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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.

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

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

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