# Pre-registration 2: Kelvin's 1874 claim about the number of molecules (written 2026-09-26, before kelvin.mjs was run)

Nature 9 (9 April 1874), p. 443, Thomson, "Kinetic Theory of the Dissipation of Energy", after
reversing every molecule of a gas whose temperature has nearly equalised:

> "The number of molecules being finite, it is clear that small finite deviations from absolute
> precision in the reversal we have supposed would not obviate the resulting disequalisation of the
> distribution of energy. But the greater the number of molecules, the shorter will be the time
> during which the disequalising will continue"

and he fixes the clock himself a few lines later: "Taking as unit of time the average interval of
free motion between consecutive collisions". Nothing about this has been run yet. What I expect is
written below, including where I expect him to be wrong, so that it can be checked against me.

## Design (fixed now)
- The same engine (engine.mjs, unchanged). Same density and same collision clock at every size:
  N in {32, 64, 128, 256, 512, 1024}; sigma = 0.05 * sqrt(128 / N); dt = 2.5e-4 * sigma / 0.05
  (so steps per collision stay about the same); speed 1. Seeds 1, 2, 3.
- "Nearly equalised": T_eq = the first forward step (checked every 10) at which the left-half
  share is <= 0.55.
- At T_eq, reverse. Perturbed copy: before reversing, every disk is moved by an independent
  random amount, uniform in [-delta*sigma, +delta*sigma] in x and in y (rounded to grid steps;
  PRNG seed = 1000 + seed). delta in {2^-8, 2^-16, 2^-24}. Exact copy: no perturbation.
- Run both back 2*T_eq steps, recording the left-half share every 10 steps.
  PEAK = highest share reached; D = back-time at which PEAK is first reached, in collisions
  (steps / 110, the measured steps per collision at N = 128, taken as the clock for all N).
  t* = first back-time at which any disk of the perturbed copy is more than sigma from the same
  disk of the exact copy.

## Predictions
- K0 (control, must hold): every exact copy reaches share 1 at back-time T_eq exactly.
- K1: PEAK falls as N grows. Spearman rho(N, PEAK) < 0 in at least 8 of the 9 (delta, seed) cells.
  (Kelvin's direction, and mine: bigger gases un-mix less when the reversal is imperfect.)
- K2, Kelvin's sentence itself: I expect it to FAIL in the form he wrote it. My reasoning: the
  error multiplies by a fixed factor per collision (measured about 3), so an imperfect reversal
  stays faithful for a fixed number of collisions whatever N is, while a bigger gas takes longer
  (in collisions) to equalise in the first place. So D should rise with N and then level off, not
  fall. Test: Spearman rho(N, D) >= 0 in at least 7 of the 9 cells means my expectation held and
  Kelvin's sentence did not; rho < 0 in at least 7 of 9 means Kelvin was right and I was wrong;
  anything between is reported as undecided.
- K3: t* in collisions is roughly independent of N: in each cell, max/min of t* over the six
  sizes <= 1.5. (If the extreme over more disks makes t* shrink with N, that is a small,
  logarithmic version of Kelvin's claim, and it will be reported as such.)

Whatever comes out is what the page prints, with this file linked.