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Observer Measurements — Examples

1 · Prerequisites

2 · Summary

These worked examples and counterexamples exercise the observer-measurement structures of sr-observer-measurements on exact rational and hyperbolic data.

The inertial measurements are computed first. A radar round trip with emission at chart time $2 T / 5$ , reflection at the event $(c T, 3 c T / 5, 0, 0)$ and reception at $8 T / 5$ has future null legs, radar time $T$ and radar distance $3 c T / 5$ , matching the chart values; a second reflection at $4 c T / 5$ gives radar time $T$ and distance $4 c T / 5$ , exhibiting radar simultaneity on the slice (Worked radar round trip: emission, reflection and reception at exact rational parameters). A clock moving at $3 c / 5$ runs slow by the exact factor $4 / 5$ , with four-velocity $U = (5 c / 4, 3 c / 4, 0, 0)$ and $η (U, U) = - c^{2}$ (Worked time dilation at speed 3c/5); a rod of rest length $5$ at $3 c / 5$ has measured length $4$ at equal times of the measuring frame, with the two frames comparing different pairs of endpoint events (Worked length contraction of a rod of rest length 5 at speed 3c/5); and the velocity-transformation formula is checked in the collinear, transverse and null cases, with exact values $5 c / 13$ , $2 c / 5$ and the null limit, verifying subluminality (Worked velocity transformation: collinear, transverse and null cases).

The accelerated and rotating constructions come next. The Rindler chart $Φ (θ, ρ, y, z) = (ρ \sinh θ, ρ \cosh θ, y, z)$ is verified to be a diffeomorphism onto the right wedge with pullback metric $- ρ^{2} d θ^{2} + d ρ^{2} + d y^{2} + d z^{2}$ , acceleration $a = ρ^{- 1} \partial_{ρ}$ of magnitude $1 / ρ$ , Born-rigid irrotational slices and a non-coordinate tetrad $[n, \partial_{ρ}] = ρ^{- 1} n$ (Worked Rindler wedge congruence: chart, metric, acceleration, rigidity and non-coordinate frame). For rigid rotation at $Ω R = 3 c / 5$ the cylinder metric $- (16 / 25) c^{2} d t^{2} + (6 / 5) c R d t d φ + R^{2} d φ^{2}$ is computed, the rest lines have slope $d t / d φ = 15 R / (16 c)$ and one-turn holonomy $Δ t = 15 π R / (8 c) \neq 0$ , while the ambient rest distribution is nowhere integrable — the distinction between the global ring obstruction and the local Frobenius obstruction (Worked rigid rotation: rest-space holonomy around the ring and failure of ring synchronization).

The Bell-type translated congruence $x_{ξ} (t) = (c t, \frac{1}{2} α t^{2} + ξ_{1}, ξ_{2}, ξ_{3})$ has exact expansion $θ (s, s) = γ^{3} α^{2} t / c^{3}$ , nonzero for every $t \neq 0$ while the labels keep constant separation $∣ ξ_{1} - ξ_{2} ∣$ ; no elasticity or breaking conclusion is drawn (Worked Bell-type translated congruence: constant launch-frame separation without Born rigidity). Along the uniformly accelerated hyperbolic worldline $x (τ) = (c^{2} / α) (\sinh (α τ / c), \cosh (α τ / c), 0, 0)$ both legs $U / c$ and $X_{1} = (\sinh, \cosh, 0, 0)$ are Fermi-Walker transported, while $X_{1}^{'} = (α / c) X_{0} \neq 0$ shows that Fermi-Walker transport is not parallel transport for accelerated motion (Worked Fermi-Walker transport along uniformly accelerated hyperbolic motion).

Finally the two counterexamples delimit the rigidity and simultaneity claims. The local construction with $ψ (λ) = λ^{2} / (2 L_{0}^{2})$ , $e_{0} = (\cosh ψ, \sinh ψ, 0, 0)$ , $e_{1} = (\sinh ψ, \cosh ψ, 0, 0)$ and $Φ (λ, ξ, y, ζ) = z (λ) + ξ e_{1} (λ) + y e_{2} + ζ e_{3}$ gives a smooth future unit field with $n^{♭} = - N d λ$ , $N = 1 + λ ξ / L_{0}^{2}$ , which is Born rigid and irrotational yet has non-closed acceleration form $a^{♭} = (k / N) d ξ$ , $d a^{♭} = (1 / L_{0}^{2} N^{2}) d λ \land d ξ \neq 0$ ; no positive smooth multiple of it is Killing, refuting the claim that every Born-rigid field is locally a positive multiple of a Killing field (Counterexample: a Born-rigid congruence need not be a Killing congruence). And the events $p = x^{- 1} (0, 0, 0, 0)$ , $q = x^{- 1} (0, 1, 0, 0)$ are simultaneous relative to $n = e_{0}$ but not relative to $n^{'} = \frac{5}{4} (e_{0} + \frac{3}{5} e_{1})$ , since $η (n^{'}, q - p) = 3 / 4 \neq 0$ ; simultaneity is therefore observer-dependent (Counterexample: simultaneity is relative to the observer vector).

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

Worked radar round trip: emission, reflection and reception at exact rational parameters

Physical scope

Exact SI radar assignments for an inertial observer in complete vacuum under the ideal-clock and ideal-light assumptions of the radar protocol, with available ideal instantaneous reflectors and a clock calibrated to zero at the chart origin. Coordinates and rest distances have unit metre, T and proper/radar time have unit second, and c has unit metre per second. No finite-apparatus, material-medium, empirical-measurement or uncertainty claim is made.

Example

Adopt the ideal-clock, ideal-light, available-reflector and SI premises of The operational radar protocol of an ideal observer, and calibrate the observer's ideal clock to read $τ = 0$ at $x^{0} = 0$ . In an inertial chart $(x^{0}, x^{1}, x^{2}, x^{3})$ of the affine Minkowski model with metric $diag (- 1, 1, 1, 1)$ let the observer be at rest at the spatial origin, so that its worldline is $x^{1} = x^{2} = x^{3} = 0$ , let $T > 0$ , and let $q : = x^{- 1} (c T, \frac{3}{5} c T, 0, 0)$ , an event of chart time $T$ and spatial coordinate $(3 c T / 5, 0, 0)$ . With $e_{-} : = x^{- 1} (\frac{2}{5} c T, 0, 0, 0)$ and $e_{+} : = x^{- 1} (\frac{8}{5} c T, 0, 0, 0)$ one has $q - e_{-} = (\frac{3}{5} c T, \frac{3}{5} c T, 0, 0)$ and $e_{+} - q = (\frac{3}{5} c T, - \frac{3}{5} c T, 0, 0)$ ; both displacements have $η$ -square $- \frac{9}{25} c^{2} T^{2} + \frac{9}{25} c^{2} T^{2} = 0$ and positive time component, hence are future null, so the radar protocol applies with emission $e_{-}$ , reflection $q$ and reception $e_{+}$ (The operational radar protocol of an ideal observer, The radar round trip of an inertial observer halves at the midpoint of the reception interval). Since the observer is inertial, its proper time between the events with temporal coordinates $λ_{-} = \frac{2}{5} c T$ and $λ_{+} = \frac{8}{5} c T$ (chart times $2 T / 5$ and $8 T / 5$ ) is $(λ_{+} - λ_{-}) / c$ with the stated clock calibration, so the radar assignments are $τ_{r a d} (q) = \frac{1}{2} (\frac{2}{5} T + \frac{8}{5} T) = T$ , the chart time of $q$ , and $ρ_{r a d} (q) = \frac{c}{2} (\frac{8}{5} T - \frac{2}{5} T) = \frac{3}{5} c T$ , the chart distance of $q$ from the observer. The same computation for $q^{'} : = x^{- 1} (c T, \frac{4}{5} c T, 0, 0)$ gives $e_{-}^{'} = x^{- 1} (\frac{1}{5} c T, 0, 0, 0)$ , $e_{+}^{'} = x^{- 1} (\frac{9}{5} c T, 0, 0, 0)$ , radar time $T$ and radar distance $\frac{4}{5} c T$ . Hence the two events $q$ and $q^{'}$ of equal chart time are radar-simultaneous, with radar distances equal to their chart distances, in agreement with The radar round trip of an inertial observer halves at the midpoint of the reception interval.

Facts & Assumptions

Given: The ideal-clock, ideal-light, reflector and SI premises of [L4], with the clock calibrated by $τ (0) = 0$ , an inertial frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ with chart $x$ (Inertial frames and inertial charts of the affine Minkowski model), $T > 0$ , the observer worldline $W = {x^{- 1} (λ, 0, 0, 0) : λ \in R}$ at rest at the spatial origin, the events $q = x^{- 1} (c T, \frac{3}{5} c T, 0, 0)$ , $e_{-} = x^{- 1} (\frac{2}{5} c T, 0, 0, 0)$ , $e_{+} = x^{- 1} (\frac{8}{5} c T, 0, 0, 0)$ and $q^{'} = x^{- 1} (c T, \frac{4}{5} c T, 0, 0)$ , $e_{-}^{'} = x^{- 1} (\frac{1}{5} c T, 0, 0, 0)$ , $e_{+}^{'} = x^{- 1} (\frac{9}{5} c T, 0, 0, 0)$ (Oriented time-oriented affine Minkowski space (the spacetime primitives)).

[L1]

Chart and frame: $x^{- 1} (λ, r_{1}, r_{2}, r_{3}) = o + λ e_{0} + r_{1} e_{1} + r_{2} e_{2} + r_{3} e_{3}$ , the coordinate basis is $η$ -orthonormal with $η (e_{a}, e_{b}) = diag (- 1, 1, 1, 1)_{a b}$ , and the spatial part of a displacement is measured by the Euclidean norm of The Euclidean spatial structure and cross product of an inertial frame (Inertial frames and inertial charts of the affine Minkowski model, Oriented time-oriented affine Minkowski space (the spacetime primitives)).

[L2]

Future null criterion: a nonzero vector $v = v^{0} e_{0} + v^{1} e_{1} + v^{2} e_{2} + v^{3} e_{3}$ is future null exactly when $η (v, v) = 0$ and $v^{0} > 0$ ; indeed the closure of the future cone is ${v : v^{0} \geq ∣ v_{v e c} ∣}$ in the coordinates of the frame, and on the null cone $v^{0} = ∣ v_{v e c} ∣$ , so for $v \neq 0$ the condition $v^{0} > 0$ is exactly $v \in cl (C^{+})$ (Timelike, null, spacelike and causal vectors of a Lorentz vector space, 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).

[L3]

Radar midpoint lemma: for a future unit $n$ , events $e_{\pm} = o + λ_{\pm} n$ with $λ_{-} < λ_{+}$ and an event $q$ , the conditions " $q - e_{-}$ future null and $e_{+} - q$ future null" hold if and only if $q = o + \frac{λ_{-} + λ_{+}}{2} n + u$ with $u \in n^{⊥}$ and $∣ u ∣ = (λ_{+} - λ_{-}) / 2$ ; in that case the midpoint is $m = \frac{e_{-} + e_{+}}{2}$ , the event $q$ lies in the slice $H (m, n)$ , and its rest distance in that slice is $(λ_{+} - λ_{-}) / 2$ (The radar round trip of an inertial observer halves at the midpoint of the reception interval, Simultaneity slices and rest distance relative to a future unit timelike vector).

[L4]

Radar protocol: an observer with proper-time function $τ$ assigns to a reflection event $q$ reached from $e_{-} = x (s_{-})$ and connected to $e_{+} = x (s_{+})$ by future null segments the values $τ_{r a d} (q) = \frac{1}{2} (τ (s_{-}) + τ (s_{+}))$ and $ρ_{r a d} (q) = \frac{c}{2} (τ (s_{+}) - τ (s_{-}))$ (The operational radar protocol of an ideal observer).

[L5]

Proper time of an inertial observer: for the straight worldline $λ \mapsto o + λ e_{0}$ the proper-time function satisfies $τ^{'} (λ) = c^{- 1} \sqrt{- η (e_{0}, e_{0})} = 1 / c$ , hence $τ (λ) = λ / c$ under the stated calibration (Proper time along a future timelike worldline, Inertial frames and inertial charts of the affine Minkowski model).

Verification

technique · direct

1.1L1L2L3algebra

The events of the statement are $e_{-} = o + \frac{2}{5} c T e_{0}$ , $e_{+} = o + \frac{8}{5} c T e_{0}$ , $q = o + c T e_{0} + \frac{3}{5} c T e_{1}$ and $m : = x^{- 1} (c T, 0, 0, 0) = o + c T e_{0}$ by [L1]; the observer direction is $n : = e_{0}$ , future unit. With $u : = \frac{3}{5} c T e_{1} \in n^{⊥}$ one has $q = m + u$ and $∣ u ∣ = \frac{3}{5} c T = \frac{1}{2} (\frac{8}{5} c T - \frac{2}{5} c T) = (λ_{+} - λ_{-}) / 2$ , so condition (ii) of [L3] holds; by the implication (ii) $\Rightarrow$ (i) of [L3] the displacements $q - e_{-}$ and $e_{+} - q$ are future null and the radar protocol applies with emission $e_{-}$ , reflection $q$ and reception $e_{+}$ . Directly, $q - e_{-} = \frac{3}{5} c T e_{0} + \frac{3}{5} c T e_{1}$ and $e_{+} - q = \frac{3}{5} c T e_{0} - \frac{3}{5} c T e_{1}$ by [L1], and $η (q - e_{-}, q - e_{-}) = - \frac{9}{25} c^{2} T^{2} + \frac{9}{25} c^{2} T^{2} = 0$ with time component $\frac{3}{5} c T > 0$ , so $q - e_{-}$ is future null by [L2], and likewise $η (e_{+} - q, e_{+} - q) = 0$ with time component $\frac{3}{5} c T > 0$ , so $e_{+} - q$ is future null.

