Manuscript V@1 · V.tex · section · e7ddf37d

The retarded coefficient sign

Sign, initial-data, and Duhamel audit › The retarded coefficient sign · lines 3953–3991

The scalar contractions obey

W_{m}^{F} (y - x) - W_{m} (y - x) = i G_{R} (x - y) .

With

R_{1, 1} = i (T_{2} - V ϕ),

the contraction in the retarded coefficient is $- G_{R}$ . After the compact vertex integration by parts,

ϕ^{(1)}_{g, h} = - G_{R} Q_{g h} ϕ .

Smearing gives

- \int f G_{R} Q_{g h} ϕ = - \int (G_{A} f) Q_{g h} ϕ = \int u_{f} Q_{g h} ϕ = ϕ (Q_{g h} u_{f}) .

This is why the compact test is

q_{g, h, f} = + Q_{g h} (- G_{A} f) .

The minus sign in $- G_{R} Q_{g h} ϕ$ has already been absorbed into $u_{f} = - G_{A} f$ ; inserting another minus in $q$ would reverse the coefficient.

For a TT tensor, $Q_{g h} = - \partial_{i} (g h^{ij} \partial_{j})$ . Thus the equivalent explicit formula is

q_{g, h, f} = - \partial_{i} (g h^{ij} \partial_{j} u_{f}) .

The leading minus here belongs to the differential operator $Q$ , not to an additional convention for the retarded product.

Displayed formulas in this passage

unlabeled · line 3956
$W_{m}^{F} (y - x) - W_{m} (y - x) = i G_{R} (x - y) .$
unlabeled · line 3960
$R_{1, 1} = i (T_{2} - V ϕ),$
unlabeled · line 3965
$ϕ^{(1)}_{g, h} = - G_{R} Q_{g h} ϕ .$
unlabeled · line 3969
$- \int f G_{R} Q_{g h} ϕ = - \int (G_{A} f) Q_{g h} ϕ = \int u_{f} Q_{g h} ϕ = ϕ (Q_{g h} u_{f}) .$
unlabeled · line 3976
$q_{g, h, f} = + Q_{g h} (- G_{A} f) .$
unlabeled · line 3985
$q_{g, h, f} = - \partial_{i} (g h^{ij} \partial_{j} u_{f}) .$