Manuscript X-A-conv@1 · X-A-conv.tex · proposition · c76fffa3

Proposition: A nonzero derived source

A nonzero forced scalar example and a symbolic diagnostic · lines 2049–2101 · proof included or linked

X-A-conv@1#prop:sourced, X-A-conv@1#eq:nonzerosource, X-A-conv@1#eq:sourcedresponse, X-A-conv@1#eq:sourcedsecond

For (eq:conformaltest),

a_{1} (h)^{μν} = ψ η^{μν}, a_{2} (h)^{μν} = 0,

and

V_{1, g} [h] ρ_{0} = \partial_{t} ψ \neq \equiv 0.

The selected zero-data correction is

ρ_{1, g} = - E_{0} \partial_{t} ψ = - \partial_{t} E_{0} ψ,

and it is not identically zero. The second correction is

ρ_{2, g} = E_{0} V_{1, g} [h] E_{0} \partial_{t} ψ .

Proof. Here $tr h = 4 ψ$ and $h_{α β} h^{α β} = 4 ψ^{2}$ . Equations (eq:aone)–(eq:atwo) give $a_{1} (h) = ψ η$ and $a_{2} (h) = 0$ . Since $\partial_{ν} t = δ_{ν}^{0}$ ,

\partial_{μ} (ψ η^{μν} \partial_{ν} t) = \partial_{t} ψ .

The chosen bump has nonconstant time dependence, proving (eq:nonzerosource).

Equation (eq:rhoone) gives the first formula in (eq:sourcedresponse). The source $ψ$ vanishes near $Σ$ , so $\partial_{t} E_{0} ψ$ has zero value and first time derivative there and solves $□_{η} (\partial_{t} E_{0} ψ) = \partial_{t} ψ$ . Uniqueness gives the second formula. If $ρ_{1, g} = 0$ , its equation would imply $\partial_{t} ψ = 0$ , a contradiction. Equation (eq:rhotwo) and $V_{2, g} [h] = 0$ give (eq:sourcedsecond).

Displayed formulas in this passage

X-A-conv@1#eq:nonzerosource · line 2057
$V_{1, g} [h] ρ_{0} = \partial_{t} ψ \neq \equiv 0.$
X-A-conv@1#eq:sourcedresponse · line 2062
$ρ_{1, g} = - E_{0} \partial_{t} ψ = - \partial_{t} E_{0} ψ,$
X-A-conv@1#eq:sourcedsecond · line 2071
$ρ_{2, g} = E_{0} V_{1, g} [h] E_{0} \partial_{t} ψ .$
unlabeled · line 2052
$a_{1} (h)^{μν} = ψ η^{μν}, a_{2} (h)^{μν} = 0,$
unlabeled · line 2085
$\partial_{μ} (ψ η^{μν} \partial_{ν} t) = \partial_{t} ψ .$

Proposition: A nonzero derived source

Displayed formulas in this passage

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