1.2L1L3L4L5algebra

Radar assignments for $q$ : the proper-time function of the observer along $W$ is $τ (λ) = λ / c$ under the stated calibration by [L5], so the emission event $e_{-}$ has parameter $λ_{-} = \frac{2}{5} c T$ and clock reading $τ (e_{-}) = \frac{2}{5} T$ , and the reception event $e_{+}$ has parameter $λ_{+} = \frac{8}{5} c T$ and reading $τ (e_{+}) = \frac{8}{5} T$ . By [L4], $τ_{r a d} (q) = \frac{1}{2} (\frac{2}{5} T + \frac{8}{5} T) = T$ , the chart time $x^{0} (q) / c$ , and $ρ_{r a d} (q) = \frac{c}{2} \cdot \frac{6}{5} T = \frac{3}{5} c T$ . The chart distance of $q$ from the observer is the Euclidean norm of the spatial coordinate difference $(3 c T / 5, 0, 0)$ , namely $\frac{3}{5} c T$ , so $ρ_{r a d} (q)$ equals the chart distance; and by [L3] the same number is the rest distance $d_{n} (m, q)$ in the slice $H (m, n)$ through the midpoint $m = x^{- 1} (c T, 0, 0, 0)$ .

1.3L1L2L3L4L5algebra

The same computation for $q^{'} = o + c T e_{0} + \frac{4}{5} c T e_{1}$ : here $q^{'} - e_{-}^{'} = \frac{4}{5} c T e_{0} + \frac{4}{5} c T e_{1}$ and $e_{+}^{'} - q^{'} = \frac{4}{5} c T e_{0} - \frac{4}{5} c T e_{1}$ by [L1], both with $η$ -square $- \frac{16}{25} c^{2} T^{2} + \frac{16}{25} c^{2} T^{2} = 0$ and time component $\frac{4}{5} c T > 0$ , hence future null by [L2]; equivalently $q^{'} = m^{'} + u^{'}$ with $m^{'} = x^{- 1} (c T, 0, 0, 0) = m$ and $u^{'} = \frac{4}{5} c T e_{1}$ , where $∣ u^{'} ∣ = \frac{4}{5} c T = \frac{1}{2} (\frac{9}{5} c T - \frac{1}{5} c T)$ satisfies condition (ii) of [L3]. Since $τ (e_{-}^{'}) = \frac{1}{5} T$ and $τ (e_{+}^{'}) = \frac{9}{5} T$ , [L4] gives $τ_{r a d} (q^{'}) = \frac{1}{2} (\frac{1}{5} T + \frac{9}{5} T) = T$ and $ρ_{r a d} (q^{'}) = \frac{c}{2} \cdot \frac{8}{5} T = \frac{4}{5} c T$ , the chart distance of $q^{'}$ ; the midpoint of the $q^{'}$ round trip is again $m$ .

2.1step 1.1step 1.2step 1.3∎

Conclusion: by steps 1.1 and 1.3 the two round trips have the same midpoint $m = m^{'} = x^{- 1} (c T, 0, 0, 0)$ , and the assigned radar times agree, $τ_{r a d} (q) = τ_{r a d} (q^{'}) = T$ ; so $q$ and $q^{'}$ are radar-simultaneous although they are distinct events with different spatial coordinates, and each lies in the slice $H (m, n)$ with the rest vector $u = \frac{3}{5} c T e_{1}$ or $u^{'} = \frac{4}{5} c T e_{1}$ by [L3]. Their radar distances $\frac{3}{5} c T$ and $\frac{4}{5} c T$ equal their chart distances and their rest distances in that slice, while the reception intervals differ, $\frac{6}{5} c T$ for $q$ and $\frac{8}{5} c T$ for $q^{'}$ , reflecting the different rest distances from the same midpoint. This is the exact rational instance of [L3]: the reflection events on one slice need not be reflections for the same emission-reception pair, and the radar assignment reproduces the chart time and chart distance for each of them.

Remarks

Exact arithmetic only. All parameters are rational multiples of $c T$ ; the nullity of the two segments and the two radar assignments are exact evaluations, and no numerical approximation or measurement model is involved.
Clock calibration. The stated calibration fixes the additive proper-time constant to zero and hence gives the displayed radar time $T$ . A different common clock offset would add that offset to both radar times while leaving their equality and both radar distances unchanged.
One midpoint, different reflection radii. The two reflection events $q$ and $q^{'}$ lie on the same slice $H (m, n)$ but at different rest distances $\frac{3}{5} c T$ and $\frac{4}{5} c T$ from the midpoint $m$ ; the lemma's statement that admissible reflections for one fixed emission-reception pair differ only in the direction of $u$ is not being applied to the pair ${q, q^{'}}$ , which has two different reception intervals, but each event separately is checked against its own pair by [L3].

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Worked time dilation at speed 3c/5

Example

In an inertial chart $x$ with metric $diag (- 1, 1, 1, 1)$ let $v = 3 c / 5$ , $γ = (1 - v^{2} / c^{2})^{- 1 / 2} = 5 / 4$ , and consider the smooth worldline $y : R \to A$ whose $x$ -coordinates are $y (t) = (c t, \frac{3 c}{5} t, 0, 0)$ .

(a) $y^{'} (t) = (c, \frac{3 c}{5}, 0, 0)$ and $η (y^{'}, y^{'}) = - c^{2} + \frac{9 c^{2}}{25} = - \frac{16}{25} c^{2}$ , so $y$ is future timelike and its proper-time rate is $τ^{'} (t) = \frac{1}{c} \sqrt{\frac{16}{25} c^{2}} = \frac{4}{5}$ : between coordinate times $t_{0} < t_{1}$ the model elapsed proper time is $\frac{4}{5} (t_{1} - t_{0}) = (t_{1} - t_{0}) / γ$ , with rate factor $4 / 5 = 1 / γ$ (Proper time along a future timelike worldline, Inertial frames and inertial charts of the affine Minkowski model).

(b) The four-velocity is $U = y^{'} / τ^{'} = (\frac{5 c}{4}, \frac{3 c}{4}, 0, 0)$ with $η (U, U) = - \frac{25 c^{2}}{16} + \frac{9 c^{2}}{16} = - c^{2}$ , and the dimensionless velocity is $u = U / c = \frac{5}{4} (1, \frac{3}{5}, 0, 0) = γ (e_{0} + \frac{3}{5} e_{1})$ for the chart's first coordinate basis vector $e_{0}$ : this is exactly the boost relation of clause (i) of Coordinate formulas of a boost: clock rates, rod lengths, simultaneity and velocity transformation (Four-velocity and four-acceleration of a future timelike worldline, Norms of four-velocity and four-acceleration, and the observer rest-space decomposition).

(c) Described in the boosted chart $x^{'}$ in which the clock is at rest (boost speed $3 c / 5$ along the first axis, The passive Lorentz boost with parameter beta), its $x^{'}$ -time rate equals its proper-time rate, and the inverse statement $Δ t = γ Δ τ$ holds with $γ = 5 / 4$ .

Facts & Assumptions

Given: An inertial chart $x$ with metric $diag (- 1, 1, 1, 1)$ , the numbers $v = 3 c / 5$ , $γ = 5 / 4$ , and the smooth worldline $y : R \to A$ whose $x$ -coordinates are $y (t) = (c t, \frac{3 c}{5} t, 0, 0)$ (Inertial frames and inertial charts of the affine Minkowski model, Proper time along a future timelike worldline).

[L1]

For a $C^{1}$ future timelike worldline the proper-time function satisfies $τ (b) - τ (a) = c^{- 1} \int_{a}^{b} \sqrt{- η (y^{'}, y^{'})} d s$ ; in an inertial chart $η (u, u) = - (u^{0})^{2} + (u^{1})^{2} + (u^{2})^{2} + (u^{3})^{2}$ (Proper time along a future timelike worldline, Inertial frames and inertial charts of the affine Minkowski model).

[L2]

The four-velocity is $U = x^{'} / τ^{'}$ , its norm is $η (U, U) = - c^{2}$ , and the dimensionless velocity $u = U / c$ is future unit (Four-velocity and four-acceleration of a future timelike worldline, Norms of four-velocity and four-acceleration, and the observer rest-space decomposition).

[L3]

Clause (i) of Coordinate formulas of a boost: clock rates, rod lengths, simultaneity and velocity transformation: the worldline $x^{- 1} (λ, \frac{v}{c} λ + a_{1}, a_{2}, a_{3})$ has proper-time rate $1 / γ$ with respect to the coordinate time of $x$ and its tangent is future timelike; in the chart $x^{'}$ of the boost with parameter $v$ the clock is at rest.

Verification

technique · direct

1.1L1L3algebra

The coordinate derivative of $y$ is $y^{'} (t) = (c, \frac{3 c}{5}, 0, 0)$ , and by [L1] $η (y^{'}, y^{'}) = - c^{2} + \frac{9}{25} c^{2} = - \frac{16}{25} c^{2} < 0$ with $y^{' 0} = c > 0$ , so $y$ is future timelike; the proper-time rate is $τ^{'} (t) = \frac{1}{c} \sqrt{\frac{16}{25} c^{2}} = \frac{4}{5}$ . Hence for $t_{0} < t_{1}$ the proper time elapsed is $\frac{4}{5} (t_{1} - t_{0}) = \frac{t_{1} - t_{0}}{5 / 4}$ , i.e. the model proper-time rate is $4 / 5 = 1 / γ$ ; equivalently $Δ t = γ Δ τ = \frac{5}{4} Δ τ$ .

1.2L2L3algebra

The four-velocity is $U = y^{'} / τ^{'} = \frac{5}{4} (c, \frac{3 c}{5}, 0, 0) = (\frac{5 c}{4}, \frac{3 c}{4}, 0, 0)$ , whose norm is $η (U, U) = - \frac{25}{16} c^{2} + \frac{9}{16} c^{2} = - c^{2}$ by [L1], in agreement with [L2]; the dimensionless velocity is $u = U / c = \frac{5}{4} (1, \frac{3}{5}, 0, 0) = γ (e_{0} + \frac{3}{5} e_{1})$ , which is the unit tangent vector of clause (i) of [L3] with $v / c = 3 / 5$ and $γ = 5 / 4$ .

2.1L3step 1.1algebra

In the chart $x^{'}$ of the boost with parameter $v = 3 c / 5$ the spatial coordinates of $y$ are constant, so the clock is at rest in $x^{'}$ ; a clock at rest in its own chart has $τ^{'} = 1$ with respect to that chart's coordinate time, so its $x^{'}$ -time rate equals its proper-time rate, and the relation $Δ t = γ Δ τ$ of step 1.1 is the inverse statement with $γ = 5 / 4$ .

3.1step 1.1step 1.2step 2.1∎

Conclusion: steps 1.1-2.1 verify the three clauses (a), (b), (c) of the Example: the worldline is future timelike with proper-time rate $4 / 5 = (1 / γ)$ , its four-velocity is $(\frac{5 c}{4}, \frac{3 c}{4}, 0, 0)$ of norm $- c^{2}$ , the dimensionless velocity is the inverse passive-boost image $γ (e_{0} + \frac{3}{5} e_{1})$ , and in the rest chart of the clock the time rate is the proper-time rate, so the inverse dilation $Δ t = γ Δ τ$ holds with $γ = 5 / 4$ .

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Worked length contraction of a rod of rest length 5 at speed 3c/5

Example

In an inertial chart $x$ let the chart $x^{'}$ be related to $x$ by a boost of speed $v = 3 c / 5$ along the first spatial axis, $γ = 5 / 4$ , so that $x^{' 0} = \frac{5}{4} (x^{0} - \frac{3}{5} x^{1})$ and $x^{' 1} = \frac{5}{4} (x^{1} - \frac{3}{5} x^{0})$ (Coordinate formulas of a boost: clock rates, rod lengths, simultaneity and velocity transformation, The passive Lorentz boost with parameter beta). Let $L_{0} = 5$ and let the rod at rest in $x^{'}$ have endpoints on the worldlines $x^{' 1} = 0$ and $x^{' 1} = 5$ , with vanishing transverse coordinates. Solving the transition for $x^{1}$ as a function of $x^{0}$ gives $x^{1} = \frac{3}{5} x^{0}$ on the first worldline and $x^{1} = \frac{3}{5} x^{0} + 4$ on the second, so at every equal $x$ -time the endpoints are separated by $4 = L_{0} / γ$ in the frame $x$ . Explicit witness: the events $p : = x^{- 1} (5, 3, 0, 0)$ and $r : = x^{- 1} (5, 7, 0, 0)$ are simultaneous in $x$ (both have $x^{0} = 5$ ) and separated by $4$ ; their $x^{'}$ -coordinates are $x^{'} (p) = (4, 0, 0, 0)$ and $x^{'} (r) = (1, 5, 0, 0)$ , so they are not simultaneous in $x^{'}$ but their $x^{'}$ -separation is $5 = L_{0}$ . Comparing events at equal $x^{'}$ -time instead: the events of the two worldlines with $x^{' 0} = 4$ are $x^{- 1} (5, 3, 0, 0)$ and $x^{- 1} (35 / 4, 37 / 4, 0, 0)$ , the second of which has $x^{' 1} = 5$ ; at equal $x^{'}$ -time $4$ the endpoints are separated by $5$ , the rest length. Thus the frame $x$ assigns length $4$ to a rod of rest length $5$ at rest in $x^{'}$ , exactly the contraction $L_{0} / γ$ of clause (ii) of Coordinate formulas of a boost: clock rates, rod lengths, simultaneity and velocity transformation, and the comparison uses the simultaneity slice of the measuring frame in the sense of Simultaneity slices and rest distance relative to a future unit timelike vector.

