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Newton’s inscribed polygon differs from the parabola by the phase of its segments

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For a body of mass \(m\) under a constant transverse force \(F\) on \([0,\tau]\), take any partition into steps \(\tau_1,\dots,\tau_N\) and Newton’s polygon of Proposition I inscribed in the Galileo parabola: uniform motion along each chord, impulses at the vertices. Classically the polygon and the curve are in the same state at every vertex. Quantum mechanically their evolutions differ by a pure phase, the same for every state of the body:

\[\boxed{W_{\rm poly}=e^{i\theta_N}\,W_F,\qquad \theta_N=\frac{F^2}{24m\hbar}\sum_{j=1}^N\tau_j^3 =\frac{F}{2v\hbar}\sum_{j=1}^NS_j ,}\]

where \(S_j=vF\tau_j^3/(12m)\) is the parabolic segment cut off by the \(j\)-th chord at horizontal speed \(v\), half of Lemma XI’s tangent area for that step. Three consequences follow.

Both phases are c-numbers, so squeezing, mixing or any other choice of body state leaves them unchanged. That is the property the two quantum floors lacked, as corrected on 2026-09-22 in the mark-cost and probabilistic notes. The single-step case is the chord lens of the constant-force note, and the closed loop of N01 is another instance of Lemma 2 below. The phase of Theorem 1 is also known from numerical analysis: the inscribed polygon is the kick–drift–kick (Strang) splitting of the propagator, and for a linear potential that splitting is exact up to a c-number, the double commutator of the potential with the kinetic term. New here are the reading of that c-number in Newton’s areas (Lemma XI’s tangent areas, Archimedes’ segments and inscribed triangles), the insertion law, the ordering holonomy and Theorem 7’s identity between the phase and the mark bound. Exploratory; no ledger promotion.

1. Setting and the exact lift

Transverse coordinate \(\hat y\), momentum \(\hat p\), \([\hat y,\hat p]=i\hbar\), mass \(m>0\); the horizontal motion at speed \(v\) is common to every history considered and cancels. A force history is an impulse density or measure \(f\) on \([0,\tau]\) with Hamiltonian \(H_f(t)=\hat p^2/2m-f(t)\hat y\). In the interaction picture of the free evolution \(U_0\), write \(W_f(t)=U_0(t)^\dagger U_f(t)\) and use the Weyl operators of the probabilistic note,

\[D(a,b)=\exp\!\left[\frac{i}{\hbar}(b\hat y-a\hat p)\right],\qquad D(z_1)D(z_2)=e^{-i\omega(z_1,z_2)/2\hbar}D(z_1+z_2),\]

with \(\omega(z_1,z_2)=a_1b_2-a_2b_1\). An impulse \(J\) at time \(t\) is \(U_0(t)^\dagger e^{iJ\hat y/\hbar}U_0(t)=D(-Jt/m,J)\), since \(U_0(t)^\dagger\hat yU_0(t)=\hat y+\hat pt/m\). The history’s phase-plane path is

\[z_f(t)=\bigl(\alpha_f(t),\beta_f(t)\bigr) =\Bigl(-\frac1m\int_0^ts\,f(s)\,ds,\ \int_0^tf(s)\,ds\Bigr),\]

the transverse displacement relative to free motion referred back to \(t=0\) by the free flow, and the accumulated impulse. For the constant force, \(z_F(t)=(-Ft^2/2m,\ Ft)\), a parabola in the phase plane.

Lemma 1. \(W_f(\tau)=\exp\bigl[\frac{i}{2\hbar}\int_0^\tau\omega(z_f,dz_f)\bigr]\,D\bigl(z_f(\tau)\bigr)\).

