Smooth source/dust scaling and test-backreaction obstructions

By [F4], exterior differentiation is real-linear, so $d{F}_{\delta }=\delta d{F}_0=0$. The connection and metric raising in [F1] are linear in the field strength, so ${\mathrm{\nabla }}_a(F_{\delta }{)}^{ab}=\delta {\mathrm{\nabla }}_a(F_0{)}^{ab}=0$. Every term in the Maxwell tensor is quadratic in $F$, hence $T_{\mathrm{E}\mathrm{M}}(g,F_{\delta })={\delta }^2{T}_{\mathrm{E}\mathrm{M}}(g,F_0)$.

The definitions in [F1] give $T_m(\eta \rho ,U)=\eta T_m(\rho ,U)$ and $j(\eta \sigma ,U)=\eta j(\sigma ,U)$. On $\{\rho >0\}$, $(\eta \sigma )\mathrm{/}(\eta \rho )=\sigma \mathrm{/}\rho$. The force equation in [F3] therefore has both sides multiplied by $\eta$ if $F$ and $U$ are held fixed.

At any point where $F_0\mathrm{\ne }0$, choose a future unit observer and its positively oriented orthonormal rest frame. Then $E$ and $B$ cannot both vanish, since the field components are $F_{0i}=-E_i\mathrm{/}c$ and $F_{ij}={ϵ}_{ijk}{B}_k$ by [F1]. With ${\mu }_0>0$, the density formula [F2] gives $u>0$. Minkowski space has $G_{ab}=0$; with $\mathrm{\Lambda }=0$ and $T_m=0$, the Einstein equation would require $0=\kappa T_{ab}^{\mathrm{E}\mathrm{M}}$, contradicted by this positive observer contraction and $\kappa >0$ from [F1].

The Coulomb energy density on ${\mathcal{A}}_{\mathrm{\ell }}$ is $\frac{{\epsilon }_0}{2}\mathrm{\mid }\mathbf{E}{\mathrm{\mid }}^2=\frac{Q^2}{32{\pi }^2{\epsilon }_0{r}^4}$ by [F2] and ${\epsilon }_0=({\mu }_0{c}^2{)}^{-1}$ from [F1], where $r=\mathrm{\mid }\mathbf{x}\mathrm{\mid }$. With $d^{3}x$ denoting Lebesgue measure in Cartesian coordinates, define the nonnegative Borel function $h_{\mathrm{\ell }}:{\mathbb{R}}^3\to [0,\mathrm{\infty })$ to equal $Q^2\mathrm{/}(32{\pi }^2{\epsilon }_0\mathrm{\mid }\mathbf{x}{\mathrm{\mid }}^4)$ when $\mathrm{\ell }<\mathrm{\mid }\mathbf{x}\mathrm{\mid }<2\mathrm{\ell }$ and $0$ otherwise. Then ${\mathcal{E}}_{\mathrm{\ell }}={\int }_{{\mathbb{R}}^3}{h}_{\mathrm{\ell }}{d}^{3}x$, and [F5] gives ${\mathcal{E}}_{\mathrm{\ell }}=4\pi {\int }_{\mathrm{\ell }}^{2\mathrm{\ell }}\frac{Q^2}{32{\pi }^2{\epsilon }_0{r}^4}{r}^{2}dr=\frac{Q^2}{8\pi {\epsilon }_0}{\int }_{\mathrm{\ell }}^{2\mathrm{\ell }}{r}^{-2}dr\mathrm{.}$ The field profile is used only on the finite annulus.

By the quadratic formula in [F1] and step 1.1, $T_{ab}^{\mathrm{E}\mathrm{M}}(g,F_{\delta })={\delta }^2{T}_{ab}^{\mathrm{E}\mathrm{M}}(g,F_0)$. Substituting this identity and the assumed vacuum equation into the displayed definition of $\mathcal{R}$ gives ${\mathcal{R}}_{ab}(g,F_{\delta };\mathrm{\Lambda })=-\kappa {\delta }^2{T}_{ab}^{\mathrm{E}\mathrm{M}}(g,F_0)$, as claimed.

The sourced Maxwell equation is ${\mathrm{\nabla }}_a{F}^{ab}=-{\mu }_0{j}^b$ by [F1]; replacing $j$ by $\eta j$ while leaving $F$ fixed changes its right-hand side to $-\eta {\mu }_0{j}^b$, so it remains the same equation only when $(\eta -1)j=0$. If instead $(F,j)\mapsto (\eta F,\eta j)$, the Maxwell equation is preserved by linearity, but step 1.1 scales $T_{\mathrm{E}\mathrm{M}}$ quadratically while step 1.2 scales $T_m$ linearly; the dust force equation then also has its left side scale by $\eta$ and its right side by ${\eta }^2$. These unequal degrees show why this scaling is not a general coupled-solution construction, while leaving special cancellations open.

Because $\mathrm{\ell }>0$, the function $r\mapsto r^{-2}$ is continuous on $[\mathrm{\ell },2\mathrm{\ell }]$ by [F6] and hence is integrable there. The primitive $-r^{-1}$ has derivative $r^{-2}$ on this interval by [F6], so the fundamental theorem of calculus gives ${\int }_{\mathrm{\ell }}^{2\mathrm{\ell }}{r}^{-2}dr={\mathrm{\ell }}^{-1}-(2\mathrm{\ell }{)}^{-1}=(2\mathrm{\ell }{)}^{-1}$. Hence ${\mathcal{E}}_{\mathrm{\ell }}=Q^2\mathrm{/}(16\pi {\epsilon }_0\mathrm{\ell })$. For every $M>0$, choosing $0<\mathrm{\ell }<Q^2\mathrm{/}(16\pi {\epsilon }_{0}M)$ gives ${\mathcal{E}}_{\mathrm{\ell }}>M$; therefore the shell energies diverge as $\mathrm{\ell }\downarrow 0$ at fixed $Q\mathrm{\ne }0$.