Proposition: First rod variations and a detected quadratic difference

Let $Y_1$ and $Z_1$ be the first original and modified rod variations. Then

and the quadratic relational lapses satisfy

A nonnegative compact spacetime test detects a strictly negative pairing.

Proof. For ${\gamma }_{ij}=a^2(1+2\epsilon D){\delta }_{ij}$, expansion of

gives ${\Delta }_{\gamma }{x}^i=\epsilon a^{-2}{\partial }_{i}D+O({\epsilon }^2)$. Hence

whose decaying solution is $b{\partial }_i\psi$.

A modified slice through background time $\tau$ has $t=\tau -\epsilon D/H+O({\epsilon }^2)$. The change in $a(t{)}^2$ cancels the first-order factor $2\epsilon D$. Terms involving spatial derivatives of the embedding are second order. Thus its induced spatial metric is $a(\tau {)}^2{\delta }_{ij}+O({\epsilon }^2)$, and the decaying first harmonic correction is ${Z_1}^i=0$.

The common raw lapse is

The unchanged time coordinate gives $u_1=u_2=0$, $n_1=A=b\psi$, and the same raw $N_2$ for both prescriptions. Equation (eq:n2) therefore gives

Choose a nonnegative compact test supported where $b\ne 0$ and $\nabla \psi \ne 0$, positive on a nonempty open set. Its pairing is strictly negative.