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Light quanta: from entropy to an energy scale

Einstein argues that light, when it is emitted or absorbed, behaves as if its energy came in separate quanta, and calls the idea a heuristic point of view. Follow all nine sections, from the trouble with classical radiation to the photoelectric effect.

Draft explanation, not yet reviewed

This is newly written explanation in modern notation; editorial and physics review are pending. The introduction and all nine numbered sections are explained below, but this is not the German source, an English translation, or a complete critical edition. The German source face holds a machine-drafted transcription with hand correction that no one has reviewed yet, and the facsimile face shows the pinned journal pages; an English translation is not ready. The headings and the way the argument is divided into passages are ours: Einstein divided his paper into an introduction and nine numbered sections, under his own titles, and a passage here does not stand for one of his paragraphs.

Show me one example before the notation →

First encounter · No algebra required

How often can everything end up in the left part?

Imagine a box divided into equal parts. Each labeled token lands in any part with the same chance. First let tokens land on their own. Then keep them together in the same part. Does adding another token change the chance that all of them land in the left part?

Authored counting examples, not measurements or movies of motion. Parts have equal size and no part is favored.

Try changing one thing

One box, one set of choices

2 tokens, 2 equal parts, each token chooses independently: 1 of 4 arrangements put all tokens in the left part.

leftright12
Arrangement 1 of 4. Labels distinguish tokens. Positions within a part are drawn only to keep the labels readable.
1 of 4 equally likely arrangements put every token in the left part.
ArrangementWhere each labeled token landsAll left?
11: left, 2: leftYes
21: right, 2: leftNo
31: left, 2: rightNo
41: right, 2: rightNo

Load this exact setup in the counting laboratory →

In the laboratory, choose Apply settings to calculate the loaded setup. Until then its prepared example stays visible.

Read every example without using the controls

For three or four tokens, try counting the arrangements before opening the table. No prediction is required.

1 token in 2 equal parts
1 of 2 equally likely arrangements put every token in the left part.
ArrangementWhere each labeled token landsAll left?
11: leftYes
21: rightNo
2 independent tokens in 2 equal parts
1 of 4 equally likely arrangements put every token in the left part.
ArrangementWhere each labeled token landsAll left?
11: left, 2: leftYes
21: right, 2: leftNo
31: left, 2: rightNo
41: right, 2: rightNo
3 independent tokens in 2 equal parts
1 of 8 equally likely arrangements put every token in the left part.
ArrangementWhere each labeled token landsAll left?
11: left, 2: left, 3: leftYes
21: right, 2: left, 3: leftNo
31: left, 2: right, 3: leftNo
41: right, 2: right, 3: leftNo
51: left, 2: left, 3: rightNo
61: right, 2: left, 3: rightNo
71: left, 2: right, 3: rightNo
81: right, 2: right, 3: rightNo
4 independent tokens in 2 equal parts
1 of 16 equally likely arrangements put every token in the left part.
ArrangementWhere each labeled token landsAll left?
11: left, 2: left, 3: left, 4: leftYes
21: right, 2: left, 3: left, 4: leftNo
31: left, 2: right, 3: left, 4: leftNo
41: right, 2: right, 3: left, 4: leftNo
51: left, 2: left, 3: right, 4: leftNo
61: right, 2: left, 3: right, 4: leftNo
71: left, 2: right, 3: right, 4: leftNo
81: right, 2: right, 3: right, 4: leftNo
91: left, 2: left, 3: left, 4: rightNo
101: right, 2: left, 3: left, 4: rightNo
111: left, 2: right, 3: left, 4: rightNo
121: right, 2: right, 3: left, 4: rightNo
131: left, 2: left, 3: right, 4: rightNo
141: right, 2: left, 3: right, 4: rightNo
151: left, 2: right, 3: right, 4: rightNo
161: right, 2: right, 3: right, 4: rightNo
2 tokens kept together in 2 equal parts
1 of 2 equally likely arrangements put every token in the left part.
ArrangementWhere each labeled token landsAll left?
11: left, 2: leftYes
21: right, 2: rightNo
10 tokens kept together in 2 equal parts
1 of 2 equally likely arrangements put every token in the left part.
ArrangementWhere each labeled token landsAll left?
11: left, 2: left, 3: left, 4: left, 5: left, 6: left, 7: left, 8: left, 9: left, 10: leftYes
21: right, 2: right, 3: right, 4: right, 5: right, 6: right, 7: right, 8: right, 9: right, 10: rightNo
1 token in 3 equal parts
1 of 3 equally likely arrangements put every token in the left part.
ArrangementWhere each labeled token landsAll left?
11: leftYes
21: middleNo
31: rightNo
2 independent tokens in 3 equal parts
1 of 9 equally likely arrangements put every token in the left part.
ArrangementWhere each labeled token landsAll left?
11: left, 2: leftYes
21: middle, 2: leftNo
31: right, 2: leftNo
41: left, 2: middleNo
51: middle, 2: middleNo
61: right, 2: middleNo
71: left, 2: rightNo
81: middle, 2: rightNo
91: right, 2: rightNo

In two equal parts, each extra independent token halves the chance again. In three equal parts, each extra independent token divides it by three. Ten tokens always kept together still have just one shared choice.

Counting tokens is not evidence that light is made of dots. The same count can help us recognize a pattern without proving what causes it.

What is getting in the way?

Are these measurements of light?

No. These are counting examples, not measurements. Counting tokens helps us understand a pattern; it is not evidence that light is made of dots.

What does keeping tokens together mean?

They are required to land in the same part every time. There is only one shared choice, however many labels the group carries. A real short string near a boundary need not meet this ideal rule.

Why label the tokens?

The first token on the left and the second on the right is a different arrangement from the first on the right and the second on the left.

Next: why this counting pattern matters

Physicists keep a bookkeeping number called entropy. It goes up by the same step each time the number of arrangements is multiplied by the same factor.

Count all the ways tokens can land, and distinguish several independent choices from a single choice shared by a group.

Section 5 turns this counting pattern into an entropy comparison. Section 6 then finds the same pattern in dilute light of a single color, as if its energy came in independent pieces.

This is an “as if” comparison in one restricted setting, not a general proof about light. The radiation calculation comes first in section 4; the interpretation follows in section 6.

More guidance: reading a fraction of the possibilities →

Show the two-token example in the instrument →

Continue to the section 5 counting argument →

Then inspect the section 6 “as if” inference →

Both links lead to our explanation of those sections, which is a draft. A reviewed German text, and an English translation aligned to it sentence by sentence, are not ready yet.

Introduction · a heuristic viewpoint

A qualification

Keep what waves explain

Why question the description of light without discarding interference?

The 4 printed paragraphs this passage explains
  • Introduction, paragraph 1: Physicists describe a gas or any other body by a finite, if very large, number of atoms and electrons, while Maxwell's theory describes light by quantities spread continuously through space. So a body's energy cannot be divided without limit, but the energy of a light ray spreading from a point is thought to spread over an ever larger volume.
  • Introduction, paragraph 2: The wave theory of light works superbly for optics and will probably never be replaced there, but optical observations measure averages over time. It is conceivable that the theory conflicts with experience when it is applied to how light is produced and transformed.
  • Introduction, paragraph 3: Einstein's proposal: black radiation, photoluminescence, the production of cathode rays by ultraviolet light and related effects seem easier to understand if the energy of light is distributed discontinuously in space. On this assumption a ray from a point carries a finite number of energy quanta, localized at points, which move without dividing and are absorbed or produced only as wholes.
  • Introduction, paragraph 4: He will set out the reasoning and the facts that led him to this view, in the hope that it may be of use to other researchers.

