Proposition: Sourced first coefficient, diagnostic only

Let $m=0$, ${\varphi }_0=t$, $h_{\mu \nu }=\psi {\eta }_{\mu \nu }$ with real $\psi \in {C_c}^{\infty }({M_T}^{\circ })$, and $k=0$. Then ${a_1}^{\mu \nu }[h]=\psi {\eta }^{\mu \nu }$, $b_1[h]=2\psi$,

and the first relational coefficient (eq:A1) is

If $\mathrm{supp}\psi \subset \{t\ge \epsilon \}\setminus \mathrm{supp}g$ and ${\partial }_t\psi \not\equiv 0$, then ${A_1}^{(g,\epsilon )}={\mathsf{E}}_0{\partial }_t\psi \ne 0$ and there is $f$ with $\int f{A}_1^{(g,\epsilon )}{\mathrm{d}}^{4}x\ne 0$. If $\mathrm{supp}\psi \subset P_g\cap \{t\ge \epsilon \}$, then ${A_1}^{(g,\epsilon )}=0$ although ${\rho }_{1,g}$ and ${\mathsf{X}}_1$ are individually nonzero whenever ${\partial }_t\psi \not\equiv 0$.

Proof. $\square t=0$, so ${\varphi }_0$ is free. From (eq:V1g) with ${\partial }_{\nu }t={{\delta }^0}_{\nu }$ and ${\eta }^{\mu 0}={{\delta }^{\mu }}_0$, ${\mathsf{V}}_{1,g}[h]{\varphi }_0={\partial }_{\mu }(g\psi {\eta }^{\mu 0})={\partial }_t(g\psi )$. From (eq:X1eq), $C^{\mu }[\psi \eta ]={\partial }^{\mu }\psi -2{\partial }^{\mu }\psi =-{\partial }^{\mu }\psi$ and ${a_1}^{0\mu }=\psi {{\delta }^{\mu }}_0$, so $\square {{\mathsf{X}}_1}^0=-{\sigma }_{\epsilon }{\partial }_t\psi -{\sigma }_{\epsilon }^{\prime }\psi =-{\partial }_t({\sigma }_{\epsilon }\psi )$ and $\square {{\mathsf{X}}_1}^i=-{\sigma }_{\epsilon }{\partial }^i\psi ={\sigma }_{\epsilon }{\partial }_i\psi$. Then $A_1={\rho }_{1,g}-{{\mathsf{X}}_1}^{\mu }{\partial }_{\mu }t={\rho }_{1,g}-{{\mathsf{X}}_1}^0$, which is (eq:A1diag). In the first case $g=0$ and ${\sigma }_{\epsilon }=1$ on $\mathrm{supp}\psi$, so the source is ${\partial }_t\psi \ne 0$ and $A_1\ne 0$ because $\square A_1\ne 0$; a bump $f$ where $A_1$ has fixed sign gives a nonzero smearing. In the second case $g={\sigma }_{\epsilon }=1$ on $\mathrm{supp}\psi$.