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

Proposition: Finite-image current and sign

An independent exact quotient check · lines 1016–1051 · proof included or linked

I-B-VP-0@1#prop:quotient, I-B-VP-0@1#eq:quotientcurrent

Direct differentiation of (eq:quotientkernel) gives

J_{φ} = - \frac{1}{8 π ^{2} N r ^{2}} k = 1 \sum^{N - 1} sin (\frac{2 π p k}{N}) cot (\frac{π k}{N}) csc^{2} (\frac{π k}{N}) .

It equals Theorem thm:current at $(α, β) = (1/ N, p / N)$ .

Proof. The image denominator is

D_{k}^{2} = r^{2} + r^{'2} - 2 r r^{'} cos (\frac{φ - φ ^{'}}{N} - \frac{2 π k}{N}) + Δ τ^{2} + Δ z^{2} .

At coincidence $D_{k}^{2} = 4 r^{2} sin^{2} (π k / N)$ . Applying $i (\partial_{φ} - \partial_{φ^{'}})$ to (eq:quotientkernel), and pairing conjugate images, gives (eq:quotientcurrent). Lemma lem:finiteidentity then yields

J_{φ} = \frac{p ( N - p ) ( 2 p - N )}{12 π ^{2} N r ^{2}},

which is precisely (eq:main). This calculation does not use the small-separation Bernoulli expansion.

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I-B-VP-0@1#eq:quotientcurrent · line 1019
$J_{φ} = - \frac{1}{8 π ^{2} N r ^{2}} k = 1 \sum^{N - 1} sin (\frac{2 π p k}{N}) cot (\frac{π k}{N}) csc^{2} (\frac{π k}{N}) .$
unlabeled · line 1034
$D_{k}^{2} = r^{2} + r^{'2} - 2 r r^{'} cos (\frac{φ - φ ^{'}}{N} - \frac{2 π k}{N}) + Δ τ^{2} + Δ z^{2} .$
unlabeled · line 1044
$J_{φ} = \frac{p ( N - p ) ( 2 p - N )}{12 π ^{2} N r ^{2}},$

Proposition: Finite-image current and sign

Displayed formulas in this passage

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