Facts & Assumptions

Given: Two inertial charts $x$ , $x^{'}$ of the affine Minkowski model related by the boost of speed $v = 3 c / 5$ along the first axis with $γ = 5 / 4$ , the number $L_{0} = 5$ , the two worldlines $x^{' 1} = 0$ and $x^{' 1} = 5$ at rest in $x^{'}$ with zero transverse coordinates, and the witness events $p = x^{- 1} (5, 3, 0, 0)$ , $r = x^{- 1} (5, 7, 0, 0)$ (Inertial frames and inertial charts of the affine Minkowski model, Oriented time-oriented affine Minkowski space (the spacetime primitives)).

[L1]

The transition is $x^{' 0} = γ (x^{0} - \frac{3}{5} x^{1})$ , $x^{' 1} = γ (x^{1} - \frac{3}{5} x^{0})$ , $x^{' 2} = x^{2}$ , $x^{' 3} = x^{3}$ with $γ = 5 / 4$ (Coordinate formulas of a boost: clock rates, rod lengths, simultaneity and velocity transformation, The passive Lorentz boost with parameter beta).

[L2]

Length comparison (clause (ii) of Coordinate formulas of a boost: clock rates, rod lengths, simultaneity and velocity transformation): two worldlines at rest in $x^{'}$ with $x^{'}$ -separation $L_{0}$ have $x$ -separation $L_{0} / γ$ at equal $x$ -times, while at equal $x^{'}$ -times their $x^{'}$ -separation is $L_{0}$ .

[L3]

Simultaneity relative to a chart is equality of the chart's time coordinate, and the events simultaneous with $p$ in the frame $x$ are the slice $H (p, e_{0})$ of the measuring frame (Simultaneity slices and rest distance relative to a future unit timelike vector, Inertial frames and inertial charts of the affine Minkowski model).

Verification

technique · direct

1.1L1L2algebra

On the worldline $x^{' 1} = ℓ$ the transition reads $γ (x^{1} - \frac{3}{5} x^{0}) = ℓ$ , so $x^{1} = \frac{3}{5} x^{0} + ℓ / γ = \frac{3}{5} x^{0} + \frac{4}{5} ℓ$ ; for $ℓ = 0$ this is $x^{1} = \frac{3}{5} x^{0}$ and for $ℓ = L_{0} = 5$ it is $x^{1} = \frac{3}{5} x^{0} + 4$ . Hence at every fixed $x^{0}$ the $x$ -separation of the two worldlines is $4 = L_{0} / γ$ as asserted.

1.2L1L3algebra

The witness events: $x^{'} (p) = (\frac{5}{4} (5 - \frac{3}{5} \cdot 3), \frac{5}{4} (3 - \frac{3}{5} \cdot 5), 0, 0) = (4, 0, 0, 0)$ and $x^{'} (r) = (\frac{5}{4} (5 - \frac{3}{5} \cdot 7), \frac{5}{4} (7 - \frac{3}{5} \cdot 5), 0, 0) = (1, 5, 0, 0)$ ; the first lies on $x^{' 1} = 0$ and the second on $x^{' 1} = 5$ , both have $x^{0} = 5$ hence are simultaneous in $x$ and separated by $7 - 3 = 4$ ; their $x^{'}$ -times $4$ and $1$ differ, their $x^{'}$ -separation is $5$ .

1.3L1L2algebra

Equal $x^{'}$ -time comparison: on the second worldline, $x^{' 0} = \frac{5}{4} (x^{0} - \frac{3}{5} (\frac{3}{5} x^{0} + 4)) = \frac{4}{5} x^{0} - 3$ ; setting $x^{' 0} = 4$ gives $x^{0} = 35 / 4$ and then $x^{1} = \frac{3}{5} \cdot \frac{35}{4} + 4 = \frac{37}{4}$ , so the two events $x^{- 1} (5, 3, 0, 0)$ and $x^{- 1} (35 / 4, 37 / 4, 0, 0)$ have equal $x^{'}$ -time $4$ ; their $x^{' 1}$ -coordinates are $0$ and $5$ , so at equal $x^{'}$ -time the separation is the rest length $5$ .

2.1step 1.1step 1.2step 1.3∎

Conclusion: the frame $x$ assigns to the rod the length $4$ at its own equal- $x$ -time slice, which is exactly $L_{0} / γ = 5 / (5 / 4) = 4$ of clause (ii) of [L2]; the same pair of worldlines has $x^{'}$ -separation $5$ at equal $x^{'}$ -times, and the witness events show explicitly that the equal- $x$ and the equal- $x^{'}$ comparisons involve different pairs of events. This verifies the length contraction formula and the role of the measuring frame's simultaneity slice.

Remarks

Which events are being compared. The two witness events of step 1.2 are simultaneous in $x$ and not in $x^{'}$ , while the two events of step 1.3 are simultaneous in $x^{'}$ and not in $x$ ; the contracted and the rest lengths therefore refer to different pairs of events on the same two worldlines, which is exactly the content of clause (ii) of Coordinate formulas of a boost: clock rates, rod lengths, simultaneity and velocity transformation.
No optical claim. The length $4$ assigned by $x$ is the distance of the rod's endpoints on a simultaneity slice of $x$ ; it is not an appearance or a photograph, and no light-propagation or measurement model is involved.

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Worked velocity transformation: collinear, transverse and null cases

Example

Let $x$ and $x^{'}$ be inertial charts related by a boost of parameter $v = 3 c / 5$ along the first axis ( $γ = 5 / 4$ , The passive Lorentz boost with parameter beta, Inertial frames and inertial charts of the affine Minkowski model).

(a) Collinear case. A worldline with coordinate velocity $w = (4 c / 5, 0, 0)$ in $x$ has, by clause (iv) of Coordinate formulas of a boost: clock rates, rod lengths, simultaneity and velocity transformation, $w_{1}^{'} = (4 c / 5 - 3 c / 5) / (1 - (3 / 5) (4 / 5)) = (c / 5) / (13 / 25) = 5 c / 13$ , so $∣ w^{'} ∣ = 5 c / 13 < c$ ; the Galilean difference $c / 5$ is not the transformed velocity.

(b) Transverse case. A worldline with $w = (0, c / 2, 0)$ has $w_{1}^{'} = (0 - 3 c / 5) / (1 - 0) = - 3 c / 5$ and $w_{2}^{'} = (c / 2) / ((5 / 4) (1 - 0)) = 2 c / 5$ , hence $∣ w^{'} ∣ = \sqrt{9 / 25 + 4 / 25} c = (\sqrt{13} / 5) c < c$ .

(c) Null limit. For $w = (c, 0, 0)$ the formula gives $w_{1}^{'} = (c - 3 c / 5) / (1 - 3 / 5) = c$ and $∣ w^{'} ∣ = c$ : a null worldline is null in every inertial chart.

In each case take the affine coordinate curve $t \mapsto (c t, w t)$ on $R$ with the stated constant $w$ ; the first two are future timelike and the third future null. The three cases verify the velocity transformation and the subluminality statement of clause (iv) of Coordinate formulas of a boost: clock rates, rod lengths, simultaneity and velocity transformation.

Facts & Assumptions

Given: Two inertial charts $x$ , $x^{'}$ related by the boost of parameter $v = 3 c / 5$ along the first axis with $γ = 5 / 4$ , and the three coordinate velocities $w = (4 c / 5, 0, 0)$ , $w = (0, c / 2, 0)$ , $w = (c, 0, 0)$ (The passive Lorentz boost with parameter beta, Inertial frames and inertial charts of the affine Minkowski model).

[L1]

Clause (iv) of Coordinate formulas of a boost: clock rates, rod lengths, simultaneity and velocity transformation: for a $C^{1}$ future causal curve with coordinate velocity $w$ , $∣ w ∣ \leq c$ , in the chart $x$ , the coordinate velocity in $x^{'}$ is $w_{1}^{'} = (w_{1} - v) / (1 - v w_{1} / c^{2})$ and $w_{⊥}^{'} = w_{⊥} / (γ (1 - v w_{1} / c^{2}))$ , with positive denominators, and $∣ w^{'} ∣ < c$ for $∣ w ∣ < c$ , with $∣ w^{'} ∣ = c$ exactly when $∣ w ∣ = c$ .

[L2]

The boost parameter is $β = v / c = 3 / 5$ and $γ = (1 - β^{2})^{- 1 / 2} = 5 / 4$ (The passive Lorentz boost with parameter beta).

Verification

technique · direct

1.1L1L2algebra

Collinear case: with $w_{1} = 4 c / 5$ , $w_{⊥} = 0$ , $v = 3 c / 5$ , the denominator of [L1] is $1 - \frac{3}{5} \cdot \frac{4}{5} = 1 - \frac{12}{25} = \frac{13}{25} > 0$ and $w_{1}^{'} = \frac{c / 5}{13 / 25} = \frac{5 c}{13}$ ; hence $w^{'} = (5 c / 13, 0, 0)$ and $∣ w^{'} ∣ = 5 c / 13 < c$ because $5 / 13 < 1$ . The Galilean difference for this passive change of frame is $4 c / 5 - 3 c / 5 = c / 5$ , which is not the transformed velocity.

1.2L1L2algebra

Transverse case: with $w_{1} = 0$ , $w_{⊥} = (c / 2, 0)$ , the denominator is $1$ and [L1] gives $w_{1}^{'} = (0 - 3 c / 5) / 1 = - 3 c / 5$ and $w_{2}^{'} = (c / 2) / (γ \cdot 1) = 2 c / 5$ , so $∣ w^{'} ∣^{2} = (9 / 25 + 4 / 25) c^{2} = 13 c^{2} / 25$ and $∣ w^{'} ∣ = (\sqrt{13} / 5) c < c$ because $\sqrt{13} < 5$ .

1.3L1L2algebra

Null limit: with $w_{1} = c$ the denominator of [L1] is $1 - 3 / 5 = 2 / 5 > 0$ and $w_{1}^{'} = (c - 3 c / 5) / (2 / 5) = (2 c / 5) / (2 / 5) = c$ , so $w^{'} = (c, 0, 0)$ and $∣ w^{'} ∣ = c$ ; this is the equality case of the subluminality statement, consistent with $∣ w ∣ = c$ .

2.1step 1.1step 1.2step 1.3∎

Conclusion: steps 1.1-1.3 evaluate the transformation formula of [L1] in the three configurations and confirm $∣ w^{'} ∣ < c$ for subluminal $w$ and $∣ w^{'} ∣ = c$ in the null case; in particular the collinear case deviates from the Galilean difference by the relativistic denominator $1 - v w_{1} / c^{2}$ , and the transverse case exhibits the factor $1 / γ$ in the perpendicular component.

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Worked Rindler wedge congruence: chart, metric, acceleration, rigidity and non-coordinate frame

Example

In the inertial chart $(x^{0}, x, y, z)$ with metric $diag (- 1, 1, 1, 1)$ consider $Φ (θ, ρ, y, z) : = (ρ \sinh θ, ρ \cosh θ, y, z)$ on $R \times (0, \infty) \times R^{2}$ , with $θ \in R$ and $ρ > 0$ . Then:

(a) the identities $x^{2} - (x^{0})^{2} = ρ^{2}$ and $x = ρ \cosh θ > 0$ show that the image of $Φ$ is the right wedge $W = {x > ∣ x^{0} ∣}$ , and the inverse $ρ = \sqrt{x^{2} - (x^{0})^{2}}$ , $θ = \frac{1}{2} \log ((x + x^{0}) / (x - x^{0}))$ is smooth on $W$ ; both maps are smooth, so $Φ$ is a diffeomorphism onto $W$ ;

(b) differentiating gives $d x^{0} = \sinh θ d ρ + ρ \cosh θ d θ$ and $d x = \cosh θ d ρ + ρ \sinh θ d θ$ , so the pullback metric is $- ρ^{2} d θ^{2} + d ρ^{2} + d y^{2} + d z^{2}$ ;

(c) the field $n : = ρ^{- 1} \partial_{θ}$ satisfies $η (n, n) = - 1$ and is future; computing with the pullback metric, the nonzero connection coefficients are $Γ_{θ θ}^{ρ} = ρ$ and $Γ_{θ ρ}^{θ} = Γ_{ρ θ}^{θ} = 1 / ρ$ , so $\nabla_{\partial_{θ}} \partial_{θ} = ρ \partial_{ρ}$ and the congruence acceleration is $a = \nabla_{n} n = ρ^{- 1} \partial_{ρ}$ with $η (a, a) = 1 / ρ^{2}$ ; for example at $ρ = 2$ one has $∣ a ∣ = 1 / 2$ in reciprocal-length units;

(d) the slices $θ = c o n s t$ are open subsets of hyperplanes through the origin with induced metric $d ρ^{2} + d y^{2} + d z^{2}$ , independent of $θ$ ; hence the congruence is Born rigid, and it is irrotational because its rest distribution is the tangent distribution of these slices;

(e) the tetrad field $(n, \partial_{ρ}, \partial_{y}, \partial_{z})$ is not a coordinate frame, since $[n, \partial_{ρ}] = [ρ^{- 1} \partial_{θ}, \partial_{ρ}] = + ρ^{- 2} \partial_{θ} = + (1 / ρ) n$ is nonzero, whereas the members of a coordinate frame commute.

