Thermal vorticity and global equilibrium obstructions

The scalar $s:=-g(\beta ,\beta )$ is smooth and strictly positive on $O$. Hence $s^{-1\mathrm{/}2}$ is smooth and positive, so $T=(k_B\sqrt{s}{)}^{-1}$ is a smooth positive kelvin-valued scalar and $u=\beta \mathrm{/}\sqrt{s}$ is a smooth future-directed vector field. Direct contraction gives $g(u,u)=g(\beta ,\beta )\mathrm{/}s=-1$.

Lowering the smooth vector field $\beta$ with $g$ gives a smooth one-form, and its covariant derivative is a smooth covariant two-tensor. Antisymmetrizing this tensor gives the smooth two-form $\varpi$; the inverse-length derivative of the inverse-joule components of $\beta$ gives its stated units. Since $\mu$ and $k_{B}T>0$ are smooth, $\zeta$ and $\alpha =-\zeta$ are defined and dimensionless on all of $O$. The only foundational assumption in this construction is the inherited ${\mathrm{A}\mathrm{C}}_{\omega }$ in [F4]; no selection or stronger choice principle is used.

The charts are smooth charts of the manifold structure in [F1]. On each chartwise $C^1$ piece, $t\mapsto {\omega }_{\gamma (t)}(\dot{\gamma }(t))$ is continuous up to the endpoints by [F2] and [F3]. The covector-vector pairing is unchanged by a chart change, so this is an intrinsically defined continuous real function; its piece integral exists by [F4].

Any two admissible finite subdivisions have a finite common refinement. Repeatedly applying [F5] shows that subdividing a piece leaves the sum unchanged; hence the defined period is a unique real number independent of the chosen subdivision.

In any smooth chart, the definitions give $({\iota }_{\xi }F{)}_a={\xi }^b({\mathrm{\partial }}_b{A}_a-{\mathrm{\partial }}_a{A}_b)$ and ${\mathrm{\partial }}_a{\mathrm{\Psi }}_{\xi }=A_b{\mathrm{\partial }}_a{\xi }^b+{\xi }^b{\mathrm{\partial }}_a{A}_b-{\mathrm{\partial }}_a{\mathrm{\Lambda }}_{\xi }$. By [A1] and [F1], ${\mathrm{\partial }}_a{\mathrm{\Lambda }}_{\xi }={\xi }^b{\mathrm{\partial }}_b{A}_a+A_b{\mathrm{\partial }}_a{\xi }^b$. Substitution yields $({\iota }_{\xi }F{)}_a=-{\mathrm{\partial }}_a{\mathrm{\Psi }}_{\xi }$. Since this calculation holds in every chart, it proves the stated one-form identity.

The scalar ${\mathrm{\Psi }}_{\xi }={\xi }^a{A}_a-{\mathrm{\Lambda }}_{\xi }$ is global and smooth because it is the contraction of globally defined smooth fields minus a global smooth scalar. Thus step 1.1 gives $\omega =-d{\mathrm{\Psi }}_{\xi }=d(-{\mathrm{\Psi }}_{\xi })$ globally, which is exact by [F2].

Fix the finite subdivision from [F1]. On each chartwise $C^1$ piece choose a chart $\kappa :U\to V\subseteq {\mathbb{R}}^n$ containing its image, and put $x(t):=\kappa (\gamma (t))$ and $\tilde{\varphi }:=\varphi \circ {\kappa }^{-1}$. By [F2] the coordinate coefficients of $\omega$ are ${\mathrm{\partial }}_i\tilde{\varphi }$; applying [F3] to $\tilde{\varphi }\circ x$ gives $\frac{d}{dt}(\varphi \circ \gamma )(t)={\omega }_{\gamma (t)}(\dot{\gamma }(t))$ in the piece interior. The same formula holds with one-sided derivatives at the endpoints: the differentiability expansion of $\tilde{\varphi }$ composed with the one-sided expansion of $x$ gives $D\tilde{\varphi }(x(t))x^{\mathrm{\prime }}(t)$. These derivatives are continuous on the closed piece by [F1] and smoothness of $\varphi$, so they are integrable by [F4].

Apply [F5] on each subdivision interval to $h=\varphi \circ \gamma$. By [F1] the period is the sum of those integrals, so it equals the sum of the increments $\varphi (\gamma (t_{j+1}))-\varphi (\gamma (t_j))$. The internal endpoint terms cancel, and the remaining increment is $\varphi (\gamma (1))-\varphi (\gamma (0))=0$ because $\gamma$ is closed. Thus the period is zero.

