Manuscript III@2 · III.tex · proposition · 82dad3e5

Proposition: Nonzero-total-source tails

What instantaneous Poisson compensation establishes › A separate source-support fact · lines 2502–2532 · proof included or linked

III@2#prop:source-support

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If $s \in C_{c}^{\infty} (R^{3})$ has $M_{s} = \int s \neq = 0$ , no smooth compact symmetric tensor solves

R_{lin} [h] = s .

A decaying solution is

h_{ij} = δ_{ij} H, H = - \frac{1}{2} N s,

with

H (x) = \frac{M _{s}}{8 π ∣ x ∣} + O (∣ x ∣^{- 2}) .

The expression $R_{lin}$ vanishes on spatial TT tensors and is unchanged by spatial pure gauges.

Proof. Integrating second derivatives of compact $h$ gives zero. For $h_{ij} = δ_{ij} H$ , $R_{lin} [h] = - 2Δ H = s$ . The tail is the leading Newton multipole. The TT statement is immediate. For a spatial gauge addition,

\partial_{i} \partial_{j} (\partial_{i} ξ_{j} + \partial_{j} ξ_{i}) - Δ (2 \partial_{i} ξ_{i}) = 0.

Displayed formulas in this passage

unlabeled · line 2506
$R_{lin} [h] = s .$
unlabeled · line 2510
$h_{ij} = δ_{ij} H, H = - \frac{1}{2} N s,$
unlabeled · line 2514
$H (x) = \frac{M _{s}}{8 π ∣ x ∣} + O (∣ x ∣^{- 2}) .$
unlabeled · line 2528
$\partial_{i} \partial_{j} (\partial_{i} ξ_{j} + \partial_{j} ξ_{i}) - Δ (2 \partial_{i} ξ_{i}) = 0.$

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section: A separate source-support fact