This realizes the Rindler congruence of Fermi-Walker transport preserves the metric, and explicit rigid and non-rigid observer congruences clause (b) with exact formulas.

Facts & Assumptions

Given: The inertial chart of the affine Minkowski model with $η = diag (- 1, 1, 1, 1)$ (Inertial frames and inertial charts of the affine Minkowski model, Oriented time-oriented affine Minkowski space (the spacetime primitives)), the map $Φ (θ, ρ, y, z) = (ρ \sinh θ, ρ \cosh θ, y, z)$ defined on $R \times (0, \infty) \times R^{2}$ , and the field $n = ρ^{- 1} \partial_{θ}$ .

[L1]

Hyperbolic functions: $\cosh^{2} s - \sinh^{2} s = 1$ , $\cosh s > ∣ \sinh s ∣$ for all real $s$ , $\sinh^{'} = \cosh$ , $\cosh^{'} = \sinh$ , and $\tanh = \sinh / \cosh$ (The six hyperbolic functions and their natural domains, Addition formulas, identities, parity, and derivatives of the hyperbolic functions).

[L2]

Future timelike criterion and units: in the inertial chart a vector with components $(u^{0}, u^{1}, u^{2}, u^{3})$ is future timelike exactly when $- (u^{0})^{2} + (u^{1})^{2} + (u^{2})^{2} + (u^{3})^{2} < 0$ and $u^{0} > 0$ (The timelike set has exactly two convex components, described in orthonormal coordinates, Timelike, null, spacelike and causal vectors of a Lorentz vector space, Time orientation, future and past timelike/null/causal vectors, and the closed cones).

[L3]

Kinematics and Frobenius: the acceleration $a$ of a congruence, its expansion $θ$ and vorticity $ω$ satisfy $\nabla_{a} n_{b} = - n_{a} a_{b} + θ_{a b} + ω_{a b}$ with $θ, ω$ horizontal; Born rigidity is $θ = 0$ ; and the rest distribution is integrable on an open set if and only if $ω = 0$ there (Kinematics of an observer congruence: acceleration, expansion, shear, vorticity and Born rigidity, Local flow boxes, adapted charts and the Frobenius criterion for rest distributions clauses (ii),(iii)).

[L4]

Lie bracket: $[f X, Y] = f [X, Y] - Y (f) X$ and coordinate frame members commute, $[\partial_{a}, \partial_{b}] = 0$ (The Lie bracket of smooth vector fields).

[L5]

The logarithm is $\exp^{- 1} : (0, \infty) \to R$ (The natural logarithm as the inverse of the exponential function); smooth Euclidean maps with invertible derivative have smooth local inverses (Choice-free smooth inverse function theorem in Euclidean space). Diffeomorphisms: a smooth bijection with smooth inverse between open subsets is a diffeomorphism (Diffeomorphisms and local diffeomorphisms of manifolds).

[L6]

The flat covariant derivative differentiates inertial components, $(\nabla_{X} Y)^{a} = X^{b} \partial_{b} Y^{a}$ (The flat covariant derivative on affine Minkowski space and fields along curves). In another coordinate basis its connection coefficients are the coefficients of these same derivatives in that basis.

Verification

technique · direct

1.1L1L5

Bijection and inverse: $Φ$ is smooth because its components are sums of smooth functions of $(θ, ρ, y, z)$ by [L1]. For $(θ, ρ, y, z)$ in the domain one has $x^{2} - (x^{0})^{2} = ρ^{2} (\cosh^{2} θ - \sinh^{2} θ) = ρ^{2}$ and $x = ρ \cosh θ > 0$ , so $Φ$ maps into $W$ . Conversely, given $(x^{0}, x, y, z)$ with $x > ∣ x^{0} ∣$ , put $ρ : = \sqrt{x^{2} - (x^{0})^{2}} > 0$ and $θ : = \frac{1}{2} \log ((x + x^{0}) / (x - x^{0})) \in R$ (the argument of the logarithm being positive because $x \pm x^{0} > 0$ ); then $e^{θ} = \sqrt{(x + x^{0}) / (x - x^{0})}$ and $e^{- θ} = \sqrt{(x - x^{0}) / (x + x^{0})}$ , so $ρ \sinh θ = ρ (e^{θ} - e^{- θ}) / 2 = \frac{(x + x^{0}) - (x - x^{0})}{2} = x^{0}$ and $ρ \cosh θ = ρ (e^{θ} + e^{- θ}) / 2 = \frac{(x + x^{0}) + (x - x^{0})}{2} = x$ by [L1], which gives a two-sided inverse. The $(θ, ρ)$ derivative block has determinant $ρ (\cosh^{2} θ - \sinh^{2} θ) = ρ > 0$ , with identity transverse block, so [L5] makes this inverse smooth locally everywhere. Hence $Φ$ is a smooth bijection onto $W$ with smooth inverse, that is, a diffeomorphism by [L5].

2.1L1step 1.1

Pullback metric: differentiating the four components of $Φ$ and using [L1] gives $d x^{0} = \sinh θ d ρ + ρ \cosh θ d θ$ , $d x = \cosh θ d ρ + ρ \sinh θ d θ$ , $d y = d y$ , $d z = d z$ . Substituting into $- (d x^{0})^{2} + (d x)^{2} + d y^{2} + d z^{2}$ : the cross terms are $- 2 ρ \sinh θ \cosh θ + 2 ρ \cosh θ \sinh θ = 0$ , the $d θ^{2}$ coefficient is $- ρ^{2} \cosh^{2} θ + ρ^{2} \sinh^{2} θ = - ρ^{2}$ and the $d ρ^{2}$ coefficient is $- \sinh^{2} θ + \cosh^{2} θ = 1$ ; hence the pullback metric is $- ρ^{2} d θ^{2} + d ρ^{2} + d y^{2} + d z^{2}$ .

3.1L1L2step 2.1

Unit future field: in these coordinates $g (\partial_{θ}, \partial_{θ}) = - ρ^{2}$ , $g (\partial_{ρ}, \partial_{ρ}) = g (\partial_{y}, \partial_{y}) = g (\partial_{z}, \partial_{z}) = 1$ and all mixed components vanish, so $n = ρ^{- 1} \partial_{θ}$ satisfies $g (n, n) = - ρ^{2} \cdot ρ^{- 2} = - 1$ . In the inertial chart, $\partial_{θ} = ρ (\cosh θ \partial_{0} + \sinh θ \partial_{1})$ by step 2.1, hence $n = \cosh θ \partial_{0} + \sinh θ \partial_{1}$ , whose time component is $\cosh θ > 0$ and whose $η$ -square is $- \cosh^{2} θ + \sinh^{2} θ = - 1$ ; by [L2] $n$ is future timelike, so $n$ is a congruence on $W$ .

4.1L1L6step 2.1step 3.1

Acceleration and connection coefficients: by the coordinate map, $\partial_{θ} = ρ (\cosh θ e_{0} + \sinh θ e_{1})$ and $\partial_{ρ} = \sinh θ e_{0} + \cosh θ e_{1}$ . Differentiating these inertial components with [L6] gives $\nabla_{\partial_{θ}} \partial_{θ} = ρ \partial_{ρ}$ , $\nabla_{\partial_{θ}} \partial_{ρ} = \nabla_{\partial_{ρ}} \partial_{θ} = ρ^{- 1} \partial_{θ}$ and $\nabla_{\partial_{ρ}} \partial_{ρ} = 0$ ; all derivatives involving the constant transverse basis fields vanish. Hence the only nonzero coordinate connection coefficients are $Γ_{θ θ}^{ρ} = ρ$ and $Γ_{θ ρ}^{θ} = Γ_{ρ θ}^{θ} = ρ^{- 1}$ , as claimed. Since $\partial_{θ} ρ = 0$ , $a = \nabla_{ρ^{- 1} \partial_{θ}} (ρ^{- 1} \partial_{θ}) = ρ^{- 2} \nabla_{\partial_{θ}} \partial_{θ} = ρ^{- 1} \partial_{ρ}$ , and $η (a, a) = ρ^{- 2}$ by step 2.1. At $ρ = 2$ , this is $1 / 4$ and $∣ a ∣ = 1 / 2$ .

4.2L1L3step 2.1step 3.1

Rigidity, irrotationality and the coframe: the $θ = c o n s t$ slices are the portions ${x^{0} = \tanh θ x} \cap W$ of hyperplanes through the origin, and their tangent space is spanned by $\partial_{ρ}, \partial_{y}, \partial_{z}$ , which are orthogonal to $n$ by the metric of step 2.1; their induced metric is $d ρ^{2} + d y^{2} + d z^{2}$ , independent of $θ$ . The rest distribution is therefore integrable, being the tangent distribution of the foliation by these slices; by [L3] its vorticity vanishes and the congruence is irrotational. Moreover $h (n, \cdot) = 0$ , the rest-basis pairings $h (\partial_{ρ}, \partial_{ρ}) = h (\partial_{y}, \partial_{y}) = h (\partial_{z}, \partial_{z}) = 1$ are constant, and $[n, \partial_{ρ}] = (1 / ρ) n$ while $[n, \partial_{y}] = [n, \partial_{z}] = 0$ . Thus $L_{n} h (X, Y) = n (h (X, Y)) - h ([n, X], Y) - h (X, [n, Y]) = 0$ on that basis and on $n$ , so the congruence is Born rigid; equivalently, the metric components on the rest slices are constant along $\partial_{θ}$ , which is the content of the supplier's rigidity computation (Fermi-Walker transport preserves the metric, and explicit rigid and non-rigid observer congruences clause (b)).

4.3L4step 3.1

Non-coordinate frame: by [L4], $[n, \partial_{ρ}] = [ρ^{- 1} \partial_{θ}, \partial_{ρ}] = ρ^{- 1} [\partial_{θ}, \partial_{ρ}] - \partial_{ρ} (ρ^{- 1}) \partial_{θ} = 0 + ρ^{- 2} \partial_{θ} = ρ^{- 1} n$ , which is nonzero since $n \neq 0$ ; here $[\partial_{θ}, \partial_{ρ}] = 0$ because $θ, ρ$ are coordinates. Hence the members of the tetrad field $(n, \partial_{ρ}, \partial_{y}, \partial_{z})$ do not commute pairwise, while the members of any coordinate frame do; so this tetrad field is not the coordinate frame of any chart.

5.1step 4.1step 4.2step 4.3∎

Conclusion: steps 1.1-2.1 verify the diffeomorphism and the exact pullback metric; step 3.1 checks that $n$ is a future unit field; step 4.1 computes the acceleration $a = ρ^{- 1} \partial_{ρ}$ with $η (a, a) = ρ^{- 2}$ ; step 4.2 verifies Born rigidity and vanishing vorticity with the explicit rest slices; and step 4.3 exhibits the nonzero commutator showing that the frame is not a coordinate frame. All identities are exact symbolic computations in the hyperbolic functions.

Remarks

The acceleration is not constant over the congruence. Each observer at fixed $ρ$ has constant congruence-acceleration magnitude $1 / ρ$ in reciprocal-length units, but the acceleration magnitude is not the same for all observers: the acceleration grows as $ρ$ decreases, and no global statement about a rigid body extending to $ρ = 0$ is made.
The metric is not curved. The pullback metric is the flat metric rewritten in curvilinear coordinates, which is why $\nabla_{\partial_{θ}} \partial_{θ} = ρ \partial_{ρ}$ has the form of a connection coefficient; the model's curvature is zero, and the example uses only the flat connection of the inertial chart.

