08-ilang-propagation / 06-mark-inertia
06mark inertiaverified
Determines the inertia of a moving body with its own mark against its cost in oscillator cells, as a function of the ratio of body speed to field speed, and whether weight over inertia is universal.
# Review: 06-mark-inertia v1
- **Subproject:** 08-ilang-propagation
- **Package:** 06-mark-inertia
- **Reviewed version:** v1
- **Date:** 2026-10-25
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VERDICT: major errors
## Issues
### I1. Items 2 and 3 are not answered on the cubic torus
- **Location:** Step 8 ("Size of R", "D=3 otherwise"), Step 9(b), Result (6.9)–(6.12), first item of Open issues; twin.
- **Severity:** major
- **Problem:** The question asks items 2 and 3 on the ring and on the cubic torus (Scope, "Class"). On the cubic torus the derivation settles them only in the regime $\varepsilon_k\gg c_0,\ \tau\sqrt{1+r^2}$ of (6.10). Outside it, $\rho_{k,\sigma}$ and the bracket of (6.12) are left as unevaluated lattice sums, and whether one speed $v$ exists, and whether $E''\Delta W$ is the same for every level and both branches, is listed as open. An item answered on a subclass is not answered. It is also not shown whether the region of (6.10) covers the whole range in which the upper branch is isolated on the cubic torus: there (U) needs $\varepsilon_k$ of the order of $\Omega_{\max}\approx2\sqrt3\,c_0$, and $R$ has a finite limit of order $1/c_0$ (Step 8), so $\varepsilon_kR$ is of order one at the edge of (U).
- **Fix:** Evaluate $\rho_{k,\sigma}$ on the cubic torus from (6.3), numerically where no closed form is available, with $L$ and $\ell$ large enough that the sums have converged (state the check). Do this for the $F_1^2$ and $F_2^2$ parts, for each branch where it is isolated under (6.4), for several levels (values of $\varepsilon_k/c_0$) and for several $r$ including $r=1$. Answer item 2 there: one $v$ or not, as a function of $r$ and of $\varepsilon_k/c_0$. Answer item 3 there: for which grips, if any, the bracket of (6.12) is the same for every level and both branches. Say how large $\varepsilon_kR$ is where (U) holds. Add the computation to the code and cite its output.
### I2. The linear-grip reading of (6.11) needs its order
- **Location:** Step 9(a), Result (6.11); twin 6.11.
- **Severity:** minor
- **Problem:** (6.11) says that $E''\Delta W$ is the same for every level and both branches iff $F_2=0$, independently of $r$. This holds in the counting $F_1\sim F_2$, where the inertia of the mark enters only at the next order. For the linear grip, $F_2=0$ exactly, so that next-order term is the first level dependence: by (6.12) it vanishes on the ring at leading order in $1/\ell$ only at $r=\sqrt2$. As written, (6.11) can be read as universality of the linear grip at every $r$.
- **Fix:** State in Step 9(a), in the Result and in the twin that for $F_2=0$ the first dependence on the level comes from $I^{\rm mark}$ at the next order, as given by (6.12).
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