} (g (λ)) \neq 0$ on $Ω$ although $G_{a b}^{(1)} (λ h) = 0$ by step 1.2, which refutes the statement. The failure is of order $λ^{2}$ , consistent with the quadratic remainder bound of First-order curvature of a flat linearized family, gauge transformations and the quadratic remainder clause (f).

Remarks

What the example does not say. It refutes the implication "linearized vacuum solution $\Rightarrow$ exact vacuum solution" for one explicit family; it does not claim that every first-order vacuum family fails to be exact, nor that any approximate solution is useless. Exact vacuum metrics may have first-order coefficients with $G^{(1)} = 0$ (any exact vacuum family has, by the expansion), and the point is only the converse failure.
Scope. The datum $h_{a b}$ here is a mathematically chosen perturbation, not a physical field; no source, matter model, measurement or uncertainty occurs.

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passOpen item page →

A flat Lorentzian four-manifold that is not isometric to Minkowski space: the rolled-up flat model R^4 / Z by a timelike translation

Statement refuted

Statement refuted: "Let $(M, g, N)$ be a connected time-oriented smooth Lorentzian four-manifold whose metric-selected connection has vanishing curvature, $R \equiv 0$ on $M$ (Metric compatibility, torsion freedom and the metric-selected connection class, The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors). Then $(M, g)$ is isometric to Minkowski space $(R^{4}, η)$ : there is a diffeomorphism $F : M \to R^{4}$ with $F^{*} η = g$ ."

The counterexample is the quotient of the flat model by the timelike translation group. Let $(\hat{M}, \hat{g}, \hat{N})$ be the flat model (The flat model spacetime: the Minkowski metric on R^4 and its inertial chart) with inertial chart $x$ , fix $L > 0$ , let $Γ : = Z$ act on $\hat{M}$ by

$n \cdot p : = x^{- 1} (x^{0} (p) + n L, x^{1} (p), x^{2} (p), x^{3} (p)),$

and let $(M, g, N)$ be the quotient time-oriented Lorentzian four-manifold constructed below with orbit map $q : \hat{M} \to M$ , descended metric $q^{*} g = \hat{g}$ and field $N_{q (p)} = d q_{p} ({\hat{N}}_{p})$ . Then:

$(M, g, N)$ is a connected smooth time-oriented Lorentzian four-manifold;
$(M, g)$ is flat: $R \equiv 0$ for the metric-selected connection of $g$ ;
$M$ contains a nonconstant closed future timelike curve $σ : [0, L] \to M$ , $σ (s) : = q (x^{- 1} (s, 0, 0, 0))$ , with $σ (L) = σ (0)$ and $g (σ^{'}, σ^{'}) = - 1$ ;
nevertheless there is no diffeomorphism $F : M \to R^{4}$ with $F^{*} η = g$ : $M$ is not isometric to Minkowski space, and in particular it admits no global chart in which the metric components are the constants $η_{a b}$ .

Facts & Assumptions

Given: The flat model $(\hat{M}, \hat{g}, \hat{N})$ with inertial chart $x$ and ${\hat{g}}_{a b} = η_{a b} = diag (- 1, 1, 1, 1)$ , the number $L > 0$ , the group $Γ = Z$ with the action above, and the orbit space $M = \hat{M} / Γ$ with its quotient topology and orbit map $q$ , with $g$ and $N$ constructed below and the curve $σ (s) = q (x^{- 1} (s, 0, 0, 0))$ .

[L1]

The flat model has ${\hat{g}}_{a b} = η_{a b}$ constant in the inertial chart, and the metric-selected connection has vanishing Christoffel symbols and curvature in that chart (The flat model spacetime: the Minkowski metric on R^4 and its inertial chart, The flat model: vanishing connection and curvature, translation Killing fields and straight geodesics).

[L2]

A smooth free proper Lie-group action has a Hausdorff second-countable smooth quotient of dimension $\dim \hat{M} - \dim Γ$ , and its orbit map is a smooth surjective submersion (Free proper action quotient manifold). A smooth map with invertible differential at a point restricts near that point to a diffeomorphism onto an open neighborhood (The smooth inverse function theorem on manifolds).

