Manuscript XVIII-B@2 · XVIII-B-v2.tex · proposition · 27ed4f4b

Proposition: Evaluated normalization checks

The once-subtracted retarded prescription › Explicit evaluation of the reduced integrals · lines 780–815 · proof included or linked

XVIII-B@2#prop:checks, XVIII-B@2#eq:lowchecks, XVIII-B@2#eq:thresholdchecks

The printed form factors obey

F_{2}^{'} (0) F_{2} (a) = \frac{1}{13440 π ^{2} m ^{2}}, = \frac{1}{2400 π ^{2}}, F_{0}^{'} (0) F_{0} (a) = \frac{17}{6720 π ^{2} m ^{2}}, = \frac{3}{80 π ^{2}} .

Their imaginary parts above threshold are

Im F_{i}^{R} (p) = π sgn (p^{0}) r_{i} (p^{2}) .

Proof. At $w = 0$ , use $J_{n} (0) = 1/ (2 n + 1)$ in (eq:F2closed)–(eq:F0closed) and $a = 4 m^{2}$ . In the scalar combination,

\frac{9}{3} - \frac{6}{5} + \frac{1}{7} = \frac{68}{35} .

At $w = 1$ , $J_{1} = 1$ , $J_{2} = 1/3$ , and $J_{3} = 1/5$ , which gives (eq:thresholdchecks).

The imaginary term in (eq:J0above), together with the recursion, gives the stated imaginary parts; equivalently use Lemma lem:dispersion.

Displayed formulas in this passage

XVIII-B@2#eq:lowchecks · line 783
$F_{2}^{'} (0) F_{2} (a) = \frac{1}{13440 π ^{2} m ^{2}}, = \frac{1}{2400 π ^{2}}, F_{0}^{'} (0) F_{0} (a) = \frac{17}{6720 π ^{2} m ^{2}}, = \frac{3}{80 π ^{2}} .$
XVIII-B@2#eq:thresholdchecks · line 783
$F_{2}^{'} (0) F_{2} (a) = \frac{1}{13440 π ^{2} m ^{2}}, = \frac{1}{2400 π ^{2}}, F_{0}^{'} (0) F_{0} (a) = \frac{17}{6720 π ^{2} m ^{2}}, = \frac{3}{80 π ^{2}} .$
unlabeled · line 794
$Im F_{i}^{R} (p) = π sgn (p^{0}) r_{i} (p^{2}) .$
unlabeled · line 805
$\frac{9}{3} - \frac{6}{5} + \frac{1}{7} = \frac{68}{35} .$