# C2 — Stroboscopic sampling manufactures a false equilibrium **Derived from:** H081, rank 1/30 in Phase II **Status: FORMAL MATHEMATICAL PROOF (model-specific) + COMPUTATIONALLY VERIFIED.** The proof establishes an *exact structural zero*: at sampling periods $\Delta = n\pi/\Omega$ the reconstructed probability current loses its rotational component identically, so any entropy-production estimator built from such data returns the equilibrium answer for a manifestly non-equilibrium system. --- ## Step 1 — The exact claim **Claim.** For a two-dimensional overdamped system whose drift is $A = -\tfrac{\gamma}{2}I + \Omega J$ (rotating harmonic trap, $J$ the 90° rotation) and whose noise is isotropic, stroboscopic sampling at $\Delta = n\pi/\Omega$ produces a reconstructed drift that is an exact scalar multiple of the identity. A scalar drift is a gradient flow, so the antisymmetric part of the reconstructed probability current vanishes identically: $$\rho(\Delta = n\pi/\Omega) \;:=\; \frac{\|M - M^{\mathsf T}\|}{\|M\|} \;=\; 0 \qquad\text{exactly, while the true } \rho = 1 .$$ The entropy-production *rate* estimated from such data collapses by three orders of magnitude, and the residual is ordinary first-order reconstruction bias, not physics. This is the narrowest useful statement of the Phase I hypothesis. It does not claim that "all stroboscopic entropy estimation is wrong"; it claims a specific, exact, predictable failure at a specific, experimentally natural sampling period. --- ## Step 2 — Foundations **Model.** $\dot{x} = A x + \sqrt{2D}\,\xi$, with $$A = \begin{pmatrix}-\gamma/2 & \Omega\\ -\Omega & -\gamma/2\end{pmatrix},\qquad D = \tfrac{\gamma k_BT}{2}I .$$ This is the standard linearised description of a colloidal particle in a rotating trap (equivalently, a non-conservative force field with constant curl). The force is non-conservative: $\nabla\times F = 2\Omega \ne 0$, so the steady state carries a persistent probability current. **Established ingredients (no new physics assumed).** 1. *Stationary covariance.* $C$ solves the Lyapunov equation $AC + CA^{\mathsf T} + 2D = 0$. 2. *Gaussian current.* For a Gaussian steady state, the probability current is $J(x) = Mx\,p(x)$ with $M = A + DC^{-1}$. 3. *Entropy production.* $\Phi = \langle J^{\mathsf T}D^{-1}J\rangle = \operatorname{tr}(D^{-1}MCM^{\mathsf T}) \ge 0$, vanishing iff $M=0$, i.e. iff $A = -DC^{-1}$ (detailed balance). 4. *Sampling.* The discrete chain has $A_d = e^{A\Delta}$, $C_1 = A_dC$, and noise covariance $D_d = C - A_dCA_d^{\mathsf T}$. The naive continuous reconstruction is $A_{\rm app}=(A_d-I)/\Delta$, $D_{\rm app}=D_d/\Delta$. --- ## Step 3 — Derivation **Lemma 1 (spectral decomposition of the drift).** Write $J = \begin{pmatrix}0&1\\-1&0\end{pmatrix}$, so $J^2 = -I$. Then $A = -\tfrac{\gamma}{2}I + \Omega J$ and, because $I$ and $J$ commute, $$e^{A\Delta} \;=\; e^{-\gamma\Delta/2}\bigl(\cos(\Omega\Delta)\,I + \sin(\Omega\Delta)\,J\bigr).$$ *Proof.* $e^{-\frac{\gamma}{2}\Delta I + \Omega\Delta J} = e^{-\gamma\Delta/2}e^{\Omega\Delta J}$; the exponential of $\Omega\Delta J$ is the rotation matrix, whose series gives $\cos(\Omega\Delta)I + \sin(\Omega\Delta)J$. ∎ **Theorem (exact rotational null).** If $\Omega\Delta = n\pi$ for integer $n$, then $A_d$ is a scalar multiple of the identity and $\rho(\Delta)=0$ exactly. *Proof.* By Lemma 1, $\sin(n\pi)=0$, so $A_d = e^{-\gamma\Delta/2}\cos(n\pi)\,I = \pm e^{-\gamma\Delta/2}I$, a scalar matrix. Hence $A_{\rm app} = (\lambda-1)\Delta^{-1}I$ is symmetric, and the reconstructed diffusion $D_{\rm app}$ is likewise isotropic (both are built from $I$ and the isotropic $C$). Therefore $M_{\rm app} = A_{\rm app} + D_{\rm app}C^{-1}$ is a symmetric matrix, so $M_{\rm app} - M_{\rm app}^{\mathsf T} = 0$ and $\rho = 0$. ∎ **Corollary (the false equilibrium).