# Section 10: Every admitted preparation-clock position is harmless for a reason

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<a id="section-10"></a>

## 10 Every admitted preparation-clock position is harmless for a reason

 The old preparation clock $C$ is retained but no longer controls the active programme. It is an ordinary free particle with $M_C=10^8$. In its co-moving coordinate, the free propagator and one-dimensional Sobolev bound give, through time three, 

$$

 e_0\le\frac{3(112+4000)}{2M_C},\quad
 e_1\le\frac{3(4000+190000)}{2M_C},\quad
 |\dot C-1|\le\frac{e_1}{M_C(1-e_0)}<3\times10^{-11}.

$$

 The displacement error is below $9\times10^{-11}$, less than the $.05$ plateau clearance. A first-exit argument proves the tube for every $C_0\in[-1.25,-.75]$. No averaging over $|\chi_C|^2$ is used. Its correlations are irrelevant to active histories by Theorem [6.1](/quantum-measurement/research/nonequilibrium-records/exact-relative-coordinate-reduction-with-finite-recoil#pa:thm:recoil).

The active writer clock is $S$. Its actual distribution is controlled by the original joint cap $f_0\le2$, not by a new Born-clock premise. A pointwise trajectory tube for every $S_0$ is neither claimed nor needed by the active-law current proof.