Light quanta · Section 6

The radiation entropy law matches the gas entropy law. The exponent identifies the light quantum.

Why does equating the volume dependence of Wien radiation entropy to Boltzmann's independent-particle entropy law suggest that monochromatic radiation behaves as independent energy quanta of magnitude $R\beta\nu / N = h\nu$ ?

The move in §6

Matching the entropy laws to find the light quantum

Static worked example

CurrentThese numbers match the current settings.

Model note

Primary outputs radiationEnergy, frequency, volumeRatio, independentPointCount: Host calculation (lq06.acceptedInputs). Owner lq06.acceptedInputs.
Primary outputs effectiveIndependentCount, quantumEnergy, quantumEnergyEv, meanQuantumEnergyWien, meanQuantumEnergyWienEv, moleculeMeanKineticEnergyEv, meanEnergyRatio, ratioAt600THz: Host calculation (radiation.quanta). Owner radiation.quanta.
Primary outputs radiationEntropy, entropyVolumeCoefficient: Host calculation (radiation.entropy). Owner radiation.entropy.
Primary outputs gasEntropy, gasEntropyVolumeCoefficient: Host calculation (radiation.configurations). Owner radiation.configurations.
Primary output correspondenceVerdict: Host calculation (lq06.correspondence). Owner lq06.correspondence.
Accepted input revision 1.
Snapshot version 1.
Not modeled: Radiation outside the Wien regime; Mechanism of emission and absorption (reserved for §§7–9); Cavity wall dynamics and boundary interactions; Wave interference patterns inside the volume.

Predict before the numbers

The radiation's entropy is S − S₀ = (R/N) ln[(V/V₀)^NE/(Rβν)], and a gas of n molecules has S − S₀ = (R/N) n ln(V/V₀). Which expression plays the part of n?

Three relations the model could have NE/(Rβν), which is E/(hν). E, the total radiant energy. E/(βν), the radiation's own entropy coefficient.

Predict before the numbers

Over a Wien spectrum, how does the mean energy of a light quantum compare with a gas molecule's mean kinetic energy at the same temperature?

Three relations the model could have They are equal, by equipartition. The quantum's is unboundedly larger, because the field has infinitely many modes. The quantum's is twice the molecule's.

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

Which expression plays the part of n, the number of things?

How to read the match

Radiation energy E (nJ)

Frequency ν (THz)

600.0 THz: each quantum carries 2.481 eV

Try

Experiment settings volume ratio, the comparison gas, temperature

Volume ratio V/V₀ (ratio)

Molecules in the comparison gas, n (count)

Temperature T (K)

Side-by-side entropy volume laws (§6, the move)

V/V₀ = 0.50

Wien monochromatic radiation (§4)

E = 9.056 nJ, ν = 600.0 THz

S − S₀ = (E/βν) · ln(V/V₀)

Written with Boltzmann’s constant, k = R/N:

= (R/N) · [ ? ] · ln(V/V₀)

Ideal gas / solute molecules (§5)

n = 10 independent particles

S − S₀ = (R/N) · ln W

Independent positions, W = (V/V₀)ⁿ:

= (R/N) · [ n ] · ln(V/V₀)

Which expression goes in the bracket? Choose one in the controls. Coefficients of ln(V/V₀) now: radiation 3.145 × 10⁻¹³ J/K, gas 1.381 × 10⁻²² J/K.

Wien spectrum mean quantum energy against a gas molecule’s kinetic energy (§6)

T = 3000 K

Integrating over a full Wien spectrum, the average energy of a light quantum is exactly twice the average translational kinetic energy of a gas molecule (at 600 THz, monochromatic h·ν is 3.20× this mean quantum energy):

Mean energy of a light quantum in a Wien spectrum, ⟨ε⟩ = 3k_BT

0.7756 eV

Mean translational kinetic energy of a gas molecule, ⟨E_kin⟩ = (3/2)k_BT

0.3878 eV

Ratio of the two: 2.0 : 1

The three logical roles of the match

1. Derivation (algebra)

The Wien radiation entropy and the Boltzmann gas entropy have the same form exactly when n = NE/(Rβν) = E/(hν).

2. Heuristic inference

Monochromatic radiation of low density, in the Wien regime, behaves thermodynamically as though it consisted of independent energy quanta of size hν.

3. Further hypothesis

Are the production of light (Stokes's rule, §7) and its transformation (the photoelectric effect, §8; ionization, §9) also exchanges in amounts of hν?

