Manuscript XI-C@1 · XI-C.tex · theorem · acec1422

Theorem: Double-retarded normalization

Normalization to the de Donder violation · lines 784–823 · proof included or linked

XI-C@1#thm:normalization, XI-C@1#eq:Dresponse, XI-C@1#eq:double

For every smooth free scalar $ϕ_{0}$ ,

D r_{g} [ϕ_{0}] X_{1} [r_{g} [ϕ_{0}]] = - \frac{1}{2} E_{0}^{R} ((\partial_{ν} g) T_{0}^{ν \cdot} [ϕ_{0}]), = - \frac{1}{2} E_{0}^{R} E_{0}^{R} ((\partial_{ν} g) T_{0}^{ν \cdot} [ϕ_{0}]) .

The complete vector in (eq:double) vanishes identically on the slab if and only if the switched free source is conserved:

(\partial_{ν} g) T_{0}^{ν μ} [ϕ_{0}] = 0 for every μ .

Proof. Trace reversal is involutive, so $P r_{g} = - \frac{1}{2} E_{0}^{R} (g T_{0})$ . Differentiation commutes with the retarded inverse on these compact interior sources. Taking the divergence and using $\partial_{ν} T_{0}^{ν μ} [ϕ_{0}] = 0$ proves (eq:Dresponse).

Retardation makes $r_{g}$ and $D r_{g}$ zero below $δ$ . Thus the layer term in (eq:coordinate) is zero, and $σ_{ε} D r_{g} = D r_{g}$ wherever this response can be nonzero. Consequently $X_{1} [r_{g}] = E_{0}^{R} D r_{g}$ , proving (eq:double).

Applying $□^{2}$ to (eq:double) recovers minus one half of the displayed source. Zero output therefore implies zero source. Conversely, zero source gives zero response by the zero-data inverse.

Displayed formulas in this passage

XI-C@1#eq:Dresponse · line 787
$D r_{g} [ϕ_{0}] X_{1} [r_{g} [ϕ_{0}]] = - \frac{1}{2} E_{0}^{R} ((\partial_{ν} g) T_{0}^{ν \cdot} [ϕ_{0}]), = - \frac{1}{2} E_{0}^{R} E_{0}^{R} ((\partial_{ν} g) T_{0}^{ν \cdot} [ϕ_{0}]) .$
XI-C@1#eq:double · line 787
$D r_{g} [ϕ_{0}] X_{1} [r_{g} [ϕ_{0}]] = - \frac{1}{2} E_{0}^{R} ((\partial_{ν} g) T_{0}^{ν \cdot} [ϕ_{0}]), = - \frac{1}{2} E_{0}^{R} E_{0}^{R} ((\partial_{ν} g) T_{0}^{ν \cdot} [ϕ_{0}]) .$
unlabeled · line 800
$(\partial_{ν} g) T_{0}^{ν μ} [ϕ_{0}] = 0 for every μ .$

Theorem: Double-retarded normalization

Displayed formulas in this passage

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