A wave can spread through space and interfere with another wave. Those successes are not withdrawn by the light-quantum proposal. The question changes: what happens when radiation is generated, absorbed, or converted into another form?

The introduction contrasts a continuously distributed field with matter described through a finite number of moving constituents. Einstein proposes considering independently moving energy elements whose energy remains localized during propagation and is exchanged as a whole. In this paper that is a heuristic viewpoint: something to investigate through consequences, not a conclusion supplied by an animation.

In the wave-optics laboratory, double the field amplitude while holding the wave shape and medium fixed. The model intensity becomes four times as large. That establishes a consequence of the wave model; it does not settle how an individual energy-transfer event occurs.

Show every step here: Keep what waves explain

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Modern qualifications

Modern photon language and later quantum theory are not premises of this reconstruction. The proposal does not say that purely optical wave phenomena have disappeared.

Assumptions and limits: Keep what waves explain

Assumed here

  • A continuous electromagnetic field describes optical propagation.
  • A model of propagation need not yet explain how matter exchanges energy with radiation.

What this does not establish

  • Drawing isolated light packets does not prove that radiation consists of them.
  • The wave description keeps its successes, interference and diffraction among them; the proposal concerns how light exchanges energy with matter.

Source context: German source · Facsimile

§1 · classical energy allocation

A derivation

A finite window cannot cure an infinite total

What happens if every radiation mode receives classical thermal energy?

The 6 printed paragraphs this passage explains
  • §1, paragraph 1: The setup for section 1: a space enclosed by perfectly reflecting walls holds gas molecules and electrons that move freely and collide like the molecules of a gas, and further electrons bound to fixed points by springlike forces. These bound electrons, called resonators, emit and absorb electromagnetic waves of definite periods.
  • §1, paragraph 2: On the current view of how light arises, the radiation in this space at equilibrium, worked out with Maxwell's theory, should be the same as black radiation, at least if resonators of every frequency that matters are present.
  • §1, paragraph 3: Setting the radiation aside, collisions with the gas require each resonator vibration along one line to have the mean energy (R/N)T, where R is the gas constant, N the number of molecules in a gram-equivalent and T the temperature. With more or less energy on average, the collisions would give energy to the gas or take it from the gas, so equilibrium needs exactly this mean energy.
  • §1, paragraph 4: Planck had found the corresponding equilibrium between a resonator and the radiation around it, assuming the radiation is as disordered as possible: the resonator's mean energy fixes the radiation's energy density at its frequency.
  • §1, footnote 2: The footnote cites Planck's paper in Annalen der Physik, volume 1, page 99, 1900.
  • §1, paragraph 5: For the radiation at each frequency to neither gain nor lose energy on balance, both conditions must hold, and together they fix its density. The result disagrees with experience, and it also means the picture allows no definite division of energy between ether and matter: the wider the range of resonator frequencies, the larger the radiation energy, without limit.

Counting cavity modes and giving each mode the classical mean energy kBTk_B T produces a spectral energy density ρν\rho_\nu. A density must be multiplied by a frequency width to give energy per unit volume.

ρν=8 π ν2c3 kB T\rho_\nu = \frac{8\,\pi\,\nu^{2}}{c^{3}}\,k_B\,T

The classical energy density per unit frequency is eight pi frequency squared times thermal energy, divided by light speed cubed.

U(νc)=∫0νcρν dνU\left(\nu_{c}\right) = \int_{0}^{\nu_{c}} \rho_\nu\,\mathrm{d}\nu
U(νc)=8 π kB T3 c3 νc3U\left(\nu_{c}\right) = \frac{8\,\pi\,k_B\,T}{3\,c^{3}}\,\nu_{c}^{3}

The energy density below the cutoff frequency is eight pi times Boltzmann's constant times temperature, divided by three times the speed of light cubed, times the cutoff frequency cubed.

At fixed positive temperature, doubling the cutoff multiplies this finite-range total by eight. Extending the model to arbitrarily high frequencies produces an unbounded total. The resonator laboratory keeps that failure distinct from the finite portion displayed on screen.

Show every step here: A finite window cannot cure an infinite total

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Modern qualifications

kBk_B and cc are modern explanatory notation. The paper writes R/NR/N for the thermal constant and LL for the speed of light. Historical names and later terminology are not additional evidence for the model.

Assumptions and limits: A finite window cannot cure an infinite total

Assumed here

  • Classical equilibrium assigns mean energy k_B T to each radiation mode.
  • The number of modes per unit volume and per unit frequency grows as the square of the frequency.

What this does not establish

  • The allocation is a classical model, not the spectrum measured across all frequencies.
  • A finite cutoff is where the calculation stops, not a property of light.

Source context: German source · Facsimile

An assumption

Two assumptions under the equilibrium of section 1

What does the equilibrium of §1 assume about the electrons and about the radiation?

The 2 printed paragraphs this passage explains
  • §1, footnote 1: The footnote says this assumption amounts to supposing that gas molecules and electrons have equal mean kinetic energies at the same temperature, which Drude assumed when he derived the ratio of heat conduction to electrical conduction in metals.
  • §1, footnote 3: The footnote defines radiation as disordered as possible: when the electric field at a point is expanded in a long Fourier series, the probabilities of the amplitudes and phases are independent of one another. The radiation here comes closer to that the more each pair of them depends on emission and absorption by particular groups of resonators.

§1 puts gas molecules, free electrons and bound electrons in a box with mirror walls, and lets the free electrons collide like the molecules of a gas. A footnote says what that amounts to: at equilibrium, gas molecules and electrons have equal mean kinetic energies. Drude, Einstein notes, used the same assumption to derive the ratio of heat conduction to electrical conduction in metals.

That equality is what fixes the mean energy of each straight-line vibration of a resonator. A resonator whose average energy were higher or lower would, through the collisions, keep giving energy to the gas or taking it, so equilibrium needs:

E=RNTE=\frac{R}{N}T

The mean energy E of one resonator vibration equals R over N times the temperature T.

The second assumption belongs to Planck's relation between a resonator and the radiation around it, which §1 uses. Planck derived it for radiation that is 'as disordered as possible', and a footnote defines the phrase. At one point of the space, expand one component Z of the electric field over a long time T in a Fourier series of sine waves with amplitudes A and phases α. Repeat the expansion from starting moments chosen at random: the amplitudes and phases come out differently each time, with a statistical probability for each combination.

Z=∑ν=1∞Aνsin⁡(2πνtT+αν)Z=\sum_{\nu=1}^{\infty}A_\nu\sin\left(2\pi\nu\frac{t}{T}+\alpha_\nu\right)

Z equals the sum over nu from one to infinity of A nu times the sine of two pi nu t over T plus alpha nu.

