The surviving area is the matched-endpoint arc–chord defect
Draft coordinator derivation for B11 review, 2026-09-07. The supplied Sol commentary correctly computes a factor of two for a fixed parabola. A genuine central-force family distinguishes two comparisons: arc versus its inertial tangent, and arc versus its matched-endpoint chord. The latter supplies the stable connection to the existing action calculation.
1. Exact signed-area identity
Put \(C_R=(0,R)\) and orient increasing \(x\) to the right and \(y\) toward the centre. For a path starting at the origin, define its signed sectorial area
\[A_R[\gamma]=\frac12\int_\gamma[x\,dy-(y-R)\,dx] =\frac R2\Delta x+\frac12\int_\gamma(x\,dy-y\,dx).\]
For two paths with the same endpoints, the term proportional to \(R\) cancels exactly. Their area difference is the signed area of the closed loop formed by the first path and the reversed second. This cancellation holds for every centre, before taking any limit.
For the perpendicular constant-force parabola \(x=vt\), \(y=Ft^2/(2m)\), \(0\le t\le\tau\), \(F,v,m,\tau>0\), the finite part is
\[A_{\rm lens}=\frac{Fv\tau^3}{12m}.\]
Its straight chord from the origin has \(\int(x\,dy-y\,dx)=0\), so this is the exact arc-minus-chord area. The inertial tangent \((vt,0)\) happens also to give that subtraction for the fixed parabola, since both have the same horizontal endpoint. Their vertical endpoints differ.
The positive ordinary areas satisfy
\[A_{\rm tangent,arc}=\frac{Fv\tau^3}{6m}=2A_{\rm lens},\qquad A_{\rm tangent,chord}=\frac{Fv\tau^3}{4m}=3A_{\rm lens},\]
and hence \(A_{\rm tangent,arc}=A_{\rm tangent,chord}-A_{\rm lens}\). The plus sign in the supplied geometrical explanation reverses this ordering.
2. A genuine receding central force
Use the explicit classical family
\[m\ddot q_R=\frac FR(C_R-q_R),\quad q_R(0)=(0,0),\quad\dot q_R(0)=(v,0),\qquad \omega_R^2=\frac{F}{mR}.\]
The force is central for every finite \(R\) and tends to the uniform field on bounded spatial sets. The exact solution is
\[x_R(t)=\frac v{\omega_R}\sin(\omega_Rt),\qquad y_R(t)=R[1-\cos(\omega_Rt)].\]
On every fixed compact time interval it tends to the parabola, while
\[x_R(\tau)=v\tau-\frac{Fv\tau^3}{6mR}+O(R^{-2}).\]
Angular momentum about \(C_R\) is \(mRv\), so the actual arc and the inertial tangent each have signed sectorial area \(Rv\tau/2\). Their difference is exactly zero for every \(R\). In the finite-part decomposition the missing term is
\[\frac R2[x_R(\tau)-v\tau]\longrightarrow-\frac{Fv\tau^3}{12m},\]
which cancels the parabola’s finite part. Substituting the limiting parabola before subtracting the divergent reference therefore changes this comparison.
Let \(\operatorname{ch}_R\) instead join the actual arc endpoints. Then
\[A_R[q_R]-A_R[\operatorname{ch}_R] =\frac{Rv}{2}\left[\tau-\frac{\sin(\omega_R\tau)}{\omega_R}\right] \longrightarrow\frac{Fv\tau^3}{12m}.\]
Thus a genuine receding-centre construction recovers the lens when the chord shares the arc endpoints. This is a precise reusable version of the supplied geometrical idea. It does not require an identification of the arc and tangent sector differences in a varying central-force family.
3. Action normalization
For the constant-force Lagrangian \(L=m|\dot q|^2/2+Fy\), the existing matched-endpoint calculation gives
\[S[q_{\rm chord}]-S[q_{\rm cl}] =\frac F{2v}A_{\rm lens} =\frac{F^2\tau^3}{24m} =\frac{\tau\Delta E_\perp}{12},\qquad \Delta E_\perp=\frac{F^2\tau^2}{2m}.\]
The supplied identity \(6(F/v)A_{\rm lens}=\tau\Delta E_\perp\) is also correct. It uses twice the matched-endpoint action difference before the factor six. Numerical normalization should be fixed rather than absorbed into an undefined threshold. \(\Delta E_\perp\) is deterministic transverse work, distinct from a quantum energy uncertainty. Comparing the actual path to its inertial tangent changes the terminal point and loses endpoint-gauge invariance.
4. Threshold versus exact quantization
With \(h=2\pi\hbar\) and equal angular steps \(\Delta\phi=2\pi/N\), the inequality \(J\Delta\phi\ge h\) gives \(J\ge N\hbar\). For any positive integer \(N\), \(J=(N+1/2)\hbar\) satisfies every one of these \(N\) cell bounds and closes the angle at \(2\pi\), while lying off the asserted action lattice. An exact tiling assumption is therefore an extra premise in the supplied Claude argument. A threshold on comparison-path defects also requires an explicit map before being applied to orbital action-angle cells.
For the supplied Kepler spectrum write \(E_N=-A/N^2\), \(A>0\). If \(\omega_N=\hbar^{-1}dE_N/dN\) and \(T_N=2\pi/\omega_N\), then
\[\frac{(E_{N+1}-E_N)T_N}{h} =\frac{N(2N+1)}{2(N+1)^2}\longrightarrow1.\]
The finite difference approaches the frequency rule; the derivative relation is exact. These algebraic checks assess the supplied conditional model and leave its physical selection premise explicit.
5. Research use
Use the arc–chord construction in A03’s nonuniform cut-point test. State the refinement law, conserved endpoint data, action normalization and physical condition proposed to exclude a vanishing remainder. Keep a supplied minimum cell axiom as a separate comparison branch. Historical attribution follows the Newton passage audit, not the modern calculation.