Manuscript XVI-B@1 · XVI-B.tex · proposition · 6a0abffd

Proposition: Evaluated correction and prepared source

The pure-trace source calculation · lines 791–861 · proof included or linked

XVI-B@1#prop:pair, XVI-B@1#eq:d1, XVI-B@1#eq:Dsymbol, XVI-B@1#eq:d-components, XVI-B@1#eq:JDeval

On (eq:witness), define

B^{μν} = η^{μν} + 2 δ^{μ}_{0} δ^{ν}_{0}, d_{1}^{μν} (t) = - i σ^{'} (t) B^{μν} .

Then

D_{σ}^{μν} (γ_{1}^{λ}, γ_{2}^{λ}) d^{μν} (t, λ) = d^{μν} (t, λ) e^{2 i λ t}, = λ d_{1}^{μν} (t) = - i λ σ^{'} (t) B^{μν} .

In components,

d^{00} = - 3 i λ σ^{'}, d^{0 i} = 0, d^{ij} = i λ σ^{'} δ^{ij} .

The prepared evaluated source is

S_{λ}^{μν} := J_{D}^{μν} (γ_{1}^{λ}, γ_{2}^{λ}) = [λ^{2} σ (t) q^{μν} + λ d_{1}^{μν} (t)] e^{2 i λ t} .

The coefficient of $σ^{''}$ in this evaluated correction is zero.

Proof. Put $ψ_{a} = e^{i λ k_{a} \cdot y}$ . Then

ψ_{1} ψ_{2} = e^{2 i λ t}, \partial_{t} (ψ_{1} ψ_{2}) = 2 i λ e^{2 i λ t}, \partial^{ν} (ψ_{1} ψ_{2}) = 2 i λ δ^{ν}_{0} e^{2 i λ t} .

Substitution in (eq:trace-identity) gives (eq:Dsymbol). The signature gives $B^{00} = 3$ , $B^{0 i} = 0$ , and $B^{ij} = - δ^{ij}$ , proving (eq:d-components).

Add $σ$ times (eq:bulk-eval) to (eq:Dsymbol), using (eq:actualsource). This gives (eq:JDeval).

Displayed formulas in this passage

XVI-B@1#eq:d1 · line 794
$B^{μν} = η^{μν} + 2 δ^{μ}_{0} δ^{ν}_{0}, d_{1}^{μν} (t) = - i σ^{'} (t) B^{μν} .$
XVI-B@1#eq:Dsymbol · line 803
$D_{σ}^{μν} (γ_{1}^{λ}, γ_{2}^{λ}) d^{μν} (t, λ) = d^{μν} (t, λ) e^{2 i λ t}, = λ d_{1}^{μν} (t) = - i λ σ^{'} (t) B^{μν} .$
XVI-B@1#eq:d-components · line 815
$d^{00} = - 3 i λ σ^{'}, d^{0 i} = 0, d^{ij} = i λ σ^{'} δ^{ij} .$
XVI-B@1#eq:JDeval · line 825
$S_{λ}^{μν} := J_{D}^{μν} (γ_{1}^{λ}, γ_{2}^{λ}) = [λ^{2} σ (t) q^{μν} + λ d_{1}^{μν} (t)] e^{2 i λ t} .$
unlabeled · line 843
$ψ_{1} ψ_{2} = e^{2 i λ t}, \partial_{t} (ψ_{1} ψ_{2}) = 2 i λ e^{2 i λ t}, \partial^{ν} (ψ_{1} ψ_{2}) = 2 i λ δ^{ν}_{0} e^{2 i λ t} .$

Proposition: Evaluated correction and prepared source

Displayed formulas in this passage

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