The radiation is as disordered as possible when that probability splits into separate factors, one for each amplitude and one for each phase:

f(A1,A2,…,α1,α2,…)=F1(A1)F2(A2)…f1(α1)f2(α2)…f(A_1,A_2,\ldots,\alpha_1,\alpha_2,\ldots)=F_1(A_1)F_2(A_2)\ldots f_1(\alpha_1)f_2(\alpha_2)\ldots

The joint probability density of all the amplitudes and phases equals a product of separate factors, one for each amplitude and one for each phase.

Then the value of any one amplitude or phase tells nothing about the others. Einstein adds that the radiation in his box comes closer to this the more each pair of an amplitude and its phase depends on emission and absorption by its own particular group of resonators. Neither assumption is derived in §1. The footnotes state them, and the equilibrium, with the unbounded total energy it leads to at the end of §1, rests on both.

Show every step here: Two assumptions under the equilibrium of section 1

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Modern qualifications

Drude's electron theory of metals (Annalen der Physik, 1900) treated the conduction electrons as a gas sharing the thermal energy of the metal, which is the use the Drude footnote points to. For Planck's relation §1 cites Ann. d. Phys. 1, p. 99, 1900; that paper derives the relation under a statistical condition on the radiation that Planck called natural radiation, and the footnote examined here is Einstein's own statement of the condition in terms of Fourier amplitudes and phases. On the plate (p. 135) the second α in the sentence after the product is set as an italic x, a misprint the sense of the sentence corrects. Whether cavity radiation meets the condition is a question §1 does not take up.

Assumptions and limits: Two assumptions under the equilibrium of section 1

Assumed here

  • Free electrons collide with the gas molecules as gas molecules collide with each other.
  • At equilibrium, gas molecules and electrons have equal mean kinetic energies (the footnote to the setup of §1).
  • Planck's relation between a resonator and the radiation around it holds for radiation that is as disordered as possible (the footnote that defines it).

What this does not establish

  • The passage explains what the footnotes state; it does not test whether real radiation in a cavity meets the disorder condition.
  • Einstein cites Drude's theory of metals in support; the passage does not assess that theory.

Earlier step: A finite window cannot cure an infinite total

Source context: German source · Facsimile

§2 · what spectral constants determine

A derivation, to an approximation

A spectral fit is not yet a free-light hypothesis

What can the low-frequency coefficient determine about molecular scale?

The 4 printed paragraphs this passage explains
  • §2, paragraph 1: Einstein will show that Planck's determination of the elementary quanta, the constants that fix the size of atoms, does not depend entirely on Planck's theory of black radiation.
  • §2, paragraph 2: Planck's formula, which fits all experience so far, reduces for long wavelengths and high densities to the form found in section 1. Matching the two coefficients gives N = 6.17 × 10²³, so a hydrogen atom weighs 1.62 × 10⁻²⁴ gram: Planck's own value, in fair agreement with other methods.
  • §2, footnote 1: The footnote cites Planck's paper in Annalen der Physik, volume 4, page 561, 1901.
  • §2, paragraph 3: So the theoretical foundations of section 1 work better the higher the energy density and the longer the wavelength, and they fail completely for short wavelengths and low densities.

Write the historical spectral constants as AA and BB in this explanatory notation. At low ν/T\nu/T, expanding the exponential denominator gives a term proportional to ν2T\nu^2 T. Its coefficient can be compared to the classical allocation from §1.

A ν3exp⁡(B νT)−1≈AB ν2 T\frac{A\,\nu^{3}}{\exp\left(\frac{B\,\nu}{T}\right) - 1} \approx \frac{A}{B}\,\nu^{2}\,T

At low frequency relative to temperature, the spectrum approaches A over B times frequency squared times temperature.

AB=8 π RN c3\frac{A}{B} = \frac{8\,\pi\,R}{N\,c^{3}}
N=BA 8 π Rc3N = \frac{B}{A}\,\frac{8\,\pi\,R}{c^{3}}

Equating the coefficients gives N equal to B over A times eight pi R over light speed cubed.

This comparison is about coefficients and their experimental provenance. In the modern SI, RR equals NAkBN_A k_B. Inserting that identity and the modern spectral constants recovers the defined NAN_A; it is not independent evidence for a molecular count.

Planck’s resonators belong to the matter that emits or absorbs radiation. Einstein’s proposed energy elements concern the radiation itself. Moving between those statements requires an argument.

Show every step here: A spectral fit is not yet a free-light hypothesis

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Modern qualifications

The paper writes the spectral constants as α\alpha and β\beta; AA and BB here are editorial names for them. In §2 Einstein derives N=6.17×1023N = 6.17 \times 10^{23} from Planck’s printed constants. The historical constant set reproduces that number and the modern set does not, so the two are kept apart.

Assumptions and limits: A spectral fit is not yet a free-light hypothesis

Assumed here

  • Both sides use the same spectral density and the same units.
  • The spectral constants and the gas constant are separate inputs to any inference drawn from them.

What this does not establish

  • Modern exact SI constants give a consistency check, not a new measurement of N.
  • A fit to the Planck spectrum does not by itself establish independently moving light elements.

Earlier step: A finite window cannot cure an infinite total

Source context: German source · Facsimile

§3 · temperature and entropy

A derivation

A spectrum can determine an entropy derivative

How does equilibrium temperature constrain radiation entropy?

The 6 printed paragraphs this passage explains
  • §2, paragraph 4: From here on, black radiation is treated from experience, without any picture of how radiation is produced or travels.
  • §3, paragraph 1: The next argument is Wien's, from a well-known paper, and is included only for completeness.
  • §3, paragraph 2: Radiation filling a volume v is taken to be fully described by its energy density at every frequency. Since radiation of different frequencies can be separated without work or heat, its entropy is the volume times a sum over frequencies of an entropy density that depends on the density and the frequency.
  • §3, footnote 1: The footnote calls that assumption arbitrary: it is the simplest one, to be kept until experiment forces it to be dropped.
  • §3, paragraph 3: For black radiation the entropy is as large as it can be for the given energy, which means the rate at which the entropy density grows with the energy density is the same at every frequency.
  • §3, paragraph 4: Warming black radiation in a unit volume, and comparing with the rule that entropy grows by the added heat divided by the temperature, shows that this common rate is one over the temperature. That is the law of black radiation, so the entropy density can be found from the radiation law, and the radiation law from it.

Let ρν\rho_\nu be energy per volume per frequency interval and sνs_\nu the corresponding entropy density. At fixed volume, equilibrium maximizes entropy while keeping total energy fixed. Moving a small amount of energy from one spectral interval to another must give no first-order entropy gain at equilibrium.

Consequently the derivatives of entropy with respect to spectral energy density have the same value across the equilibrium spectrum. The thermodynamic identity dS/dE = 1/T, with other constraints fixed, identifies that common value.

(∂sν∂ρν)ν=1T\left(\frac{\partial s_\nu}{\partial \rho_\nu}\right)_{\nu} = \frac{1}{T}

At fixed frequency, the partial derivative of spectral entropy density with respect to spectral energy density is inverse temperature.

