Manuscript I-B-VP-0@1 · I-B-VP-0-v1.tex · section · 8a9d0682

A differentiated image–diffraction identity

A differentiated image–diffraction identity · lines 1398–1493

I-B-VP-0@1#eq:contour, I-B-VP-0@1#eq:current-contour

This appendix records a second representation, with its differentiation restriction explicit. It is not needed for Theorem thm:current.

For a chosen angular lift define

K (ϑ) = {k \in Z : ∣ ϑ - 2 π α k ∣ < π},

R_{k}^{2} = r^{2} + r^{'2} - 2 r r^{'} cos (ϑ - 2 π α k) + Δ z^{2}, ρ (u)^{2} = r^{2} + r^{'2} + 2 r r^{'} cosh u + Δ z^{2} .

The diffraction weight is

Φ_{β} (u, ϑ) = n \in Z \sum sin (π ν_{n}) e^{- ν_{n} u} e^{i (n + β) ϑ / α} .

I-B-G's angular summation and the Euclidean heat integral give CheungIBG (:closed) CheungIBVP (:contour)

G_{E, β} = k \in K (ϑ) \sum \frac{e ^{2 π i β k}}{4 π ^{2} ( R _{k} ^{2} + Δ τ ^{2} )} - \frac{1}{π α} \int_{0}^{\infty} \frac{Φ _{β} ( u , ϑ )}{4 π ^{2} ( ρ ( u ) ^{2} + Δ τ ^{2} )} d u .

Assume here that $\frac{1}{2 α} \in / N$ , so zero angular separation is not an image boundary. Define the real weight

Ξ_{β} (u) = - i α \partial_{ϑ} Φ_{β} (u, 0) = n \sum (n + β) sin (π ν_{n}) e^{- ν_{n} u}, u > 0.

Then

J_{φ} = - \frac{α}{4 π ^{2} r ^{2}} 0 < k < 1/ (2 α) \sum \frac{sin ( 2 π β k ) cos ( π α k )}{sin ^{3} ( π α k )} + \frac{1}{8 π ^{3} α r ^{2}} \int_{0}^{\infty} \frac{Ξ _{β} ( u )}{cosh ^{2} ( u /2 )} d u .

Its value is the polynomial in (eq:main).

To justify (eq:current-contour), choose a small angular neighbourhood of zero containing no image boundary. The image set is constant there. The closed geometric-series expression for $Φ_{β}$ in I-B-G :Phi has denominators bounded away from zero at $u = 0$ , also after one angular derivative. At infinity, the differentiated series decays exponentially. The diffraction denominator grows as $e^{u}$ , uniformly on compact regular spatial sets. Differentiation under the integral is therefore justified on this neighbourhood. Apply $2 i α \partial_{ϑ}$ , set coincidence, pair the $k, - k$ image terms, and use $ρ (u)^{2} = 4 r^{2} cosh^{2} (u /2)$ .

When $α > 1/2$ , the image sum is empty, so the full current is represented by the differentiated diffraction term. This is a representation of a stationary spectral observable, not a claim about an emitted retarded signal.

joint integrability alone is insufficient. At an image boundary, local integrability of $Φ_{β} (u, ϑ)$ does not by itself justify pointwise differentiation under its integral. Angular derivatives can worsen the boundary singularity, and changing image indicators also contribute distributionally. For such parameters, use the differentiated total distribution in (eq:contour) and its analytic remainder from Proposition prop:kernel, or use the mode proof. No separate pointwise boundary prescription is introduced.

Displayed formulas in this passage

I-B-VP-0@1#eq:contour · line 1426
$G_{E, β} = k \in K (ϑ) \sum \frac{e ^{2 π i β k}}{4 π ^{2} ( R _{k} ^{2} + Δ τ ^{2} )} - \frac{1}{π α} \int_{0}^{\infty} \frac{Φ _{β} ( u , ϑ )}{4 π ^{2} ( ρ ( u ) ^{2} + Δ τ ^{2} )} d u .$
I-B-VP-0@1#eq:current-contour · line 1448
$J_{φ} = - \frac{α}{4 π ^{2} r ^{2}} 0 < k < 1/ (2 α) \sum \frac{sin ( 2 π β k ) cos ( π α k )}{sin ^{3} ( π α k )} + \frac{1}{8 π ^{3} α r ^{2}} \int_{0}^{\infty} \frac{Ξ _{β} ( u )}{cosh ^{2} ( u /2 )} d u .$
unlabeled · line 1406
$K (ϑ) = {k \in Z : ∣ ϑ - 2 π α k ∣ < π},$
unlabeled · line 1410
$R_{k}^{2} = r^{2} + r^{'2} - 2 r r^{'} cos (ϑ - 2 π α k) + Δ z^{2}, ρ (u)^{2} = r^{2} + r^{'2} + 2 r r^{'} cosh u + Δ z^{2} .$
unlabeled · line 1417
$Φ_{β} (u, ϑ) = n \in Z \sum sin (π ν_{n}) e^{- ν_{n} u} e^{i (n + β) ϑ / α} .$
unlabeled · line 1441
$Ξ_{β} (u) = - i α \partial_{ϑ} Φ_{β} (u, 0) = n \sum (n + β) sin (π ν_{n}) e^{- ν_{n} u}, u > 0.$

Nearby in this section

bibliography: References