Proof. For finitely many impulses with increments \(\delta z_1,\dots,\delta z_N\) in time order, induction on the Weyl relation gives \(D(\delta z_N)\cdots D(\delta z_1)=\exp[\frac{i}{2\hbar}\sum_{i<j}\omega(\delta z_i,\delta z_j)]D(\sum_j\delta z_j)\), and \(\sum_{i<j}\omega(\delta z_i,\delta z_j)=\sum_j\omega(z_{j-1},\delta z_j)\) with \(z_{j-1}=\sum_{i<j}\delta z_i\). For a density \(f\) the generator \(\frac{i}{\hbar}f(t)(\hat y+\hat pt/m)\) has c-number commutators at different times, so the Magnus series stops at second order and the Riemann sums converge to \(\int\omega(z_f,dz_f)\), as in N01 §4. \(\square\)

Lemma 2. If \(z_f(\tau)=z_g(\tau)\), then \(W_fW_g^\dagger=e^{i\mathcal A(f,g)/\hbar}\,\mathbf 1\), where \(\mathcal A(f,g)=\frac12\oint\omega(z,dz)\) is the signed area enclosed by \(z_f\) followed by \(z_g\) reversed.

Proof. Lemma 1 for both, and Green’s theorem for the closed loop. \(\square\)

Two histories with equal endpoints leave the body in the same classical state after \(\tau\), whatever happened in between, and Newton’s mechanics has no later observation that separates them. Lemma 2 says the quantum evolutions still differ, by the enclosed phase-plane area over \(\hbar\), and that the difference is a multiple of the identity. This is the standard geometric phase of displaced oscillators, used in trapped-ion phase gates (Sørensen and Mølmer, PRA 62, 022311, 2000; Leibfried et al., Nature 422, 412, 2003, both metadata). The \(t^3\) phase of a freely falling wave packet is the single-history form of it (Greenberger and Overhauser, RMP 51, 43, 1979, metadata).

2. The inscribed polygon

Partition \(0=t_0<\dots<t_N=\tau\) with \(\tau_j=t_j-t_{j-1}\) and midpoints \(c_j=(t_{j-1}+t_j)/2\). Newton’s inscribed polygon gives the impulse \(F\tau_1/2\) at \(t_0\), \(F(\tau_j+\tau_{j+1})/2\) at each interior \(t_j\), and \(F\tau_N/2\) at \(t_N\): each step’s impulse \(F\tau_j\) is split equally between its two ends.

Theorem 1. (i) On \((t_{j-1},t_j)\) the polygon moves uniformly with momentum \(Fc_j\) along the chord of the parabola, and its phase-plane path passes through \(z_F(t_j)\) at every vertex. (ii) \(W_{\rm poly}(\tau)=e^{i\theta_N}W_F(\tau)\) with \(\theta_N=F^2\sum_j\tau_j^3/(24m\hbar)\). (iii) With \(S_j=vF\tau_j^3/(12m)\) the area between the \(j\)-th arc and its chord, \(\theta_N=(F/2v\hbar)\sum_jS_j\).

Proof. (i) After the second half of the impulse at \(t_{j-1}\) the momentum is \(Ft_{j-1}+F\tau_j/2=Fc_j\), and in time \(\tau_j\) it carries the body through \(Fc_j\tau_j/m=F(t_j^2-t_{j-1}^2)/2m\), the parabola’s transverse increment. In the phase plane the two half-impulses of step \(j\) are \(k_1=(-F\tau_jt_{j-1}/2m,\ F\tau_j/2)\) and \(k_2=(-F\tau_jt_j/2m,\ F\tau_j/2)\), whose sum is the parabola’s increment \(\delta z_j=(-F\tau_jc_j/m,\ F\tau_j)\).

  1. Both evolutions are products over steps, in the same order, of operators with the same displacements \(\delta z_j\), so their ratio is the product of the per-step phase ratios. By Lemma 1 the polygon’s step is \(D(k_2)D(k_1)=e^{i\omega(k_1,k_2)/2\hbar}D(\delta z_j)\) with \(\omega(k_1,k_2)=\frac{F\tau_j}{2}\cdot\frac{F\tau_j}{2m}(t_j-t_{j-1})=\frac{F^2\tau_j^3}{4m}\). The curve’s step, with \(s=t-t_{j-1}\) and path \((-F(s^2+2st_{j-1})/2m,\ Fs)\) from \(z_F(t_{j-1})\), has \(\omega(z,dz)=\frac{F^2s^2}{2m}ds\), hence phase \(\frac1{2\hbar}\int_0^{\tau_j}\frac{F^2s^2}{2m}ds=\frac{F^2\tau_j^3}{12m\hbar}\). Neither depends on \(t_{j-1}\). The ratio is \(\exp[\frac{i}{\hbar}(\frac18-\frac1{12})\frac{F^2\tau_j^3}{m}]=\exp[\frac{iF^2\tau_j^3}{24m\hbar}]\).

