Metric-free Maxwell equations and metric-dependent vacuum closure

Expand in the fixed chart. The spatial part of $dF$ is $(\mathrm{div}B)vo{l}_3$ and its $dt$ part is $dt\wedge ({\mathrm{\partial }}_t\mathcal{B}+d_{\mathrm{\Omega }}\mathcal{E})$, where $d_{\mathrm{\Omega }}\mathcal{E}=({\mathrm{\partial }}_2{E}_3-{\mathrm{\partial }}_3{E}_2)d{x}^2\wedge d{x}^3+({\mathrm{\partial }}_3{E}_1-{\mathrm{\partial }}_1{E}_3)d{x}^3\wedge d{x}^1+({\mathrm{\partial }}_1{E}_2-{\mathrm{\partial }}_2{E}_1)d{x}^1\wedge d{x}^2$. Thus $d_{\mathrm{\Omega }}\mathcal{E}$ has the curl components in the ordered spatial 2-form basis. Likewise, the spatial part of $dG$ is $(\mathrm{div}D)vo{l}_3$ and its $dt$ part is $dt\wedge ({\mathrm{\partial }}_t\mathcal{D}-d_{\mathrm{\Omega }}\mathcal{H})$, where $d_{\mathrm{\Omega }}\mathcal{H}$ is given by the same curl formula with $H$ in place of $E$; $\mathcal{J}={\rho }_{f}vo{l}_3-dt\wedge {\mathcal{J}}_{\mathrm{s}\mathrm{p}}$. These coefficient identities follow by applying the $C^1$ component formula of [F1] to each displayed wedge term. Equality of the independent spatial and $dt$ coefficients therefore gives both directions of the claimed equivalence.

In vacuum $P=M=0$, so $D={ϵ}_{0}E$ and $H={\mu }_0^{-1}B$ by [A2]. Applying the basis formula [F3] to the six terms of $F$ gives ${\star }_{\eta }F=cdt\wedge (B_{1}d{x}^1+B_{2}d{x}^2+B_{3}d{x}^3)+c^{-1}(E_{1}d{x}^2\wedge d{x}^3+E_{2}d{x}^3\wedge d{x}^1+E_{3}d{x}^1\wedge d{x}^2)$. Using ${ϵ}_0{\mu }_0{c}^2=1$, this equals ${\mu }_{0}cG$. The Lorentzian sign check ${\star }_{\eta }^2=-\mathrm{id}$ on 2-forms is [F4], so the closure has the stated signature convention.

From [A1] and [A2], $\mathrm{div}D={ϵ}_0\mathrm{div}E+\mathrm{div}P=\rho +\mathrm{div}P=\rho -{\rho }_b={\rho }_f$. Also, $\mathrm{curl}H={\mu }_0^{-1}\mathrm{curl}B-\mathrm{curl}M=J+({\mu }_0{c}^2{)}^{-1}{\mathrm{\partial }}_{t}E-\mathrm{curl}M=J+{ϵ}_0{\mathrm{\partial }}_{t}E-\mathrm{curl}M$. Since $J_f=J-{\mathrm{\partial }}_{t}P-\mathrm{curl}M$ and ${\mathrm{\partial }}_{t}D={ϵ}_0{\mathrm{\partial }}_{t}E+{\mathrm{\partial }}_{t}P$, this is $\mathrm{curl}H=J_f+{\mathrm{\partial }}_{t}D$. The two remaining component equations, $\mathrm{div}B=0$ and $\mathrm{curl}E=-{\mathrm{\partial }}_{t}B$, are unchanged from [A1]. Thus the component equations hold, and step 1.1 gives $dF=0$ and $dG=\mathcal{J}$.

Under the additional $C^2$ hypothesis, $\mathcal{J}=dG$ is $C^1$ and $d\mathcal{J}$ is defined by [F1]. In repeated-index notation, $d(dG)=\frac{1}{2}{\mathrm{\partial }}_{\kappa }{\mathrm{\partial }}_{\lambda }{G}_{\mu \nu }d{x}^{\kappa }\wedge d{x}^{\lambda }\wedge d{x}^{\mu }\wedge d{x}^{\nu }$. Terms with $\kappa =\lambda$ vanish by alternation; for each $\kappa \mathrm{\ne }\lambda$, exchanging the derivative indices leaves the coefficient unchanged by [F6] and reverses the wedge sign. Thus all terms cancel and $d\mathcal{J}=d(dG)=0$. From the definition $\mathcal{J}={\rho }_{f}vo{l}_3-dt\wedge {\mathcal{J}}_{\mathrm{s}\mathrm{p}}$ in [F1], its coefficient along $dt\wedge vo{l}_3$ is ${\mathrm{\partial }}_t{\rho }_f+\mathrm{div}{J}_f$; setting this coefficient to zero gives the stated continuity equation.

