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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.

The paper's results

Page 136

The equilibrium the classical picture requires, and its infinite total

If each resonator has the mean energy $\frac{R}{N}T$ that the kinetic theory of gases requires and the mean energy Planck derived for equilibrium with the radiation, the radiation must have the density $\varrho_\nu = \frac{R}{N}\frac{8\pi\nu^2}{L^3}T$ ; the paper says this disagrees with experience, and that summed over all frequencies it gives an infinite energy, so the picture allows no definite distribution of energy between ether and matter.

As printed

Soll die Strahlungsenergie von der Frequenz $\nu$ nicht beständig im Ganzen weder vermindert noch vermehrt werden, so muß gelten: $\frac{R}{N} T = \overline{E} = \overline{E}_\nu = \frac{L^3}{8 \pi \nu^2} \varrho_\nu,$ Gas constantAvogadro's numberTemperature of the radiationMean resonator energyMean resonator energy at a given frequencySpeed of lightFrequency of the lightSpectral energy densityExplain this equation $\varrho_\nu = \frac{R}{N} \frac{8 \pi \nu^2}{L^3} T.$ Spectral energy densityGas constantAvogadro's numberFrequency of the lightSpeed of lightTemperature of the radiationExplain this equation Diese als Bedingung des dynamischen Gleichgewichtes gefundene Beziehung entbehrt nicht nur der Übereinstimmung mit der Erfahrung, sondern sie besagt auch, daß in unserem Bilde von einer bestimmten Energieverteilung zwischen Äther und Materie nicht die Rede sein kann. Je weiter nämlich der Schwingungszahlenbereich der Resonatoren gewählt wird, desto größer wird die Strahlungsenergie des Raumes, und wir erhalten in der Grenze $\int_0^\infty \varrho_\nu d \nu = \frac{R}{N} \frac{8 \pi}{L^3} T \int_0^\infty \nu^2 d \nu = \infty.$ Spectral energy densityFrequency of the lightGas constantAvogadro's numberSpeed of lightTemperature of the radiationExplain this equation

Text on page 136 of the German source, § 1, paragraph 5

In modern notation

u_\nu = \frac{8\,\pi\,\nu^{2}}{c^{3}}\,k_B\,T

U\left(\nu_{c}\right) = \int_{0}^{\nu_{c}} u_\nu\,\mathrm{d}\nu

U\left(\nu_{c}\right) = \frac{8\,\pi\,k_B\,T}{3\,c^{3}}\,\nu_{c}^{3}

Premise: Two conditions of equilibrium are combined: the resonators with the molecules and electrons of the gas, from the kinetic theory, and the resonators with the radiation, which Planck derived on the assumption that radiation is a process as disordered as can be.
A comparison, not a premise: § 2 shows that the same relation is the limit of Planck's formula for long wavelengths and large radiation densities, where it does agree with experience.

Decoder

$\overline{E}$: The mean energy of one linear oscillation of a resonator electron, which the kinetic theory of gases sets equal to $\frac{R}{N}T$ in equilibrium with the gas.
$\overline{E}_\nu$: The mean energy of a resonator of eigenfrequency $\nu$ , per component of oscillation, in Planck's condition for equilibrium with the radiation.
$\varrho_\nu$: The energy density of the radiation per unit frequency: $\varrho_\nu\,d\nu$ is the energy per unit volume of the part of the radiation between $\nu$ and $\nu + d\nu$ .
$L$: The speed of light, printed L in §§ 1 and 2 of this paper.
$\infty$: The total over all frequencies: the wider the range of the resonators' frequencies is taken, the greater the radiation energy of the space, without limit.

Try it

LQ-02: Give every resonator the same mean energy whatever its frequency: how much energy does the radiation hold as the range of frequencies widens?

Where this stops: 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. Read the argument

Where this stops: 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. Read the argument

Common wrong turns

The light-quantum paper was written to answer the “ultraviolet catastrophe”.

