Can the ordering temperature be estimated cheaply from known ordered structures in the databases?

EXPERIMENTS.md · lines 5633–5700

E97 — An ordering temperature the screen can afford: prediction

E96 showed rung 0 rewards the ordering hazard because the on-lattice ground states were taken out of its hull. The repair is to give rung 0 the ordering question, and it is nearly free: the 401 bcc orderings excluded in E86 are already relaxed with MACE and on disk, and the driving force is linear in temperature (E95, departure 0.000 meV/atom). So the temperature at which the random alloy stops beating its own ordered competitor is one division, not a Monte Carlo.

T_cross = T_lo - d(T_lo) * (T_hi - T_lo) / (d(T_hi) - d(T_lo))

where d is the driving force against the hull WITH the orderings in it. Below T_cross the ordered state is lower and the alloy is not a random solid solution. This is a mean-field quantity in the sense that it compares two fixed states rather than sampling configurations, but it is built on measured MACE energies rather than a Bragg-Williams estimate from the ECIs.

Validated before deployment against the four ordering temperatures rung 1 measured in E82 - the only four in this project measured under the current expansion.

Predicted:

1. It overestimates the Monte Carlo T_od, because comparing the random state against a single fully ordered state ignores the partial order that a real transition passes through, and because mean-field treatments of bcc ordering run high by roughly a quarter. Expect +10 to +50 per cent.
2. It ranks the four correctly. This is the part that matters: rung 0 is a gate and a ranking, and a quantity that is uniformly high but correctly ordered is usable where one that scrambles the order is not.
3. It costs under 5 ms, since the two hull queries it needs are already made.

Falsified as a gate if the ranking is wrong. If it cannot order four compositions whose true answers are known, it cannot be trusted to gate thousands that are not, and rung 0 must instead carry the cost of a short Monte Carlo or the ordering question stays at rung 1 with the generator blind to it.

Outcome. All three predictions falsified, and the reason is that I substituted my own idea for the advice I was given.

It underestimates by 38 to 72 per cent where an overestimate was predicted, the ranking of the middle two is wrong, and it costs 6.7 ms rather than under 5.

Cause, and it is decisive: all 401 bcc orderings in the hull are binary. Not one is ternary or higher. A three- or four-element solid solution can therefore only be compared against mixtures of binary orderings, which are a strictly worse competitor than its own ternary ground state, so its crossover comes out low - and the error grows with element count, which is exactly the ratio column. Two-element compositions score 0.51 and 0.62; the three-element one scores 0.28.

The databases hold binary superstructures because binaries are what people compute and observe. They will not hold the multi-element ones, and no amount of fetching will change that.

This is the mistake worth recording. Fable's suggestion was a Bragg-Williams spinodal computed from the expansion's own ECIs - enumerate two-sublattice partitions at the composition, evaluate each analytically, take the lowest. I substituted a hull of measured structures, reasoning that real MACE energies beat a mean-field estimate. They do, for structures that exist. The expansion knows every composition continuously and the database knows only the ones somebody has made, and that is precisely the property this problem needs. The advice was better than my improvement on it, for the reason that decided the result.