Lemma: Scalar–lapse separation

The harmonic relational scalar ${\mathcal{R}}_{\mathrm{H}}$ and the scalar-clock relational lapse ${\mathcal{N}}_{\mathrm{E}}$ are different functional prescriptions.

More concretely, on a common flat, homogeneous comparison domain:

1. ${\mathcal{R}}_{\mathrm{H}}$ and ${\mathcal{N}}_{\mathrm{E}}$ are not equal as functionals;

2. at a fixed point, ${\mathcal{N}}_{\mathrm{E}}$ does not factor through the scalar value ${\mathcal{R}}_{\mathrm{H}}$ alone;

3. original spatial rods are not the relational scalar.

These statements do not exclude identities on specially restricted families or after additional, explicitly specified constructions.

Proof. Take $q=\eta$. The harmonic map of Definition def:harmonic is ${\mathsf{H}}_q=\mathrm{id}$. For any homogeneous clock $T(t)$ with $\dot{T}>0$, the original spatial rods are ${Y_T}^i=x^i$. Choose the clock itself as the background reference, so the relational map is the identity. Then

For $T(t)=ct+b$ with $c>0$, these are $ct+b$ and $1/c$. They are not the same functional.

For the factorization claim, fix a background clock $T_0(t)=ct+b$, a point $p=(t_c,{\mathbf{x}}_c)$, and a smooth time bump $\chi$ equal to one near $t_c$. Compare

For sufficiently small $e$, these are timelike clocks. At $p$ they all have the same scalar value,

The relational inverse sends $p$ to $p$ in time because the clock values agree there. Spatial rods remain $x^i$. Thus

The lapse varies with $a$ on a fiber of the scalar-value map. Lemma lem:factor excludes factorization through that value.

Finally, on the homogeneous flat family the scalar is independent of $\mathbf{x}$, whereas ${Y_T}^i=x^i$ is not. Even componentwise, they cannot be equal as fields on an open spatial set.