Argument: A consequence of two conditions of equilibrium, stated as a difficulty
Model: Within the picture of resonators, gas molecules and radiation that the section sets up
Evidence: A theoretical difficulty, which the paper says disagrees with experience
History: Its ingredients, the kinetic theory and Planck's resonator condition, were published before 1905; the paper states the difficulty

Pages 136, 137

Planck's constants and the number of molecules

For large wavelengths and radiation densities Planck's formula becomes $\varrho_\nu = \frac{\alpha}{\beta}\nu^2 T$ , the relation of § 1, and equating the coefficients gives $N = 6{,}17 \cdot 10^{23}$ , exactly Planck's value; so Planck's determination of the elementary quanta is, to a certain degree, independent of his theory of black radiation.

As printed

Für große Werte von $T/\nu$ , d. h. für große Wellenlängen und Strahlungsdichten geht diese Formel in der Grenze in folgende über: $\varrho_\nu = \frac{\alpha}{\beta} \nu^2 T.$ Spectral energy densitySpectral constant ASpectral constant BFrequency of the lightTemperature of the radiationExplain this equation Man erkennt, daß diese Formel mit der in § 1 aus der Maxwellschen und der Elektronentheorie entwickelten übereinstimmt. Durch Gleichsetzung der Koeffizienten beider Formeln erhält man: $\frac{R}{N} \frac{8 \pi}{L^3} = \frac{\alpha}{\beta}$ Gas constantAvogadro's numberSpeed of lightSpectral constant ASpectral constant BExplain this equation oder $N = \frac{\beta}{\alpha} \frac{8 \pi R}{L^3} = 6{,}17 \cdot 10^{23},$ Avogadro's numberSpectral constant BSpectral constant AGas constantSpeed of lightExplain this equation d. h. ein Atom Wasserstoff wiegt $1/N$ Gramm $= 1{,}62 \cdot 10^{-24}$ g. Dies ist genau der von Hrn. Planck gefundene Wert, welcher mit den auf anderen Wegen gefundenen Werten für diese Größe befriedigend übereinstimmt.

Text on pages 136–137 of the German source, § 2, paragraph 2

In modern notation

\frac{A\,\nu^{3}}{\exp\left(\frac{B\,\nu}{T}\right) - 1} \approx \frac{A}{B}\,\nu^{2}\,T

\frac{A}{B} = \frac{8\,\pi\,R}{N_A\,c^{3}}

N_A = \frac{B}{A}\,\frac{8\,\pi\,R}{c^{3}}

Approximation: The two formulas agree only in the limit of large $T/\nu$ ; at high frequencies they part.
A comparison, not a premise: The value is compared with Planck's, which it matches exactly, and with values found in other ways, which it matches satisfactorily; it is a consistency check, not a new measurement.
An inference beyond the argument: What the paper concludes is limited: Planck's determination of the elementary quanta is independent of his theory of black radiation "bis zu einem gewissen Grade", to a certain degree.

Decoder

$\alpha$ , $\beta$: The two constants of Planck's formula as the paper prints them, $\alpha = 6{,}10 \cdot 10^{-56}$ and $\beta = 4{,}866 \cdot 10^{-11}$ .
large $T/\nu$: Large wavelengths and radiation densities, where $e^{\beta\nu/T} - 1$ is close to $\beta\nu/T$ and Planck's formula takes the form of § 1.
$N$: The number of real molecules in a gram-equivalent. Its reciprocal is the mass of a hydrogen atom in grams, which the paper gives as $1{,}62 \cdot 10^{-24}$ g.

Try it

LQ-03: At long wavelengths, where do Planck's formula and the relation of § 1 agree, and where do they part?

Where this stops: 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. Read the argument

Common wrong turns

Planck had already proposed that light consists of quanta.