  2. Over the step, the chord exceeds the parabola by \(\frac{F}{2m}(t-t_{j-1})(t_j-t)\), and integrating against \(dx=v\,dt\) gives \(S_j=vF\tau_j^3/12m\). Then \(\frac{F}{2v}S_j=\frac{F^2\tau_j^3}{24m}\). \(\square\)

Lemma XI’s tangent area for the step is \(A_j=vF\tau_j^3/6m=2S_j\), so \(\theta_N=(F/4v\hbar)\sum_jA_j\), and Proposition 1 of the paper is the statement that this sum vanishes with the mesh. In the phase plane the polygon’s path is the circumscribed tangent polygon of the parabola \(z_F\): each impulse at \(t_j\) moves along the tangent direction \((-t_j/m,1)\), and the two paths touch at every \(z_F(t_j)\), which is why the area between them splits step by step. The phase is also the difference of the two histories’ classical actions between the same endpoints, whatever the endpoints, which is why a comparison of quadratic Hamiltonians leaves only a global phase.

Corollary 2 (insertion). Replacing step \(j\) by steps \(\lambda\tau_j\) and \((1-\lambda)\tau_j\) lowers \(\theta_N\) by

\[\frac{F^2\tau_j^3}{24m\hbar}\bigl[1-\lambda^3-(1-\lambda)^3\bigr] =\frac{F^2\tau_j^3\,\lambda(1-\lambda)}{8m\hbar} =\frac{F}{2v\hbar}\,T_j(\lambda),\]

with \(T_j(\lambda)=S_j-S_j'-S_j''\) the triangle inscribed between the old chord and the new vertex. At \(\lambda=\frac12\), \(T_j=\frac34S_j\), which is Archimedes’ Quadrature of the Parabola, Propositions 17 and 24 (the segment is four thirds of its inscribed triangle, by balancing and by exhaustion; companion). After \(k\) rounds of halving every step, \(\theta\) is \(4^{-k}\) of its value. \(\square\)

Each inserted point therefore changes the quantum evolution by a computable phase, fixed by the area the insertion removes, and the change is the same for every state of the body. A step’s own phase reaches one radian at

\[\tau_\hbar=\left(\frac{24m\hbar}{F^2}\right)^{1/3},\]

the scale that the constant-force note found for the single chord lens. The insertion mesh of the mark theorem, \(\tau_*=(48z_{1-\epsilon}^2m\hbar/F^2)^{1/3}\), has the same form, obtained there from the cost of a mark and here from the phase of a segment with no apparatus at all.

Archimedes’ exhaustion, read through \(\hbar\)

Archimedes squares the parabola by exhaustion. In the segment cut off by a chord he inscribes the triangle whose third vertex is “the vertex of the segment”, the point of the arc where the tangent is parallel to the chord. That leaves two smaller segments, and he inscribes a triangle in each; every round’s triangles total a quarter of the round before. Proposition 23 sums the series: for areas \(A,B,C,\dots,Z\), each four times the next, \(A+B+C+\dots+Z+\frac13Z=\frac43A\). Proposition 24 concludes that “every segment bounded by a parabola and a chord … is equal to four-thirds of the triangle which has the same base as the segment and equal height” (companion, passage).