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Worked rigid rotation: rest-space holonomy around the ring and failure of ring synchronization

Example

Fix $R > 0$ and let $Ω$ satisfy $Ω R = 3 c / 5$ , so that $γ = (1 - 9 / 25)^{- 1 / 2} = 5 / 4$ and $c^{2} - Ω^{2} R^{2} = (16 / 25) c^{2}$ . In the inertial chart the ring observer at angular position $ϕ = ϕ_{0} + Ω t$ (chart time $t$ ) has unit tangent

$n = γ (1, - \frac{3}{5} \sin ϕ, \frac{3}{5} \cos ϕ, 0) = \frac{5}{4} (1, - \frac{3}{5} \sin ϕ, \frac{3}{5} \cos ϕ, 0),$

which is future unit because $γ^{2} (- 1 + \frac{9}{25}) = - \frac{25}{16} \cdot \frac{16}{25} = - 1$ . Writing the cylinder $r = R$ in the rotating angular coordinate $φ : = ϕ - Ω t$ , so that the point has inertial coordinates $(c t, R \cos (φ + Ω t), R \sin (φ + Ω t))$ , the induced metric is

$g = - (c^{2} - Ω^{2} R^{2}) d t^{2} + 2 Ω R^{2} d t d φ + R^{2} d φ^{2} = - \frac{16}{25} c^{2} d t^{2} + \frac{6}{5} c R d t d φ + R^{2} d φ^{2} .$

The rest-space covector of $n$ restricted to the cylinder is proportional to $- (c^{2} - Ω^{2} R^{2}) d t + Ω R^{2} d φ = - \frac{16}{25} c^{2} d t + \frac{3}{5} c R d φ$ , so a curve tangent to the rest spaces satisfies $d t / d φ = Ω R^{2} / (c^{2} - Ω^{2} R^{2}) = 15 R / (16 c)$ ; integrating over one turn $φ \in [0, 2 π]$ gives the nonzero chart-time offset

$Δ t = \frac{2 π Ω R^{2}}{c^{2} - Ω^{2} R^{2}} = \frac{15 π R}{8 c} \neq 0.$

Hence the curve tangent to the rest spaces starting at the observer returns to the same rotating label at a strictly later chart time: the rest simultaneity of the rotating congruence does not close up around the ring, and no globally single-valued real rest-synchronization time with time-oriented nonzero differential exists on the whole cylinder. The ambient congruence extends to the smooth future unit rotating field $n = γ (r) [\partial_{0} + (Ω / c) (- x^{2} \partial_{1} + x^{1} \partial_{2})]$ on $r < c / ∣ Ω ∣$ , which is Born rigid, and the three-dimensional ambient rest distribution is nowhere integrable because its Frobenius form is nonzero; the one-dimensional rest distribution on the cylinder is locally integrable, so its nonzero global holonomy checked above is a different obstruction. This preserves the distinction between ambient local nonintegrability and global ring synchronization failure.

Facts & Assumptions

Given: An inertial chart $(x^{0} = x^{0}, x^{1}, x^{2}, x^{3})$ of the affine Minkowski model with $η = diag (- 1, 1, 1, 1)$ (Inertial frames and inertial charts of the affine Minkowski model, Oriented time-oriented affine Minkowski space (the spacetime primitives)), real $Ω$ and $R > 0$ with $Ω R = 3 c / 5$ , the ambient rotating field $n$ and the ring of the Example (Fermi-Walker transport preserves the metric, and explicit rigid and non-rigid observer congruences clause (c)).

[L1]

Supplier clause (c): on the open region $r^{2} = (x^{1})^{2} + (x^{2})^{2} < c^{2} / Ω^{2}$ the field $K = \partial_{0} + (Ω / c) (- x^{2} \partial_{1} + x^{1} \partial_{2})$ satisfies $η (K, K) = - 1 + Ω^{2} r^{2} / c^{2}$ , the field $n = γ (r) K$ with $γ (r) = (1 - Ω^{2} r^{2} / c^{2})^{- 1 / 2}$ is a smooth future unit timelike Born-rigid congruence; $κ : = η (K, \cdot) = (- 1, - (Ω / c) x^{2}, (Ω / c) x^{1}, 0)$ has $d κ = 2 (Ω / c) d x^{1} \land d x^{2}$ and $κ \land d κ = - 2 (Ω / c) d x^{0} \land d x^{1} \land d x^{2}$ , so $n^{♭} \land d (n^{♭}) \neq 0$ when $Ω \neq 0$ ; on the cylinder $r = R$ , $x^{3} = c o n s t$ with rotating angle $φ$ , the ring observer $ϕ = ϕ_{0} + Ω t$ has $n = γ (R) (1, - (Ω R / c) \sin ϕ, (Ω R / c) \cos ϕ, 0)$ , the induced metric is $- (c^{2} - Ω^{2} R^{2}) d t^{2} + 2 Ω R^{2} d t d φ + R^{2} d φ^{2}$ , the ambient rest covector restricted to the cylinder is proportional to $- (c^{2} - Ω^{2} R^{2}) d t + Ω R^{2} d φ$ , the rest lines satisfy $d t / d φ = Ω R^{2} / (c^{2} - Ω^{2} R^{2})$ , the one-turn offset is $2 π Ω R^{2} / (c^{2} - Ω^{2} R^{2})$ , and no globally single-valued real rest-synchronization time on the cylinder exists when $Ω \neq 0$ (Fermi-Walker transport preserves the metric, and explicit rigid and non-rigid observer congruences clause (c), Kinematics of an observer congruence: acceleration, expansion, shear, vorticity and Born rigidity).

[L2]

Frobenius criterion: for a smooth nowhere-zero one-form $α$ on an open set, the rest distribution $\ker α$ is integrable if and only if $α \land d α = 0$ ; for the rest form $α = n^{♭}$ this holds if and only if the vorticity vanishes, $ω = 0$ (Local flow boxes, adapted charts and the Frobenius criterion for rest distributions clauses (ii),(iii)(b)).

[L3]

A future timelike unit vector field is a congruence, and a vector of the inertial chart with components $(u^{0}, u^{1}, u^{2}, u^{3})$ has $η (u, u) = - (u^{0})^{2} + (u^{1})^{2} + (u^{2})^{2} + (u^{3})^{2}$ (Kinematics of an observer congruence: acceleration, expansion, shear, vorticity and Born rigidity, Inertial frames and inertial charts of the affine Minkowski model, Timelike, null, spacelike and causal vectors of a Lorentz vector space).

[L4]

Sine and cosine are the everywhere-convergent power series of Sine and cosine defined by their real power series, with derivatives $\cos$ and $- \sin$ (The derivatives of sine and cosine are cosine and minus sine), Pythagorean identity $\sin^{2} ϕ + \cos^{2} ϕ = 1$ (Parity and the Pythagorean identity for sine and cosine), and period $2 π$ (The zero sets of sine and cosine and the least positive common period 2 pi); the resulting angular parametrization modulo $2 π$ is the unit circle ( $t \mapsto (\cos t, \sin t)$ is a bijection from $[0, 2 π)$ onto the real unit circle).

Verification

technique · direct

1.1L1L3algebra

Numerical data: $Ω R = 3 c / 5$ gives $Ω^{2} R^{2} = 9 c^{2} / 25$ , hence $c^{2} - Ω^{2} R^{2} = 16 c^{2} / 25$ , $2 Ω R^{2} = 2 (Ω R) R = 6 c R / 5$ , $Ω R^{2} / (c^{2} - Ω^{2} R^{2}) = (3 c R / 5) / (16 c^{2} / 25) = 15 R / (16 c)$ and $γ (R) = (1 - 9 / 25)^{- 1 / 2} = 5 / 4$ ; the ring field is $n = γ (R) (1, - (3 / 5) \sin ϕ, (3 / 5) \cos ϕ, 0) = \frac{5}{4} (1, - \frac{3}{5} \sin ϕ, \frac{3}{5} \cos ϕ, 0)$ .

2.1L3step 1.1L4

Future unit property: by [L3], $η (n, n) = \frac{25}{16} (- 1 + \frac{9}{25} (\sin^{2} ϕ + \cos^{2} ϕ)) = \frac{25}{16} (- 1 + \frac{9}{25}) = - \frac{25}{16} \cdot \frac{16}{25} = - 1$ , and the time component $\frac{5}{4} > 0$ makes $n$ future; this verifies the formula of the Example.

2.2L1step 1.1algebraL4

Cylinder metric and rest covector: substituting $ϕ = φ + Ω t$ , hence $d ϕ = d φ + Ω d t$ , into $- c^{2} d t^{2} + R^{2} d ϕ^{2}$ (the transverse coordinates being fixed) gives $- (c^{2} - Ω^{2} R^{2}) d t^{2} + 2 Ω R^{2} d t d φ + R^{2} d φ^{2}$ ; with the values of step 1.1 this is $- \frac{16}{25} c^{2} d t^{2} + \frac{6}{5} c R d t d φ + R^{2} d φ^{2}$ . The covector $κ$ of [L1] restricts as $- c d t + (Ω R^{2} / c) d ϕ = - c d t + (Ω R^{2} / c) (d φ + Ω d t) = - \frac{c^{2} - Ω^{2} R^{2}}{c} d t + \frac{Ω R^{2}}{c} d φ$ , so $n^{♭} = γ (R) κ$ is proportional to $- (c^{2} - Ω^{2} R^{2}) d t + Ω R^{2} d φ = - \frac{16}{25} c^{2} d t + \frac{3}{5} c R d φ$ ; solving the kernel condition $- (c^{2} - Ω^{2} R^{2}) \dot{t} + Ω R^{2} \dot{φ} = 0$ gives $d t / d φ = Ω R^{2} / (c^{2} - Ω^{2} R^{2}) = \frac{15 R}{16 c}$ .

3.1L1step 2.2algebraL4

One-turn holonomy: along the rest lines of step 2.2 the chart time changes by $Δ t = \int_{0}^{2 π} \frac{15 R}{16 c} d φ = \frac{15 π R}{8 c}$ , which is nonzero; hence the rest line starting at $φ = 0$ returns to the same rotating label $φ = 2 π$ at the chart time $Δ t > 0$ , and it is not closed as a curve of simultaneity. The no-global-synchronization conclusion of [L1] applies to these ring rest lines with their nonzero time-oriented slope; its proof tests a real synchronization function on a closed angular lift and uses the nonzero fixed-sign angular derivative to contradict periodicity. Thus no such global synchronization time exists.

4.1L1L2step 3.1

Ambient nonintegrability: for $Ω \neq 0$ the Frobenius form of the ambient congruence is $n^{♭} \land d (n^{♭}) = γ^{2} κ \land d κ = - 2 γ^{2} (Ω / c) d x^{0} \land d x^{1} \land d x^{2}$ , which is nonzero at every point of the region $r < c / ∣ Ω ∣$ ; by [L2] the three-dimensional ambient rest distribution is nowhere integrable, so no local ambient rest-simultaneity hypersurface exists there. On the cylinder the rest distribution is one-dimensional: it is locally integrable because any nowhere-zero vector field has local flow-box coordinates, and its global nontrivial holonomy is the distinct obstruction computed in step 3.1, not an instance of the ambient pointwise nonintegrability. Both the ambient congruence and the ring congruence are Born rigid by [L1].

5.1step 2.1step 3.1step 4.1∎

Conclusion: steps 1.1-2.1 verify the future unit ring field; step 2.2 computes the induced cylinder metric, the rest covector and the slope $d t / d φ = 15 R / (16 c)$ ; step 3.1 computes the exact nonzero holonomy $Δ t = 15 π R / (8 c)$ and the failure of global ring synchronization; and step 4.1 exhibits the ambient nonintegrability of the three-dimensional rest distribution. The example therefore realizes the distinction, promised by [L1], between a locally integrable one-dimensional ring distribution with nonzero global holonomy and a nowhere-integrable ambient distribution.

Remarks

Why the numbers are exact. The choice $Ω R = 3 c / 5$ makes $γ = 5 / 4$ and $c^{2} - Ω^{2} R^{2} = (16 / 25) c^{2}$ , so the slope is exactly $15 R / (16 c)$ and the one-turn offset is exactly $15 π R / (8 c)$ . The ambient Frobenius three-form is exactly $- 2 γ (r)^{2} (Ω / c) d x^{0} \land d x^{1} \land d x^{2}$ , as computed in step 4.1; no approximation enters.
Two distinct obstructions. The ambient three-dimensional rest distribution is nowhere integrable for $Ω \neq 0$ (nowhere-zero vorticity by [L2]); the one-dimensional distribution on the ring cylinder is locally integrable but has nonzero holonomy around the ring. The second is not a consequence of the first, and no locally defined synchronization function on the cylinder can be single-valued around the ring.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6.1-sol)Open item page →

Worked Bell-type translated congruence: constant launch-frame separation without Born rigidity

Example

In an inertial chart let $q (t) = (\frac{1}{2} α t^{2}, 0, 0)$ with $α > 0$ , and consider the congruence of translated worldlines $x_{ξ} (t) = (c t, q (t) + ξ)$ for $ξ$ in an open subset of $R^{3}$ , defined while $∣ q_{1}^{'} (t) ∣ = α ∣ t ∣ < c$ (Fermi-Walker transport preserves the metric, and explicit rigid and non-rigid observer congruences clause (d), Inertial frames and inertial charts of the affine Minkowski model). Then $n = γ (1, α t / c, 0, 0)$ with $γ = (1 - α^{2} t^{2} / c^{2})^{- 1 / 2}$ is a smooth future unit timelike field, and $s = γ (α t / c, 1, 0, 0)$ is a unit vector orthogonal to $n$ . Direct computation of the expansion on $s$ gives

$θ (s, s) = s^{a} s^{b} \partial_{a} n_{b} = γ^{3} \frac{α^{2} t}{c^{3}},$

which is nonzero for every $t \neq 0$ on the subluminal domain; hence the congruence is not Born rigid on any neighbourhood of a time $t \neq 0$ , and the failure is already present at $t = 0$ on the two sides, where $θ (s, s)$ changes sign. Nevertheless, for every fixed chart time $t$ the separation between the member labelled $ξ_{1}$ and the member labelled $ξ_{2}$ is $∣ ξ_{1} - ξ_{2} ∣$ , constant in $t$ : the congruence has constant launch-frame separation while failing Born rigidity. This is the Bell-type translated congruence of the cited theorem; a physical conclusion about a breaking thread or rod would additionally require an elasticity model and a failure criterion and is not drawn here.