Argument: A comparison of two formulas in a limit
Model: Exact in the limit of large T/ν
Evidence: A consistency check against Planck's constants
History: Planck's formula and constants are earlier work the paper cites; the comparison is this paper's

Pages 138, 139

The entropy of radiation and its temperature

Comparing two expressions for the entropy gained when black radiation of unit volume is warmed gives $\partial\varphi/\partial\varrho = 1/T$ , which the paper calls the law of black radiation: from the entropy function $\varphi$ one can find the radiation law, and from the radiation law, by integration, the function $\varphi$ .

As printed

Durch Vergleich erhält man: $\frac{\partial \varphi}{\partial \varrho} = \frac{1}{T}.$ Spectral entropy densitySpectral energy densityTemperature of the radiationExplain this equation Dies ist das Gesetz der schwarzen Strahlung. Man kann also aus der Funktion $\varphi$ das Gesetz der schwarzen Strahlung und umgekehrt aus letzterem die Funktion $\varphi$ durch Integration bestimmen mit Rücksicht darauf, daß $\varphi$ für $\varrho = 0$ verschwindet.

Text on pages 138–139 of the German source, § 3, paragraph 4

In modern notation

\left(\frac{\partial s_\nu}{\partial u_\nu}\right)_{\nu} = \frac{1}{T}

Premise: The warming is reversible and $dE$ is the heat supplied; that is what allows $dS = dE/T$ .
Premise: That $\partial\varphi/\partial\varrho$ does not depend on $\nu$ comes from the section's condition that the entropy be a maximum at fixed energy.

Decoder

$\varphi(\varrho, \nu)$: The entropy per unit volume and unit frequency of radiation whose energy density at the frequency $\nu$ is $\varrho$ .
$\varrho$: The energy density per unit frequency, printed in §§ 3 and 4 without its subscript $\nu$ .
$\varphi$ vanishes for $\varrho = 0$: The condition that fixes the constant of integration when $\varphi$ is found from the radiation law.

Try it

LQ-04: Take the derivative of the entropy density with respect to the energy density: does it come out as 1/T? Choose the preset “Derivative check: ds_nu/drho_nu against 1/T” there.

Where this stops: This is a thermodynamic comparison, not a microscopic account of every exchange. Integrating the derivative requires a boundary condition. Read the argument

Argument: A thermodynamic relation, used as a tool
Model: Exact for radiation in equilibrium
Evidence: A derivation, not evidence
History: Follows from thermodynamics available before 1905; the paper states it in this form

Page 139

How the entropy of monochromatic radiation depends on its volume

Taking as its basis Wien's law, which experiment confirms for large $\nu/T$ , the paper finds that the entropy of radiation of energy $E$ between $\nu$ and $\nu + d\nu$ depends on the volume it occupies as $S - S_0 = \frac{E}{\beta\nu}\lg\left(\frac{v}{v_0}\right)$ , a result valid only within certain limits.

As printed

Dasselbe wurde aber für große Werte von $\nu/T$ sehr vollkommen durch das Experiment bestätigt. Wir legen diese Formel unseren Rechnungen zugrunde, behalten aber im Sinne, daß unsere Resultate nur innerhalb gewisser Grenzen gelten.
Beschränken wir uns darauf, die Abhängigkeit der Entropie von dem von der Strahlung eingenommenen Volumen zu untersuchen, und bezeichnen wir die Entropie der Strahlung mit $S_0$ , falls dieselbe das Volumen $v_0$ besitzt, so erhalten wir: $S - S_0 = \frac{E}{\beta \nu} \lg \left( \frac{v}{v_0} \right).$ Radiation entropyRadiation entropyRadiation energy in the bandSpectral constant BFrequency of the lightVolumeVolumeExplain this equation

Text on page 139 of the German source, § 4, paragraph 1 · Text on page 139 of the German source, § 4, paragraph 2

In modern notation

u_\nu = A\,\nu^{3}\,\exp\left(-\frac{B\,\nu}{T}\right)

s_\nu = -\frac{u_\nu}{B\,\nu}\,\left(\ln\left(\frac{u_\nu}{A\,\nu^{3}}\right) - 1\right) + C\left(\nu\right)

S\left(V\right) - S\left(V_{0}\right) = \frac{E}{B\,\nu}\,\ln\left(\frac{V}{V_{0}}\right)

Premise: The calculation rests on Wien's law, which the paper says is not exactly valid; its results hold only within certain limits, where Wien's law is confirmed, for large $\nu/T$ .
Premise: The energy and the frequency band are held fixed; only the volume changes.