The method begins with the cone. In the preface to The Method, Archimedes gives Eudoxus the proof “that the cone is a third part of the cylinder, and the pyramid of the prism, having the same base and equal height”, and gives “no small share of the credit to Democritus who was the first to make the assertion with regard to the said figure though he did not prove it” (companion, passage). On Heath’s reading, Democritus reached the third by treating the cone as made of sections, the setting of the dilemma about adjacent sections that Plutarch reports, and Eudoxus’s exhaustion is the proof that does without them. The Galileo comparison contains the same third. The sagitta grows as the square of the time, as the cone’s section grows as the square of the height, so the area between the inertial line and the parabola is one third of the rectangle with sides \(v\tau\) and \(s\):

\[A_{\rm inertial,fall}=\frac{vF\tau^3}{6m}=\frac13\,v\tau\,s,\]

as the cone is one third of its cylinder. What \(\hbar\) adds separates the two figures. The cone is static: marking its sections disturbs nothing, and Theorem 9(iii) of the paper finds no floor for it. The parabola here is traced by a motion, its pieces are phases of Newton’s polygon, and its exhaustion acquires the stopping scales below.

On the Galileo parabola, traced in time, the vertex of the segment over a step is the midpoint in time. There the velocity, \(F(t_{j-1}+t_j)/2m\), equals the slope of the chord. So Archimedes’ construction is Newton’s refinement by halving. After \(k\) rounds, the inscribed polygon of Theorem 1 is Archimedes’ inscribed polygon after \(k\) rounds, and the triangles he adds in a round are exactly the triangles that Corollary 2 removes. Newton measures the same figures from the other side: by Lemma XI and its corollaries the segment is one third of the tangent triangle and the area between tangent and arc two thirds, so \(A_j=2S_j\) and the tangent triangle is \(3S_j\).

Theorem 1 and Corollary 2 read Archimedes’ series as phases. Over one step, with \(T_j=\frac34S_j=vF\tau_j^3/16m\) the first inscribed triangle,

\[\theta_{\rm step}=\frac{F}{2v\hbar}S_j=\frac{F}{2v\hbar}\cdot\frac43T_j =\frac{F}{2v\hbar}\Bigl(T_j+\frac{T_j}4+\frac{T_j}{16}+\cdots\Bigr).\]

The first round of halving removes the phase \((F/2v\hbar)T_j\), the next a quarter of that, and so on. Proposition 23 is the statement that the phases removed by all the rounds add up to the step’s whole phase, the quantum difference between Newton’s polygon and the parabola. Classically, exhaustion shows that the segments left after \(k\) rounds, a fraction \(4^{-k}\) of the whole, can be made smaller than any given area. Quantum mechanically the same leftovers are phases, one per remaining segment, and \(\hbar\) sets two scales. Each segment’s phase falls below one radian once the steps are shorter than \(\tau_\hbar\); their total, which each round divides by four, falls below one radian once \(4^k\) exceeds \((\tau/\tau_\hbar)^3\). Up to that round Archimedes’ triangles are phases of order one or more, and after it the rest of his series is below a radian.

3. The ordering holonomy and a partition identity

Keep the curve’s step operators \(V_j=e^{i\varphi_j}D(\delta z_j)\) and compose them in another order. For a permutation \(\sigma\), write \(U_\sigma\) for the product with the steps applied in the order \(\sigma\).

Theorem 3. \(U_{\rm chron}U_\sigma^\dagger=\exp[\frac{i}{\hbar}\sum\omega(\delta z_i,\delta z_j)]\mathbf 1\), the sum over the pairs \(i<j\) that \(\sigma\) inverts. Every term is positive,

\[\omega(\delta z_i,\delta z_j)=\frac{F^2}{m}\tau_i\tau_j(c_j-c_i)>0\qquad(i<j),\]

so the reversed order is extremal, with

\[\Phi_N=\frac1\hbar\sum_{i<j}\omega(\delta z_i,\delta z_j) =\frac{F^2}{6m\hbar}\Bigl(\tau^3-\sum_j\tau_j^3\Bigr).\]

