Exact finite-horizon phase recovery
Known initial position and momentum give three momentum-risk regimes under a complete noisy position history: free acceleration, a two-arc transient, and record-limited saturation. Position risk saturates earlier. Composition can therefore lose position information as well as momentum information during the transient.
R13, 2026-09-11. Let m,F,epsilon be positive, T >= 0, with known initial state, arbitrary measurable forces |f| <= F, and records y(t)=q(t)+e(t) for every t in [0,T], |e(t)| <= epsilon. There is no speed ceiling, stochastic law or measurement disturbance. Subtract the known inertial trajectory. The separate scalar minimax absolute errors are Q for position and P for canonical momentum. H=QP has action units, with no 2 pi normalization. Proof and literature status are recorded in C082–C083 and B44.
1. The exact radii
Put a=F/m, tau=sqrt(epsilon/a), and s=T/tau. Then
\[Q(T)=\varepsilon\min\{s^2/2,1\},\qquad P(T)=\sqrt{mF\varepsilon}\,V(s),\]
\[\boxed{V(s)=\begin{cases} s,&0\le s\le\sqrt2,\\ \sqrt{2s^2+4}-s,&\sqrt2\le s\le4,\\ 2,&s\ge4. \end{cases}}\]
Both coordinate extrema are attained by the same central-fibre motion. Consequently a blind interval b >= 0 after the last record has exact risks
\[P_b=P(T)+Fb,\qquad Q_b=Q(T)+bP(T)/m+Fb^2/(2m).\]
Here T is the observed horizon and T+b the prediction time. These products are products of global coordinate risks, not lower bounds on both errors for every individual motion.
2. Written proof through bounded-curvature paths
R11’s symmetric-fibre argument reduces each scalar minimax risk to the largest endpoint deviation of a path hidden by the zero record. Scale time by tau and position by epsilon. The admissible paths satisfy x(0)=x’(0)=0, |x’‘|<=1 almost everywhere, and |x(t)|<=1 on [0,s]. Let v=x’(s)>=0; symmetry covers the other sign.
First v<=s. For fixed v the acceleration u obeys integral u=v. Write u=2w-1 with 0<=w<=1 and integral w=(s+v)/2. The endpoint position is integral (s-t)u(t)dt. Its minimum places w=1 on the final interval of length (s+v)/2, since the weight s-t is decreasing. For a direct proof, let w_* be that final-interval indicator and c its left endpoint. Since integral (w-w_*)=0 and \(((s-t)-(s-c))(w-w_*)\ge0\) pointwise, integrating proves the minimum. Hence
\[x(s)\ge\frac{v^2+2sv-s^2}{4},\qquad v\le\sqrt{2s^2+4}-s.\]
Finally, for t in [s-v,s], x’(t)>=v-(s-t). This interval exists because v<=s, and integration gives
\[2\ge x(s)-x(s-v)\ge v^2/2,\qquad v\le2.\]
These three upper bounds have lower envelope V(s): the first and second meet at sqrt(2), the second and third at 4; the middle expression increases from sqrt(2) to 2 on that interval. The endpoint position separately obeys x(s)<=min(s^2/2,1).
For s<=sqrt(2), take u=+1 throughout. For sqrt(2)<=s<=4, set h=(s-V(s))/2=s-sqrt(s^2/2+1). Take u=-1 for h time and then u=+1 until s. The endpoint has x(s)=1 and x’(s)=V(s). Its minimum occurs at time 2h and equals -h^2. Since 0<=h<=1 and 2h<=s, the entire path stays in [-1,1]. For s>=4, wait at zero for s-4, apply u=-1 for one unit, then u=+1 for three units. This reaches -1 with zero velocity after the first two active units, then ends at +1 with velocity 2. Thus every upper bound is attained, including both regime junctions and s=0.
Opposite paths admit the same inertial record and share the exact initial state. Compatible-interval midpoint estimators give matching upper minimax bounds by R11. Because one extremizer maximizes position and momentum with the same sign, appending force +F for the blind interval attains both displayed prediction bounds. The triangle inequality and the force bound give the matching upper bounds on the central fibre.
3. Composition at every horizon
Let constituent i have (m_i,F_i,epsilon_i) and the same observed T. Allow the Cartesian product of force/error classes, as in R12. Set M=sum m_i, F_Sigma=sum F_i, E=sum m_i epsilon_i/M. With every constituent record retained the exact centre and total-momentum radii are
\[Q_A=\frac{\sum_i m_iQ_i(T)}M,\qquad P_A=\sum_iP_i(T).\]
With only the centre record retained, R12’s proportional force/error lift works at every T. Thus Q_B,P_B are exactly section 1’s single-body formulas with (m,F,epsilon)=(M,F_Sigma,E). Information inclusion gives Q_B>=Q_A and P_B>=P_A without any common-regime assumption.
There is a precise transient position-loss criterion. Put A_i=F_iT^2/2 and B_i=m_i epsilon_i. Then
\[M Q_A=\sum_i\min(A_i,B_i),\qquad M Q_B=\min(\sum_iA_i,\sum_iB_i).\]
The difference is strictly positive exactly when at least one A_i
For a concrete mixed-regime example, choose equal masses m and precision epsilon, F_1=F and F_2=16F, at T=sqrt(m epsilon/F). Then s_1=1, s_2=4, while the aggregate s_B=sqrt(17/2). The exact radii are
\[Q_A=\tfrac34\varepsilon,\qquad Q_B=\varepsilon,\] \[P_A=9\sqrt{mF\varepsilon},\qquad P_B=(\sqrt{714}-17)\sqrt{mF\varepsilon}>P_A.\]
Indeed sqrt(714)>26 since 714>676. The products are respectively (27/4)sqrt(mF) epsilon^(3/2) and (sqrt(714)-17)sqrt(mF) epsilon^(3/2). The second exceeds the first. For identical copies H_A=H_B=n H_1(T) at every horizon.
4. Limits and next physical premise
At fixed positive m,F,epsilon and T down to zero, H=F2T3/(2m). At fixed T>0 and epsilon down to zero, H=2 sqrt(mF) epsilon^(3/2) once T>=4 sqrt(m epsilon/F). The gap closes in either limit. The precision and force parameters set the scale; the finite preparation time sets when it is reached. R11’s sufficient time 4 tau is the exact earliest momentum saturation time for this known-initial-state experiment.
The source capsule is Seeber–Haimovich’s prepared bounded-curvature pair (B42) plus R12’s product-fibre and aggregate-image proofs. The finite-horizon rearrangement supplies the missing transient. The next bounded task R14 should restrict the record errors by one explicit shared apparatus constraint, for example sum_i (e_i/epsilon_i)^2<=1 pointwise, and derive the changed central fibre and aggregate image for two identical constituents. This tests which composition conclusion depended on freely aligned full-width errors.