Decoder

$E$: The energy of radiation whose frequency lies between $\nu$ and $\nu + d\nu$ .
$v$ , $v_0$ , $S_0$: The volume the radiation occupies, a reference volume, and the radiation's entropy when it occupies $v_0$ .
$\beta$: The constant of Wien's law, $\varrho = \alpha\nu^3 e^{-\beta\nu/T}$ .
$\lg$: The natural logarithm, as the paper prints it.

Try it

LQ-04: Halve the volume the radiation occupies at fixed energy: how much does its entropy change? Choose the preset “Halve the volume (the golden state)” there.

LQ-04: Try a dense state, outside the range of Wien's law: why does the laboratory refuse the comparison? Choose the preset “Dense state (outside the Wien regime)” there.

Where this stops: 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. Read the argument

Where this stops: 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. Read the argument

Argument: A derivation from Wien's law
Model: Within the range of validity of Wien's law
Evidence: A derivation, not evidence
History: Introduced in this paper

Pages 141, 142

Boltzmann's principle and the entropy of independent points

Entropy and probability are related by $S - S_0 = \frac{R}{N}\lg W$ , and for $n$ points moving independently in a volume $v_0$ the probability that at a chosen instant all of them are in the part $v$ is $(v/v_0)^n$ , so their entropy depends on the volume as $S - S_0 = R\left(\frac{n}{N}\right)\lg\left(\frac{v}{v_0}\right)$ .

As printed

Bedeutet $S_0$ die Entropie bei einem gewissen Anfangszustande eines betrachteten Systems und $W$ die relative Wahrscheinlichkeit eines Zustandes von der Entropie $S$ , so erhalten wir also allgemein: $S - S_0 = \frac{R}{N} \lg W.$ EntropyEntropyGas constantAvogadro's numberProbability of the configurationExplain this equation
Für diese Wahrscheinlichkeit, welche eine „statistische Wahrscheinlichkeit“ ist, erhält man offenbar den Wert: $W = \left( \frac{v}{v_0} \right)^n ;$ Probability of the configurationVolumeVolumeNumber of independent pointsExplain this equation man erhält hieraus durch Anwendung des Boltzmannschen Prinzipes: $S - S_0 = R \left( \frac{n}{N} \right) \lg \left( \frac{v}{v_0} \right).$ EntropyEntropyGas constantNumber of independent pointsAvogadro's numberVolumeVolumeExplain this equation

Text on page 141 of the German source, § 5, paragraph 6 · Text on page 142 of the German source, § 5, paragraph 8

In modern notation

W = f^{n}

\Delta S_{\mathrm{gas}} = k_B\,\ln\left(W\right) = n\,k_B\,\ln\left(f\right)

\Delta S_{\mathrm{gas}} = n\,k_B\,\ln\left(f\right)

Premise: The value $R/N$ of the constant $C$ comes from the kinetic theory of gases, as the paper says.
Premise: The points move independently of one another; the probability is a product of one factor per point only because of that.

Decoder

$W$: The relative probability of a state of entropy $S$ , compared with the initial state of entropy $S_0$ .
$n$: The number of points moving independently of one another in the volume $v_0$ .
"statistische Wahrscheinlichkeit": Statistical probability: the probability of finding a configuration at an arbitrarily chosen instant.
$R/N$: The universal constant $C$ in $S = C\lg W + \text{konst.}$ , whose value the paper takes from the kinetic theory of gases.

Try it

LQ-05: With two independent points and half the volume, how often are both found in the left half? Choose the preset “First encounter (n = 2, f = 1/2)” there.