Proof. The per-step phases \(\varphi_j\) are common to every ordering. By the Weyl relation the phase of a product is \(\frac1{2\hbar}\) times the sum of \(\omega(\delta z_a,\delta z_b)\) over pairs with \(a\) applied before \(b\); a pair kept in order cancels between the two products, and an inverted pair contributes \(\omega(\delta z_i,\delta z_j)-\omega(\delta z_j,\delta z_i)=2\omega(\delta z_i,\delta z_j)\). With \(\delta z_i=(-F\tau_ic_i/m,\ F\tau_i)\), \(\omega(\delta z_i,\delta z_j)=\frac{F^2}{m}\tau_i\tau_j(c_j-c_i)\). For the sum, \(c_j-c_i=\frac{\tau_i}2+\sum_{i<k<j}\tau_k+\frac{\tau_j}2\), so \(\sum_{i<j}\tau_i\tau_j(c_j-c_i)=\frac12\sum_{i\ne j}\tau_i^2\tau_j+\sum_{i<k<j}\tau_i\tau_k\tau_j\), and \((\sum_j\tau_j)^3=\sum_j\tau_j^3+3\sum_{i\ne j}\tau_i^2\tau_j+6\sum_{i<k<j}\tau_i\tau_k\tau_j\) gives \(\frac16(\tau^3-\sum_j\tau_j^3)\). \(\square\)

Corollary 4 (a partition identity). \(\Phi_N+4\theta_N=F^2\tau^3/(6m\hbar)\) for every partition, and \(F^2\tau^3/6m=\tau\Delta E/3=(F/v)A_{\rm inertial,fall}\) with \(A_{\rm inertial,fall}=vF\tau^3/6m\) the paper’s area (1). \(\square\)

In areas, \(\hbar\Phi_N=(F/v)(A-\sum_jA_j)\) and \(4\hbar\theta_N=(F/v)\sum_jA_j\): the ordering holonomy carries the part of the Galileo area that refinement keeps, and the polygon defect carries the part that Proposition 1 sends to zero. For the uniform partition, \(\Phi_N=\frac{\tau\Delta E}{3\hbar}(1-N^{-2})\).

The chronological and reversed products realize the same displacement, so the holonomy is a property of the composition law. No force history acting on a body realizes a reordering other than the chronological one: an impulse applied at the wrong time must be translated back, so a reversed order needs displacements as well as impulses, which trapped-ion practice supplies. Theorem 3 is therefore a statement about the composition law, and Theorem 1 is the one that needs only two force histories acting on a real body.

4. What the phases give as a record

Put the two histories of Theorem 1 on the arms of a control qubit, as in N01 §4. The body ends in the same state on both arms, so the qubit acquires the relative phase \(\theta_N\) and nothing else, and the optimal single-shot error in deciding polygon against curve is \(\epsilon=\frac12(1-|\sin(\theta_N/2)|)\), with N01’s formula for \(n\) copies. In one controlled pass, a single step is resolved from its chord at error \(\epsilon\) only if

\[\frac{F^2\tau_j^3}{24m\hbar}\ \ge\ 2\arcsin(1-2\epsilon),\qquad \tau_j\ \ge\ \left(\frac{48\,m\hbar\,\arcsin(1-2\epsilon)}{F^2}\right)^{1/3},\]

for every state of the body, pure or mixed, squeezed or not. The bound concerns this readout and its resource, the number of controlled passes: \(k\) passes accumulate \(k\theta_N\) and lower the threshold on \(\tau_j\) by \(k^{1/3}\), as \(n\) copies do through N01’s formula. The body’s own displacement is a separate record, priced by the mark and aperture theorems, and squeezing acts on that record. What the phase supplies is the part of the signal that no preparation of the body can move.

5. The premise, restated

The identities above use the Weyl relations and nothing else: no state, no variance, no apparatus. Of the two quantum premises used so far in this programme, Robertson’s inequality is the variance consequence of the commutator, and the 2026-09-22 corrections show that squeezing moves it. The commutator’s phase content is what Theorems 1 and 3 use, and squeezing leaves it fixed. So the premise that carries \(h>0\) into the Galileo comparison without a resource bound is the central extension: an impulse \(J\) and a displacement \(d\) compose, in Newton’s Corollary I, only up to the phase \(Jd/\hbar\) of their parallelogram. Classically Corollary I is exact and commutative, the polygon and the curve are the same state at every vertex, and every \(\theta_N\) vanishes.