Values at these settings

Quantity	Value
Radiation energy E	9.056 nJ
Frequency ν	600.0 THz
Volume ratio V/V₀	0.5
Number of independent quanta, n_eff	2.2778 × 10¹⁰
Energy of each, hν	3.9756 × 10⁻¹⁹ J
Energy of each, hν	2.4814 eV
Radiation: coefficient of ln(V/V₀)	3.1448 × 10⁻¹³ J/K
Gas: coefficient of ln(V/V₀)	1.3806 × 10⁻²² J/K
Radiation: entropy change	−2.1798 × 10⁻¹³ J/K
Gas: entropy change	−9.5699 × 10⁻²³ J/K
Mean quantum energy over a Wien spectrum, 3k_BT	0.77556 eV
Mean kinetic energy of a gas molecule, (3/2)k_BT	0.38778 eV
Ratio of the two	2

When faint light of one colour is squeezed into half the room, its entropy falls by the same rule as a gas of independent particles does. Matching the two rules says how many particles the light would have to contain, and so how much energy each one carries, an amount set by the colour of the light.

Section 4 found that radiation of energy E in a narrow band at frequency ν, where Wien's law holds, changes its entropy with volume as S − S₀ = (E/βν) ln(v/v₀). Section 5 used Boltzmann's principle, S − S₀ = (R/N) ln W: for n independent moving points the probability that all of them are in a part v of the volume v₀ is W = (v/v₀)ⁿ, so S − S₀ = (R/N) n ln(v/v₀). Section 6 writes the radiation result in the same form, S − S₀ = (R/N) ln[(v/v₀)^{(N/R)(E/βν)}], and reads off the probability that all the radiation energy is in v. The exponent plays the part of n: the energy behaves as if it consisted of n = NE/(Rβν) independent quanta, each of size Rβν/N, which is hν. At the defaults, 9.06 × 10⁻⁹ J at 600 THz, that is 2.28 × 10¹⁰ quanta of 2.48 eV. Halving the volume lowers the radiation's entropy by 2.18 × 10⁻¹³ J/K, as it would for a gas of that many points; the instrument's gas of 10 points loses 9.57 × 10⁻²³ J/K. Einstein then compared the mean quantum of a whole Wien spectrum, 3(R/N)T, with a molecule's mean kinetic energy, (3/2)(R/N)T: at 3000 K, 0.776 eV against 0.388 eV, a factor of 2.

Boltzmann's principle ties entropy to probability: S − S₀ = (R/N) ln W, where R/N, the gas constant divided by the number of molecules in a mole, is 1.381 × 10⁻²³ J/K. Take n independent points moving about a box of volume v₀. The chance that one of them is, at a given moment, in a part v of the box is v/v₀. Because they move independently, the chance that all n are there together is the product, (v/v₀)ⁿ. With 10 points and half the box, that is (1/2)¹⁰ = 1/1024. The entropy difference is then (R/N) ln[(v/v₀)ⁿ] = (R/N) n ln(v/v₀) = 1.381 × 10⁻²³ × 10 × ln 0.5 = −9.57 × 10⁻²³ J/K. Now the radiation. Section 4 gives S − S₀ = (E/βν) ln(v/v₀). Pull R/N out in front, E/βν = (R/N) × (N/R)(E/βν), and use the rule that a number times a logarithm is the logarithm of a power: S − S₀ = (R/N) ln[(v/v₀)^{(N/R)(E/βν)}]. Set this beside the gas. The two have the same form if n = (N/R)(E/βν). Divide the energy by that count to find the energy of one part: E/n = Rβν/N. With today's constants Rβ/N is Planck's h, 6.626 × 10⁻³⁴ J s, so each part carries hν = 6.626 × 10⁻³⁴ × 6.00 × 10¹⁴ = 3.976 × 10⁻¹⁹ J, or 2.48 eV. The defaults hold E = 9.056 × 10⁻⁹ J, so n = 9.056 × 10⁻⁹/(3.976 × 10⁻¹⁹) = 2.28 × 10¹⁰. The radiation's coefficient E/βν is n times R/N, 2.28 × 10¹⁰ × 1.381 × 10⁻²³ = 3.145 × 10⁻¹³ J/K, and halving the volume changes the entropy by 3.145 × 10⁻¹³ × ln 0.5 = −2.18 × 10⁻¹³ J/K. The probability that all that light is in one half at once is 1/2 raised to the power 2.28 × 10¹⁰, too small to write out. Last, the mean quantum. Over a whole Wien spectrum at temperature T, the energy per unit frequency is αν³e^−βν/T, and the number of quanta per unit frequency is that divided by Rβν/N; the ratio of the two integrals is 3(R/N)T. At 3000 K that is 3 × 1.381 × 10⁻²³ × 3000 = 1.243 × 10⁻¹⁹ J, or 0.776 eV, twice a gas molecule's (3/2)(R/N)T = 0.388 eV. A 600 THz quantum, at 2.48 eV, is 3.20 times the mean.