[L3]

The countable group $Z$ with its discrete topology has a Hausdorff second-countable zero-dimensional smooth structure given by singleton charts, and addition and negation are smooth in these charts, so it is a Lie group. Its action on a four-manifold is smooth when each $p \mapsto n \cdot p$ is smooth; a smooth action and freeness and properness are as in Left group actions, transitive actions, and faithful actions and Free and proper Lie-group actions: free means trivial stabilizers, proper means the action-graph map $Θ (n, p) = (n \cdot p, p)$ has compact preimages of compact sets.

[L4]

Chart formulas (Christoffel symbols of an affine connection, Tensoriality of the curvature, its coordinate formula and its point-normal metric formula clause (b), Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields): the Christoffel symbols of the metric-selected connection in a chart are $Γ^{a}_{b c} = \frac{1}{2} g^{a d} (\partial_{b} g_{c d} + \partial_{c} g_{b d} - \partial_{d} g_{b c})$ and the curvature components are $R^{a}_{b c d} = \partial_{c} Γ^{a}_{d b} - \partial_{d} Γ^{a}_{c b} + Γ^{a}_{c e} Γ^{e}_{d b} - Γ^{a}_{d e} Γ^{e}_{c b}$ ; a tensor vanishes at a point as soon as all its components in one chart around that point vanish.

[L5]

A local diffeomorphism $s$ with $s^{*} \hat{g} = g$ has components $g_{a b} = \partial_{a} u^{c} \partial_{b} u^{d} ({\hat{g}}_{c d} \circ s)$ for the chart $u = x \circ s$ , and $d q_{p}$ is a linear isomorphism where $q$ is a local diffeomorphism (The differential of a smooth map, $C^{r}$ and smooth maps between smooth manifolds).

[L6]

For a chart with constant metric components $η_{a b}$ , the symbols of [L4] vanish identically in that chart, hence so does the curvature there (Metric compatibility, torsion freedom and the metric-selected connection class).

[L7]

Mean value theorem (Cauchy's mean value theorem: for $f, g$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $(f (b) - f (a)) g^{'} (c) = (g (b) - g (a)) f^{'} (c)$ ; no hypothesis on $g^{'}$ is needed in this product form with $g (s) = s$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length): if $f : [a, b] \to R$ is continuous and differentiable on $(a, b)$ , there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ .

[L8]

A compact subset of $R^{4}$ has bounded coordinate functions (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right), and for $F : M \to R^{4}$ with $F^{*} η = g$ one has $η (d F_{p} V, d F_{p} W) = g_{p} (V, W)$ for all $p, V, W$ (The differential of a smooth map, Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model).

Counterexample

technique · direct; verify the action hypotheses, construct the quotient and descend its metric and time orientation, exhibit constant-component charts for flatness, then exclude a global isometry with the mean value theorem applied to the time coordinate of the closed timelike curve

1.1L1L3L8

(The action.) In the chart $x$ the action reads $(x^{0}, x^{1}, x^{2}, x^{3}) \mapsto (x^{0} + n L, x^{1}, x^{2}, x^{3})$ ; each $p \mapsto n \cdot p$ is a smooth translation, so the action is smooth; it is an action of the group $Z$ on $\hat{M}$ with $0 \cdot p = p$ and $(n + m) \cdot p = n \cdot (m \cdot p)$ . It is free: $n \cdot p = p$ forces $n L = 0$ , hence $n = 0$ . It preserves $\hat{g}$ : the differential of a translation in the chart is the identity, so ${\hat{g}}_{n \cdot p} (d (n \cdot)_{p} V, d (n \cdot)_{p} W) = {\hat{g}}_{p} (V, W)$ by [L1]. It is proper: if $K_{1}, K_{2} \subseteq \hat{M}$ are compact and $p \in K_{1}$ with $n \cdot p \in K_{2}$ , then $∣ n ∣ L \leq \sup_{K_{1}} ∣ x^{0} ∣ + \sup_{K_{2}} ∣ x^{0} ∣ < \infty$ by [L8], so only finitely many $n$ occur; hence $Θ^{- 1} (K_{1} \times K_{2})$ is a finite union of compact sets, one for each such $n$ , and is compact. Finally $\hat{N} = \partial_{0}$ is $Γ$ -invariant with $\hat{g} (\hat{N}, \hat{N}) = - 1 < 0$ , since it is invariant under translations.

