Manuscript X-B-ST@1 · X-B-ST.tex · section · d03c150c

The one inherited scalar vertex

The frozen classical handoff › The one inherited scalar vertex · lines 519–557

X-B-ST@1#eq:vertex, X-B-ST@1#eq:Lone, X-B-ST@1#eq:switchterms, X-B-ST@1#eq:Hessians

The only interaction vertex used in the response construction is

V_{1, g} [h; ψ] L_{1, g} [h] u = \frac{1}{2} \int g (a_{1}^{μν} [h] \partial_{μ} ψ \partial_{ν} ψ - m^{2} b_{1} [h] ψ^{2}) d^{4} x = - \frac{1}{2} ⟨ ψ, L_{1, g} [h] ψ ⟩, = \partial_{μ} (g a_{1}^{μν} [h] \partial_{ν} u) + g m^{2} b_{1} [h] u .

In particular,

L_{1, g} [h] u = g (\partial_{μ} (a_{1}^{μν} [h] \partial_{ν} u) + m^{2} b_{1} [h] u) + (\partial_{μ} g) a_{1}^{μν} [h] \partial_{ν} u .

The derivative of $g$ is part of the operator.

Formal self-adjointness follows by compactly supported integration by parts. The sign ledger is

S_{0}^{''} = - P_{m}, V_{1, g}^{'} = - L_{1, g} [h] ψ, V_{1, g}^{''} = - L_{1, g} [h] .

The positive Euler operator $P_{m}$ is not the variational Hessian $S_{0}^{''}$ .

Displayed formulas in this passage

X-B-ST@1#eq:vertex · line 523
$V_{1, g} [h; ψ] L_{1, g} [h] u = \frac{1}{2} \int g (a_{1}^{μν} [h] \partial_{μ} ψ \partial_{ν} ψ - m^{2} b_{1} [h] ψ^{2}) d^{4} x = - \frac{1}{2} ⟨ ψ, L_{1, g} [h] ψ ⟩, = \partial_{μ} (g a_{1}^{μν} [h] \partial_{ν} u) + g m^{2} b_{1} [h] u .$
X-B-ST@1#eq:Lone · line 523
$V_{1, g} [h; ψ] L_{1, g} [h] u = \frac{1}{2} \int g (a_{1}^{μν} [h] \partial_{μ} ψ \partial_{ν} ψ - m^{2} b_{1} [h] ψ^{2}) d^{4} x = - \frac{1}{2} ⟨ ψ, L_{1, g} [h] ψ ⟩, = \partial_{μ} (g a_{1}^{μν} [h] \partial_{ν} u) + g m^{2} b_{1} [h] u .$
X-B-ST@1#eq:switchterms · line 537
$L_{1, g} [h] u = g (\partial_{μ} (a_{1}^{μν} [h] \partial_{ν} u) + m^{2} b_{1} [h] u) + (\partial_{μ} g) a_{1}^{μν} [h] \partial_{ν} u .$
X-B-ST@1#eq:Hessians · line 548
$S_{0}^{''} = - P_{m}, V_{1, g}^{'} = - L_{1, g} [h] ψ, V_{1, g}^{''} = - L_{1, g} [h] .$