What this model leaves out

Radiation outside the Wien regime
Mechanism of emission and absorption (reserved for §§7–9)
Cavity wall dynamics and boundary interactions
Wave interference patterns inside the volume

The rule this laboratory evaluates

From src/physics/reference/radiation/quanta.ts, the audited TypeScript reference evaluator.

// Paper 1, §6: matching the entropy coefficients
// Radiation entropy:     S - S_0 = (E / (beta * nu)) * ln(V / V_0)
// Boltzmann gas entropy: S - S_0 = (R / N) * n * ln(V / V_0)
//
// Equating the exponents in W = (V / V_0)^n:
// n_eff = (N / R) * (E / (beta * nu)) = E / (h * nu)
// Energy per quantum: epsilon = E / n_eff = (R * beta * nu) / N = h * nu
//
// Mean quantum energy over a Wien spectrum:
// <epsilon> = 3 * (R / N) * T = 3 * k_B * T, twice a molecule's 1.5 * k_B * T

The physical argument

The entropy volume laws placed side by side

In §4, Einstein showed that for monochromatic radiation of energy $E$ and frequency $\nu$ in the Wien regime, changing the enclosing volume from $V_0$ to $V$ changes the entropy by:

S - S_0 = \frac{E}{\beta\nu}\ln\frac{V}{V_0}

In §5, Einstein evaluated Boltzmann's principle $S - S_0 = \frac{R}{N}\ln W$ for a system of $n$ independent particles in a container, finding that the statistical probability of finding all $n$ particles in a subvolume $V$ is $W = (V/V_0)^n$ , leading to:

S - S_0 = \frac{R}{N}\,n\ln\frac{V}{V_0} = k_B\,n\ln\frac{V}{V_0}

The move: equating the functional forms

To make the two equations directly comparable, Einstein rewrites the radiation entropy formula with Boltzmann's constant factor $R/N$ outside the logarithm:

S - S_0 = \frac{R}{N}\ln\left[\left(\frac{V}{V_0}\right)^{\frac{N}{R}\frac{E}{\beta\nu}}\right]

Comparing this with the gas probability law reveals that the statistical probability that all the monochromatic radiation energy $E$ is found in subvolume $V$ is:

W = \left(\frac{V}{V_0}\right)^{\frac{N}{R}\frac{E}{\beta\nu}}

The exponent $n_{\text{eff}} = \frac{N E}{R\beta\nu} = \frac{E}{h\nu}$ plays precisely the role of the particle count $n$ .

Energy per element and historical constants

If a total energy $E$ is composed of $n_{\text{eff}}$ independent quanta, each quantum carries an energy:

\epsilon = \frac{E}{n_{\text{eff}}} = \frac{R\beta\nu}{N} = h\nu

Using the 1905 experimental values for the gas constant $R = 8{,}31\cdot 10^7\text{ erg/K}$ , Wien's constant $\beta = 4{,}866\cdot 10^{-11}\text{ K}\cdot\text{s}$ , and Avogadro's number $N = 6{,}17\cdot 10^{23}$ , the product is:

\frac{R\beta}{N} = 6{,}5537\cdot 10^{-27}\text{ erg}\cdot\text{s}

Einstein does not print this product. It is the constant Planck called $h$ , computed from the values Einstein takes from Planck, and it lies about 1.1% below the modern value, 6.626 × 10⁻²⁷ erg·s. So the energy of the packets found from the entropy of radiation alone is Planck's quantum of action, to that precision.

Mean quantum energy over a Wien spectrum

Einstein further calculated the average energy of light quanta in thermal radiation at temperature $T$ by integrating over the full Wien spectrum:

\begin{aligned}\langle \epsilon \rangle &= \frac{\int_0^\infty \alpha\nu^3 e^{-\beta\nu/T} d\nu}{\int_0^\infty \frac{N}{R\beta\nu}\alpha\nu^3 e^{-\beta\nu/T} d\nu} \\ &= 3\frac{R}{N}T \\ &= 3 k_B T\end{aligned}

This is exactly twice the average translational kinetic energy of a monoatomic gas molecule, $\langle E_{\text{kin}} \rangle = \frac{3}{2} k_B T$ .

The three logical roles

Derivation (Mathematical Identity): The radiation entropy volume law and the ideal gas entropy volume law agree identically for all volume ratios $V/V_0$ if and only if $n = N E / (R\beta\nu) = E / (h\nu)$ .
Heuristic Inference (Thermodynamic Analogy): In the Wien regime of low radiation density, monochromatic radiation behaves thermodynamically as though it consisted of $n_{\text{eff}}$ mutually independent energy quanta $h\nu$ .
Further Physical Hypothesis (Emission and Absorption): This analogy suggests investigating whether the processes of emission and absorption of light also proceed by discrete quanta of size $h\nu$ (demonstrated in §7 for Stokes' rule, §8 for photoelectricity, and §9 for gas ionization).

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