2.1step 1.1L2L3

(The smooth quotient.) By step 1.1, [L2] and the zero-dimensional Lie-group structure of [L3], $M$ is a Hausdorff second-countable smooth four-manifold and $q$ is a smooth surjective submersion. Thus $d q_{p}$ is onto between four-dimensional tangent spaces and hence invertible for every $p$ . The inverse function theorem in [L2] gives a local diffeomorphism $q ∣_{\hat{U}} : \hat{U} \to V$ at each point, with smooth inverse $s : V \to \hat{U}$ . The quotient is connected as the image of connected $\hat{M}$ under the continuous surjection $q$ .

3.1step 1.1step 2.1L5L8

(Metric and time orientation.) For $r = q (p)$ and $v, w \in T_{r} M$ define $g_{r} (v, w) : = {\hat{g}}_{p} ((d q_{p})^{- 1} v, (d q_{p})^{- 1} w)$ and $N_{r} : = d q_{p} ({\hat{N}}_{p})$ . These definitions are independent of the representative: if $p^{'} = n \cdot p$ , differentiating $q \circ (n \cdot) = q$ gives $d q_{p^{'}} \circ d (n \cdot)_{p} = d q_{p}$ , so the inverse images of $v, w$ at $p^{'}$ are the translates of those at $p$ . Metric invariance and field invariance from step 1.1 therefore give the same values for $g_{r}$ and $N_{r}$ . The form $g_{r}$ is symmetric and nondegenerate with signature $(-, +, +, +)$ because $d q_{p}$ is invertible and ${\hat{g}}_{p}$ has that signature. Over every inverse section $s$ of step 2.1 one has $g ∣_{V} = s^{*} \hat{g}$ and $N_{r} = d q_{s (r)} ({\hat{N}}_{s (r)})$ , both smooth by their chart formulas. Thus $g$ is a smooth Lorentzian metric with $q^{*} g = \hat{g}$ ; this identity also uniquely determines $g$ , since $q$ is onto and $d q_{p}$ is invertible. Furthermore $g (N, N)_{q (p)} = \hat{g} (\hat{N}, \hat{N})_{p} = - 1$ , so $N$ is a time orientation.

4.1step 2.1step 3.1L1L5L8

(The closed curve.) The curve $\hat{σ} (s) : = x^{- 1} (s, 0, 0, 0)$ on $[0, L]$ is smooth and satisfies $\hat{σ} (L) = x^{- 1} (L, 0, 0, 0) = 1 \cdot \hat{σ} (0)$ . Hence $σ = q \circ \hat{σ}$ is smooth and $σ (L) = q (1 \cdot \hat{σ} (0)) = σ (0)$ . Its tangent is $σ^{'} (s) = d q_{\hat{σ} (s)} (\partial_{0}) \neq 0$ by step 2.1, and step 3.1 gives $g (σ^{'}, σ^{'}) = - 1$ and $g (N, σ^{'}) = - 1$ . Thus $σ$ is future timelike and nonconstant with coincident endpoints.

4.2step 2.1step 3.1L1L4L5L6

(Flatness.) Let $r \in M$ and $p \in q^{- 1} (r)$ . Since $q$ is a local diffeomorphism there are open neighborhoods $\hat{U}$ of $p$ and $V$ of $r$ with $q ∣_{\hat{U}} : \hat{U} \to V$ a diffeomorphism; let $s : V \to \hat{U}$ be its inverse and $u : = x \circ s : V \to R^{4}$ the induced chart (Smooth manifolds and their smooth charts). For $r^{'} \in V$ and $v, w \in T_{r^{'}} M$ one has $g_{r^{'}} (v, w) = {\hat{g}}_{s (r^{'})} (d s_{r^{'}} v, d s_{r^{'}} w)$ by $q^{*} g = \hat{g}$ from step 3.1. In the induced chart $u = x \circ s$ , $d s$ takes each coordinate basis vector to the corresponding $x$ -coordinate basis vector; hence $g_{a b} = {\hat{g}}_{a b} \circ s = η_{a b}$ , constant by [L1] and [L5]. By [L6] the Christoffel symbols of the metric-selected connection vanish identically in this chart, and then the curvature formula of [L4] gives $R^{a}_{b c d} (r) = 0$ . As $r \in M$ was arbitrary and vanishing of a tensor at a point is chart-independent, $R \equiv 0$ on $M$ . The same charts exhibit for each point a neighborhood isometric to an open subset of Minkowski space, in accordance with clause (d) of Vanishing curvature on a normal neighborhood is a local isometry to Minkowski space.

