Manuscript III@2 · III.tex · section · 82dad3e5

Distributional kernel checks

Sign, inverse, boundary, and kernel audit › Distributional kernel checks · lines 3075–3125

For the temporal deformation, let

T_{ij} (d) = \frac{d _{i} d _{j}}{d ^{3}} + \frac{δ _{ij}}{d} .

Away from zero,

\partial_{i} (\frac{d _{i} d _{j}}{d ^{3}}) = \frac{d _{j}}{d ^{3}}, \partial_{i} (\frac{δ _{ij}}{d}) = - \frac{d _{j}}{d ^{3}} .

The shrinking-sphere flux of this degree- $- 1$ kernel tends to zero. Thus $\partial_{i} T_{ij} = 0$ distributionally.

Only the $j, k$ -symmetric part of the spatial kernel contributes:

S_{ij k} (d) = \frac{3 d _{i} d _{j} d _{k}}{d ^{5}} + \frac{d _{i} δ _{j k}}{d ^{3}} - \frac{δ _{ij} d _{k} + δ _{ik} d _{j}}{d ^{3}} .

It satisfies $\partial_{j} S_{ij k} = 0$ away from zero. Its flux is

\int_{∣ d ∣ = ε} S_{ij k} n_{j} d S = \int_{S^{2}} (3 n_{i} n_{k} - δ_{ik}) d Ω = 3 \frac{4 π}{3} δ_{ik} - 4 π δ_{ik} = 0.

Replacing a smooth test on the sphere by its value at zero leaves an error tending to zero. There is no delta contribution, and the divergence identity holds distributionally.

Integration by parts against compact pure gauges therefore gives

D_{C}^{0} [K ξ] = 0, D_{C}^{i} [K ξ] = 0.

Together with the kernel comparison and Poisson compensation, this also checks the Coulomb transformation sign in our convention.

The general kernel comparison requires spatial Schwartz behavior. No boundary identity for arbitrary $1/ r$ metric tails is asserted.

Displayed formulas in this passage

unlabeled · line 3078
$T_{ij} (d) = \frac{d _{i} d _{j}}{d ^{3}} + \frac{δ _{ij}}{d} .$
unlabeled · line 3083
$\partial_{i} (\frac{d _{i} d _{j}}{d ^{3}}) = \frac{d _{j}}{d ^{3}}, \partial_{i} (\frac{δ _{ij}}{d}) = - \frac{d _{j}}{d ^{3}} .$
unlabeled · line 3094
$S_{ij k} (d) = \frac{3 d _{i} d _{j} d _{k}}{d ^{5}} + \frac{d _{i} δ _{j k}}{d ^{3}} - \frac{δ _{ij} d _{k} + δ _{ik} d _{j}}{d ^{3}} .$
unlabeled · line 3103
$\int_{∣ d ∣ = ε} S_{ij k} n_{j} d S = \int_{S^{2}} (3 n_{i} n_{k} - δ_{ik}) d Ω = 3 \frac{4 π}{3} δ_{ik} - 4 π δ_{ik} = 0.$
unlabeled · line 3117
$D_{C}^{0} [K ξ] = 0, D_{C}^{i} [K ξ] = 0.$