Manuscript VI@1 · VI.tex · section · b5b60afb

The compactly switched minimal scalar vertex

Bare dressing, solution coefficients, and external response › The compactly switched minimal scalar vertex · lines 1247–1297

VI@1#eq:matter-vertex, VI@1#eq:Q, VI@1#eq:QL, VI@1#eq:Q-switch, VI@1#eq:external-response

Fix

g \in C_{c}^{\infty} ((0, T) \times R^{3}; R) .

For a smooth external tensor $h$ , the matter vertex is

V_{1}^{m} (g; h, φ) = \frac{1}{2} \int g h^{μν} T_{μν} (η, φ) d^{4} x .

For $H = g h$ , define

Q_{H} = - \partial_{μ} (H ˉ^{μν} \partial_{ν}) - \frac{1}{2} m^{2} tr_{η} H .

It is the scalar Hessian contribution of (eq:matter-vertex), and it is formally self-adjoint. Direct expansion gives

Q_{H} = L_{H} + \frac{1}{2} (tr_{η} H) P .

Hence, on a free scalar,

Q_{g h} ϕ_{0} = g L_{h} ϕ_{0} - (\partial_{μ} g) h ˉ^{μν} \partial_{ν} ϕ_{0} .

Let $G_{R}$ be the retarded inverse of $P$ , with time kernel

G_{R} (t) = - Θ (t) \frac{sin ( t Ω _{m} )}{Ω _{m}}, Ω_{m} = - Δ + m^{2} .

Since $g$ is compactly supported after $t = 0$ , the first scalar response with zero past data is

ϕ_{1, g} [h, ϕ_{0}] = - G_{R} Q_{g h} ϕ_{0} .

Displayed formulas in this passage

VI@1#eq:matter-vertex · line 1254
$V_{1}^{m} (g; h, φ) = \frac{1}{2} \int g h^{μν} T_{μν} (η, φ) d^{4} x .$
VI@1#eq:Q · line 1261
$Q_{H} = - \partial_{μ} (H ˉ^{μν} \partial_{ν}) - \frac{1}{2} m^{2} tr_{η} H .$
VI@1#eq:QL · line 1271
$Q_{H} = L_{H} + \frac{1}{2} (tr_{η} H) P .$
VI@1#eq:Q-switch · line 1276
$Q_{g h} ϕ_{0} = g L_{h} ϕ_{0} - (\partial_{μ} g) h ˉ^{μν} \partial_{ν} ϕ_{0} .$
VI@1#eq:external-response · line 1291
$ϕ_{1, g} [h, ϕ_{0}] = - G_{R} Q_{g h} ϕ_{0} .$
unlabeled · line 1250
$g \in C_{c}^{\infty} ((0, T) \times R^{3}; R) .$
unlabeled · line 1285
$G_{R} (t) = - Θ (t) \frac{sin ( t Ω _{m} )}{Ω _{m}}, Ω_{m} = - Δ + m^{2} .$