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Positive-parameter normalized time and spatial ends

Statement

Use the same family, including the contraction $W_{q} (T) = d_{q} - D_{q}$ , and let $q > 0$ . The exact coordinate change $t_{q} = \frac{d_{q}}{4 q} t, x_{q} = q x, r_{q} = q r$ gives $(g_{q})_{t_{q} t_{q}} = - 1 + O_{5} (r_{q}^{- 1}), (g_{q})_{t_{q} x_{q}^{i}} = O_{5} (r_{q}^{- 2}), (g_{q})_{x_{q}^{i} x_{q}^{j}} = δ_{i j} + O_{5} (r_{q}^{- 1}) .$ The constants and a common end radius may be chosen uniformly on $q \in [δ, 1]$ for any fixed $δ > 0$ . No uniform normalized end at $q = 0$ is asserted. Since $d_{q}$ is bounded above and away from zero, $t_{q}$ has the scale $t / q$ (meaning that its coefficient relative to $t / q$ is bounded above and below by positive constants). A fixed normalized radius $r_{q} = R$ corresponds to original radius $r = R / q$ , which tends to infinity as $q ↓ 0$ .

Facts & Assumptions

[F1]

Differentiated asymptotic bounds and parameter uniformity defines fixed-chart differentiated bounds and parameter uniformity. Mass-one Ernst decay and uniform Harrison denominator bounds gives the denominator expansion, the pole constant bounds and the mass-one potential and metric estimates.

[F2]

The differentiated spatial decay supplied by the Harrison lift and Uniform spatial components and the cylinder scale give the spatial one-form and metric estimates.

[F3]

Stationary Harrison one-form and metric gives the tensor and the contraction condition; Mass-one convention and Harrison denominator fixes the pole-evaluation normalization. The time coordinate is normalized using that same $d_{q}$ , not an independently chosen constant.

Proof

**Given:**The fixed-parameter family with the normalized contraction.

Goal: Establish the exact time scaling and differentiated normalized end estimates.

1.1F1F2F3givenalgebra

In the original chart put $a_{i} = g_{0 i}$ and $W_{t} = d_{q} - D_{q}$ . Expanding the Harrison tensor gives $g_{q} (T, T) = d_{q}^{2} g_{00} / (4 D_{q})$ and $(g_{q})_{0 i} = \frac{d_{q}}{4 D_{q}} (D_{q} a_{i} - f (W_{q})_{i})$ . This is also obtained from the pullback formula and $A_{q} T = (d_{q} / D_{q}) T$ . For $q \in [δ, 1]$ , [F1] yields $D_{q} = 4 q^{2} + O_{5} (r^{- 1})$ and $D_{q}^{- 1} = (4 q^{2})^{- 1} + O_{5} (r^{- 1})$ , with constants depending on $δ$ . Hence $(g_{q})_{00} = - (d_{q} / (4 q))^{2} + O_{5} (r^{- 1})$ and $(g_{q})_{0 i} = O_{5} (r^{- 2})$ . The spatial block is $q^{2} δ_{i j} + O_{5} (r^{- 1})$ by [F2].

2.1F1step 1.1algebra∎

The coordinate vector factors are $\partial_{t_{q}} = (4 q / d_{q}) \partial_{t}$ and $\partial_{x_{q}^{i}} = q^{- 1} \partial_{x^{i}}$ . Insert these into the three blocks from step 1.1. The leading temporal and spatial coefficients become $- 1$ and $δ_{i j}$ . A derivative of order $k$ in $x_{q}$ gives $q^{- k}$ times the original spatial derivative, and $r = r_{q} / q$ ; thus a remainder of order $r^{- b - k}$ becomes $q^{b} r_{q}^{- b - k}$ before the block's constant coordinate factors. All these factors are controlled on $[δ, 1]$ . This proves the stated $O_{5}$ estimates. The scale statement follows from the $d_{q}$ bounds in [F1], and the radius identity follows directly from $r_{q} = q r$ . Neither computation permits putting $q = 0$ into these coordinates.

Depends on

Used by · 1

Introductory Harrison asymptotic claims

Dependency tree · 2 levels

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