Reasoning laboratory · Preview

The same average can hide different arrangements.

Seeing half the points inside on average does not tell you whether they were placed independently. Keep the point count and region fixed, and compare points placed one by one with points locked to a single shared position.

Keep the mean. Change the dependence.

Static worked example
Hold these settings fixed for both models

Which measurements would you keep?

Underdetermined: the selected measurements have identical predictions. These choices cannot distinguish the two candidates.

Predictions at n = 4, f = 2/4. Counts are dimensionless.
ObservableIndependentLockedKept?
Probability that one specified point is inside0.50.5No
Mean number of points inside22Yes
Probability that all points are inside0.06250.5No
Variance of the number inside14No

Independent positions

Each labeled point is placed uniformly and independently. Several, all or none may land inside.

Perfectly locked positions

All labeled points share one uniformly placed, coincident position. Only all-in or all-out is possible. This is not a finite-size rigid cluster straddling a boundary.

Inspect the full count distributions

Each bar uses the same probability scale, zero to one. The numbers are model probabilities, not a sampled histogram.

Probability of exactly K points inside
KIndependentLocked
0 0.0625 0.5
1 0.25 0
2 0.375 0
3 0.25 0
4 0.0625 0.5
Connect the joint event to the entropy argument

The all-inside constraint has probability W = fn for independent points, but W = f for perfectly locked points. Applying Boltzmann’s logarithm gives ΔS/kB = ln W, with the same reference volume and point count.

Independent: -2.7725887. Locked: -0.69314718.

This is the log weight of a constraint, not the Shannon entropy of the count distribution. Agreeing with a volume law does not uniquely prove independence or establish the light-quantum hypothesis.

Test a count record against both models

Record how often K = 0, 1, …, 4 points were inside across independent repeat placements at the displayed settings. Enter one frequency for each K, including zeros. Consecutive frames of one correlated trajectory are not independent trials.

No file is uploaded and nothing is stored automatically. Analyzing asserts exact counting, known n and f, and identical independently repeated trials. Counting errors, finite cluster geometry and partially correlated models are not included.

Or replace the settings and record with an explicitly constructed example:

The link shares settings and selected measurements only. Your count record and explanation stay out of it. Copy them separately before leaving this tab.

Four points placed in half a box end up with two inside on average, whether each is placed on its own or all four are locked together. The average cannot tell the two apart; how often all four land inside, one time in 16 against one time in 2, can.

The light-quanta paper's §5 asks how likely it is that n points moving independently in a volume v0 are all found, at a chosen moment, in a part v of it. For independent points the answer is (v/v0)n, and through Boltzmann's principle the logarithm of that probability gives the entropy change (R/N) n ln(v/v0), with the number of points as the coefficient. §6 then reads radiation's entropy in the same form and concludes that, where Wien's law holds, radiation behaves as if it consisted of independent quanta. The workbench tests the premise that makes the exponent n rather than 1. It keeps four points and a region of half the volume, f = 1/2, and compares two ideal candidates: points placed one by one, and all four locked to one shared position. Both give a mean of 2 points inside, and both give each point a one-half chance of being inside, so neither measurement distinguishes them. The chance that all four are inside does: 1/16 = 0.0625 for independent points and 1/2 for locked ones, which is fn against f. So does the variance of the number inside, 1 against 4. A count record can be entered as how often 0, 1, 2, 3 or 4 points were inside. A single placement with 1, 2 or 3 inside is impossible for the locked candidate, and the workbench reports its likelihood as zero, not as a small number. With one point, or a region of none or all of the volume, the candidates agree on everything and no measurement separates them.

Which premise earns the exponent?

These are two explicit ideal candidates, not every possible form of correlation. The comparison calculates predictions and can analyze a count record you enter. It does not produce historical observations or claim a reviewed edition.

The light-quanta paper’s §5 counts independently placed points. The chance that every one lies in a fraction f of the original volume is fn. One perfectly locked group has only one placement to make, so its corresponding probability is f. The logarithm turns this difference into an entropy coefficient.

In the Brownian configuration argument, independently placed units likewise determine the power of available volume. Constituents locked into one ideal point-like unit do not acquire independent placements merely because you can count several labels.

Try keeping only the mean, then add the all-inside probability or the variance. Finally set n to one, or choose the empty or whole region: a useful measurement can cease to distinguish the candidates in a degenerate case.

Rejecting perfect locking does not establish every independence assumption. Finite-size clusters, partial correlations, measurement errors and time dependence need other models. Real measurements must specify those conditions and their uncertainty.

Inspect the numerical owners

The same binomial and locked-probability functions serve this comparison and LQ-05. The workbench projects their results; its controls do not implement a second probability law.

Configuration-count source · Prediction and likelihood source