Manuscript Em01-D@2 · Em01-D-v2.tex · lemma · d3b49151

Lemma: Exact diagnostic difference

Exact nonphysical eight-variable diagnostic › Difference of complete families · lines 1957–2001 · proof included or linked

Em01-D@2#lem:diagnosticdifference, Em01-D@2#eq:diagnosticdifference

Back to manuscript

Put

s = κ (y - a), t = κ (b - a) .

Then

u_{δ} (Φ) = i ⟨ R (b - a) δ_{D}^{(4)} (a - z), \partial_{b^{0}}^{2} [δ^{- 3} p (s / δ) p^{'} ((s - t) / δ) Φ_{I_{*}} (a, b, z, y)] ⟩ .

Proof. The kernel difference is

K_{6}^{I_{*}, I_{*}} (D_{1}) - K_{6}^{I_{*}, I_{*}} (D_{0}) = - \partial_{b^{0}}^{2} R (b - a) δ_{D}^{(4)} (a - z) .

The derivative $\partial_{a^{0}}$ in $K_{6}$ acts on $R$ alone, not on the incidence.

For (eq:diagnosticC), the kinetic/trace-kinetic combination is

Z_{I_{*}}^{ν} (C_{δ}) = - ℓ^{ν} δ^{- 3} p (s / δ) p^{'} ((s - t) / δ) .

The kinetic tensor contraction gives $ℓ^{ν}$ , while the trace contribution subtracts $2 ℓ^{ν}$ . The derivative along $ℓ$ annihilates both profile arguments and the $y$ -independent internal kernel. Since $ℓ^{ν} \partial_{ν} g = 1$ , the output divergence leaves exactly the stated product. Transposing the two $b^{0}$ -derivatives proves (eq:diagnosticdifference).

Displayed formulas in this passage

Em01-D@2#eq:diagnosticdifference · line 1964
$u_{δ} (Φ) = i ⟨ R (b - a) δ_{D}^{(4)} (a - z), \partial_{b^{0}}^{2} [δ^{- 3} p (s / δ) p^{'} ((s - t) / δ) Φ_{I_{*}} (a, b, z, y)] ⟩ .$
unlabeled · line 1960
$s = κ (y - a), t = κ (b - a) .$
unlabeled · line 1979
$K_{6}^{I_{*}, I_{*}} (D_{1}) - K_{6}^{I_{*}, I_{*}} (D_{0}) = - \partial_{b^{0}}^{2} R (b - a) δ_{D}^{(4)} (a - z) .$
unlabeled · line 1989
$Z_{I_{*}}^{ν} (C_{δ}) = - ℓ^{ν} δ^{- 3} p (s / δ) p^{'} ((s - t) / δ) .$

Nearby in this section

section: Difference of complete families