Manuscript VII-J-ST@1 · VII-J-ST.tex · section · f1f1d82c

Persistent spatial marks and exact times

Exact regular-part benchmark through the incidence parameter › Persistent spatial marks and exact times · lines 1421–1462

VII-J-ST@1#eq:benchmark-times

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In sector zero choose the dimensionless transverse mark

x_{⊥} = (1/64, 1/128) .

At that mark,

h (x_{⊥}) = 1/64, k (x_{⊥}) = 3/640.

Choose two regular persistent ports

A = (x_{⊥}, ζ = 0), B = (x_{⊥}, ζ = L), L = \frac{1}{64},

and slice parameters

s_{-} = - \frac{1}{4}, s_{+} = \frac{1}{4} .

Both ports have identical time coordinates on each marking. Their input and output times are

τ_{in} = - \frac{15}{64}, τ_{out} = \frac{323}{640} .

Take a convex developed spatial tube containing

[3/256, 5/256] \times [1/256, 3/256] \times [- 1/256, 5/256] .

Its closure is entirely regular and lies inside the first sector. Choose a finite time interval containing $[τ_{in}, τ_{out}]$ . All operational paths and gates remain in this tube.

The marks $A, B$ are spatial locations, not evolving abstract cell labels. The local branch at $x_{⊥}$ changes at $s = h - k = 7/640$ , but the mark and its quantum register do not cease to exist.

Displayed formulas in this passage

VII-J-ST@1#eq:benchmark-times · line 1442
$τ_{in} = - \frac{15}{64}, τ_{out} = \frac{323}{640} .$
unlabeled · line 1424
$x_{⊥} = (1/64, 1/128) .$
unlabeled · line 1428
$h (x_{⊥}) = 1/64, k (x_{⊥}) = 3/640.$
unlabeled · line 1432
$A = (x_{⊥}, ζ = 0), B = (x_{⊥}, ζ = L), L = \frac{1}{64},$
unlabeled · line 1437
$s_{-} = - \frac{1}{4}, s_{+} = \frac{1}{4} .$
unlabeled · line 1449
$[3/256, 5/256] \times [1/256, 3/256] \times [- 1/256, 5/256] .$