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

Leading instantaneous slice inversion

What instantaneous Poisson compensation establishes › Leading instantaneous slice inversion · lines 2397–2435

Use

{h_{ij} (x), π^{k l} (y)} = δ_{i}^{(k} δ_{j}^{l)} δ^{3} (x - y),

and

R_{lin} [h] = \partial_{i} \partial_{j} h_{ij} - Δ h_{ii} .

The generator

G_{0} [ξ] = \int_{R^{3}} (- 2 ξ^{0} R_{lin} [h] - 2 ξ^{i} \partial_{j} π^{ij}) d^{3} x

gives

δ h_{ij} = \partial_{i} ξ_{j} + \partial_{j} ξ_{i},

δ π^{ij} = 2 (\partial_{i} \partial_{j} - δ_{ij} Δ) ξ^{0} .

For $χ_{0} = Π_{tr}$ , $χ_{i} = A_{i}$ ,

δ χ = M ξ, M = (- 4Δ 0 0 Δ δ_{ij}) .

Displayed formulas in this passage

unlabeled · line 2400
${h_{ij} (x), π^{k l} (y)} = δ_{i}^{(k} δ_{j}^{l)} δ^{3} (x - y),$
unlabeled · line 2405
$R_{lin} [h] = \partial_{i} \partial_{j} h_{ij} - Δ h_{ii} .$
unlabeled · line 2410
$G_{0} [ξ] = \int_{R^{3}} (- 2 ξ^{0} R_{lin} [h] - 2 ξ^{i} \partial_{j} π^{ij}) d^{3} x$
unlabeled · line 2419
$δ h_{ij} = \partial_{i} ξ_{j} + \partial_{j} ξ_{i},$
unlabeled · line 2422
$δ π^{ij} = 2 (\partial_{i} \partial_{j} - δ_{ij} Δ) ξ^{0} .$
unlabeled · line 2427
$δ χ = M ξ, M = (- 4Δ 0 0 Δ δ_{ij}) .$