Facts & Assumptions

Given: An inertial chart of the affine Minkowski model with $η = diag (- 1, 1, 1, 1)$ (Inertial frames and inertial charts of the affine Minkowski model, Oriented time-oriented affine Minkowski space (the spacetime primitives)), the number $α > 0$ , the worldline $t \mapsto (c t, \frac{1}{2} α t^{2}, 0, 0)$ and its translates $x_{ξ} (t) = (c t, q (t) + ξ)$ for $ξ$ in an open subset of $R^{3}$ , and the field of the Example (Fermi-Walker transport preserves the metric, and explicit rigid and non-rigid observer congruences clause (d)).

[L1]

Supplier clause (d): for a $C^{2}$ function $q$ with $∣ q_{1}^{'} ∣ < c$ the translated congruence has $n = γ (1, q_{1}^{'} / c, 0, 0)$ , $γ = (1 - q_{1}^{' 2} / c^{2})^{- 1 / 2}$ , unit longitudinal rest vector $s = γ (q_{1}^{'} / c, 1, 0, 0)$ , and first-order projected deformation $θ (s, s) = s^{a} s^{b} \partial_{a} n_{b} = γ^{3} q_{1}^{'} (t) q_{1}^{''} (t) / c^{3}$ , nonzero exactly when $q_{1}^{'} q_{1}^{''} \neq 0$ ; when $q$ is smooth these formulas agree with the smooth congruence definitions of Kinematics of an observer congruence: acceleration, expansion, shear, vorticity and Born rigidity. For every fixed chart time the members labelled $ξ_{1}, ξ_{2}$ have separation $∣ ξ_{1} - ξ_{2} ∣$ (Fermi-Walker transport preserves the metric, and explicit rigid and non-rigid observer congruences clause (d)).

[L2]

Born rigidity of a congruence means the vanishing of the expansion, $θ = 0$ (Kinematics of an observer congruence: acceleration, expansion, shear, vorticity and Born rigidity).

[L3]

The four-velocity of a future timelike worldline $t \mapsto (c t, r (t))$ with $∣ r^{'} ∣ < c$ is $U = γ_{w} (c, r^{'})$ with $γ_{w} = (1 - ∣ r^{'} ∣^{2} / c^{2})^{- 1 / 2}$ , and $n = U / c$ (Four-velocity and four-acceleration of a future timelike worldline, Fermi-Walker transport preserves the metric, and explicit rigid and non-rigid observer congruences clause (d)).

[L4]

Chain rule and one-variable algebra: with $β : = α t / c$ and $γ = (1 - β^{2})^{- 1 / 2}$ one has $\frac{d γ}{d t} = γ^{3} β β^{'} = γ^{3} α^{2} t / c^{2}$ and $\frac{d (γ β)}{d t} = γ^{3} β^{'} = γ^{3} α / c$ , and $1 - β^{2} = γ^{- 2}$ (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

Verification

technique · direct

1.1L1L3algebra

With $q_{1} (t) = \frac{1}{2} α t^{2}$ one has $q_{1}^{'} (t) = α t$ and $q_{1}^{''} (t) = α$ , so the field of the Example is $n = γ (1, α t / c, 0, 0)$ and $s = γ (α t / c, 1, 0, 0)$ , where $γ = (1 - α^{2} t^{2} / c^{2})^{- 1 / 2}$ is defined and positive on the subluminal domain $∣ t ∣ < c / α$ . Evaluating the diagonal form gives $η (n, n) = γ^{2} (- 1 + α^{2} t^{2} / c^{2}) = - 1$ , $η (s, s) = γ^{2} (- α^{2} t^{2} / c^{2} + 1) = 1$ and $η (n, s) = γ^{2} (- α t / c + α t / c) = 0$ ; hence $n$ is a future unit timelike field and $s$ a unit rest vector.

2.1L1L2L4step 1.1algebra

Direct computation of the contraction: with $n_{b} = η_{b c} n^{c} = (- γ, γ β, 0, 0)$ and $s = γ (β, 1, 0, 0)$ , and with $n$ depending only on $t = x^{0} / c$ , one has $\partial_{1} n_{b} = \partial_{2} n_{b} = \partial_{3} n_{b} = 0$ and $\partial_{0} = \frac{1}{c} \frac{d}{d t}$ ; by [L4], $\partial_{0} n_{0} = - γ^{3} β β^{'} / c = - γ^{3} α^{2} t / c^{3}$ and $\partial_{0} n_{1} = \frac{1}{c} \frac{d}{d t} (γ β) = γ^{3} β^{'} / c = γ^{3} α / c^{2}$ . Hence $s^{a} s^{b} \partial_{a} n_{b} = (γ β)^{2} \partial_{0} n_{0} + (γ β) (γ) \partial_{0} n_{1} = γ^{5} β^{2} (- β β^{'}) / c + γ^{5} β β^{'} / c = γ^{5} β β^{'} (1 - β^{2}) / c = γ^{3} β β^{'} / c = γ^{3} α^{2} t / c^{3}$ , where $1 - β^{2} = γ^{- 2}$ and only the cross term $(a, b) = (0, 1)$ occurs because $\partial_{1} n_{0} = 0$ ; this reproduces the supplier formula of [L1]. Since $α > 0$ and $γ \geq 1$ on the domain, $θ (s, s) = 0$ exactly at $t = 0$ and is nonzero for every $t \neq 0$ , with opposite signs on the two sides of $0$ . By [L2] the congruence is therefore not Born rigid on any neighbourhood of any $t \neq 0$ , and at $t = 0$ the rigidity condition fails immediately on both sides.

2.2L1step 1.1

The translated members at a fixed chart time $t$ have positions $q (t) + ξ_{1}$ and $q (t) + ξ_{2}$ in the spatial coordinates, so their Euclidean separation is $∣ ξ_{1} - ξ_{2} ∣$ , independent of $t$ ; equivalently, in the three-dimensional spatial coordinates the congruence is a translation family at each instant even though $θ \neq 0$ off $t = 0$ . Thus constant instantaneous separation of the labels and Born rigidity are different requirements, which is the content of [L1].

3.1step 2.1step 2.2∎

Conclusion and scope: the computation exhibits the exact expansion $γ^{3} α^{2} t / c^{3}$ , nonzero for $t \neq 0$ and zero at $t = 0$ , so the translated congruence is not Born rigid while its labels keep constant separation $∣ ξ_{1} - ξ_{2} ∣$ . No statement is made about a material rod, a thread or a fracture: a breaking or stress conclusion would require an elasticity model, a constitutive law and a failure criterion, none of which is part of the kinematic model, and no such conclusion is drawn.

Remarks

Momentary rigidity does not persist. The expansion is linear in $t$ near $t = 0$ and vanishes exactly there, so the congruence is momentarily rigid at the launch time but not rigid on any time neighbourhood; this is the kinematic counterpart of the Bell-type setup, and the constant separation of the labels is a statement about the labelling, not about the metric distances in the rest spaces.
Exact data. Only algebraic functions of $t$ , $α$ and $c$ arise; at $c = 1$ , $β = α t$ , $β^{'} = α$ the formula reads $θ (s, s) = γ^{3} β β^{'}$ , an exact algebraic expression with no approximation.

CounterexampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

Counterexample: a Born-rigid congruence need not be a Killing congruence

Statement refuted

False claim: every smooth Born-rigid future unit timelike field $n$ (Kinematics of an observer congruence: acceleration, expansion, shear, vorticity and Born rigidity, Charts, tetrads, observer congruences and adapted charts in affine Minkowski space) is locally a positive multiple of a Killing field: there is a positive smooth function $f$ on a neighbourhood of every point with $L_{f n} η = 0$ (The smooth local Herglotz-Noether classification of Born-rigid motions).

Facts & Assumptions

Given: The affine Minkowski space with an inertial chart $(x^{0}, x^{1}, x^{2}, x^{3})$ in which $η = diag (- 1, 1, 1, 1)$ and the flat covariant derivative differentiates components (Oriented time-oriented affine Minkowski space (the spacetime primitives), Inertial frames and inertial charts of the affine Minkowski model, The flat covariant derivative on affine Minkowski space and fields along curves), the kinematics of a congruence with acceleration $a$ , expansion $θ$ and vorticity $ω$ (Kinematics of an observer congruence: acceleration, expansion, shear, vorticity and Born rigidity), a length scale $L_{0} > 0$ , and the functions and curves of the Construction below.

[L1]

Killing equation (The smooth local Herglotz-Noether classification of Born-rigid motions, [F10] and steps 10.1, 11.1 there): for a smooth future unit timelike field $n$ with expansion $θ$ , acceleration one-form $a^{♭}$ and the Lie derivative of the metric, $L_{n} η = 2 θ - (n^{♭} \otimes a^{♭} + a^{♭} \otimes n^{♭})$ ; and for a positive smooth $f$ , $L_{f n} η = f L_{n} η + d f \otimes n^{♭} + n^{♭} \otimes d f$ , so $L_{f n} η = 0$ if and only if $\nabla_{a} (f n)_{b} + \nabla_{b} (f n)_{a} = 0$ .

[L2]

Hyperbolic functions: $\cosh^{2} s - \sinh^{2} s = 1$ , $\cosh s > 0$ , $\cosh^{'} ⁣ = \sinh$ , $\sinh^{'} ⁣ = \cosh$ , $\sinh 0 = 0$ , $\cosh 0 = 1$ (The six hyperbolic functions and their natural domains, Addition formulas, identities, parity, and derivatives of the hyperbolic functions).

[L3]

Fundamental theorem of calculus: if $F (λ) = \int_{0}^{λ} g (s) d s$ for a continuous $g$ , then $F$ is differentiable with $F^{'} = g$ (The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive).

[L4]

Choice-free inverse function theorem: a smooth map on an open subset of $R^{n}$ whose derivative at a point is invertible restricts to a diffeomorphism between open neighbourhoods (Choice-free smooth inverse function theorem in Euclidean space).

[L5]

Exterior derivative: $d^{2} = 0$ on smooth differential forms (The exterior derivative squares to zero).

[L6]

Future timelike coordinate criterion: in an inertial chart a vector with components $(u^{0}, u^{1}, u^{2}, u^{3})$ is future timelike exactly when $- u^{0}^{2} + u^{1}^{2} + u^{2}^{2} + u^{3}^{2} < 0$ and $u^{0} > 0$ (The timelike set has exactly two convex components, described in orthonormal coordinates, Timelike, null, spacelike and causal vectors of a Lorentz vector space, Time orientation, future and past timelike/null/causal vectors, and the closed cones).

Counterexample

technique · direct

1.1L2L3L6

Construction of the data: fix $L_{0} > 0$ and put $ψ (λ) : = λ^{2} / (2 L_{0}^{2})$ , $k (λ) : = ψ^{'} (λ) = λ / L_{0}^{2}$ , $e_{0} (λ) : = (\cosh ψ (λ), \sinh ψ (λ), 0, 0)$ and $e_{1} (λ) : = (\sinh ψ (λ), \cosh ψ (λ), 0, 0)$ in the inertial coordinates, and let $z (λ) : = (\int_{0}^{λ} \cosh ψ (s) d s, \int_{0}^{λ} \sinh ψ (s) d s, 0, 0)$ ; these are smooth in $λ$ , $z (0) = 0$ , and $z^{'} = e_{0}$ by [L3]. By [L2], $η (e_{0}, e_{0}) = - \cosh^{2} ψ + \sinh^{2} ψ = - 1$ , $η (e_{1}, e_{1}) = - \sinh^{2} ψ + \cosh^{2} ψ = 1$ and $η (e_{0}, e_{1}) = - \cosh ψ \sinh ψ + \sinh ψ \cosh ψ = 0$ , so $(e_{0}, e_{1}, e_{2}, e_{3})$ is an orthonormal frame at every $λ$ ; moreover $e_{0}$ is future by [L6] because its time component $\cosh ψ > 0$ and $η (e_{0}, e_{0}) = - 1$ .

1.2L2L4

Local chart: define $Φ (λ, ξ, y, ζ) : = z (λ) + ξ e_{1} (λ) + y e_{2} + ζ e_{3}$ on $R^{4}$ . Differentiating in $λ$ gives $\partial_{λ} Φ = z^{'} + ξ e_{1}^{'} = e_{0} + ξ k e_{0} = N e_{0}$ with $N : = 1 + k (λ) ξ$ , using $e_{1}^{'} = k e_{0}$ from [L2]; further $\partial_{ξ} Φ = e_{1}$ , $\partial_{y} Φ = e_{2}$ , $\partial_{ζ} Φ = e_{3}$ . At $(λ, ξ, y, ζ) = 0$ these four vectors are $e_{0} (0) = (1, 0, 0, 0)$ , $e_{1} (0) = (0, 1, 0, 0)$ , $e_{2} = (0, 0, 1, 0)$ , $e_{3} = (0, 0, 0, 1)$ , so the derivative matrix is the identity and is invertible; by [L4] $Φ$ restricts to a diffeomorphism from an open neighbourhood of $0$ onto an open neighbourhood $W$ of $Φ (0) = 0$ . Shrinking to a coordinate box $B = {(λ, ξ, y, ζ) : ∣ λ ∣ < ε L_{0}, ∣ ξ ∣ < ε L_{0}, ∣ y ∣ < ε L_{0}, ∣ ζ ∣ < ε L_{0}}$ with $0 < ε < 1$ (intersected with the neighbourhood supplied by [L4]) keeps a diffeomorphism onto its image and gives $∣ k ξ ∣ = ∣ λ ξ ∣ / L_{0}^{2} < ε^{2}$ , hence $N = 1 + k ξ \geq 1 - ε^{2} > 0$ on $B$ . We write again $W : = Φ (B)$ and use $(λ, ξ, y, ζ)$ as coordinates on $W$ .

