Manuscript VII-F@1 · VII-F.tex · proposition · 2d81c03f

Proposition: Complete finite-edge length table

Cellulation and the isolated particle event › Edges, lengths, and areas · lines 708–768 · proof included or linked

VII-F@1#prop:lengths

The induced lengths of all finite edges in $K_{s}$ are:

edge type e_{i} c_{i} t_{i} o_{i} u_{i} v_{i}^{edge} w_{i}^{-} w_{i}^{+} multiplicity 33333333 s < 0 532142 inactive inactive inactive inactive s > 0 5 (1 - s /2) 32142 \frac{3}{2} s \frac{3 41}{11} s \frac{3 65}{22} s \frac{3 65}{22} s

Every active edge has positive length on its open parameter interval. Every new edge vanishes linearly at $s = 0$ .

Proof. For an edge in a developed chart,

ℓ^{2} = ∣Δ x ∣^{2} - (Δ T)^{2} .

Along $e_{i}$ , the spatial displacement is $3 v_{i}$ before the event and $(3 - \frac{3}{2} s) v_{i}$ after it, while $Δ T = \frac{2}{3} ∣Δ x ∣$ . This gives the first row.

For $u_{i}$ , both endpoints have time coordinate $2 s$ , so its length is $3 s /2$ .

For $v_{i}^{edge}$ , the spatial displacement is $(u_{s}, u_{s})$ and the time displacement is $3 u_{s} /5$ . Hence

ℓ^{2} = u_{s}^{2} (2 - \frac{9}{25}) = (\frac{3 41}{11} s)^{2} .

For $w_{i}^{-}$ , the spacetime displacement from $B_{i}$ to $C_{i}$ is

(Δ T, Δ ξ, Δ η) = \frac{s}{22} (18, - 3, 30) .

Therefore

ℓ^{2} = \frac{s ^{2}}{2 2 ^{2}} (9 + 900 - 324) = (\frac{3 65}{22} s)^{2} .

The other perimeter edge is symmetric. The remaining rows have $Δ T = 0$ and follow immediately.

Displayed formulas in this passage

unlabeled · line 711
$edge type e_{i} c_{i} t_{i} o_{i} u_{i} v_{i}^{edge} w_{i}^{-} w_{i}^{+} multiplicity 33333333 s < 0 532142 inactive inactive inactive inactive s > 0 5 (1 - s /2) 32142 \frac{3}{2} s \frac{3 41}{11} s \frac{3 65}{22} s \frac{3 65}{22} s$
unlabeled · line 732
$ℓ^{2} = ∣Δ x ∣^{2} - (Δ T)^{2} .$
unlabeled · line 744
$ℓ^{2} = u_{s}^{2} (2 - \frac{9}{25}) = (\frac{3 41}{11} s)^{2} .$
unlabeled · line 753
$(Δ T, Δ ξ, Δ η) = \frac{s}{22} (18, - 3, 30) .$
unlabeled · line 759
$ℓ^{2} = \frac{s ^{2}}{2 2 ^{2}} (9 + 900 - 324) = (\frac{3 65}{22} s)^{2} .$

Proposition: Complete finite-edge length table

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