Take $t_0<t_1$ in $I$ and let $e_0,e_1,e_2$ be the ordered vertices of ${\mathrm{\Delta }}^2$. The product prism has the oriented triangulation $P={\tau }_0-{\tau }_1+{\tau }_2$, with ${\tau }_0=[(t_0,e_0),(t_1,e_0),(t_1,e_1),(t_1,e_2)]$, ${\tau }_1=[(t_0,e_0),(t_0,e_1),(t_1,e_1),(t_1,e_2)]$, and ${\tau }_2=[(t_0,e_0),(t_0,e_1),(t_0,e_2),(t_1,e_2)]$. In coordinates $(t,u^1,u^2)$, their ordered vertex determinants are $t_1-t_0,-(t_1-t_0),t_1-t_0$, so the coefficients orient all three simplices consistently. Applying the signed-face formula [F7] cancels their internal faces and gives $\mathrm{\partial }P={\iota }_{t_1\mathrm{\#}}{\mathrm{\Delta }}^2-{\iota }_{t_0\mathrm{\#}}{\mathrm{\Delta }}^2-P(\mathrm{\partial }{\mathrm{\Delta }}^2)$, where ${\iota }_t(u)=(t,u)$ and $P(\mathrm{\partial }{\mathrm{\Delta }}^2)=P([e_1,e_2])-P([e_0,e_2])+P([e_0,e_1])$, with $P([e_a,e_b])=[(t_0,e_a),(t_1,e_a),(t_1,e_b)]-[(t_0,e_a),(t_0,e_b),(t_1,e_b)]$. For each singular simplex $\sigma$ in $z$, compose these parameter maps with $q_{\sigma }(t,u)=(t,\sigma (u))$. The top and bottom chains become the time-slice copies of $z$; after summing with the real coefficients of $z$, side terms cancel because $\mathrm{\partial }z=0$. Each composite $q=q_{\sigma }\circ {\tau }_i$ is a smooth singular 3-simplex into $U$: the affine parameter maps and the smooth-simplex extensions of $\sigma$ give an extension to a neighbourhood, which may be restricted so its image lies in the open set $U$. Hence $q^{\ast }F$ is a $C^1$ two-form on a neighbourhood of its standard parameter simplex. Write $F=\frac{1}{2}{F}_{\mu \nu }(y)d{y}^{\mu }\wedge d{y}^{\nu }$ in laboratory coordinates. By [F10], the pullback coefficients are ${\eta }_{ij}=F_{\mu \nu }(q){\mathrm{\partial }}_i{q}^{\mu }{\mathrm{\partial }}_j{q}^{\nu }$, so $(d\eta {)}_{kij}={\mathrm{\partial }}_k{\eta }_{ij}+{\mathrm{\partial }}_i{\eta }_{jk}+{\mathrm{\partial }}_j{\eta }_{ki}$. By [F9], the terms differentiating $F_{\mu \nu }$ are $({\mathrm{\partial }}_{\lambda }{F}_{\mu \nu })(q)({\mathrm{\partial }}_k{q}^{\lambda }{\mathrm{\partial }}_i{q}^{\mu }{\mathrm{\partial }}_j{q}^{\nu }+{\mathrm{\partial }}_i{q}^{\lambda }{\mathrm{\partial }}_j{q}^{\mu }{\mathrm{\partial }}_k{q}^{\nu }+{\mathrm{\partial }}_j{q}^{\lambda }{\mathrm{\partial }}_k{q}^{\mu }{\mathrm{\partial }}_i{q}^{\nu })$, which is exactly $(dF{)}_{\lambda \mu \nu }(q){\mathrm{\partial }}_k{q}^{\lambda }{\mathrm{\partial }}_i{q}^{\mu }{\mathrm{\partial }}_j{q}^{\nu }$. The remaining terms are $F_{\mu \nu }(q)({\mathrm{\partial }}_{ki}{q}^{\mu }{\mathrm{\partial }}_j{q}^{\nu }+{\mathrm{\partial }}_i{q}^{\mu }{\mathrm{\partial }}_{kj}{q}^{\nu }+{\mathrm{\partial }}_{ij}{q}^{\mu }{\mathrm{\partial }}_k{q}^{\nu }+{\mathrm{\partial }}_j{q}^{\mu }{\mathrm{\partial }}_{ik}{q}^{\nu }+{\mathrm{\partial }}_{jk}{q}^{\mu }{\mathrm{\partial }}_i{q}^{\nu }+{\mathrm{\partial }}_k{q}^{\mu }{\mathrm{\partial }}_{ji}{q}^{\nu })$. The first and fourth cancel after interchanging $\mu ,\nu$ and using ${\mathrm{\partial }}_{ki}q={\mathrm{\partial }}_{ik}q$; the second and fifth cancel by ${\mathrm{\partial }}_{kj}q={\mathrm{\partial }}_{jk}q$; the third and sixth cancel by ${\mathrm{\partial }}_{ij}q={\mathrm{\partial }}_{ji}q$. Each cancellation uses $F_{\mu \nu }=-F_{\nu \mu }$ and [F6]. Thus $d(q^{\ast }F)=q^{\ast }(dF)=0$ by step 2.1. The $C^1$ Stokes identity [F8] applies to each pullback on the standard parameter 3-simplex, so the boundary integrals sum to zero. The top and bottom terms are ${\mathrm{\Phi }}_z(t_1)$ and $-{\mathrm{\Phi }}_z(t_0)$ because the time-slice pullback of $F=\mathcal{B}-dt\wedge \mathcal{E}$ is $\mathcal{B}$ by [F1]. Finite-chain linearity in [F1] gives ${\mathrm{\Phi }}_z(t_1)-{\mathrm{\Phi }}_z(t_0)=0$. Hence the cycle flux is constant.