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

Corollary: Source-level traceless diagnostic

The pure-trace source calculation › Spatial traceless channel · lines 899–941 · proof included or linked

XVI-B@1#cor:tf, XVI-B@1#eq:tf, XVI-B@1#eq:tf-contractions, XVI-B@1#eq:source-contraction

The spatial traceless projections and contractions satisfy

q_{TF}^{ij} h_{ij} q^{ij} = n^{i} n^{j} - \frac{1}{3} δ^{ij}, = \frac{2}{3}, (d_{1})_{TF}^{ij} h_{ij} d_{1}^{ij} = 0, = 0.

Consequently

h_{ij} S_{λ}^{ij} = \frac{2}{3} λ^{2} σ (t) e^{2 i λ t} .

Proof. Equation (eq:qcounter) gives $δ_{ij} q^{ij} = 1 - 9 = - 8$ . Subtracting one third of that trace gives

q_{TF}^{ij} = n^{i} n^{j} - 3 δ^{ij} + \frac{8}{3} δ^{ij} = n^{i} n^{j} - \frac{1}{3} δ^{ij} .

Equation (eq:d-components) shows that the correction is spatially proportional to $δ^{ij}$ , so its traceless projection is zero.

Also $h_{ij} δ^{ij} = 0$ and

h_{ij} n^{i} n^{j} = ∣ n ∣^{4} - \frac{1}{3} ∣ n ∣^{2} = \frac{2}{3} .

The contractions and (eq:source-contraction) follow.

Displayed formulas in this passage

XVI-B@1#eq:tf · line 902
$q_{TF}^{ij} h_{ij} q^{ij} = n^{i} n^{j} - \frac{1}{3} δ^{ij}, = \frac{2}{3}, (d_{1})_{TF}^{ij} h_{ij} d_{1}^{ij} = 0, = 0.$
XVI-B@1#eq:tf-contractions · line 902
$q_{TF}^{ij} h_{ij} q^{ij} = n^{i} n^{j} - \frac{1}{3} δ^{ij}, = \frac{2}{3}, (d_{1})_{TF}^{ij} h_{ij} d_{1}^{ij} = 0, = 0.$
XVI-B@1#eq:source-contraction · line 914
$h_{ij} S_{λ}^{ij} = \frac{2}{3} λ^{2} σ (t) e^{2 i λ t} .$
unlabeled · line 925
$q_{TF}^{ij} = n^{i} n^{j} - 3 δ^{ij} + \frac{8}{3} δ^{ij} = n^{i} n^{j} - \frac{1}{3} δ^{ij} .$
unlabeled · line 935
$h_{ij} n^{i} n^{j} = ∣ n ∣^{4} - \frac{1}{3} ∣ n ∣^{2} = \frac{2}{3} .$