Manuscript XX-C@1 · XX-C.tex · lemma · 7a8cab37

Lemma: Rod-source normalization at zero spatial frequency

An intrinsic inverse-distance obstruction for two rods › The nonzero leading coefficient · lines 1504–1546 · proof included or linked

XX-C@1#lem:sourcecoefficient, XX-C@1#eq:rodmatrix

For $k \neq = 0$ ,

k_{i} k_{j} C^{ia} A^{- 1} (C^{j b})^{†} = \frac{2}{M ^{2}} ∣ k ∣^{2} δ_{ab} + \frac{p + 1}{2 M ^{2}} k_{a} k_{b} .

Proof. The metric functional contracted with $k_{i}$ has the spatial symmetric coefficient

B^{a}_{k l} = \frac{1}{2} k_{a} δ_{k l} - k_{(k} δ_{l) a} .

Its spatial trace is $k_{a} /2$ , and

B^{a} : B^{b} = \frac{1}{2} ∣ k ∣^{2} δ_{ab} + \frac{1}{4} k_{a} k_{b} .

Four-dimensional trace reversal in the metric inverse subtracts one half of the product of these traces. Therefore the metric contribution is

\frac{4}{M ^{2}} (\frac{1}{2} ∣ k ∣^{2} δ_{ab} + \frac{1}{8} k_{a} k_{b}) .

The scalar contribution is

\frac{p ^{2}}{A ^{2}} k_{a} k_{b} = \frac{p}{2 M ^{2}} k_{a} k_{b} .

Adding them gives (eq:rodmatrix). The formal adjoint here uses the field pairing, not an independently substituted positive metric.

Displayed formulas in this passage

XX-C@1#eq:rodmatrix · line 1507
$k_{i} k_{j} C^{ia} A^{- 1} (C^{j b})^{†} = \frac{2}{M ^{2}} ∣ k ∣^{2} δ_{ab} + \frac{p + 1}{2 M ^{2}} k_{a} k_{b} .$
unlabeled · line 1519
$B^{a}_{k l} = \frac{1}{2} k_{a} δ_{k l} - k_{(k} δ_{l) a} .$
unlabeled · line 1525
$B^{a} : B^{b} = \frac{1}{2} ∣ k ∣^{2} δ_{ab} + \frac{1}{4} k_{a} k_{b} .$
unlabeled · line 1533
$\frac{4}{M ^{2}} (\frac{1}{2} ∣ k ∣^{2} δ_{ab} + \frac{1}{8} k_{a} k_{b}) .$
unlabeled · line 1539
$\frac{p ^{2}}{A ^{2}} k_{a} k_{b} = \frac{p}{2 M ^{2}} k_{a} k_{b} .$