The displayed definitions give $u=k_{B}T\beta$ and $k_{B}T=(-g(\beta ,\beta ){)}^{-1\mathrm{/}2}$, so $u=\beta \mathrm{/}\sqrt{-g(\beta ,\beta )}$ is future unit timelike and $\beta =u\mathrm{/}(k_{B}T)$. Applying the exact equivalence [F2] to this $u$ and $T$ gives the four displayed kinematic conditions if and only if $\beta$ is Killing. No hypersurface-orthogonality or zero-vorticity condition is added.

In the Killing subcase of the Statement, fix any constant $T_{\ast }>0$ kelvins and set $\xi :=k_B{T}_{\ast }\beta$. This is a smooth future-timelike Killing field with dimensionless components. Its lapse is $N_{\xi }=k_B{T}_{\ast }\sqrt{-g(\beta ,\beta )}=T_{\ast }\mathrm{/}T$, so $\xi \mathrm{/}{N}_{\xi }=u$ and $N_{\xi }T=T_{\ast }$. To meet the orientation hypothesis of [F3] and [F4], fix any point of $O$ and a connected chart domain $U\subseteq O$ containing it, and orient $U$ by its ordered coordinates. The restricted fields satisfy all stationary thermal-carrier hypotheses of [F4], and [A3] supplies its massive-shell ansatz on $U$. The future shell is defined by $g$ and the time orientation, and the charged transport by its covariant equation in [F3]; neither depends on this auxiliary chart orientation. Thus the restricted scalar ${\mathcal{L}}_{F,q_{\mathrm{S}\mathrm{I}}}f$ and the one-form $d\zeta +c{q}_{\mathrm{S}\mathrm{I}}\omega$ agree with their global definitions on overlaps. Applying [F4] on $U$ gives stationarity there exactly when $d\zeta =-c{q}_{\mathrm{S}\mathrm{I}}\omega$ there, since $\beta$ is already Killing. As the point was arbitrary, the same equivalence holds globally on $O$, without a global orientation or a simultaneous choice of oriented charts. By [A1] this is exactly $d\alpha =q_{\mathrm{e}\mathrm{f}\mathrm{f}}\omega$. This identifies the charged stationarity condition without assuming stationarity in the separate solvability argument below. The only inherited set-theoretic premise is ${\mathrm{A}\mathrm{C}}_{\omega }$ in [F9]; no stronger choice principle is invoked.

Since $q_{\mathrm{e}\mathrm{f}\mathrm{f}}\mathrm{\ne }0$, any global solution of $d\alpha =q_{\mathrm{e}\mathrm{f}\mathrm{f}}\omega$ gives $\omega =d(\alpha \mathrm{/}{q}_{\mathrm{e}\mathrm{f}\mathrm{f}})$, which is exact by [F5]. Conversely, if $\omega =d\psi$ for a global smooth $\psi$, choosing $\alpha =q_{\mathrm{e}\mathrm{f}\mathrm{f}}\psi$ solves the equation, and $\mu =-k_{B}T\alpha$ is a smooth global chemical-potential scalar with the stated units. Thus the scalar transport equation has a global solution exactly when $\omega$ is exact. This is a solvability result and does not itself assert that the supplied fields are stationary.

If $\xi =k_B{T}_{\ast }\beta$ has the global compensator in the Statement, [F8] gives ${\iota }_{\xi }F=-d{\mathrm{\Psi }}_{\xi }$. Since $\beta =\xi \mathrm{/}(k_B{T}_{\ast })$, we get $\omega =-(k_B{T}_{\ast }{)}^{-1}d{\mathrm{\Psi }}_{\xi }$, an exact one-form; [F7] then forces every period to vanish. This stronger potential-symmetry condition is separate from the obstruction hypotheses.

If any closed curve has ${\mathrm{Per}}_{\gamma }(\omega )\mathrm{\ne }0$, [F7] contradicts the exactness forced by step 1.3. Hence no global smooth charged chemical scalar can solve the transport equation.

In the rotating Minkowski model, ${\mathrm{\partial }}_{x^0}$ and the smooth rotation generator ${\mathrm{\partial }}_{\varphi }=-y{\mathrm{\partial }}_x+x{\mathrm{\partial }}_y$ are Killing because the metric is invariant under time translations and rotations. Their constant linear combination $\beta$ is Killing and $g(\beta ,\beta )=-b^2+b^2{\mathrm{\Omega }}^2{r}^2=-b^2(1-{\mathrm{\Omega }}^2{r}^2)$. Also $g(\beta ,{\mathrm{\partial }}_{x^0})=-b<0$, so it is future-directed in the stated time orientation wherever it is timelike. Thus its future-timelike domain is exactly $1-{\mathrm{\Omega }}^2{r}^2>0$. Substituting this norm into [F1] gives the stated temperature; when $\mathrm{\Omega }=0$ the norm is $-b^2$ for every $r$.