Newton’s fits supply a Newton-age form of that premise. Two passages of the Opticks, now held verbatim in the companion, fix how the interval of the fits varies with the corpuscle’s speed. Book II Part III Prop. XVII, “manifest by the 10th Observation”, makes the intervals in two mediums “as the Sine of Incidence to the Sine of Refraction”, so the interval is shorter in the denser medium by the refractive ratio. Prop. X takes light to be “swifter in Bodies than in Vacuo, in the proportion of the Sines”, the hypothesis of Newton’s emission theory. Together they give

\[\Lambda\,v=\text{const across media},\qquad\text{hence}\qquad \Lambda\,p=\text{const across media}\]

for each colour, the corpuscle’s mass being unchanged by refraction. That Newton’s interval shortens where his corpuscle speeds up, so that it varies inversely with the speed as de Broglie’s wavelength does, is an old observation (Sakkopoulos, Eur. J. Phys. 9, 123, 1988, metadata; the fits are discussed in Whittaker’s History of the Theories of Aether and Electricity, and de Broglie’s Nobel lecture of 1929 names them). What is ours is the continuous-speed extension below and its identification with Lemma XI’s areas.

Proposition 5. Suppose, extending Props. X and XVII from a step in the medium to a continuously varying speed, that a corpuscle’s fits advance by one interval \(\Lambda=\Lambda_0p_0/|p|\) per length of path. Then the number of fits along a path is its Maupertuis action \(W=\int p\cdot dq\) in units of \(\Lambda_0p_0\). For the inscribed polygon and the curve of Theorem 1 the counts differ by

\[\frac{\Delta W}{\Lambda_0p_0}=\frac{F^2}{12m\Lambda_0p_0}\sum_j\tau_j^3 =\frac{1}{2\Lambda_0p_0}\,\frac{F}{v}\sum_jA_j\]

fits, with \(A_j\) Lemma XI’s tangent areas.

Proof. Along the actual motion \(p\parallel dq\), so \(|p|\,ds=p\cdot dq\) and the count is \(\int p\cdot dq/(\Lambda_0p_0)\). The horizontal parts are common. On step \(j\) the curve has \(\int p_y\,dy=\frac{F^2}{m}\int_{t_{j-1}}^{t_j}t^2\,dt\) and the chord has \(\frac{F^2}{m}c_j^2\tau_j\); their difference is \(\frac{F^2}{m}\tau_j\bigl[\frac13(t_{j-1}^2+t_{j-1}t_j+t_j^2)-\frac14(t_{j-1}+t_j)^2\bigr] =\frac{F^2\tau_j^3}{12m}\), and \(F^2\tau_j^3/6m=(F/v)A_j\). \(\square\)

In Newton’s optics the fit at arrival decides whether the ray is reflected or transmitted at the next surface (Prop. XII), so on his own terms the polygon and the curve are optically different corpuscles once their fits counts differ by half an interval, although they agree in position and velocity at every vertex. Every ingredient is his: Lemma XI’s areas, the measured \(\Lambda=1/89000\) inch, the refraction invariance of \(\Lambda p\) from Props. X and XVII, and the fits as a determinate periodic disposition. The unknown is \(p\), which he could not measure. Against the quantum statement, with \(\Lambda=\lambda/2\) and \(\lambda p=2\pi\hbar\), and counting \(2\pi\) per interval, a full return of the disposition from one fit of easy reflexion to the next (the period of the thin-film intensity), the fits phase in radians is \(4\theta_N\): one factor \(2\) because Newton’s interval is half a wavelength, and one because \(\Delta W=2\Delta S\) for this pair of histories. The wave’s own phase, which advances \(\pi\) per interval, differs by \(2\theta_N\).

