Sharing Energy at Random

Teach me

Speed

Guided Inquiry

1 · The grid and the ledger

Let’s see what we’re working with. On the left: 100 atoms, with all 100 quanta piled on ONE atom. On the right: a live histogram counting how many atoms hold each amount. Each step picks a random atom; if it holds a quantum, one hops to another random atom (empty picks count as steps).

2 · Predict the settled shape

Here’s your first prediction. Suppose you let it shuffle for a long time from this all-on-one start. What do you think the histogram becomes?

3 · Press ▶ Play and watch

Let’s get it going. Press ▶ Play — the shuffle now runs at 3,000 steps/s. Let the steps counter pass ~10,000 while you watch the total-quanta and average-energy panels. What did they do while the bars churned?

4 · The pile-up at zero wins

Here’s the surprise. The most common condition is holding nothing, and the average-energy atom is NOT the typical atom. The total and the average never changed — yet the shape is lopsided.

5 · Does the start matter?

Now let’s change the starting point. Suppose you restart from spread-evenly — every atom begins with exactly one quantum. Take a guess: after a long shuffle, the settled shape is…

6 · Your turn: restart evenly

Let’s test it. Click "Spread evenly" under Starting arrangement, press ▶ Play, and let the bars settle. You’ll see a thin curve — the equilibrium N·P(n) — now overlays them. Compared to the shape the all-on-one start settled into, the even start settles…

7 · Read it as a straight line

Now for a neat trick. Flip the histogram to a log axis (button in the controls). The falling shape becomes a near-straight staircase, and one number — the fitted slope panel — now describes the whole distribution.

8 · What sets the slope

One last step. Raise quanta-per-atom toward 4 and watch the log line flatten — you can read the fitted-slope panel climb toward zero. Average energy alone sets that ratio. And here’s the big reveal: this is the Boltzmann factor — ⟨E⟩ plays the role of kT.

Listen

Starting arrangement

Number of atoms10×10 = 100

3×3 = 920×20 = 400

Quanta per atom1 quantum/atom

Shuffle speed20 /s

20 /s100,000 /s

Histogram vertical axis

q 100

⟨n⟩ = q/N 1.000

log-histogram —

ln(q/(q+N)) → —

Multiplicity: the number of ways (microstates) to place q quanta on N atoms.

Probability one atom holds n quanta — the Einstein-solid marginal.

Each level is a near-fixed fraction of the one below (finite-N steepening is small).

The log-slope; equality with ln(q/(q+N)) holds for large N. The average energy ⟨E⟩ plays the role of temperature — ⟨E⟩ → kT for many quanta per atom.

The Physics

Energy sharing · what this sim shows, the physics behind it, and every control

At a glance

Share a fixed amount of energy among many atoms by trading it at random, and the energy ends up spread in a definite, reproducible way. This sim lets you watch the sharing happen, then measure the result against the counting argument that predicts it.

What you'll learn

How to count the arrangements of energy among atoms, and why equal weight on each one is a natural rule.
How the distribution of energy among atoms follows from that counting.
How the settled shape depends on the starting arrangement and on the energy per atom.
How the sharing connects to temperature.

Each atom holds a whole number of quanta (orange). A trade moves one quantum between two atoms. The histogram counts atoms at each n.

What's on screen

Atom grid (left): each grey disc is one atom. Orange dots and the ×n label show how many quanta it holds; the steps counter sits at the bottom.
Histogram (right): bars count how many atoms hold n quanta, with n along the bottom.
Gold curve, "equilibrium N·P(n)": the counting prediction for the atoms at each n. The inquiry reveals it; Free Exploration shows it always.
Dashed teal line, "fitted line": a straight-line fit to the bars on the log axis.
Readouts: Total quanta q, Average energy ⟨n⟩ = q/N, and the Fitted slope.

See itPut all 100 quanta on one atom and shuffle. Watch for: the one loaded atom handing quanta to the others, and the bars on the right moving off their starting spike.

The controls

Press ⌖ on a row to see where that control sits in the sim.