Einstein attributes this thermodynamic reasoning to Wien. The next step is to express 1/T as a function of density by using Wien’s spectral law, not by assuming particles of light in advance.

Show every step here: A spectrum can determine an entropy derivative

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Modern qualifications

The variational calculation assumes that the radiation entropy can be represented by a function of spectral density and frequency. That is a stated thermodynamic assumption, not a derivation of a microscopic state-counting model.

Assumptions and limits: A spectrum can determine an entropy derivative

Assumed here

  • Equilibrium thermodynamics holds, and energy and entropy add up over the spectrum.
  • Vary the energy distribution at fixed volume and fixed total energy.

What this does not establish

  • This is a thermodynamic comparison, not a microscopic account of every exchange.
  • Integrating the derivative requires a boundary condition.

Source context: German source · Facsimile

§4 · dilute radiation and the volume law

A derivation, to an approximation

The constant cannot simply be dropped

Why does the zero-radiation boundary condition matter?

The 2 printed paragraphs this passage explains
  • §4, paragraph 1: Wien's radiation law is known not to hold exactly, but experiment confirmed it very well when the frequency is large compared with the temperature. Einstein builds on it while keeping in mind that his results hold only within those limits.
  • §4, paragraph 2: Putting Wien's law into the relation from section 3 gives the entropy density. For radiation of energy E in a narrow band of frequencies, the entropy changes by E/(βν) times the logarithm of v/v₀ when its volume changes from v₀ to v.
ρν=A ν3 exp⁡(−B νT)\rho_\nu = A\,\nu^{3}\,\exp\left(-\frac{B\,\nu}{T}\right)

Wien spectral energy density is A times frequency cubed times the exponential of minus B frequency over temperature.

Divide by Aν3A\nu^3 and take the natural logarithm. Its argument is dimensionless. Solving for inverse temperature turns the entropy derivative from §3 into an explicit function of density.

1T=−1B ν ln⁡(ρνA ν3)\frac{1}{T} = -\frac{1}{B\,\nu}\,\ln\left(\frac{\rho_\nu}{A\,\nu^{3}}\right)

Inverse temperature is minus one over B frequency times the natural logarithm of density over A frequency cubed.

sν=−ρνB ν (ln⁡(ρνA ν3)−1)+C(ν)s_\nu = -\frac{\rho_\nu}{B\,\nu}\,\left(\ln\left(\frac{\rho_\nu}{A\,\nu^{3}}\right) - 1\right) + C\left(\nu\right)

The integrated entropy density is minus density over B frequency times the bracketed logarithm minus one, plus a frequency-dependent constant.

As positive density approaches zero, ρln⁡ρ\rho\ln\rho tends to zero. The remaining limit is C(ν)C(\nu). The boundary condition sν→0s_\nu \to 0 therefore sets C(ν)=0C(\nu) = 0. Without this condition, the volume comparison in the next step has an extra term.

Show every step here: The constant cannot simply be dropped

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Modern qualifications

This explanation uses natural logarithms and editorial AA, BB notation. The source prints lg⁡\lg. The interpretation of BB as h/kBh/k_B belongs to the later comparison of coefficients.

Assumptions and limits: The constant cannot simply be dropped

Assumed here

  • The narrow spectral region obeys the Wien approximation.
  • At zero radiation density, radiation entropy density vanishes.

What this does not establish

  • A positive inferred temperature alone is not a guarantee that the Wien approximation is accurate.
  • The spectral constants are not initially interpreted as particle energies.

Earlier step: A spectrum can determine an entropy derivative

Source context: German source · Facsimile

A derivation, to an approximation

Compare two states, not a compression movie

What is held fixed when volume changes?

The 2 printed paragraphs this passage explains
  • §4, paragraph 2: Putting Wien's law into the relation from section 3 gives the entropy density. For radiation of energy E in a narrow band of frequencies, the entropy changes by E/(βν) times the logarithm of v/v₀ when its volume changes from v₀ to v.
  • §4, paragraph 3: So the entropy of dilute monochromatic radiation depends on its volume by the same law as the entropy of an ideal gas or a dilute solution. Einstein will interpret this with Boltzmann's principle, that the entropy of a system is a function of the probability of its state.

Within a narrow band, write E=V Δν ρνE = V\,\Delta\nu\,\rho_\nu and S=V Δν sνS = V\,\Delta\nu\,s_\nu. Substitute ρν=E/(V Δν)\rho_\nu = E/(V\,\Delta\nu) into the integrated entropy density. Comparing VV with V0V_0 at the same EE and band removes terms independent of volume.

S(V)−S(V0)=EB ν ln⁡(VV0)S\left(V\right) - S\left(V_{0}\right) = \frac{E}{B\,\nu}\,\ln\left(\frac{V}{V_{0}}\right)

The entropy difference is E over B frequency times the natural logarithm of the volume ratio.

E/(Bν)E/(B\nu) has entropy units. Only after identifying B=h/kBB = h/k_B can it be written kBk_B times the dimensionless coefficient E/(hν)E/(h\nu). This distinction matters when interpreting a displayed coefficient as a count.

ΔSC=Δν C(ν) (V−V0)\Delta S_C = \Delta\nu\,C\left(\nu\right)\,\left(V - V_{0}\right)

An unfixed entropy-density constant would add bandwidth times C of frequency times the volume difference.

Doubling the accessible volume adds (E/(Bν))ln⁡2(E/(B\nu))\ln 2 under the stated constraints. Doubling it again adds the same amount. The radiation-entropy laboratory compares those constrained states and exposes the otherwise hidden constant.

Show every step here: Compare two states, not a compression movie

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Modern qualifications

A modern pointwise comparison gives relative spectral-density error exp⁡(−hν/(kBT))\exp(-h\nu/(k_B T)) for Wien versus Planck. That diagnostic is not, by itself, an error bound for the entire integrated entropy argument.

Assumptions and limits: Compare two states, not a compression movie

Assumed here

  • Energy E, central frequency ν, and bandwidth Δν are fixed in both states.
  • The entropy-density constant has been fixed by the zero-radiation condition.
  • Both states satisfy the dilute Wien-regime and narrow-band assumptions.

What this does not establish

  • A moving mirror generally changes frequency and energy, so it is not this held-fixed comparison.
  • The result is not valid merely because one of the two endpoint states is dilute.

Earlier step: The constant cannot simply be dropped

Source context: German source · Facsimile

§5 · independent configurations

A definition

Which probability the entropy uses

What does Einstein mean by the probability of a state, and why does he insist on it?

The printed paragraph this passage explains
  • §5, paragraph 1: Einstein notes that molecular theory often uses the word probability in a sense that differs from its mathematical definition, and says he will show in another paper that statistical probability is enough; here he gives only the general principle and some special cases.

§5 opens with a warning about a word. When entropy is calculated from molecular theory, Einstein writes, 'probability' is often used in a sense that does not match its definition in the theory of probability. In particular, which cases count as equally probable is often fixed by hypothesis, even where the theoretical picture is definite enough to deduce them instead.

He announces that he will show in a separate paper that the statistical probability is all that thermal questions need, and hopes this removes a logical difficulty that still stands in the way of applying Boltzmann's principle. Here he states the principle only in general and applies it to special cases. The first special case shows what statistical probability means: the probability that, at a moment chosen at random, the system is found in the state considered.

