
of the lightest neutrino:
𝐿
𝜈
=
ℏ𝑐
4 𝑚
1
𝑐
2
= 62.1 𝜇m. (8)
Nothing in the published model supplies this scale. It is hypothesis H
𝜈
1. At this spacing one light-tick,
𝑐𝜏 = 4𝐿
𝜈
, equals the reduced Compton wavelength of the lightest neutrino.
A structure that carries the lattice at its own spacing, set by its rest energy in bond quanta, has
a counterpart at the nuclear scale. With sixteen quanta, 𝑚
𝑝
𝑐
2
= 16 ℏ𝑐/(4𝐿), the rule gives 𝐿 =
4ℏ/(𝑚
𝑝
𝑐) = 0.8412 fm. This is within 0.1% of the measured proton charge radius, 0.84075(64) fm [22].
Sixteen is the number of two-step walks from one node of the complete graph on five nodes. These five
nodes are the four vertices of a tetrahedral void and the node it traps. A single bond, the graph of one
step along a worldline, gives one. We use this comparison only as motivation. One quantum at 𝐿
𝜈
plays
the role for the neutrino that sixteen quanta at 0.84 fm would play for the proton. We test the choice of
one quantum below.
4.5 A worldline defect, and the one cube that needs time
In the extension of the model to 𝐷
4
[15], a neutrino is not a defect of the spatial lattice. It is a time-only
worldline. It is an operator that wraps the time direction. It has no presence in any equal-time slice. It
commutes with every stabilizer, so it fires no local syndrome. It is labeled by one of the three 16-cells of
the 𝐷
4
shell. Its energy per unit of spatial volume is therefore not defined at any instant. Its action per
unit of four-volume is defined.
The geometry of 𝐷
4
fixes the four-cell. No three of the twelve spatial bond vectors are mutually
perpendicular. So no cube of space has bond-length edges. The 24 bond vectors of 𝐷
4
do fall into
exactly three frames of four mutually perpendicular vectors. Each frame pairs the time axis with one
spatial axis. One example is
(1, 1, 0, 0), (1, −1, 0, 0), (0, 0, 1, 1), (0, 0, 1, −1) (9)
in coordinates (𝑡, 𝑥, 𝑦, 𝑧). The other two frames pair 𝑡 with 𝑦 and with 𝑧. So every cube of edge 𝐿 in
𝐷
4
has an edge that leans into time. The eight vectors ±𝑣 of one frame are the vertices of one of the
three inscribed 16-cells of the 𝐷
4
shell. Ref. [15] assigns the three flavors to these 16-cells through its
hypothesis H2 (scripts frame_cube.py, lorentz_cell.py).
Each frame spans a four-cell of edge 𝐿 and four-volume 𝐿
4
. The cell contains two 𝐷
4
sites. Its two
time-bearing edges are null in Lorentzian signature. Build a three-dimensional cube on one of them. Its
three-volume is 𝐿
3
in Euclidean signature, 𝐿
3
/
√
2 when projected onto an equal-time slice, and zero in
Lorentzian signature. The three-dimensional count is therefore ambiguous. The four-volume of the cell
is 𝐿
4
in either signature. The vacuum side has the same property. An energy density with 𝑤 = −1 has a
stress tensor proportional to the metric, so it is the same for every observer. An action per four-volume
needs no choice of time slicing. We therefore count both sides of the relation per orthogonal-frame
four-cell. The four-volume of the cell is fixed. The duration of the neutrino’s quantum within it is not
fixed. This duration is where the model’s two clocks disagree (Section 4.7).
4.6 The count as an impedance-matching condition
On a lattice, a compact 𝑈 (1) field has a charging energy 𝑈 and a bond stiffness 𝐽. In a common normal-
ization its coupling is 𝛼 =
𝑈/𝐽/4𝜋. Its wave impedance is proportional to
𝑈/𝐽 [23]. Two regions
of lattice with different spacings join without reflection when their ratios 𝑈/𝐽 agree. A mismatched
boundary reflects part of what crosses it and stores it. We apply the same condition to the channel that a
time-only worldline uses. This channel does not propagate in space, so the speed of its waves does not
fix its impedance. We compare the two stiffnesses as actions over the same four-cell.
5