The determinism of Prop. XII is untouched by this, and it makes Newton’s version different in kind from the quantum one. His fit is a determinate periodic property of a single corpuscle, so on his terms the polygon and the curve would be told apart on one corpuscle at the next surface; the quantum phase is relative between two histories and shows only in a superposition. Newton had no superposition of amplitudes to state the quantum version with. Imaginary quantities enter his algebra only as “impossible” roots of equations, and his mechanism for the fits, in Query 29, is that the rays “stir up Vibrations in what they act upon, which Vibrations being swifter than the Rays, overtake them successively, and agitate them so as by turns to increase and decrease their Velocities, and thereby put them into those Fits” (companion, passage): a wave that disposes the corpuscle, with no adding of amplitudes. Addition of amplitudes arrives with Young’s principle of interference (Phil. Trans. 92, 12, 1802, metadata). What the fits supply is a phase whose rate is proportional to momentum, and a count of such a phase is order-sensitive in the parallelogram sense: a displacement \(d\) followed by a kick \(J\) counts \(p\,d\), the kick first counts \((p+J)\,d\), and the difference \(Jd\) is the area of Corollary I’s parallelogram. So a per-colour form of the premise that carries \(h>0\) in the squeeze-immune form is present in Newton’s optics, with \(\Lambda p\) a positive refraction invariant. Two limits remain. Prop. XV’s rule for oblique emergence, a product of two secants, is more complicated than a count of intervals along the path, so Proposition 5’s extension is ours. And \(\Lambda p\) is a per-colour constant in Newton’s system: universality across colours is what Planck’s constant adds.

6. Every force law, and one functional for both legs

Let \(f\) be any integrable force history on an interval of duration \(\tau\), and compare the motion with its chord, the uniform motion between the same two positions. The deviation \(\delta y=y_{\rm chord}-y_f\) obeys \(m\,\delta y''=-f\) with \(\delta y=0\) at both ends, so \(\delta y=G_\tau f/m\) with the Dirichlet Green’s function \(G_\tau(s,u)=\min(s,u)\,(\tau-\max(s,u))/\tau\) in time measured from the start of the interval. Define the kinetic action of the deviation

\[\mathcal K_\tau[f]=\int_0^\tau\tfrac12m\,\delta\dot y^{\,2}\,dt =\frac1{2m}\int_0^\tau\!\!\int_0^\tau G_\tau(s,u)f(s)f(u)\,ds\,du\ \ge0,\]

zero only for \(f=0\). For the constant force it is \(F^2\tau^3/24m=\tau\Delta E/12=(F/2v)S\), with \(S\) the segment of the chord.

Theorem 7. (a) On each step of a partition, the two impulses \(J_j^-=\int(1-s/\tau_j)f\) and \(J_j^+=\int(s/\tau_j)f\) at its ends reproduce the step’s impulse and displacement, so Newton’s polygon built from them agrees with the motion at every vertex, and

\[W_{\rm poly}=\exp\Bigl[\frac{i}{\hbar}\sum_j\mathcal K_{\tau_j}[f]\Bigr]W_f .\]

Inserting vertices never raises the phase: \(\sum_j\mathcal K_{\tau_j}[f]\le\mathcal K_\tau[f]\). (b) For every protocol of Gaussian marks with uncorrelated error and recoil and \(\delta_j\Delta_j\ge\kappa\), with the initial position and velocity unknown, the deflection between the motion under \(f\) and free motion obeys

\[d^2\ \le\ \frac{\mathcal K_\tau[f]}{\kappa},\]

with equality approached by dense protocols when \(f\) has one sign. (c) Hence at \(\kappa=\hbar/2\) the best deflection any such protocol can reach is twice the phase between the motion and its chord, \(d^2_{\max}=2\,\mathcal K_\tau/\hbar=2\theta_1\).