S−S0=RNlg⁡WS-S_0=\frac{R}{N}\lg W

The entropy difference S minus S zero equals R over N times the logarithm of W.

Here W is the probability of the state with entropy S relative to a starting state with entropy S0S_0, R is the gas constant, N the number of real molecules in a gram-equivalent, and lg is the natural logarithm as the paper writes it. Read with a statistical W, the formula needs no list of equally likely cases: a state the system is found in at 1 moment in 16, compared with a state it is always in, has W = 1/16, and its entropy is lower by R/N times ln 16.

The reading matters for everything after it. The probability that n independent points are all in part of a volume, and in §6 the probability that all the energy of monochromatic radiation is in part of its volume, are statistical probabilities in exactly this sense. The step from entropy to energy quanta is only as good as that reading of W.

Show every step here: Which probability the entropy uses

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Modern qualifications

Boltzmann had tied entropy to the probability of a state in the kinetic theory of gases; Einstein calls the relation Boltzmann's principle and writes its constant as R/N. Planck's 1901 paper, which §2 cites (Ann. d. Phys. 4, p. 561), writes that constant as k, the letter later written kBk_B. Before 1905 Einstein had published three papers on the molecular foundations of thermodynamics in the Annalen der Physik (1902, 1903 and 1904). The separate paper promised in §5 is not named, and this passage does not identify it; nor does §5 say whose use of probability it criticizes.

Assumptions and limits: Which probability the entropy uses

Assumed here

  • The probability of a state is its statistical probability: the chance of finding the system in that state at a moment chosen at random.
  • Boltzmann's principle, as Einstein uses it: the entropy of a system is a function of the probability of its state.
  • The argument compares entropy differences between states; it needs no absolute entropy.

What this does not establish

  • Einstein says he will show in a separate paper that statistical probability is enough; this paper does not give that proof, and this passage does not supply it.
  • A moment chosen at random presupposes a system in a steady condition, watched over a long time; the passage does not treat systems that are not.

Source context: German source · Facsimile

A derivation

Independence supplies the exponent

Why is the probability f to the power n, rather than just f?

The 8 printed paragraphs this passage explains
  • §5, paragraph 2: If entropy is a function of the probability of a state, two systems that do not act on each other have entropies that add while their probabilities multiply, because their states are independent events.
  • §5, paragraph 3: It follows that entropy is a universal constant times the logarithm of the probability, plus a constant. Kinetic theory gives that constant as R/N, so the entropy difference between two states is R/N times the logarithm of their relative probability.
  • §5, paragraph 4: A special case: n moving points, such as molecules, 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 of them that they do not act on each other.
  • §5, paragraph 5: This system, an ideal gas or a dilute solution for example, has some entropy. Imagine all n points moved into a part v of the volume with nothing else changed: the new state has a different entropy, to be found from Boltzmann's principle.
  • §5, paragraph 6: The question: how probable is it that, at a moment chosen at random, all n independently moving points happen to be in the part v?
  • §5, paragraph 7: That statistical probability is v/v₀ raised to the power n, and Boltzmann's principle turns it into an entropy difference of R(n/N) times the logarithm of v/v₀.
  • §5, paragraph 8: Remarkably, this equation, from which Boyle and Gay-Lussac's law and the matching law of osmotic pressure follow, needs no assumption about the law by which the molecules move.
  • §5, footnote 1: The footnote derives the gas law from that entropy difference: the work p dv equals T dS, which gives pv = R(n/N)T.

Each independent point has probability f of being inside the chosen fraction. For all n to be inside, multiply n independent probabilities. Taking the logarithm makes the exponent n a coefficient.

W=fnW = f^{n}
ΔSgas=kB ln⁡(W)=n kB ln⁡(f)\Delta S_{\mathrm{gas}} = k_B\,\ln\left(W\right) = n\,k_B\,\ln\left(f\right)

The probability is f to the n; the entropy difference is Boltzmann's constant times its logarithm, or n times Boltzmann's constant times the logarithm of f.

With four independent points and half the volume, the probability is 1/16. If all four positions are exactly the same uniformly distributed position, the probability is 1/2 instead. Having four labels is not enough to justify four independent factors.

The concentration from V₀ to V smaller than V₀ has a negative entropy difference. Reversing the constrained comparison reverses the sign. The independent-configurations laboratory makes the probability and entropy readouts separate.

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Modern qualifications

This is Boltzmann’s principle read backwards. Einstein computes how the entropy of dilute radiation changes with volume, compares it with the entropy of nn independent points, kln⁡Wk\ln W with W=fnW = f^n, and concludes that such radiation behaves thermodynamically as if it consisted of nn independent energy quanta. For large nn, fnf^n is astronomically small, so the laboratory reports its logarithm.

Assumptions and limits: Independence supplies the exponent

Assumed here

  • Each point is uniformly distributed over the initial volume.
  • The point positions are statistically independent for the independent model.
  • Entropy differences are related to logarithms of relative configuration probabilities.

What this does not establish

  • The locked-position example is an authored mathematical counterexample, not an established alternative theory of radiation.
  • Spontaneous concentration probability and constrained-state entropy are related but are not the same observable.
What is getting in the way?
  • An unfamiliar word or symbol

    f is a fraction of the volume: the chosen part divided by the whole, one half when the part is half of it. n is the number of points. W is a probability, the chance that at one moment all n points are inside the chosen part, and ln is the natural logarithm.

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  • An algebraic move

    Multiplying n equal factors f is written fⁿ, and the logarithm turns that power into a product: ln fⁿ = n ln f. So k_B ln W becomes n k_B ln f, the number of points times the change for one. Because f is less than 1, ln f is negative, and so is the entropy change for crowding the points into the smaller part.

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  • The physical reason for a step

    In the model each point moves over the whole volume without regard to the others, as the molecules of a dilute gas do. So the chance that any one point is in the chosen part is f, whatever the other points are doing. That independence is a premise of the model; the counting does not prove it.

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  • The connection to the picture

    Split the volume into the chosen part and the rest, and take one snapshot. One point is in the part a fraction f of the time. For two, the part must catch the first and also the second: f of the time, and then f of those times, which is f × f. Each further point multiplies by f again.

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  • The purpose of the calculation

    Section 6 uses the result in reverse. The entropy of dilute monochromatic radiation changes with volume the way n k_B ln f does for n independent points, so in that respect such radiation behaves as if it consisted of n independent energy quanta. The exponent is what carries the comparison, which is why the passage says it comes from independence.

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  • Too much at once

    One example: four points, half the volume. One point is in the left half with chance 1/2. Two are, with chance 1/2 × 1/2 = 1/4. All four are, with chance 1/16. If the four were locked together and moved as one, the chance would stay 1/2. That difference is the whole passage.

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Source context: German source · Facsimile

§6 · the heuristic correspondence

A heuristic step, to an approximation

A coefficient suggests an energy element

What plays the role of the number of independent things?