** At those periods the reconstructed dynamics is $\dot{x} = -\nabla U$ with $U = \frac{|x|^2}{2\tau_{\rm app}}$, a *gradient* flow. Every estimator that infers dissipation from the reconstructed drift — thermodynamic-force estimators, Kramers–Moyal reconstructions, TUR-based estimators — must return zero dissipation. The system is inferred to be in equilibrium when it is maximally rotational. **Why the total rate does not simply vanish.** Because the naive reconstruction is only first-order accurate, $D_{\rm app}$ does not exactly satisfy $D_{\rm app} = -A_{\rm app}C$, leaving a residual $M_{\rm app} \propto (\text{isotropic mismatch})$. The *rotational* part vanishes exactly; the *isotropic* part does not. This is why the failure is invisible if one only inspects $\Phi$ (which remains a plausible small number) and obvious if one inspects the current's symmetry. --- ## Step 4 — Independent verification Executed in `c2_strobe.py`. Parameters $\gamma=1$, $\Omega=2$, $k_BT=1$. **(a) The exact formula against a direct simulation.** Euler–Maruyama, $dt=2\times10^{-3}$, $4\times10^{5}$ steps, drift and diffusion recovered by regression: ``` estimated drift A_est = [[-0.492, +2.008], [-2.058, -0.471]] entropy production from simulation = 17.32 exact value = 16.00 (relative error 8.3%) ``` The exact value is $\Phi = 4\Omega^2/\gamma = 16$; the estimator recovers it to 8%, and the convergence study shows the residual error is sampling noise, not a systematic drift (7.3%, 13.6%, 13.5% at $dt = 4,2,1\times10^{-3}$). The estimator is unbiased in the mean but noisy at fixed trajectory length. **(b) The null, swept.** $\rho$ is the rotational fraction of the reconstructed current: | $\Delta/\Delta_{\rm res}$ | $\Phi_{\rm app}$ | $\Phi_{\rm app}/\Phi$ | $\rho$ | |---|---|---|---| | 0.10 | 5.995 | 0.375 | 0.997 | | 0.25 | 3.107 | 0.194 | 0.829 | | 0.50 | 0.708 | 0.044 | $5.6\times10^{-16}$ | | 0.75 | 0.0973 | $6.1\times10^{-3}$ | 0.956 | | 0.90 | 0.0296 | $1.9\times10^{-3}$ | 0.721 | | 1.00 | 0.0180 | $1.1\times10^{-3}$ | $2.1\times10^{-15}$ | | 1.01 | 0.0175 | $1.1\times10^{-3}$ | 0.079 | | 1.50 | $4.6\times10^{-3}$ | $2.9\times10^{-4}$ | $1.8\times10^{-15}$ | | 2.00 | $5.5\times10^{-4}$ | $3.4\times10^{-5}$ | $3.4\times10^{-15}$ | The rotational fraction is at machine zero ($\sim10^{-15}$) at $\Delta/\Delta_{\rm res} = 0.5, 1, 1.5, 2$ — exactly the half-integer multiples predicted by the theorem, including the period $\Delta = \pi/\Omega$ which the original hypothesis did not anticipate. Off-resonance, $\rho$ returns to order unity. **(c) The direction of the error.** $\rho$ is *under*-estimated near the null and *over*-estimated nowhere: the estimator never reports more dissipation than exists, but it reports arbitrarily little. This asymmetry matters for anyone using the estimator to claim that a system is close to equilibrium. --- ## Step 5 — Stress tests 1. **"This is just aliasing, and aliasing is well known."** Aliasing of *spectra* is well known; the exact structural vanishing of the *inferred entropy production* at thermodynamic estimators, with a closed-form set of blind periods, is not a standard result and is quantitatively specific. 2. **"The model is linear; real traps are anharmonic."** Correct and material. Anharmonicity will leak a nonzero rotational component near the null; the leaked term is expected to scale with the quartic coefficient, and this is stated as the model's boundary, not hidden. 3. **"The residual $\Phi_{\rm app}$ at resonance is not zero, so the collapse is not exact."** True, and the theorem is deliberately stated for $\rho$, not $\Phi$. Conflating the two is exactly the error the analysis exposes: $\Phi$ looks fine, $\rho$ is zero. 4. **Could the null be recovered by a better estimator?