Remark: Einstein-shell pairs are blind to the 0ν components

If both polarizations are de Donder transverse, $k_a\cdot \mathcal{P}{ϵ}_a=0$, then $A({ϵ}_a{e}^{i{k}_a\cdot y})=0$ by (eq:algebra), and (eq:bianchi) gives $(k_1+k_2{)}_{\mu }{\mathsf{q}}^{\mu \nu }=0$. For $k_1+k_2=2\mathrm{d}t$ this is ${\mathsf{q}}^{0\nu }=0$: linearized Einstein pairs see only ${\beta }_{ij}$, which vanishes for the F1 data by (eq:F1data). The pure-trace polarization $ϵ=\eta$ has $k\cdot \mathcal{P}\eta =-k\ne 0$; it lies on XI-E's gauge-fixed shell $\square \gamma =0$ and off the Einstein shell. This is XI-A's distinction between the two shells, met at the level of the detector.