The Clocks Beneath Collision Energy

The secondary term identifies a deficit beneath the leading collision energy. Its continuous part grows like the square of a logarithm, with coefficient $1/\pi^2$.

The deficit also has an exact description at each resolution. A fixed arithmetic mass is distributed among remainder clocks, and the deficit measures their weighted positions. Every increase in resolution advances the clocks together. The divisors of the new integer determine which ones reset.

The mass totals one, with a fixed share assigned to each clock.

Remainders at five and six

A remainder clock of length $k$ runs through $0,1,\ldots,k-1$ and returns to zero. Its position at resolution $N$ is the remainder when $N$ is divided by $k$.

The first five clocks show the effect at consecutive resolutions:

The clocks of lengths two, three and six reset together because all three divide six. Four and five advance without resetting. Every clock longer than six advances from five to six as well.

Each clock carries a particular positive weight. For period $k$, add the reciprocals of the smaller positive integers coprime to $k$, then divide by $k^2$:

$$w_k=\frac1{k^2}\sum_{\substack{1\le a<k\\(a,k)=1}}\frac1a.$$

For example, the smaller integers coprime to four are one and three. Its weight is $(1+1/3)/16=1/12$. The first five weights are

$$w_2=\frac14,\quad w_3=\frac16,\quad w_4=\frac1{12}, \quad w_5=\frac1{12},\quad w_6=\frac1{30}.$$

Reducing the pairs in the collision energy by their common factor gives these weights directly.

The weighted remainder identity

Let $J(N)$ be the continuous collision capacity and let $D_N=N-J(N)$ be its deficit below the leading value $N$. The capacity is one third of the gcd/lcm square-table sum used in the secondary-term calculation.

In terms of the remainder clocks,

$$\sum_{k\ge2}w_k=1, \qquad D_N=\frac23\sum_{k\ge2}w_k(N\bmod k).$$

This is an identity at every integer resolution. It includes the long clocks whose periods exceed $N$; their position is still $N$.

At resolution five the deficit is $11/6$. At six it is $17/10$. The simultaneous resets therefore make the deficit fall by $2/15$, even though the resolution increased. The exact increment equation accounts for that fall:

$$D_N-D_{N-1} =\frac23\left(1-\sum_{\substack{k\mid N\\k\ge2}}kw_k\right).$$

Advancing all clocks adds two thirds to the deficit. Each reset subtracts the corresponding weighted period. The divisor sum accounts for those subtractions.

At a large prime resolution there is only one nontrivial resetting period, and the increment approaches $2/3$. Integers with sufficiently many small divisors produce arbitrarily large downward increments.

The same accounting explains the leading capacity law. For each fixed clock, its position stays below its period while $N$ grows. Its position divided by $N$ tends to zero. Since the positive weights total one, the whole deficit divided by $N$ tends to zero. Thus $J(N)/N$ tends to one. The separate sampling estimate carries that continuous leading law into the cubic digit energy.

Stationary mean age

If a clock of length $k$ is observed uniformly through a complete cycle, its average position is $(k-1)/2$.

Keep the actual collision weights and add those average positions through period $M$. With the same factor of two thirds, the result is

$$D_*(M)=\frac13\sum_{2\le k\le M}w_k(k-1) =\frac{\log^2 M}{\pi^2}+O(\log M).$$

The secondary coefficient appears in the truncated average age of the conserved mass. Coprimality supplies the factor involving $\pi$, just as it did in the original secondary-term proof.

Clocks of arbitrarily large period have infinite stationary mean age in total, despite their finite total probability. Subtracting the average age at each finite cutoff gives a centered observable with finite limiting variance.

The clocks also have to remain compatible. The position modulo two must agree with the position modulo four after reduction. It is impossible to choose all the clock positions independently.

The space of compatible remainder choices is the classical space of profinite integers. Adding one advances every clock. A uniform state means that, on any fixed finite collection of clocks, every compatible position has its appropriate equal share. This supplies a precise stationary ensemble for the arithmetic dynamics.

The potential and its increments

Subtract each clock’s average position before adding its contribution. For a finite cutoff this gives

$$Y_M(x)=\frac23\sum_{2\le k\le M}w_k \left(r_k(x)-\frac{k-1}{2}\right).$$

Under the stationary ensemble, these functions converge in mean square to a mean-zero potential $Y$ with positive finite variance. The omitted part has mean-square norm at most a constant times $\log(2M)/\sqrt M$.

The increment process is the difference between the potential at consecutive phases. If $g$ denotes the deficit increment, then

$$\sum_{j=1}^{H}g(x+j)=Y(x+H)-Y(x).$$

All intermediate values cancel. Consequently the variance of a stationary block sum stays bounded, however long the block becomes. For independent increments with a fixed positive variance, that variance would grow with the block length. Here the arithmetic dependence prevents that accumulation.

Adjacent increments have negative covariance. Consecutive integers cannot share a divisor greater than one, which constrains their reset patterns. At shifts divisible by many small periods, the increment process returns close to itself in mean square.

The profinite space and its periodic harmonic modes are classical. The collision calculation specifies the weights of those modes. The manuscript determines their amplitudes and spectral tail, solves the finite alignment problem, and proves that the stationary distribution has no atoms and extends arbitrarily far in both directions.

Cutoffs on the integer orbit

Fix a nonnegative integer $n$ and keep admitting longer clocks. Every newly admitted clock has position $n$, but its stationary midpoint is roughly half its period. The centered cutoff $Y_M(n)$ therefore tends to minus infinity as $M$ grows, even though $Y_M$ converges in mean square over the stationary ensemble.

The nonnegative integer orbit is dense in the profinite space and has stationary measure zero. Mean-square convergence therefore permits this divergence on the integer orbit.

The actual capacity calculation increases the observation and the cutoff together. It asks for $Y_n(n)$: the state at integer $n$, seen through all clocks of length at most $n$.

Transferring the stationary law to this moving observation requires estimates uniform in the cutoff. The finite-grid question from the secondary-term calculation, whether its sampling defect tends to one, also remains open.