Manuscript I-B-VP@3 · I-B-VP-v3.tex · section · 93c04eb4

Exact Friedrichs observables and analytical checks

Exact Friedrichs observables and analytical checks · lines 1220–1238

At conformal coupling the neutral energy density is negative:

ε = - \frac{α ^{- 4} - 1}{1440 π ^{2} r ^{4}} .

At $α = 1/2$ , the single nontrivial image gives

H_{E} (s) = \frac{1}{4 π ^{2} ( 4 r ^{2} + s ^{2} )} .

Then

F = \frac{1}{16 π ^{2} r ^{2}}, H_{E}^{''} (0) = - \frac{1}{32 π ^{2} r ^{4}}, \frac{1}{2} Δ_{E} F = \frac{1}{8 π ^{2} r ^{4}},

so $T^{t}_{t} = 1/ (96 π^{2} r^{4})$ . This is an independent sign check.

Displayed formulas in this passage

unlabeled · line 1223
$ε = - \frac{α ^{- 4} - 1}{1440 π ^{2} r ^{4}} .$
unlabeled · line 1227
$H_{E} (s) = \frac{1}{4 π ^{2} ( 4 r ^{2} + s ^{2} )} .$
unlabeled · line 1231
$F = \frac{1}{16 π ^{2} r ^{2}}, H_{E}^{''} (0) = - \frac{1}{32 π ^{2} r ^{4}}, \frac{1}{2} Δ_{E} F = \frac{1}{8 π ^{2} r ^{4}},$