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What happens after one more step?

Put a total share of one across three bins. Each step redistributes it using the fixed rule. The shares settle toward equality while their total stays one.

Step 1 of 6

Put one total across three places.

Begin with 80%, 15% and 5%. These are example weights, and together they sum to one. We will follow this same arrangement through the specified transfer.

1. A supplied three component distribution. Choose a concrete starting example. p₀ = (0.80, 0.15, 0.05). 2. The total weight one. Separate the equal distribution and its deviations. Equal part = (1/3, 1/3, 1/3). 3. The initial deviation and the specified transfer basis. Project onto the two unit directions. a₂ ≈ 0.459619, a₃ ≈ 0.347011. 4. The two deviation amplitudes and fixed rates. Multiply by the two transfer eigenvalues. The equal part stays; both deviations shrink. 5. The unchanged equal part and shrunken deviations. Reconstruct p₁ from its three modes. p₁ ≈ (0.362060, 0.304995, 0.332945). 6. The same transfer applied repeatedly. Raise the two shrink factors to n. pₙ approaches (1/3, 1/3, 1/3).
Arrows carry the named inputs into later calculations. Added inputs stay marked. Reduced motion is on. Use Next step or choose a stage below.
Carry forward
A supplied three component distribution
Added at this step
Example initial distribution
Operation
Choose a concrete starting example
Result
p₀ = (0.80, 0.15, 0.05)
p0=(0.8,0.15,0.05),∑ip0,i=1\begin{gathered}p_0=(0.8,0.15,0.05),\\[.4em]\sum_i p_{0,i}=1\end{gathered}

In words: Choose a concrete starting example

Why this step works and what it assumes

This starting distribution is an illustration. The three component transfer is the existing finite construction; its selection and physical time interpretation remain separate source questions.

Read the original calculation
See where this fits in the whole theory
Try the individual controls

Try this. Choose a starting preset. Press One step, then Run 8 steps, and compare the three shares with their unchanged total.

Redistribute a fixed total

Choose a starting arrangement, then take one step. The specified transfer brings the three weights closer together while their total stays one.

Bin 180.00%Bin 215.00%Bin 35.00%
Filled bars show the current picture. Dashed outlines show the values before this step. Animation connects the two calculated arrangements.
Step 0 · total 1.000000Before: 0.8000 · 0.1500 · 0.0500
After: 0.8000 · 0.1500 · 0.0500

These are mathematical weights updated by the original three-component transfer. A step here does not specify a number of seconds or the full physical dynamics.

Choose a starting distribution

A preset divides the total share among three bins. Read the three starting amounts and check that they add to one.

Apply the rule once

One step redistributes the shares using the specified rule. Compare the new bin amounts with the starting ones.

Repeat the same rule

Run eight steps. The differences shrink toward equal thirds while the total remains one.

Explore the full interactive calculation

Watch a distribution relax

Prepare three starting weights, then apply the existing family transfer. The uniform mode stays fixed. Two deviation modes shrink at different rates.

Starting distribution

0.8 · 0.15 · 0.05

After 0 transfer steps

0.8 · 0.15 · 0.05

110−610−1210−1810−2402468
Mode magnitudes on a logarithmic scale. The continuous lines interpolate the integer step values for reading. A zero initial mode stays zero and lies below the plotted range.
Stationary mode
0.577350eigenvalue 1
Slower deviation
4.59619e-1eigenvalue 64/729
Faster deviation
3.47011e-1eigenvalue 1/729
Inspect the actual transfer and its basis
T=1311T+(23)6u2u2T+(13)6u3u3TT=\frac13\mathbf1\mathbf1^T+\left(\frac23\right)^6u_2u_2^T+\left(\frac13\right)^6u_3u_3^T
u2=(1,−1,0)2,u3=(1,1,−2)6,pn=Tnp0u_2=\frac{(1,-1,0)}{\sqrt2},\qquad u_3=\frac{(1,1,-2)}{\sqrt6},\qquad p_n=T^np_0

The symmetric three component transfer is the one specified in v221. This apparatus uses its exact spectral decomposition. All three displayed amplitudes use unit length eigenvectors. The stationary amplitude is 1/√3 for a distribution with total weight one; signed deviation amplitudes use the basis shown here.

Read the original finite transfer

These are dimensionless steps of a specified relaxation operator. The screen’s playback speed is only a reading aid. Coherent quantum evolution, physical seconds and the original interacting field source require their separate maps.

How this fits into the whole theory

This example shows what the specified redistribution rule does. The clock example instead follows positions around a returning cycle.

What this experiment establishes

The stated transfer spectrum establishes relaxation for that operator. It neither supplies the full interacting physical Hamiltonian nor assigns seconds to an internal step; the physical source, spectrum and dimensional calibration must support that interpretation.