Control	What it changes	Range
Starting arrangement	How the quanta are dealt out before any trading: all on one atom, one each, or scattered at random. Choosing one re-deals the grid.	All on one / Spread evenly / Random opens at All on one
Number of atoms	The grid size N. Total quanta follow from it: q = N × quanta per atom.	3×3 = 9 to 20×20 = 400 opens at 10×10 = 100
Quanta per atom	The average energy per atom, ⟨n⟩ = q/N.	1–6 opens at 1
Shuffle speed	Trades per second. Below about 60 /s you see each quantum hop; above it the bars update in bulk. It never restarts the run.	20–100,000 /s opens at 20 /s; 3,000 /s once the inquiry starts the run
Histogram vertical axis	Switches the bar heights between a linear and a log scale. The run is not restarted.	linear / log opens linear
Readouts	Total quanta q, Average energy ⟨n⟩, and the Fitted slope with its prediction.	panel
▶ Play / ⏸ Pause	Runs or freezes the shuffle.	top bar
Speed	Multiplies the Shuffle speed.	0.25×–4× opens at 1×
↻ Reset	Deals the grid again with the current start and sliders, and pauses.	top bar

Fixed: one trade picks a donor atom, then a different recipient atom, both at random, and moves one quantum if the donor has one. An empty pick still counts as a step. Energy is counted in quanta of size ε.

Why trading does not even things out

A microstate is the full ledger: which atom holds how many quanta. Each trade moves one quantum, and the reverse trade is exactly as likely as the forward one. That symmetry makes every microstate equally likely over a long run. The number of microstates is a count:

Ways to place q identical quanta on N distinguishable atoms.

At the defaults, N = 100 and q = 100, there are about 4.5 × 10⁵⁸ microstates, and exactly one has every atom holding one quantum.
Nothing pulls the energy toward uneven arrangements. They are simply far more numerous than even ones.
Trading conserves q and N, so the counting is over a fixed total energy (the microcanonical set).

Key idea. Random trading does not aim at fairness. It wanders among equally likely microstates, and almost none of them are even.

See itStart with one quantum on every atom, the single even microstate, and shuffle slowly. Watch for: the tall bar at n = 1 breaking up, with the atoms at n = 1 falling from 100 to about 25.

The staircase it settles into

One atom holds n quanta when the other N − 1 atoms share the remaining q − n. Counting those cases gives the settled shape, and the ratio between neighbouring levels:

Chance that one given atom holds n quanta.

Each level is nearly the same fraction of the one below.

The fullest bin is n = 0. At the defaults the levels n = 0, 1, 2, 3 hold about 50, 25, 13 and 6 atoms: each is about 0.50 of the one below.
A constant factor per level is a straight line on a log axis. Switch the Histogram vertical axis to log to see it.
The start is forgotten. Every start reaches the same curve; only the time differs. Measured here at N = 100, q = 100: about 1,000 trades from Spread evenly, about 30,000 from All on one.

Equilibrium N·P(n) at the defaults on a log axis, with the straight line through it.

See itOn the log axis, run from All on one, then from Spread evenly. Watch for: both runs ending with about 50 atoms at n = 0, close to the gold curve.

What sets the slope

For many atoms the exact ratio simplifies to a single number x, and the populations fall as a geometric series:

Large N: a constant factor x per level.

N ⟨n⟩

These are the sim's own Number of atoms and Quanta per atom sliders; moving them re-deals the grid.

Only ⟨n⟩ = q/N matters. At ⟨n⟩ = 1, growing the grid from 3×3 to 20×20 moves the predicted slope from −0.672 to −0.693. Raising ⟨n⟩ from 1 to 4 moves it to −0.223.
The Fitted slope is only a snapshot. It is a fit to the bars as they are right now, so it jumps from frame to frame and reads flatter than the prediction: by roughly 10–20% at N = 100 and about 5% at N = 400. The prediction is the reference.

See itOn the log axis, raise ⟨n⟩ from 1 to 4 with 100 atoms. Watch for: the line tilting from steep to shallow, with the Prediction going from −0.693 to −0.223 and the Fitted slope following it, a little flatter.