The 7 printed paragraphs this passage explains
  • §4, paragraph 3: So the entropy of dilute monochromatic radiation depends on its volume by the same law as the entropy of an ideal gas or a dilute solution. Einstein will interpret this with Boltzmann's principle, that the entropy of a system is a function of the probability of its state.
  • §6, paragraph 1: Section 4 found how the entropy of monochromatic radiation depends on its volume. Rewritten as R/N times the logarithm of a power of the volume ratio, and compared with Boltzmann's principle, it leads to a conclusion.
  • §6, paragraph 2: For monochromatic radiation of energy E shut into a volume v₀ by mirrors, the probability that at a moment chosen at random all its energy is in the part v is the volume ratio raised to the power (N/R)(E/βν).
  • §6, paragraph 3: A connecting line: from that probability Einstein draws a further conclusion, stated next.
  • §6, paragraph 4: Monochromatic radiation of low density, where Wien's law holds, behaves in its thermal properties as if it consisted of independent energy quanta of size Rβν/N.
  • §6, paragraph 5: Einstein compares the mean size of these energy quanta in black radiation, which Wien's formula gives as 3(R/N)T, with the mean kinetic energy of a molecule's motion at the same temperature, (3/2)(R/N)T.
  • §6, paragraph 6: If radiation behaves, in how its entropy depends on volume, like a medium made of such energy quanta, it is natural to ask whether light is also produced and transformed as if it consisted of them. The rest of the paper takes up that question.
ΔSgas=n kB ln⁡(f)\Delta S_{\mathrm{gas}} = n\,k_B\,\ln\left(f\right)
ΔSrad=EB ν ln⁡(f)\Delta S_{\mathrm{rad}} = \frac{E}{B\,\nu}\,\ln\left(f\right)

The gas entropy change has coefficient n times Boltzmann's constant; the radiation change has coefficient E divided by B times frequency.

If the radiation is interpreted through the independent-element analogy, matching the coefficients suggests neff=E/(kBBν)n_\text{eff} = E/(k_B B\nu). Its corresponding energy per element is kBBνk_B B\nu. The algebra identifies a scale conditional on the analogy; it does not prove the analogy.

neff=EkB B νn_{\mathrm{eff}} = \frac{E}{k_B\,B\,\nu}
Eneff=kB B ν=h ν\frac{E}{n_{\mathrm{eff}}} = k_B\,B\,\nu = h\,\nu

The effective coefficient is E divided by Boltzmann's constant times B times frequency. The implied energy scale is Boltzmann's constant times B times frequency, written h times frequency in modern notation.

Keep a non-integer effective coefficient as it stands. The entropy comparison concerns a macroscopic relation, not a directly observed list of individual packets. In the entropy-comparison laboratory, compare the coefficients before revealing their interpretation.

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Modern qualifications

Einstein calls his proposal a heuristic point of view, and the paper’s title says so. A matching mathematical form can suggest a model without forcing it: Planck’s energy elements of 1900 belong to the oscillators in the walls, Einstein’s quanta to the radiation itself, and the two agree on some numbers while saying different things about light.

Assumptions and limits: A coefficient suggests an energy element

Assumed here

  • Radiation in a narrow band obeys the entropy law derived for dilute (Wien) radiation.
  • Independent points obey the entropy law for n points, with k_B identified as R/N.
  • A matching dependence on volume is read as suggesting independent energy elements.

What this does not establish

  • E/(hν) is a pure number that need not be a whole number, and it is never rounded to one.
  • This does not derive Planck’s law or establish a general theory outside the Wien regime.
  • Extending the picture to individual emission and absorption processes is a further hypothesis.

Earlier step: Compare two states, not a compression movieIndependence supplies the exponent

Source context: German source · Facsimile

§7 · fluorescence and its conditions

A derivation

An energy budget has conditions

When is emitted fluorescent light restricted to a lower frequency?

The 4 printed paragraphs this passage explains
  • §7, paragraph 1: In photoluminescence, light of one frequency is turned into light of another. If both consist of energy quanta, each absorbed quantum gives rise on its own to an emitted one, and unless the substance supplies energy, the emitted quantum cannot carry more energy than the absorbed one: the emitted frequency is at most the exciting one, which is Stokes's rule.
  • §7, paragraph 2: On this view, under weak illumination the light produced is proportional to the exciting light, since each quantum acts independently, and there is no lower limit of intensity below which light fails to excite.
  • §7, paragraph 3: Exceptions to Stokes's rule are conceivable in two cases: when so many quanta are transformed at once that one emitted quantum takes energy from several exciting ones, and when the light is not like black radiation within the range of Wien's law, as with light from a very hot source.
  • §7, paragraph 4: The second case is of special interest: radiation that does not follow Wien's law might behave differently in its energy even when very dilute.

If an event receives only hνinh\nu_\text{in}, emits hνouth\nu_\text{out}, and retains a nonnegative remainder, energy conservation gives a bound on the outgoing frequency. It is a consequence of the assumed event budget, not evidence that all fluorescent events have that budget.

h νout≤h νinh\,\nu_{\mathrm{out}} \le h\,\nu_{\mathrm{in}}

The emitted quantum energy is at most the absorbed quantum energy in the restricted event budget.

Supply extra thermal energy or allow more than one absorbed quantum and the bound changes. The fluorescence laboratory's controls expose those extra-energy and channel assumptions rather than quietly creating energy.

Proportional fluorescence intensity requires further assumptions about absorption and conversion yield. Holding those assumptions fixed can make a weak-illumination rate proportional to the incident rate; the energy bound alone does not determine the yield.

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Modern qualifications

In §7 Einstein applies the picture to Stokes’s rule, that fluorescent light does not have a higher frequency than the light that excites it. He states when the rule may fail, for instance when so many quanta act at once that one emitted quantum draws its energy from several. The energy budget here is a modern way of drawing the same account; it says nothing about the inner workings of a particular fluorescent material.

Assumptions and limits: An energy budget has conditions

Assumed here

  • An event absorbs one light quantum and produces at most one emitted quantum in this idealized budget.
  • No additional energy reservoir contributes in the restricted case.
  • The light-quantum hypothesis is being applied to a transformation process.

What this does not establish

  • The limit on frequency holds under these conditions; it does not forbid higher-frequency emission in general.
  • The budget does not predict actual material rates or spectra.
  • A dense field need not satisfy the independent-quantum assumptions used in the analogy.

Earlier step: A coefficient suggests an energy element

Source context: German source · Facsimile

§8 · photoelectric energy and counts

A derivation

More electrons is not more energy per electron

What changes when frequency rises, and what changes when power rises?

