Manuscript XX-E@1 · XX-E.tex · lemma · cf7a8f40

Lemma: Gauge displacement of the original first rod

The evaluated classical Ward identity and cutoff breaking › The actual compact geometric gauge group · lines 514–553 · proof included or linked

XX-E@1#lem:rodgauge, XX-E@1#eq:rodgauge

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For every admitted compact $ξ$ ,

ℓ (K ξ) = 0, R (K ξ)^{a} = ξ^{a} .

Proof. The metric and scalar gauge coefficients are

(K ξ)_{00} = - 2 ξ ˙^{0}, (K ξ)_{ϕ} = \frac{A}{t} ξ^{0},

(K ξ)_{ij} = 2 a^{2} H ξ^{0} δ_{ij} + a^{2} (\partial_{i} ξ^{j} + \partial_{j} ξ^{i}) .

Substitution in (eq:ell) gives zero.

The clock displacement $u_{1} = - ξ^{0}$ cancels the $2 a^{2} H ξ^{0}$ term in the induced metric. Consequently

(C K ξ)^{ia} = δ^{ia} \partial_{j} ξ^{j} - \partial_{i} ξ^{a} - \partial_{a} ξ^{i},

and

\partial_{i} (C K ξ)^{ia} = - Δ ξ^{a} .

Equation (eq:R) gives $R (K ξ)^{a} = G_{sp} Δ ξ^{a} = ξ^{a}$ . The equality uses compact spatial support and the unchanged decaying Newton boundary condition.

Displayed formulas in this passage

XX-E@1#eq:rodgauge · line 517
$ℓ (K ξ) = 0, R (K ξ)^{a} = ξ^{a} .$
unlabeled · line 526
$(K ξ)_{00} = - 2 ξ ˙^{0}, (K ξ)_{ϕ} = \frac{A}{t} ξ^{0},$
unlabeled · line 530
$(K ξ)_{ij} = 2 a^{2} H ξ^{0} δ_{ij} + a^{2} (\partial_{i} ξ^{j} + \partial_{j} ξ^{i}) .$
unlabeled · line 540
$(C K ξ)^{ia} = δ^{ia} \partial_{j} ξ^{j} - \partial_{i} ξ^{a} - \partial_{a} ξ^{i},$
unlabeled · line 546
$\partial_{i} (C K ξ)^{ia} = - Δ ξ^{a} .$

Lemma: Gauge displacement of the original first rod

Displayed formulas in this passage

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