Independent configurations and the gas analogy

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 k_B. 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.

Take one point first. It moves about the whole volume v₀, and no part of the volume is preferred, so at a moment chosen at random it is in the part v with probability v/v₀, the fraction of the volume that part occupies. Call that fraction f; for the left half, f = 1/2. Now take a second point that moves independently of the first. Independently means that where the first point is tells you nothing about where the second is, and for independent events the probability that both happen is the product of their probabilities: f × f = f². Each further point multiplies by f again, so for n points W = fⁿ. With four points and half the volume, W = (1/2)⁴ = 1/16 = 0.0625. A sampler checks this: it places the four points at random 10 000 times and counts the placements with all four on the left, expecting about 10 000/16 = 625. With seed 12345 it finds 606, a fraction of 0.0606, within the ordinary scatter of such a count, whose standard deviation is √(10 000 × 0.0625 × 0.9375) ≈ 24. The enumeration view reaches the same number by counting. Divide the box into two equal cells; each of the four points can be in either, which makes 2 × 2 × 2 × 2 = 16 equally likely arrangements, and exactly one has all four on the left. Now the entropy. Boltzmann's principle says S − S₀ = (R/N) lg W, where lg is Einstein's natural logarithm, R the gas constant and N the number of molecules in a gram-molecule, so R/N is Boltzmann's constant k_B. The logarithm of a power brings the power down in front, lg fⁿ = n lg f, so S − S₀ = (R/N) n lg(v/v₀). For four points and half the volume that is 4 × lg(1/2) = −2.773 in units of R/N. Squeezing the points into a smaller part lowers the entropy, as compressing a gas does. The same arithmetic for n = 60 gives 60 × lg(1/2) = −41.59, so W = e^−41.59 = 8.67 × 10⁻¹⁹. The sampler finds nothing in 10 000 tries because the expected number of successes, 10 000 × 8.67 × 10⁻¹⁹, is far below one, which is why the logarithmic view states log₁₀ W = −18.06 instead of waiting for an event. Last, lock the points together so that they move as one. Now one placement decides all of them, W = f = 1/2 for any n, and the entropy change is lg(1/2), as for a single point. The coefficient in front of the logarithm counts independent placements, and §6 uses exactly that: radiation's entropy has E/(Rβν/N) in that place, so it behaves as that many independent quanta.

/** Independent point configuration probability W = (V/V0)^n = f^n. */
export function independentPointsProbability(
  n: number,
  f: number | { p: bigint; q: bigint },
): IndependentPointsProbabilityResult {
  let fNum: number;
  let exact: Rational | undefined;

  if (typeof f === "object") {
    fNum = Number(f.p) / Number(f.q);
    exact = {
      numerator: bigIntPow(f.p, n),
      denominator: bigIntPow(f.q, n),
    };
  } else {
    fNum = f;
  }

  const lnW = n * Math.log(fNum);
  const packed = packLogRepresentation(lnW, "probability");

  return {
    status: "value",
    quantityId: "configurationProbability",
    unit: "",
    value: packed.value,
    linearRepresentable: packed.linearRepresentable,
    lnW: packed.logFields.lnW ?? lnW,
    log10W: packed.logFields.log10W ?? lnW / Math.LN10,
    ...(exact ? { exactRational: exact } : {}),
    deltaSOverKb: lnW,
  };
}