The 9 printed paragraphs this passage explains
  • §8, paragraph 1: The usual view, that the energy of light is spread continuously, meets great difficulties in explaining the photoelectric effects, which Lenard described in detail.
  • §8, footnote 1: The footnote cites Lenard, Annalen der Physik, volume 8, pages 169 and 170, 1902.
  • §8, paragraph 2: If the exciting light consists of energy quanta, the production of cathode rays by light can be pictured simply: quanta enter the surface layer and give their energy, at least in part, to electrons, the simplest case being one quantum to one electron. An electron loses some energy on its way out and must do a work P to leave, so the fastest electrons leave with the energy (R/N)βν − P.
  • §8, paragraph 4: With E = 9.6 × 10³, Π × 10⁻⁸ is the potential in volts that the body reaches when irradiated in a vacuum.
  • §8, paragraph 5: As an order-of-magnitude check, with P′ = 0 and the frequency at the ultraviolet end of the solar spectrum, the formula gives about 4.3 volts, the same order as Lenard's results.
  • §8, footnote 3: The footnote cites Lenard, Annalen der Physik, volume 8, pages 165 and 184, plate I, figure 2, 1902.
  • §8, paragraph 6: If the formula is right, Π plotted against the frequency of the exciting light must be a straight line whose slope does not depend on the substance.
  • §8, footnote 4: The footnote cites the same paper by Lenard, pages 150 and 166 to 168.
  • §8, paragraph 7: As far as Einstein can see, the view does not contradict what Lenard observed: if each quantum acts independently, the speeds of the electrons do not depend on the intensity of the light, while the number of electrons leaving is proportional to it.

The most energetic electron receives a complete quantum and escapes with the least modeled energy loss. Subtracting the escape work Φ\Phi gives the maximum kinetic energy, provided emission is energetically allowed.

Kmax=h ν−ΦK_{\mathrm{max}} = h\,\nu - \Phi
ν0=Φh\nu_0 = \frac{\Phi}{h}

Maximum kinetic energy equals h frequency minus escape work, with threshold frequency Phi over h.

For a hypothetical work function of 2 electron volts and a frequency of 600 terahertz, the maximum is about 0.4814 electron volts using modern SI constants. Doubling optical power while retaining that frequency and escape work leaves the maximum unchanged.

Power is energy arriving per second. At fixed frequency, dividing it by hνh\nu gives the incoming quantum rate. Under a fixed yield, twice the power can therefore produce twice the electron count per second. Below threshold, negative hν−Φh\nu - \Phi is an energy deficit, not a negative kinetic energy of an emitted electron.

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Modern qualifications

The source discusses the potential attained by an illuminated body; a later collector apparatus is a distinct arrangement. Later measurements such as Millikan’s 1916 work are not premises available in 1905, and agreement with the energy relation alone does not prove the complete light-quantum picture.

Assumptions and limits: More electrons is not more energy per electron

Assumed here

  • One electron can receive one quantum of energy hν.
  • Escape requires positive work Φ, with any additional loss nonnegative.
  • Frequency, work function and yield stay fixed when the power changes.

What this does not establish

  • In the laboratory the threshold and the straight line follow from the model’s assumptions; they are not experimental proof.
  • The work function used in a hypothetical example is not a sourced value for a named metal.
  • Real collection and emission rates require additional material and apparatus assumptions.

Earlier step: A coefficient suggests an energy element

Source context: German source · Facsimile

A qualification

Einstein's own limits on the photoelectric laws

Where does Einstein expect his photoelectric regularities might stop holding?

The printed paragraph this passage explains
  • §8, paragraph 8: Einstein adds that the likely limits of validity of these laws call for remarks like those he made about deviations from Stokes's rule.

After working out how the escaping electrons should behave, Einstein adds one sentence of caution: about the presumable limits of validity of these regularities, remarks would have to be made like those about the presumable deviations from Stokes's rule. His word, 'mutmaßlich', means presumable: he expects limits, and he does not locate them.

The regularities are those of the paragraph before. If each energy quantum gives its energy to electrons independently of all the others, the speeds of the escaping electrons do not depend on the intensity of the exciting light, while their number is proportional to it. The energy relation behind them assumes one quantum for one electron:

ΠE=Rβν−P′ΠE=R\beta\nu-P'

Pi times E equals R beta nu minus P prime.

The remarks on Stokes's rule, in §7, named two conditions. First, so many quanta are being transformed at once in each unit of volume that one produced quantum can take its energy from several exciting quanta. Second, the light is not of the kind black radiation has within the range of Wien's law, for instance light from a body so hot that Wien's law no longer holds at the wavelength concerned.

Carried over, as the sentence invites: in very intense light an electron might gather energy from more than one quantum, and then its energy is no longer bounded by one quantum's (R/N)βν less the work P it spends leaving. For light outside Wien's range, the entropy argument of §6 that gave quanta of size Rβν/N does not apply, so it does not secure the size of the energy portions. Einstein spells out neither case for the photoelectric effect; the caution stays a caution.

Show every step here: Einstein's own limits on the photoelectric laws

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Modern qualifications

In 1905 the photoelectric data were Lenard's, and the paper claims only that its view does not contradict them, as far as Einstein can see. Millikan's measurements of 1916 later tested the straight-line relation between stopping potential and frequency; that is later evidence, not a premise available in 1905. Emission in which one electron takes the energy of two light quanta at once was observed only after lasers made such intensities available, decades later; it is one concrete form of the first condition, and the paper does not anticipate it in any detail.

Assumptions and limits: Einstein's own limits on the photoelectric laws

Assumed here

  • The regularities in question are those of the paragraph before: the speeds of the escaping electrons do not depend on the intensity of the light, and their number is proportional to it.
  • In §7 Einstein named two conditions under which Stokes's rule might fail.
  • §8 says only that similar remarks would apply to the photoelectric laws; it does not spell them out.

What this does not establish

  • The two conditions are carried over from §7 by the analogy Einstein's sentence invites; §8 does not state them itself, so this passage marks them as a reading of his remark.
  • The paper gives no intensity and no temperature at which the laws fail, and neither does this passage.

Earlier step: More electrons is not more energy per electronAn energy budget has conditions

Source context: German source · Facsimile

A qualification

A stopping voltage is a magnitude with a sign convention

Why does retarding an electron not make its charge positive?

The 6 printed paragraphs this passage explains
  • §8, paragraph 2: If the exciting light consists of energy quanta, the production of cathode rays by light can be pictured simply: quanta enter the surface layer and give their energy, at least in part, to electrons, the simplest case being one quantum to one electron. An electron loses some energy on its way out and must do a work P to leave, so the fastest electrons leave with the energy (R/N)βν − P.
  • §8, paragraph 3: If the body is charged to the positive potential Π that just stops it losing charge, Π times the electron's charge equals (R/N)βν − P; for a gram-equivalent of electrons this reads ΠE = Rβν − P′.
  • §8, footnote 2: The footnote says the relation is unchanged if the electron must first be freed from a neutral molecule; P′ is then the sum of two works.
  • §8, paragraph 9: The argument assumed that at least some quanta give all their energy to a single electron; without that assumption the equation becomes an inequality, ΠE + P′ ≤ Rβν.
  • §8, paragraph 10: For cathode luminescence, the reverse process, a similar argument gives ΠE + P′ ≥ Rβν. In Lenard's experiments the electrons had fallen through hundreds or thousands of volts to make visible light, so one electron's energy must go into many light quanta.
  • §8, footnote 5: The footnote cites Lenard, Annalen der Physik, volume 12, page 469, 1903.
e Vs=Kmax=h ν−Φe\,V_s = K_{\mathrm{max}} = h\,\nu - \Phi

The positive charge magnitude times the stopping-voltage magnitude equals maximum kinetic energy, h frequency minus escape work.