5.1step 4.1L5L7L8algebra

(No global isometry.) Suppose $F : M \to R^{4}$ satisfies $F^{*} η = g$ . Let $f : = (F \circ σ)^{0} = x^{0} \circ F \circ σ$ ; $f$ is smooth by [L5], so continuous on $[0, L]$ and differentiable on $(0, L)$ , with $f (L) - f (0) = 0$ because $σ (L) = σ (0)$ . By [L7] there is $c \in (0, L)$ with $f^{'} (c) = 0$ . On the other hand, for every $s$ , [L8] and step 4.1 give $η ((F \circ σ)^{'} (s), (F \circ σ)^{'} (s)) = g (σ^{'} (s), σ^{'} (s)) = - 1$ , that is $- (f^{'} (s))^{2} + \sum_{i = 1}^{3} ((F \circ σ)^{i})^{'} (s)^{2} = - 1,$ so $(f^{'} (s))^{2} = 1 + \sum_{i = 1}^{3} ((F \circ σ)^{i})^{'} (s)^{2} \geq 1$ and $f^{'} (s) \neq 0$ ; in particular $f^{'} (c) \neq 0$ , a contradiction. Hence no such $F$ exists. Since a global chart $u$ with $g_{a b} = η_{a b}$ constant would make $u$ an isometry onto its image (with $η$ given by those constants), no such chart exists either.

6.1step 2.1step 3.1step 4.1step 4.2step 5.1∎

(Conclusion.) Steps 2.1, 3.1, 4.1 and 4.2 produce a connected time-oriented smooth Lorentzian four-manifold $(M, g, N)$ with $R \equiv 0$ and a nonconstant closed future timelike curve; step 5.1 shows it is not isometric to Minkowski space and admits no global chart with constant components $η_{a b}$ . This refutes the statement; the local isometry statement of Vanishing curvature on a normal neighborhood is a local isometry to Minkowski space is not contradicted but is precisely local, as its clause (d) asserts.

Local versus global. The example isolates the topological obstruction to the converse of the local flatness theorem: vanishing curvature gives locally constant-component charts but does not prevent a compact timelike direction, which produces closed timelike curves and defeats the global argument in step 5.1 (the real-valued function $f$ on the circle direction cannot be strictly increasing). The construction is purely mathematical: no matter, observer, clock or measurement occurs.
Related models. Other identification patterns of the flat model give flat manifolds with different global properties (for example identifications by spacelike translations, as in the flat torus). The specific timelike identification here is the one for which the closed curve is timelike; a spacelike identification instead yields closed spacelike curves and no closed causal curve.

Flat And Newtonian Bridges — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

First-order curvature of the diagonal perturbation h_00 = -2 phi, h_ij = -2 phi delta_ij for the explicit potential phi = |x|^2/2

Example

Facts & Assumptions

Verification

Straight timelike geodesics of the flat model: constant coordinate velocity, four-velocity and the clock-reading identity

Example

Facts & Assumptions

Verification

Remarks

An exact observer-relative energy and momentum computation at relative speed c/2, with the expansion bounds of the low-velocity lemma checked by exact rational comparison

Example

Facts & Assumptions

Verification

Explicit stationary weak-field data with a dimensionless quadratic potential: exact Poisson source and the first-order Einstein coefficient

Example

Facts & Assumptions

Verification

Remarks

The scaled Newtonian comparison system for the quadratic potential: explicit harmonic solutions, a compact convex containing ball and the hypotheses of the scaled trajectory theorem

Example

Facts & Assumptions

Verification

A first-order vacuum family whose exact metric has nonzero Ricci tensor at order epsilon^2: a harmonic diagonal perturbation with phi = x^1

Statement refuted

Facts & Assumptions

Counterexample

Remarks

A flat Lorentzian four-manifold that is not isometric to Minkowski space: the rolled-up flat model R^4 / Z by a timelike translation

Statement refuted

Facts & Assumptions

Counterexample

Sources