Global electrochemical redshift from charged Liouville stationarity

Since $f>0$, ${\mathcal{L}}_{F,q}f=0$ is equivalent to ${\mathcal{L}}_{F,q}(\zeta +g(\beta ,P))=0$. In a chart, the connection term in the vertical part of ${\mathcal{L}}_{F,q}$ combines with $P^a{\mathrm{\partial }}_a{\beta }_b$ to give $P^a{P}^b{\mathrm{\nabla }}_a{\beta }_b$; its electromagnetic part gives $cq{\beta }^a{F}_{ab}{P}^b$. Thus on the future shell $\frac{{\mathcal{L}}_{F,q}f}{f}=P^a{P}^b{\mathrm{\nabla }}_{(a}{\beta }_{b)}+P^b({\mathrm{\nabla }}_b\zeta +cq{\beta }^a{F}_{ab})\mathrm{.}$ The stationarity equation sets this quadratic-plus-linear expression to zero for every future massive shell vector.

If ${\mathcal{L}}_{F,q}f=0$, apply [F5] pointwise to $A_{ab}:={\mathrm{\nabla }}_{(a}{\beta }_{b)}$ and $b_b:={\mathrm{\nabla }}_b\zeta +cq{\beta }^a{F}_{ab}$. It gives ${\mathrm{\nabla }}_{(a}{\beta }_{b)}=0$ and $d\zeta =-cq{\iota }_{\beta }F$. Conversely these equations make both terms in the expression of step 1.1 vanish for every shell vector, so ${\mathcal{L}}_{F,q}f=0$. This proves the first equivalence using the exact massive-shell clause of [F5].

Now impose the additional global condition [A2]. In the stationary setting $\beta =u\mathrm{/}(k_{B}T)$ is collinear with $\xi$. When the first condition holds, the thermal-normalization interface [F4] gives a constant $T_K>0$ with $NT=T_K$ and $\beta =\xi \mathrm{/}(k_B{T}_K)$. By [F3], ${\iota }_{\xi }F=-d{\mathrm{\Psi }}_{\xi }$ and ${\mathrm{\Phi }}_{\xi }=-c{\mathrm{\Psi }}_{\xi }$, hence ${\iota }_{\xi }F=d{\mathrm{\Phi }}_{\xi }\mathrm{/}c$. Substitution into the second condition gives $d\zeta =-cq{\iota }_{\beta }F=-\frac{cq}{k_B{T}_K}{\iota }_{\xi }F=-\frac{q}{k_B{T}_K}d{\mathrm{\Phi }}_{\xi }\mathrm{.}$ Therefore $d(\zeta +q{\mathrm{\Phi }}_{\xi }\mathrm{/}(k_B{T}_K))=0$.

Conversely, under [A2], suppose the three displayed redshift equations hold. Then $\beta =\xi \mathrm{/}(k_B{T}_K)$ is Killing since $T_K$ is constant and $\xi$ is Killing. Also $\zeta +q{\mathrm{\Phi }}_{\xi }\mathrm{/}(k_B{T}_K)={\mu }_K\mathrm{/}(k_B{T}_K)$ is constant, so $d\zeta =-qd{\mathrm{\Phi }}_{\xi }\mathrm{/}(k_B{T}_K)=-cq{\iota }_{\beta }F$ by [F3]. Step 2.1 then gives ${\mathcal{L}}_{F,q}f=0$.

Around every point choose a chart whose image contains a convex open ball. The chain rule and $d(\zeta +q{\mathrm{\Phi }}_{\xi }\mathrm{/}(k_B{T}_K))=0$ give zero total derivative for the coordinate representative on that ball, so [F6] makes it constant there. It is therefore locally constant on $O$. The level set of this function through any chosen point is nonempty and open, and its complement is also open because the function is locally constant. Connectedness of $O$ forces that level set to be all of $O$. Write its constant value as ${\mu }_K\mathrm{/}(k_B{T}_K)$. Multiplying $\zeta +q{\mathrm{\Phi }}_{\xi }\mathrm{/}(k_B{T}_K)={\mu }_K\mathrm{/}(k_B{T}_K)$ by $k_B{T}_K=k_{B}NT$ yields $N\mu +q{\mathrm{\Phi }}_{\xi }={\mu }_K$.

Finally [F3] gives $H_{\xi }=-g(P,\xi )+q{\mathrm{\Phi }}_{\xi }$. Since $\xi =Nu$ and $E_{\mathrm{l}\mathrm{o}\mathrm{c}}=-g(P,u)$, $-g(P,\xi )=N{E}_{\mathrm{l}\mathrm{o}\mathrm{c}}$. Using $NT=T_K$ and $N\mu +q{\mathrm{\Phi }}_{\xi }={\mu }_K$, $-\frac{H_{\xi }-{\mu }_K}{k_B{T}_K}=-\frac{N{E}_{\mathrm{l}\mathrm{o}\mathrm{c}}+q{\mathrm{\Phi }}_{\xi }-{\mu }_K}{k_{B}NT}=-\frac{E_{\mathrm{l}\mathrm{o}\mathrm{c}}-\mu }{k_{B}T}=\zeta +g(\beta ,P)\mathrm{.}$ This is an algebraic equality of dimensionless exponents; it asserts no statistical preparation or normalization.