2.1step 1.1step 1.2L2

Metric and unit field: by step 1.2, $d Φ = N e_{0} d λ + e_{1} d ξ + e_{2} d y + e_{3} d ζ$ ; since $(e_{0}, e_{1}, e_{2}, e_{3})$ is $η$ -orthonormal with $e_{0}$ timelike, the pullback metric on $W$ is $g = - N^{2} d λ^{2} + d ξ^{2} + d y^{2} + d ζ^{2}$ . Hence the vector field $n : = N^{- 1} \partial_{λ}$ satisfies $g (n, n) = - N^{2} \cdot N^{- 2} = - 1$ , and $n = e_{0} (λ)$ is future by step 1.1; $n$ is smooth on $W$ because $N$ and the coordinates are, so $n$ is a smooth future unit timelike field, that is, a congruence on $W$ . The rest space $n^{⊥}$ at each point is spanned by $\partial_{ξ}, \partial_{y}, \partial_{ζ}$ , since these are $g$ -orthogonal to $\partial_{λ}$ and have positive squares.

3.1step 1.1step 2.1algebra

Rigidity and vanishing vorticity: the inertial components of $n$ are $e_{0} (λ)$ , so its directional derivative in a vector $v$ is $d λ (v) k e_{1}$ by step 1.1 and the flat component derivative. For a rest vector $r$ spanned by $\partial_{ξ}, \partial_{y}, \partial_{ζ}$ , one has $d λ (r) = 0$ , hence $\nabla_{r} n = 0$ . Thus both the symmetric and antisymmetric projected derivatives vanish on pairs of rest vectors. Since $θ$ and $ω$ are horizontal in both slots, $θ = ω = 0$ everywhere. The field is Born rigid and irrotational.

4.1step 2.1step 3.1L2

Acceleration one-form: since $d λ (n) = 1 / N$ , the derivative from step 3.1 gives $a = \nabla_{n} n = (k / N) e_{1} = (k / N) \partial_{ξ}$ . Therefore $a^{♭} = (k / N) d ξ$ by the metric of step 2.1. As $N = 1 + k (λ) ξ$ and $\partial_{λ} N = k^{'} ξ$ , one has $\partial_{λ} (k / N) = (k^{'} N - k k^{'} ξ) / N^{2} = k^{'} (N - k ξ) / N^{2} = 1 / (L_{0}^{2} N^{2})$ . The remaining derivatives contribute no term to $d (k / N) \land d ξ$ , so $d a^{♭} = (1 / (L_{0}^{2} N^{2})) d λ \land d ξ$ , nowhere zero on $W$ because $N > 0$ .

5.1step 4.1L1L5

No positive multiple is Killing: let $f > 0$ be smooth on $W$ and suppose $L_{f n} η = 0$ . By [L1] and $θ = 0$ , $0 = L_{f n} η = f L_{n} η + d f \otimes n^{♭} + n^{♭} \otimes d f = n^{♭} \otimes β + β \otimes n^{♭}$ with $β : = d f - f a^{♭}$ . Evaluating at $(n, r)$ for a rest vector $r$ gives $n^{♭} (n) β (r) + β (n) n^{♭} (r) = - β (r) = 0$ , and evaluating at $(n, n)$ gives $- 2 β (n) = 0$ ; since every vector is a sum of a multiple of $n$ and a rest vector by step 2.1, $β = 0$ , that is, $d f = f a^{♭}$ and hence $d \log f = a^{♭}$ . Applying $d$ and [L5] gives $0 = d^{2} \log f = d a^{♭} = (1 / (L_{0}^{2} N^{2})) d λ \land d ξ \neq 0$ by step 4.1, a contradiction. Therefore no positive smooth $f$ on any nonempty open subset of $W$ makes $f n$ a Killing field: the constructed congruence is Born rigid and future unit, but it is not locally a positive multiple of a Killing field.

6.1step 5.1∎

Conclusion and clarifications: steps 1.1-5.1 construct a smooth future unit timelike field $n$ on the open set $W$ that is Born rigid ( $θ = 0$ ) and irrotational ( $ω = 0$ ), and step 5.1 shows that for every positive smooth $f$ the field $f n$ fails the Killing equation on every nonempty open subset; hence the claim of the Statement refuted section is false. The Rindler congruence is not a counterexample of this kind: its metric $- ρ^{2} d θ^{2} + d ρ^{2} + d y^{2} + d z^{2}$ is independent of $θ$ , so $\partial_{θ}$ is a Killing field and the unit field $ρ^{- 1} \partial_{θ}$ is a positive multiple of it (Fermi-Walker transport preserves the metric, and explicit rigid and non-rigid observer congruences clause (b)). The rotational classification of The smooth local Herglotz-Noether classification of Born-rigid motions is untouched, because the present congruence is irrotational: the counterexample concerns the irrotational branch of Born-rigid motions.

Remarks

What fails and what survives. The false claim would make every Born-rigid motion Killing up to a positive rescaling on a neighbourhood. Here the obstruction is exact: the acceleration one-form is closed precisely for Killing rescalings (by [L1] the Killing condition forces $d \log f = a^{♭}$ ), and $a^{♭}$ is not closed on any nonempty open subset of $W$ because $d a^{♭} = (1 / L_{0}^{2} N^{2}) d λ \land d ξ$ is nowhere zero. The rotational branch of the Herglotz-Noether classification, where $d a^{♭} = 0$ is a theorem, is not affected.
The construction is local and choice-free. The primitive $z$ is defined through the fundamental theorem of calculus rather than by a closed-form integral; no global quotient, tube completeness or material model is introduced, and the only nonlinear-analytic input is the choice-free inverse function theorem of [L4].

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Counterexample: simultaneity is relative to the observer vector

Statement refuted

False claim: the simultaneity relation $η (n, q - p) = 0$ between events of the affine model is independent of the future unit timelike vector $n$ (Simultaneity slices and rest distance relative to a future unit timelike vector).

Counterexample. In an inertial chart let $p : = x^{- 1} (0, 0, 0, 0)$ and $q : = x^{- 1} (0, 1, 0, 0)$ , so that $d : = q - p$ is spacelike with $η (d, d) = 1$ , and let $n : = e_{0} = (1, 0, 0, 0)$ and $n^{'} : = γ (e_{0} + \frac{3}{5} e_{1}) = \frac{5}{4} (1, \frac{3}{5}, 0, 0)$ , a boost of speed $3 c / 5$ along the first axis. Then $η (n^{'}, n^{'}) = \frac{25}{16} (- 1 + \frac{9}{25}) = - 1$ , so $n^{'}$ is also a future unit timelike vector; $η (n, d) = 0$ , so $p$ and $q$ are simultaneous relative to $n$ and lie in the slice $H (p, n)$ ; and $η (n^{'}, d) = \frac{5}{4} \cdot \frac{3}{5} = \frac{3}{4} \neq 0$ , so $p$ and $q$ are not simultaneous relative to $n^{'}$ and lie in different slices for $n^{'}$ . The two slices $H (p, n)$ and $H (p, n^{'})$ are distinct three-dimensional affine hyperplanes through $p$ . Hence simultaneity in the sense of Simultaneity slices and rest distance relative to a future unit timelike vector depends on the observer vector and is not an observer-independent relation.

Facts & Assumptions

Given: The affine Minkowski model with an inertial frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ and its chart $x$ (Inertial frames and inertial charts of the affine Minkowski model, Timelike, null, spacelike and causal vectors of a Lorentz vector space), the events $p = x^{- 1} (0, 0, 0, 0) = o$ and $q = x^{- 1} (0, 1, 0, 0) = o + e_{1}$ , the displacement $d = q - p = e_{1}$ , the future unit vectors $n = e_{0}$ and $n^{'} = γ (e_{0} + \frac{3}{5} e_{1})$ with $γ = (1 - \frac{9}{25})^{- 1 / 2} = \frac{5}{4}$ , and the slices $H (p, n) = p + n^{⊥}$ , $H (p, n^{'}) = p + n^{' ⊥}$ (Simultaneity slices and rest distance relative to a future unit timelike vector).

[L1]

$η (e_{a}, e_{b}) = diag (- 1, 1, 1, 1)_{a b}$ for the inertial frame's basis, so $η$ is evaluated on coordinate vectors by $η (u, v) = - u^{0} v^{0} + u^{1} v^{1} + u^{2} v^{2} + u^{3} v^{3}$ (Inertial frames and inertial charts of the affine Minkowski model, Oriented time-oriented affine Minkowski space (the spacetime primitives)).

[L2]

The future unit timelike vectors are the vectors $n$ with $η (n, n) = - 1$ and $n \in C^{+}$ ; the first basis vector $e_{0}$ of an inertial frame is future unit, and proper orthochronous Lorentz maps preserve the causal classification and the time orientation (Timelike, null, spacelike and causal vectors of a Lorentz vector space, Lorentz maps are isomorphisms of determinant ±1 that preserve causal character and permute the cones; the Lorentz sets are groups, Linear and affine Lorentz maps, and the Lorentz group).

[L3]

The passive boost with parameter $β$ , $∣ β ∣ < 1$ , is a proper orthochronous Lorentz map with inverse the boost with parameter $- β$ , and its matrix in the inertial basis has entries $B^{0}_{0} = B^{1}_{1} = γ$ and $B^{0}_{1} = B^{1}_{0} = - γ β$ on the first two coordinate directions, so $B (- β) e_{0} = γ (e_{0} + β e_{1})$ for the first basis vector, with $γ = (1 - β^{2})^{- 1 / 2}$ (The passive Lorentz boost with parameter beta, The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)).

[L4]

For a future unit $n$ , $n^{⊥} = {v : η (n, v) = 0}$ and $H (p, n) = p + n^{⊥} = {z : η (n, z - p) = 0}$ ; the slice contains $p$ and is a three-dimensional affine hyperplane (Simultaneity slices and rest distance relative to a future unit timelike vector, Norms of four-velocity and four-acceleration, and the observer rest-space decomposition clause (c), The timelike set has exactly two convex components, described in orthonormal coordinates).

Counterexample

technique · direct

1.1L1L2L3

The displacement is $d = q - p = e_{1}$ by [L1], and $η (d, d) = η (e_{1}, e_{1}) = 1$ , so $d$ is spacelike; the vector $n = e_{0}$ is future unit by [L2]; the vector $n^{'} = \frac{5}{4} (e_{0} + \frac{3}{5} e_{1})$ is the image $B (- 3 / 5) e_{0}$ of $e_{0}$ under the boost with parameter $β = 3 / 5$ by [L3], hence $n^{'}$ is future timelike by [L2], and its norm is $η (n^{'}, n^{'}) = \frac{25}{16} (η (e_{0}, e_{0}) + \frac{6}{5} η (e_{0}, e_{1}) + \frac{9}{25} η (e_{1}, e_{1})) = \frac{25}{16} (- 1 + \frac{9}{25}) = - 1$ by [L1], so $n^{'}$ is future unit; in components $n^{'} = \frac{5}{4} e_{0} + \frac{3}{4} e_{1}$ .

2.1L1L4step 1.1

Simultaneity relative to $n$ : $η (n, d) = η (e_{0}, e_{1}) = 0$ by [L1], so $q - p \in n^{⊥}$ and $q \in H (p, n)$ by [L4]; the events $p, q$ are simultaneous relative to $n$ .

2.2L1L4step 1.1

Non-simultaneity relative to $n^{'}$ : $η (n^{'}, d) = \frac{5}{4} η (e_{0}, e_{1}) + \frac{3}{4} η (e_{1}, e_{1}) = 0 + \frac{3}{4} = \frac{3}{4} \neq 0$ by [L1] and step 1.1, so $q - p \notin n^{' ⊥}$ and $q \notin H (p, n^{'})$ by [L4]; the events $p, q$ are not simultaneous relative to $n^{'}$ .

2.3L1L4step 1.1

The two slices are distinct: the vector $e_{1}$ satisfies $η (n, e_{1}) = 0$ , so $e_{1} \in n^{⊥}$ , while $η (n^{'}, e_{1}) = \frac{3}{4} \neq 0$ , so $e_{1} \notin n^{' ⊥}$ ; hence $n^{⊥} \neq n^{' ⊥}$ as subspaces of $V$ . Both $H (p, n) = p + n^{⊥}$ and $H (p, n^{'}) = p + n^{' ⊥}$ are three-dimensional affine hyperplanes through $p$ by [L4], and since their direction subspaces differ they are distinct sets.