VsV_s is a nonnegative magnitude. The sign of a laboratory electrode voltage must also specify which electrode is the reference. Using ee for a positive magnitude does not change the negative charge of the electron.

If the electron receives only part of the available energy or loses energy inside matter, its outgoing kinetic energy is at most hν−Φh\nu - \Phi. These losses do not reverse the frequency dependence or turn extra intensity into extra energy for one electron.

A footnote to the stopping condition allows that the light may also have to free each electron from a neutral molecule, at the cost of some work, before the electron can leave. The paper says nothing in the derived relation changes: its P′P', which it defines as the potential of a gram-equivalent of negative electricity with respect to the body, is then read as a sum of two terms, the work to free the electron and the work to leave the body. In the letters used here, Φ\Phi stands for that sum, and the stopping potential still rises in a straight line with frequency, with the same slope.

An inverse energy-budget question asks for the smallest accelerating-potential magnitude capable of supplying a quantum hνh\nu. Within that idealized conversion, eVeV must be at least hνh\nu. This is an energetic lower bound, not a promise that every collision produces light.

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Modern qualifications

This modern voltage-magnitude convention does not silently replace the paper’s printed charge and potential symbols. The historical 4.3-volt check belongs to its separately identified constant and unit convention in the photoelectric laboratory. The footnote on P′P' leaves room for electrons bound in neutral molecules without giving that work a value; §9 applies the same one-quantum budget to freeing an electron from a gas molecule, which is ionization.

Assumptions and limits: A stopping voltage is a magnitude with a sign convention

Assumed here

  • e denotes a positive charge magnitude; the electron charge is −e.
  • A retarding potential removes kinetic energy, with the sign convention of the electrodes stated.

What this does not establish

  • No stopping endpoint for emitted electrons is supplied when emission is not allowed.
  • A negative energy deficit and a signed collector voltage are different quantities.
  • The footnote gives no size for the work of freeing an electron from a molecule; the passage's figures for it are hypothetical.

Earlier step: More electrons is not more energy per electron

Source context: German source · Facsimile

§9 · ionization bounds and closing scope

A derivation

A threshold does not specify a yield

What can energy conservation say about gas ionization?

The 4 printed paragraphs this passage explains
  • §9, paragraph 1: When ultraviolet light ionizes a gas, each absorbed quantum should ionize one molecule, so the work to ionize a molecule cannot exceed the energy of one quantum. Lenard's longest effective wavelength for air, about 1.9 × 10⁻⁵ cm, gives an upper limit of 6.4 × 10¹² erg per gram-equivalent.
  • §9, paragraph 2: Stark's smallest measured ionizing voltage for air, about 10 volts, gives a nearly equal upper limit of 9.6 × 10¹². A further consequence, worth testing: the number of gram-molecules ionized should equal the absorbed light energy divided by Rβν, for any gas with no noticeable absorption at that frequency that is not accompanied by ionization.
  • §9, footnote 1: The footnote cites Stark, Die Elektrizität in Gasen, page 57, Leipzig 1902.
  • §9, footnote 2: The footnote notes that inside the gas the ionizing voltage for negative ions is five times larger.

Let II be the required energy for the specified ionization channel. A single quantum can pay this cost only when hνh\nu is at least II. A smaller quantum leaves an energy deficit; a larger quantum makes the event energetically possible, not certain.

h ν≥Ih\,\nu \ge I

One-quantum ionization requires quantum energy at least equal to the specified ionization energy.

For monochromatic absorbed energy EabsE_\text{abs}, the number of absorbed quanta is represented by Eabs/(hν)E_\text{abs}/(h\nu). With at most one counted ionization per quantum this gives an upper bound. It becomes an equality only when all absorbed quanta each produce one counted ionization.

nions≤Eabsh νn_{\mathrm{ions}} \le \frac{E_{\mathrm{abs}}}{h\,\nu}

The ionization count is at most absorbed energy divided by quantum energy, under the at-most-one-ionization assumption.

The paper compares two upper limits for the work of ionizing a gram-equivalent of air. Lenard's longest ionizing wavelength gives 6.4×10126.4\times10^{12} erg, and Stark's smallest measured ionizing voltage, about 10 volts at platinum anodes, gives 9.6×10129.6\times10^{12}, nearly the same. A footnote concedes that inside the gas the ionizing voltage for negative ions is five times larger. Each measured voltage caps the work, so the smallest gives the tightest cap; the fivefold figure caps it only at about 4.8×10134.8\times10^{13} erg, and the near agreement the paper notes holds for the smallest value.

The ionization laboratory separates incident energy, absorbed fraction, threshold, and conversion assumptions. Its historical checks use their own stated units; this explanation does not infer a new gas species or a real-material yield from a threshold slider.

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Modern qualifications

§9 applies the same bound to gases ionized by ultraviolet light: at most one molecule ionized for each absorbed quantum, and none unless the quantum carries at least the ionization energy. The paper ends after this section, with its date-line. The footnote to Stark's 10 volts (J. Stark, Die Elektrizität in Gasen, p. 57, Leipzig 1902) adds, without a source of its own, that inside the gas the ionizing voltage for negative ions is five times larger; the comparison of upper limits rests on the smallest measured value.

Assumptions and limits: A threshold does not specify a yield

Assumed here

  • One quantum supplies the ionization energy in the stated idealization.
  • Only the absorbed energy counts, not all the energy that arrives.
  • A count equality requires one counted ionization per absorbed quantum.

What this does not establish

  • A threshold is a necessary energy condition, not a material cross-section or a guaranteed event.
  • An unknown absorbed fraction or yield is never silently taken to be 100 percent.
  • The energy budget does not predict recombination, collisions, or avalanche multiplication.
  • The footnote gives no source and no reason for the fivefold ionizing voltage inside the gas, and the passage supplies neither.

Earlier step: A coefficient suggests an energy element

Source context: German source · Facsimile

References and source status

This is newly written explanation in modern notation; editorial and physics review are pending. The introduction and all nine numbered sections are explained below, but this is not the German source, an English translation, or a complete critical edition. The German source face holds a machine-drafted transcription with hand correction that no one has reviewed yet, and the facsimile face shows the pinned journal pages; an English translation is not ready. The headings and the way the argument is divided into passages are ours: Einstein divided his paper into an introduction and nine numbered sections, under his own titles, and a passage here does not stand for one of his paragraphs.

A. Einstein, Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt. Annalen der Physik (4), 17, 132–148 (1905).

A. Einstein, On the motion of particles suspended in liquids at rest required by the molecular-kinetic theory of heat. Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.

A. Einstein, Zur Elektrodynamik bewegter Körper. Annalen der Physik (4), 17, 891–921 (1905).

A. Einstein, Does the inertia of a body depend upon its energy content? Annalen der Physik (4), 18, 639–641 (1905). External 1923 Perrett–Jeffery translation, electronically transcribed by John Walker; its notation was modernized. A reference for this explanatory preview, not this edition’s reviewed translation or pinned facsimile.

R. A. Millikan, A Direct Photoelectric Determination of Planck's "h". Physical Review (2), 7 (3), 355–388 (1916).

26 foundation readings sit behind this argument.

The question we were answering:

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