What actually forces the potential

Layer II · Boundary mechanics

The real question on this page is not what the potential looks like. It is what forces it. A two-term cost function is not a result. Anyone can write one down. What matters is which part of it is a consequence of the axioms, which part needs one more axiom, and which part is provably free.

So this page is not an optimisation exercise, and it should not be read as one. It is an accounting of what the axiom system does and does not determine. Half of that accounting is positive. Half of it is a negative theorem, and the negative half is the more informative one.

The potential

V (κ) = maintenance price c_{0} κ + uncertainty price \frac{α C _{F}}{κ}

The first term is what it costs to hold a sharp boundary. The second is what it costs not to know. $C_{F}$ is the core constant — the only datum that depends on the shape of the kernel. In the robotic anchor these two terms are literally the two halves of a controller: what it costs to be stiff, and what it costs to be uncertain.

The entropic term is forced Machine-Checked

The $1/ κ$ law is not a modelling choice. It is what the rescaling law of the kernel forces, and that is a theorem, not a preference:

H (κ) = \int_{R} h (F (κ z)) d z = \frac{C _{F}}{κ}, H is antitone in κ .

The sharper the boundary, the less entropy it carries. Stated precisely: the $1/ κ$ law is the weight- $(- 1)$ equivariance of the entropy functional.

The maintenance term is not forced — until one axiom is added Machine-Checked

Under the categorical axioms alone, the maintenance side is entirely unconstrained: two distinct admissible models exist. Nothing in Layer I touches it. The claim that "the form is forced" was, as stated, false.

A7 — additive maintenance cost. Maintenance cost is additive over disjoint subsystems, and continuous in sharpness. Assumption
A7 is not a technical convenience. It is falsifiable: in the robotic reading, the torques needed to hold two independent joints are the sum of the torques needed to hold each. If some system's maintenance cost is not additive, A7 fails — and the form is no longer forced.

Under A7 the maintenance cost becomes linear, $M (κ) = c_{0} κ$ with $c_{0} = M (1)$ . The mathematical kernel is the standard fact that a continuous additive map between real vector spaces is linear. Nothing exotic — and machine-checkable.

The scientific core of Layer II is a negative result Machine-Checked

Read this part twice. Two models can both satisfy A7 and still differ:

M_{1} : c_{0} = 1, M_{2} : c_{0} = 2, M_{1} . maint (1) \neq = M_{2} . maint (1)

⟹ M_{1} . V (1) \neq = M_{2} . V (1)

So the axiom system is not categorical. The potential is determined only up to the coefficient $c_{0}$ . That is a theorem, not a caveat, and it is the formal closure of the objection that "the axioms do not produce the formula".

The answer is not a denial. The axioms do produce the form — linear maintenance plus an entropic

1/ κ

term — and they provably do not produce the coefficient. That coefficient is fixed by the normalisation

σ^{*} = 1

, which is a gauge condition, not a consequence.

A theory that claimed to derive it from nothing would be claiming to derive a unit.

Every element, and what forces it

Statement	Status	Forced by
Entropic term $α C_{F} / κ$	Machine-Checked	The rescaling law of the kernel (weight $- 1$ )
Maintenance term, without A7	Free	Nothing — two admissible models exist
A7 additivity	Assumption	Chosen, and falsifiable
Linearity $M (κ) = c_{0} κ$	Machine-Checked	A7 plus continuity
The coefficient $c_{0}$	Not determined	Only the gauge condition $σ^{*} = 1$
A7's legitimacy for real systems	Empirical prediction	Untested

What this page does not claim. That the potential describes any real system. A7, and the identification of

κ

with a measurable stiffness, are a modelling choice and a mapping hypothesis respectively. Their legitimacy is an empirical question — and it is the question the predictions are built to fail on. Mapping hypothesis