The slope is a temperature

The trading rule contains no energy scale and no temperature. Yet a constant factor per level is the Boltzmann factor. Give one quantum the energy ε:

Reading the slope gives kT. The sim counts quanta, so it works in units of ε.

kT/ε = −1/slope = 1/ln(1 + 1/⟨n⟩). ⟨n⟩ = 1 gives kT = 1.44 ε, and ⟨n⟩ = 4 gives 4.48 ε. More energy per atom means a higher temperature and a flatter slope.
⟨E⟩ = kT only for ⟨n⟩ ≫ 1. The sim has no zero-point ε/2, so ⟨E⟩ falls short of kT by about ε/2. At the slider's maximum, ⟨n⟩ = 6, kT = 6.49 ε and ⟨E⟩ = 6 ε is 7.5% below it.
One atom follows the Boltzmann law because the other N − 1 act as its reservoir. That reservoir is finite, so P(n) is only approximately geometric and ends at n = q.

See itSet ⟨n⟩ = 6, the hottest setting, on the log axis. Watch for: a much flatter line, with the prediction near −0.15, which means kT ≈ 6.5 ε.

Where it matters

Einstein's 1907 solid. Each atom in a crystal is three harmonic oscillators, each with levels spaced by ε = ħω. One "atom" in the sim stands for one such oscillator.
Hot crystals. When ⟨n⟩ ≫ 1, each oscillator carries ⟨E⟩ → kT, so a mole holds 3RT and the heat capacity is C = 3R ≈ 24.9 J mol⁻¹ K⁻¹, the Dulong–Petit value.
Cold crystals. When ⟨n⟩ ≪ 1, nearly every oscillator sits at n = 0 and C falls toward zero. With ε/k ≈ 1300 K for diamond, Einstein's model gives about a quarter of 3R at 300 K.
Photons. The same law, written as an occupancy, is Planck's: ⟨n⟩ = 1/(e^ε/kT − 1) for a mode of light in a cavity.

The sim's coldest setting, ⟨n⟩ = 1, is kT = 1.44 ε with C ≈ 0.96 × 3R, already close to classical. The freeze-out needs ⟨n⟩ ≪ 1, below the slider's range.

Go deeper: the same law at three temperatures

Populations of the levels at ⟨n⟩ = 0.1, 1 and 4. C is Einstein's heat capacity per mole of atoms, 3R · u²eᵘ/(eᵘ − 1)² with u = ε/kT.

As the temperature rises, the n = 0 level empties and the higher levels fill, but the geometric shape stays.
C approaches 3R from below: 0.63 × 3R at ⟨n⟩ = 0.1, 0.96 at ⟨n⟩ = 1 and 1.00 at ⟨n⟩ = 4.

Summary

In one line. Random trading gives every arrangement equal weight, and counting them gives populations that fall by a constant factor per quantum, set only by the average energy per atom: the Boltzmann factor.

Key points

There are Ω = C(N + q − 1, q) microstates, all equally likely.
Populations fall from n = 0 by about q/(q + N) per level, whatever the start.
That factor depends only on ⟨n⟩ = q/N, and kT = ε/ln(1 + 1/⟨n⟩).

Common mix-ups

"Random sharing evens the energy out." The even arrangement is one microstate among 4.5 × 10⁵⁸, and uneven ones far outnumber it.
"The average atom is the typical atom." At ⟨n⟩ = 1 the mean is one quantum, yet the most common value is zero and about half the atoms hold nothing.
"Once settled, the bars stop moving." They keep fluctuating: at the defaults the n = 0 bar wanders by about ±4 atoms around 49.7. The gold curve is the average, not any one snapshot.

Things to try on your own

At ⟨n⟩ = 1, run from All on one and from Spread evenly, and note the step count at which the n = 0 bar first reaches about 50.
At ⟨n⟩ = 3 on the log axis, compare the Fitted slope with its prediction at 5×5 atoms, then at 20×20.
Read the prediction at ⟨n⟩ = 2 and compute kT/ε = −1/slope. Check it against 1/ln(1.5) ≈ 2.47.

Predict before reveal	Commit a histogram shape before the shuffle settles; feedback names the "randomness levels out" intuition.
Invariant made visible	Total quanta and average energy per atom stay fixed on screen while the histogram churns.
Microstate vs macrostate	Equal microstates, unequal macrostates: the falling shape comes from counting arrangements.
Start-independence	All-on-one, spread-evenly and random all settle onto the same reference curve N·P(n).
Operational numbers	The fitted log-slope settles near ln(q/(q+N)); average energy per atom alone sets it — ⟨E⟩ plays the role of kT for many quanta per atom.