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Thermal vorticity and global equilibrium obstructions — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Rigid bath and nonexact electric period

Example

Use global Cartesian length coordinates $(x^{0}, x, y, z)$ on Minkowski space $M = R^{4}$ , with $x^{0} = c t$ , metric $- (d x^{0})^{2} + d x^{2} + d y^{2} + d z^{2}$ , and future direction selected by $\partial_{x^{0}}$ . Set $r : = \sqrt{x^{2} + y^{2}}$ and let $ϕ$ be the polar angle away from $r = 0$ ; the rotation field $\partial_{ϕ} : = - y \partial_{x} + x \partial_{y}$ is defined smoothly on all of $M$ by its Cartesian expression. Let

$O : = R_{x^{0}} \times {(x, y) \in R^{2} : \frac{1}{4} m < \sqrt{x^{2} + y^{2}} < \frac{3}{4} m} \times R_{z},$

so the spatial section is an annulus and all fields below are defined on this four-dimensional product domain. Adopt the exact SI defining values $k_{B} = 1.380649 \times 10^{- 23} {J K}^{- 1}$ and $c = 299792458 {m s}^{- 1}$ . For $b = 1 J^{- 1}$ , $Ω = 1 m^{- 1}$ , and $r = \frac{1}{2} m$ , the rotating field $β = b (\partial_{x^{0}} + Ω \partial_{ϕ})$ has $T = 2 J / (\sqrt{3} k_{B})$ . Define the smooth one-form $η : = (- y d x + x d y) / (x^{2} + y^{2})$ on $O$ , set $A : = C_{0} x^{0} η$ and $F : = d A = C_{0} d x^{0} \land η$ for a fixed nonzero $C_{0}$ in $T m$ . For any fixed nonzero SI charge $q_{S I}$ in coulombs, define $q_{e f f} : = c q_{S I} \neq 0$ in ${C m s}^{- 1}$ . Then $ω : = ι_{β} F = b C_{0} (η - Ω d x^{0})$ is closed and has period $2 π b C_{0} \neq 0$ on every positively oriented circle of fixed $x^{0}, z, r$ in the spatial annulus. Hence no global smooth dimensionless scalar $α : O \to R$ satisfies $d α = q_{e f f} ω$ .

Statement references

Rotation and charged equilibrium obstruction

Period of a smooth one-form along a closed curve

An exact one-form has zero period on every closed piecewise-C¹ curve

Facts & Assumptions

Given: The length-coordinate Minkowski metric, $b = 1 J^{- 1}$ , $Ω = 1 m^{- 1}$ , $r = \frac{1}{2} m$ , $C_{0} \neq 0$ in $T m$ , and $q_{e f f} \neq 0$ .

[F1]

Rotation and charged equilibrium obstruction: the rotating temperature is $T = 1 / (k_{B} b \sqrt{1 - Ω^{2} r^{2}})$ on $r < 1 / ∣ Ω ∣$ , and a nonzero period of $ω = ι_{β} F$ forbids a global solution of $d α = q_{e f f} ω$ .

[F2]

Period of a smooth one-form along a closed curve: on a smooth manifold, for a smooth one-form and a closed continuous piecewise- $C^{1}$ curve with finitely many chartwise pieces and one-sided endpoint derivatives, the period is the sum of the integrals of the one-form evaluated on the curve tangent over those pieces.

[F3]

An exact one-form has zero period on every closed piecewise-C¹ curve: every exact smooth one-form has zero period on every closed piecewise- $C^{1}$ curve.

[A1]

The domain is $O = R_{x^{0}} \times {(x, y) : \frac{1}{4} m < \sqrt{x^{2} + y^{2}} < \frac{3}{4} m} \times R_{z}$ , with $x^{0} = c t$ , the global Cartesian Minkowski metric, and the globally smooth rotation generator $\partial_{ϕ} = - y \partial_{x} + x \partial_{y}$ .

[A2]

The global one-form $η = (- y d x + x d y) / (x^{2} + y^{2})$ and the fields $A = C_{0} x^{0} η$ , $F = d A = C_{0} d x^{0} \land η$ are smooth on $O$ .

[A3]

The adopted exact SI values are $k_{B} = 1.380649 \times 10^{- 23} {J K}^{- 1}$ and $c = 299792458 {m s}^{- 1}$ ; $q_{S I} \neq 0$ coulombs and $q_{e f f} : = c q_{S I} \neq 0$ has units ${C m s}^{- 1}$ .

Verification

Given: [A1], [A2], and [F1]–[F3].

1.1givenF1A1algebra

The chosen radius satisfies $Ω^{2} r^{2} = (1 m^{- 1})^{2} (\frac{1}{2} m)^{2} = 1 / 4$ , so $\sqrt{1 - Ω^{2} r^{2}} = \sqrt{3} / 2$ and the rotating temperature formula in [F1] gives $T = 2 J / (\sqrt{3} k_{B})$ . The annulus has $r < 3 / 4 m < 1 / ∣ Ω ∣$ , so every point lies inside the timelike domain.

1.2givenA1A2algebra

The form $η$ is smooth because $r^{2} = x^{2} + y^{2} > 0$ . Its coefficients are $η_{x} = - y / r^{2}$ and $η_{y} = x / r^{2}$ , with $\partial_{x} η_{y} = (y^{2} - x^{2}) / r^{4} = \partial_{y} η_{x}$ , so $d η = 0$ . Also $η (\partial_{ϕ}) = (- y / r^{2}) (- y) + (x / r^{2}) x = 1$ . Hence $d A = C_{0} d x^{0} \land η = F$ , and contraction gives $ω = ι_{β} F = b C_{0} (η - Ω d x^{0})$ . Since $d η = 0$ and $d (d x^{0}) = 0$ , $d ω = 0$ .

1.3givenF2A1A2algebra

For $γ (s) = (x_{0}^{0}, r \cos (2 π s), r \sin (2 π s), z_{0})$ , $0 \leq s \leq 1$ , its tangent satisfies $d x^{0} (\dot{γ}) = 0$ and direct substitution into $η = (- y d x + x d y) / r^{2}$ gives $η_{γ (s)} (\dot{γ} (s)) = 2 π$ . Hence ${Per}_{γ} (ω) = \int_{0}^{1} 2 π b C_{0} d s = 2 π b C_{0} \neq 0$ .

2.1givenF3step 1.3algebra∎

If a global smooth $α$ solved $d α = q_{e f f} ω$ , then $ω = d (α / q_{e f f})$ because $q_{e f f} \neq 0$ . This would be exact, so [F3] would force ${Per}_{γ} (ω) = 0$ , contradicting step 1.3. Hence no such global scalar exists.