Light quanta · Paper 1, §5 heuristic foundation

Independent configurations and the gas analogy

How does counting independent configurations produce an entropy that depends on volume as n ln(V/V₀), and what changes when the configurations are not independent? Wien-regime radiation follows the same volume law; the counting is a premise of the light-quantum analogy and does not by itself demonstrate light quanta.

Light quanta · §5 statistical microstate counting

Independent configurations and Boltzmann entropy

How counting independent configurations gives an entropy that depends on the volume, and what changes when the positions are locked together.

Static worked example

CurrentThese numbers match the current settings.

Model note
  • Primary outputs configurationProbability, lnW, log10W, deltaSOverKb, expectedTrialsToOne: Host calculation (radiation.independentPointsProbability). Owner radiation.independentPointsProbability.
  • Primary outputs sampleFraction, successCount, drawCountAfter: Host calculation (radiation.sampleIndependentPoints). Owner radiation.sampleIndependentPoints.
  • Primary output lockedProbability: Host calculation (radiation.lockedPositionsProbability). Owner radiation.lockedPositionsProbability.
  • Seed 12345.
  • Accepted input revision 1.
  • Snapshot version 1.
  • Not modeled: Interactions between points.; Gas dynamics or time evolution.; Radiation itself (this is the gas and dilute-solution analogy, not a model of light).; Correlations other than the fully locked case..
Total volume V₀ (Full box)Subvolume V = 0.5 V₀ (50%)
Subvolume VV₀ − V
Mode: Independent pointsW = fn = (0.50)4 = 0.062500

Binomial distribution P(k): the chance that k of the 4 points lie inside

0
1
2
3
4

The last bar, k = 4, is every point inside: W = P(4) = f4.

Microstate enumeration: Total microstates: 16. Favorable: 1. Exact ratio: 1 / 16.

Predict before the numbers

Ten independent points move in the box. At a moment picked at random, what is the chance that all ten are in the left half?

Three relations the model could have

Predict before the numbers

Now the ten points are locked together and move as one. What is the chance that all ten are in the left half?

Three relations the model could have

The result appears when you choose, say you have one in mind, or skip.

Try
Experiment settings the display view, the locked-positions counterexample
Display view:

These two apply at once.

Worked example: 4 independent points all sit in a fraction 0.5 of the volume with probability 0.5 to the power 4, 0.0625.

Calculated microstate and entropy outputs

QuantitySymbolic formValueMeaning
Relative state probabilityW = (V/V₀)n = fn0.06250000Probability that all n independent points are found in V
Natural logarithm ln Wn ln f−2.772589Proportional to the entropy difference ΔS / kB
Dimensionless entropy change ΔS/kBn ln(V/V₀)−2.772589Matches Wien-regime radiation entropy S − S₀ = (E / hν) kB ln(V/V₀)
Base-10 logarithm log₁₀ Wn log₁₀ f−1.20412Order of magnitude (for example 10⁻¹⁸ at n = 60)

Paper assumptions (§5 as printed)

  • No favored part of the space or direction in volume V₀.
  • Negligible interactions among the n movable points.
  • Other movable points may also be present without altering the independent distribution.
  • No assumption is needed about the laws of motion of the points.

What this model leaves out (not modeled)

  • Interactions between points.
  • Gas dynamics or time evolution.
  • Radiation itself (this is the gas and dilute-solution analogy, not a model of light).
  • Correlations other than the fully locked case.

If points wander independently through a box, the chance of finding all of them in its left half at the same moment is one half multiplied by itself once for each point: one in 16 for four points. Einstein turned that chance, through Boltzmann's principle, into the way a gas's entropy depends on its volume, and then read radiation the same way.

§5 of the light-quanta paper takes n points moving in a volume v0, with nothing assumed about how they move except that no part of the space and no direction is preferred, and so few that they do not act on one another. It asks for the probability that, at a moment picked at random, all n are in a part v of the volume, and answers W = (v/v0)n. Boltzmann's principle, S − S0 = (R/N) lg W, then gives S − S0 = R(n/N) lg(v/v0), from which the gas law and the law of osmotic pressure follow. The instrument sets the fraction f = v/v0 and the number n. At its defaults, n = 4 and f = 1/2, W = 1/16 = 0.0625 and the entropy difference is ln W = −2.773 in units of R/N, which is kB. A seeded run of 10 000 random placements, seed 12345, finds all four inside 606 times, a fraction of 0.0606. The enumeration view lists every arrangement of the points among equal cells, 24 = 16 of them here with one favourable, and stops at 220 arrangements rather than freeze the page. For large n the chance is too small to see by sampling: at n = 60 it is 8.67 × 10−19, one success in about 1.2 × 1018 tries, so the logarithmic view states log10 W = −18.06 instead. Locking the points into one group makes W = f = 1/2 whatever n is, which shows that the exponent n comes from the independence of the points, not from how many labels there are. In §6 Einstein reads radiation's entropy in the same form, with E/(Rβν/N) in the place of n.

The independence argument in Einstein 1905 §5

In §5 of the 1905 light-quanta paper, Einstein applies Boltzmann's principle S − S₀ = (R/N) lg W to an ideal gas of n movable points in volume V₀. If the points move independently with no favored position or direction, the statistical probability that all n points are found in a subvolume V is simply:

W = (V / V₀)n, so S − S₀ = (R / N) n ln(V / V₀)

Comparing this gas entropy with the monochromatic radiation entropy found in §4, S − S₀ = (E / hν) kB ln(V / V₀), leads directly to the conclusion: monochromatic radiation behaves energetically as if it consists of E / (hν) independent energy quanta of magnitude hν.

Read the original German source text and translation for §5

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