3.1step 2.1step 2.2step 2.3∎

Conclusion: the single pair of events $p = x^{- 1} (0, 0, 0, 0)$ , $q = x^{- 1} (0, 1, 0, 0)$ is simultaneous relative to the future unit vector $n = e_{0}$ by step 2.1 and not simultaneous relative to the future unit vector $n^{'} = \frac{5}{4} (e_{0} + \frac{3}{5} e_{1})$ by step 2.2, and the corresponding slices through $p$ are distinct by step 2.3. Therefore the simultaneity relation defined by $η (n, q - p) = 0$ is not independent of the observer vector $n$ , and the claim of the Statement refuted section is false.

Remarks

Exact rational data. The boost speed $3 c / 5$ was chosen so that $γ = 5 / 4$ and the two pairings $η (n, d) = 0$ , $η (n^{'}, d) = 3 / 4$ are exact rational numbers; no numerical approximation is involved.
The witness for distinctness of the direction subspaces is $e_{1}$ . The vector $e_{1}$ lies in $n^{⊥}$ and not in $n^{' ⊥}$ ; the two rest spaces of distinct future unit vectors are in fact always distinct, indeed, if $n^{⊥} = n^{' ⊥}$ , then $h_{n} (n^{'}) \in n^{⊥}$ is orthogonal to every vector of $n^{⊥}$ (because both $n$ and $n^{'}$ are); positive definiteness forces $h_{n} (n^{'}) = 0$ . The decomposition theorem then makes $n^{'}$ a multiple of $n$ , and future unit normalization forces that multiple to be $1$ .

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Worked Fermi-Walker transport along uniformly accelerated hyperbolic motion

Example

In an inertial chart with metric $diag (- 1, 1, 1, 1)$ let $α > 0$ and consider the smooth worldline $x : R \to A$ with inertial coordinates $x (τ) = \frac{c^{2}}{α} (\sinh (α τ / c), \cosh (α τ / c), 0, 0)$ , parametrized by $τ$ ; write $θ : = α τ / c$ . Its derivatives are $U : = x^{'} = c (\cosh θ, \sinh θ, 0, 0)$ and $A : = U^{'} = α (\sinh θ, \cosh θ, 0, 0)$ , and one computes $η (U, U) = c^{2} (- \cosh^{2} θ + \sinh^{2} θ) = - c^{2}$ , $η (A, A) = α^{2} (- \sinh^{2} θ + \cosh^{2} θ) = α^{2}$ and $η (U, A) = c α (- \cosh θ \sinh θ + \sinh θ \cosh θ) = 0$ ; thus $τ$ is a proper-time parameter (up to an additive constant) and $U$ is the four-velocity of a uniformly accelerated worldline. Put $X_{0} : = U / c = (\cosh θ, \sinh θ, 0, 0)$ and $X_{1} : = (\sinh θ, \cosh θ, 0, 0)$ ; then $η (X_{0}, X_{0}) = - 1$ , $η (X_{0}, X_{1}) = 0$ and $η (X_{1}, X_{1}) = 1$ , so $(X_{0}, X_{1})$ is an orthonormal pair. A direct substitution into the Fermi-Walker rule shows that both are Fermi-Walker transported: $X_{0}^{'} = A / c$ equals $\frac{1}{c^{2}} (η (A, X_{0}) U - η (U, X_{0}) A) = \frac{1}{c^{2}} (0 + c A) = A / c$ , and $X_{1}^{'} = \frac{α}{c} X_{0}$ equals $\frac{1}{c^{2}} (η (A, X_{1}) U - η (U, X_{1}) A) = \frac{1}{c^{2}} (α U - 0) = \frac{α}{c} X_{0}$ . Moreover $X_{1}$ is not parallel transported, since $X_{1}^{'} = \frac{α}{c} X_{0} \neq 0$ for $α > 0$ ; Fermi-Walker transport therefore differs from parallel transport for accelerated motion. The norm $η (X_{1}, X_{1}) = 1$ is constant along the worldline, an instance of the preservation clause of Fermi-Walker transport preserves the metric, and explicit rigid and non-rigid observer congruences, and adjoining the constant fields $\partial_{y}, \partial_{z}$ gives a Fermi-Walker transported orthonormal tetrad $(U / c, X_{1}, \partial_{y}, \partial_{z})$ along the hyperbolic worldline.

Facts & Assumptions

Given: An inertial chart of the affine Minkowski model with $η = diag (- 1, 1, 1, 1)$ (Inertial frames and inertial charts of the affine Minkowski model, Oriented time-oriented affine Minkowski space (the spacetime primitives)), the number $α > 0$ , the curve $x (τ) = \frac{c^{2}}{α} (\sinh θ, \cosh θ, 0, 0)$ with $θ = α τ / c$ , and the fields $U = x^{'}$ , $A = U^{'}$ , $X_{0} = U / c$ , $X_{1} = (\sinh θ, \cosh θ, 0, 0)$ of the Example (Four-velocity and four-acceleration of a future timelike worldline).

[L1]

Fermi-Walker transport (Fermi-Walker transport along a timelike worldline clauses (a),(b),(c)): in a proper-time parametrization (so the derivative of the proper-time function is $1$ ), a $C^{1}$ field $X$ along a $C^{2}$ future timelike worldline is Fermi-Walker transported exactly when $X^{'} = \frac{1}{c^{2}} (η (A, X) U - η (U, X) A)$ ; the four-velocity $U$ is always Fermi-Walker transported; and on an interval where $A = 0$ the rule coincides with parallel transport, which in the flat model means $X^{'} = 0$ .

[L2]

Preservation and tetrads (Fermi-Walker transport preserves the metric, and explicit rigid and non-rigid observer congruences clause (a), Existence, uniqueness and linearity of Fermi-Walker transport): Fermi-Walker transport preserves $η (X, Y)$ for transported pairs; and the unique transported family with initial values a tetrad whose first vector is $U / c$ is an orthonormal observer-adapted tetrad at every parameter.

[L3]

Four-velocity and proper time (Four-velocity and four-acceleration of a future timelike worldline, Proper time along a future timelike worldline, Norms of four-velocity and four-acceleration, and the observer rest-space decomposition clauses (a),(b)): for a $C^{1}$ future timelike worldline parametrized by proper time, $U = x^{'}$ satisfies $η (U, U) = - c^{2}$ and, when $x$ is $C^{2}$ , $η (U, A) = 0$ ; conversely a parameter is a proper-time parameter exactly when $\sqrt{- η (x^{'}, x^{'})} = c$ along the curve.

[L4]

Hyperbolic functions and the chain rule: $\cosh^{2} s - \sinh^{2} s = 1$ , $\cosh s > 0$ , $\sinh^{'} = \cosh$ , $\cosh^{'} = \sinh$ (The six hyperbolic functions and their natural domains, Addition formulas, identities, parity, and derivatives of the hyperbolic functions), and differentiation of the composite $τ \mapsto (\sinh θ (τ), \cosh θ (τ))$ with $θ = α τ / c$ uses the chain rule (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

Verification

technique · direct

1.1L3L4algebra

Derivatives and causal character: by [L4], $x^{'} (τ) = \frac{c^{2}}{α} \cdot \frac{α}{c} (\cosh θ, \sinh θ, 0, 0) = c (\cosh θ, \sinh θ, 0, 0) = U$ , so $U$ is $C^{1}$ and $x$ is $C^{2}$ ; further $U^{'} = c \cdot \frac{α}{c} (\sinh θ, \cosh θ, 0, 0) = α (\sinh θ, \cosh θ, 0, 0) = A$ . Evaluating the diagonal form, $η (U, U) = c^{2} (- \cosh^{2} θ + \sinh^{2} θ) = - c^{2}$ , so $x$ is future timelike with $U$ future (time component $c \cosh θ > 0$ ) and $τ$ is a proper-time parameter by [L3]; $η (A, A) = α^{2} (\cosh^{2} θ - \sinh^{2} θ) = α^{2}$ and $η (U, A) = c α (- \cosh θ \sinh θ + \sinh θ \cosh θ) = 0$ ; in particular $A \neq 0$ because $α > 0$ .

2.1step 1.1L4

Orthonormal pair: $X_{0} = U / c = (\cosh θ, \sinh θ, 0, 0)$ and $X_{1} = (\sinh θ, \cosh θ, 0, 0)$ satisfy $η (X_{0}, X_{0}) = - \cosh^{2} θ + \sinh^{2} θ = - 1$ , $η (X_{0}, X_{1}) = - \cosh θ \sinh θ + \sinh θ \cosh θ = 0$ and $η (X_{1}, X_{1}) = - \sinh^{2} θ + \cosh^{2} θ = 1$ by [L4]; and $X_{0}^{'} = U^{'} / c = A / c$ .

2.2L1step 1.1L4

$X_{1}$ is Fermi-Walker transported: $X_{1}^{'} = \frac{α}{c} (\cosh θ, \sinh θ, 0, 0) = \frac{α}{c} X_{0}$ , and $\frac{1}{c^{2}} (η (A, X_{1}) U - η (U, X_{1}) A) = \frac{1}{c^{2}} (α U - 0) = \frac{α}{c} X_{0}$ , because $η (A, X_{1}) = α (- \sinh^{2} θ + \cosh^{2} θ) = α$ and $η (U, X_{1}) = c (- \cosh θ \sinh θ + \sinh θ \cosh θ) = 0$ by [L4]; the right-hand side equals $X_{1}^{'}$ , so $X_{1}$ is transported.

3.1L1step 1.1step 2.1

$X_{0}$ is Fermi-Walker transported: by [L1] and steps 1.1-2.1, $\frac{1}{c^{2}} (η (A, X_{0}) U - η (U, X_{0}) A) = \frac{1}{c^{2}} (0 - (- c) A) = A / c = X_{0}^{'}$ , because $η (A, X_{0}) = η (A, U) / c = 0$ and $η (U, X_{0}) = η (U, U) / c = - c$ .

3.2L1L2step 2.2

Difference from parallel transport and norm preservation: $X_{1}^{'} = \frac{α}{c} X_{0} \neq 0$ since $α > 0$ and $X_{0} \neq 0$ , so $X_{1}$ is not parallel transported by [L1]; Fermi-Walker transport therefore differs from parallel transport along this accelerated worldline. The norm is preserved: $η (X_{1}, X_{1}) = 1$ is constant, in agreement with the preservation clause [L2].

4.1L1L2step 2.1step 3.1step 2.2

Tetrad: the constant fields $\partial_{y} = (0, 0, 1, 0)$ and $\partial_{z} = (0, 0, 0, 1)$ have $\partial_{y}^{'} = \partial_{z}^{'} = 0$ and $η (A, \partial_{y}) = η (U, \partial_{y}) = 0$ , likewise for $\partial_{z}$ , so the right-hand side of the Fermi-Walker rule vanishes and both are transported by [L1]. Hence $(U / c, X_{1}, \partial_{y}, \partial_{z}) = (X_{0}, X_{1}, \partial_{y}, \partial_{z})$ is an orthonormal tetrad every member of which is Fermi-Walker transported, as asserted in the Example; this also realizes the observer-adapted tetrad of [L2].

5.1step 3.1step 2.2step 3.2step 4.1∎

Conclusion: steps 1.1-2.1 verify the kinematic data, steps 3.1 and 2.2 verify that both legs $X_{0} = U / c$ and $X_{1}$ satisfy the Fermi-Walker rule, step 3.2 exhibits the nonzero derivative $X_{1}^{'} = \frac{α}{c} X_{0}$ showing that Fermi-Walker transport is not parallel transport for $α > 0$ , and step 4.1 completes the transported orthonormal tetrad. All identities are exact evaluations in the hyperbolic functions and the diagonal form.

Remarks

Why the correction term matters. The Fermi-Walker rule subtracts exactly the acceleration-dependent term $c^{- 2} (η (A, X) U - η (U, X) A)$ ; along this worldline the term rotates $X_{1}$ at the rate $α / c$ while keeping $X_{0}$ transported, which is what makes the pair $(X_{0}, X_{1})$ a nonrotating orthonormal frame in the Fermi-Walker sense.
Uniform acceleration. The worldline has constant acceleration magnitude $\sqrt{η (A, A)} = α$ and lies in the region $x^{0}^{2} < x^{1}^{2}$ ; the example uses only the local differential identities and no horizon or coordinate-singularity statement, which belongs to the accelerated-observers page.

Observer Measurements — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

Worked radar round trip: emission, reflection and reception at exact rational parameters

Example

Facts & Assumptions

Verification

Remarks

Worked time dilation at speed 3c/5

Example

Facts & Assumptions

Verification

Worked length contraction of a rod of rest length 5 at speed 3c/5

Example

Facts & Assumptions

Verification

Remarks

Worked velocity transformation: collinear, transverse and null cases

Example

Facts & Assumptions

Verification

Worked Rindler wedge congruence: chart, metric, acceleration, rigidity and non-coordinate frame

Example

Facts & Assumptions

Verification

Remarks

Worked rigid rotation: rest-space holonomy around the ring and failure of ring synchronization

Example

Facts & Assumptions

Verification

Remarks

Worked Bell-type translated congruence: constant launch-frame separation without Born rigidity

Example

Facts & Assumptions

Verification

Remarks

Counterexample: a Born-rigid congruence need not be a Killing congruence

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Counterexample: simultaneity is relative to the observer vector

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Worked Fermi-Walker transport along uniformly accelerated hyperbolic motion

Example

Facts & Assumptions

Verification

Remarks

Sources