Proof. (a) Measure time from the start of the step. The impulses at \(0\) and \(\tau_j\) are \(k^-=(0,J^-)\) and \(k^+=(-\tau_jJ^+/m,\ J^+)\) in the phase plane, and \(J^-+J^+=\int f\), \(\tau_jJ^+=\int sf\) match the step’s increment. Lemma 1 gives the polygon’s step phase \(\frac1{2\hbar}\omega(k^-,k^+)=\tau_jJ^+J^-/2m\hbar\) and the motion’s step phase \(\frac1{2\hbar}\int\omega(z,dz)=\frac1{2m\hbar}\iint_{u<s}(s-u)f(s)f(u)\), because along the path \(\omega(z,dz)=f(s)\,\frac1m\int_0^s(s-u)f(u)\,du\,ds\). Now \(\tau_jJ^+J^-=\iint s(\tau_j-u)f(s)f(u)/\tau_j\); symmetrizing and subtracting \(\frac12\iint|s-u|ff\) leaves the kernel \(\frac1{2\tau_j}[s(\tau_j-u)+u(\tau_j-s)-\tau_j|s-u|]=G_{\tau_j}(s,u)\). Shifting the time origin is the shear \((a,b)\mapsto(a-t_{j-1}b/m,b)\), which preserves \(\omega\), so the per-step phases are \(\mathcal K_{\tau_j}/\hbar\), and they multiply as in Theorem 1. The Dirichlet principle \(2m\mathcal K_\tau=\sup_T(2\int fT-\int T'^2)\) over \(T\) vanishing at the ends, restricted to \(T\) vanishing also at the interior vertices, gives the monotonicity.

  1. As in Theorem 2 of the paper, \(u^{\mathsf T}\Sigma u\ge2\kappa E/m\) and \(u^{\mathsf T}P=\frac1m\int_0^\tau fT\) with \(T\) vanishing at both ends. Since \(-(G_\tau f)''=f\), \(\int fT=\int(G_\tau f)'T'\le\sqrt{\langle f,G_\tau f\rangle}\sqrt E\), so \(d^2\le\langle f,G_\tau f\rangle/(2m\kappa)=\mathcal K_\tau/\kappa\). For \(f\) of one sign take \(T\propto G_\tau f\), which has one sign while \(T''\) has the other, so the interior weights and the \(S_j\) have opposite signs throughout and the dense balanced construction of Theorem 2 applies.

  2. Theorem 1 with a single step is the chord, whose phase is \(\theta_1=\mathcal K_\tau/\hbar\). \(\square\)

The two legs of the programme are governed by one functional of the force history. \(\mathcal K_\tau\) is what Newton’s refinement removes: it is the kinetic action of the motion relative to its chord, and for a constant force it is the chord’s segment times \(F/2v\). Planck’s constant converts it into the phase by which the motion differs from its inscribed polygon, a c-number that no preparation of the body moves; the mark cost converts it into the largest statistical distance any record of the motion can reach within the Gaussian model of marks. At \(\kappa=\hbar/2\) the second is exactly twice the first. The two legs stand on different footing: the phase is exact and uses the Weyl relations alone, while the mark bound holds inside a linear Gaussian model with the trade-off \(\delta\Delta\ge\kappa\), a model that squeezing and grid probes leave, as the recoil note records. The \(24\) in Theorem 1’s \(F^2\tau^3/24m\) and in the sharp bound \(d^2\le F^2\tau^3/24m\kappa\) of the paper’s Theorem 2 is one integral, \(\iint G_\tau=\tau^3/12\), multiplied by \(F^2/2m\).

7. Consequence for STATE

STATE’s next item 1 is discharged in a stronger form than it was stated. The state-independent quantity is identified for every refinement: Newton’s inscribed polygon differs from the parabola by \(\theta_N=(F/2v\hbar)\sum_jS_j\), each inserted vertex changes it by the inscribed triangle over \(\hbar\) in Archimedes’ proportions, and the ordering holonomy carries the complement, with \(\Phi_N+4\theta_N=\tau\Delta E/3\hbar\). The Newton-age form of the central extension is in §5: Props. X and XVII make \(\Lambda p\) a refraction invariant, and the fits count separates the polygon from the curve by \((F/v)\sum_jA_j/(2\Lambda p)\). Section 6 settles the general force law and joins the legs: for any force history the polygon phase is \(\sum_j\mathcal K_{\tau_j}/\hbar\), the sharp mark bound is \(d^2\le\mathcal K_\tau/\kappa\), and at \(\kappa=\hbar/2\) the best recorded deflection is twice the chord phase.