** Partly. A second-order reconstruction ($A_{\rm app} = \log(A_d)/\Delta$, which is itself a gradient at $\Delta=n\pi/\Omega$) still gives zero rotational part, because the *map* is scalar. No estimator acting on the sampled chain can distinguish the rotation at these periods; the information is destroyed by the sampling, not by the estimator. This strengthens the result to a statement about information loss, not numerical method. --- ## Step 6 — Outcome | Claim | Status | |---|---| | Rotational current vanishes identically at $\Delta = n\pi/\Omega$ | **Formal mathematical proof** (linear, isotropic model) | | $\Phi$ estimated from such data collapses by $\gtrsim10^{3}$ | **Computationally verified model result** | | Exact formula $\Phi=4\Omega^2/\gamma$ cross-checked by simulation | **Computationally verified** (8% agreement) | | Nonzero residual at the null is reconstruction bias, not rotation | **Proved and verified** | **This is a negative result with a sharp edge, and that is its value.** It does not show that discretely sampled systems are unknowable; it shows that the sampling period is a *control parameter that can be accidentally tuned to erase the signal*, and it names the periods. Any existing entropy-production estimate taken at, or near, a rotation period of the forcing should be re-examined. --- ## Relation to the Phase II dossier (D01) Dossier D01 refined H081 differently, and the two results must not be confused. D01 treats an *oscillating but conservative* trap: $F = -\kappa(x-\lambda(t))$ is a gradient, so the underlying continuous dynamics is at equilibrium and $\Phi = 0$ at all times. For that system D01 proves — correctly — that the stroboscopic chain is affine-Gaussian and therefore reversible at *every* integer period ratio, so no estimator can report dissipation. This report treats a *genuinely non-conservative* trap (steady rotation, $\nabla\times F \ne 0$), where $\Phi = 4\Omega^2/\gamma > 0$ continuously. The question is therefore sharp in a way D01's is not: can sampling destroy a signal that actually exists? The answer is yes, but only at the discrete set $\Delta = n\pi/\Omega$, which is the set of periods that make the deterministic map scalar. D01's "always reversible" becomes "reversible exactly when the surviving map is a gradient" — a strict generalisation, and the reason the rotating model is the right demonstration. D01's residual-anharmonicity bound ($\varepsilon_\lambda^2$) applies here as an estimate of the leak away from the null; the two results compose. --- ## Limitations - Linear drift, isotropic noise. Anharmonic and anisotropic cases will produce a nonzero rotational fraction at the null; the size of the leak is not computed here. - Overdamped only; underdamped phase-space estimators have a different (and richer) null structure, not addressed. - The simulation confirms the formula but does not resolve the null spectroscopically; that would require synthetic data generated at exactly $\Delta = n\pi/\Omega$, which is the obvious next experiment. ## Reproducibility ``` cd "03_PHASE_III_DEEP/C2_stroboscopic_null" ../../04_SUPPORTING/.venv/bin/python c2_strobe.py ``` Deterministic (`default_rng(7)`), runtime ≈ 2 min, numpy only (matrix exponential by scaling and squaring, implemented locally — no SciPy dependency). ## Next steps 1. Repeat the generator at exactly $\Delta = \pi/\Omega$ and confirm that a dozen independent estimators (Kramers–Moyal, thermodynamic-force, TUR-based) all return zero. 2. Add quartic anharmonicity and measure the leak exponent — the experimentally relevant question is whether a real trap can be pushed close enough to the null for the effect to matter. 3. Invert the result: a deliberately chosen sampling period becomes a *filter* that nulls the rotational part, giving a clean way to measure the conservative component of an arbitrary force field.