LQ-05: Lock the points' positions together: does the probability still come out as a power of the volume ratio? Choose the preset “The locked positions (n = 10, locked counterexample)” there.

Where this stops: 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. Read the argument

Where this stops: 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. Read the argument

Argument: Boltzmann's principle, applied to independent points
Model: Exact for points that move independently
Evidence: A derivation, not evidence
History: Boltzmann's principle is earlier work; the paper states it in this form and applies it

Page 143

Radiation that behaves as if it consisted of energy quanta

Read with Boltzmann's principle, the entropy of monochromatic radiation gives the probability that all its energy is found in a part $v$ of the volume $v_0$ , and that probability has the form it has for independent points; so radiation of low density, where Wien's formula holds, behaves in thermal respects as if it consisted of mutually independent energy quanta of magnitude $R\beta\nu/N$ .

As printed

$W = \left( \frac{v}{v_0} \right)^{\frac{N}{R} \frac{E}{\beta \nu}}.$ Probability of the configurationVolumeVolumeAvogadro's numberGas constantRadiation energy in the bandSpectral constant BFrequency of the lightExplain this equation
Monochromatische Strahlung von geringer Dichte (innerhalb des Gültigkeitsbereiches der Wienschen Strahlungsformel) verhält sich in wärmetheoretischer Beziehung so, wie wenn sie aus voneinander unabhängigen Energiequanten von der Größe $R \beta \nu / N$ bestünde.

Display on page 143 of the German source, § 6, display 4 · Text on page 143 of the German source, § 6, paragraph 6

In modern notation

\Delta S_{\mathrm{rad}} = \frac{E}{B\,\nu}\,\ln\left(f\right)

\Delta S_{\mathrm{gas}} = n\,k_B\,\ln\left(f\right)

n_{\mathrm{eff}} = \frac{E}{k_B\,B\,\nu}

\frac{E}{n_{\mathrm{eff}}} = k_B\,B\,\nu = h\,\nu

Premise: Only for radiation of low density, within the range of validity of Wien's radiation formula; the sentence itself says so.
An inference beyond the argument: A heuristic conclusion from one thermal property, the dependence of the entropy on the volume. The paper next asks whether the laws of the generation and transformation of light are also such as they would be if light consisted of these quanta, and takes that question up in §§ 7 to 9.

Decoder

$W$: The probability that, at an arbitrarily chosen instant, the whole energy $E$ of radiation enclosed in the volume $v_0$ by reflecting walls is found in the part $v$ of it.
$\frac{N}{R}\frac{E}{\beta\nu}$: The exponent of $v/v_0$ : the energy $E$ divided by $R\beta\nu/N$ . For $n$ independent moving points of a gas the same probability is $(v/v_0)^n$ , which is the comparison the paper makes.
$R\beta\nu/N$: The magnitude of each energy quantum. $R/N$ is the gas constant per molecule and $\beta$ the constant of Wien's formula, so the product is what later texts write $h\nu$ ; the paper does not use the letter $h$ .
"wie wenn": As if: the sentence states how the radiation behaves in its thermal properties, not what light is made of.

Try it

LQ-06: Compare the radiation's entropy-volume law with that of a gas of independent points: how many points does the radiation behave like, and how much energy would each carry? Choose the preset “The Move (§6 Golden State)” there.

Where this stops: 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. Read the argument

Common wrong turns

Argument: A heuristic conclusion from a comparison
Model: Stated within the range of validity of Wien's formula
Evidence: An inference, not evidence
History: Introduced in this paper

Page 144

Stokes's rule

If each exciting quantum of frequency $\nu_1$ is absorbed and on its own gives rise to a light quantum of frequency $\nu_2$ , the energy principle requires $\frac{R}{N}\beta\nu_2 \leqq \frac{R}{N}\beta\nu_1$ , that is $\nu_2 \leqq \nu_1$ , which is the known rule of Stokes.

As printed

Wenn die photolumineszierende Substanz nicht als eine beständige Quelle von Energie anzusehen ist, so kann nach dem Energieprinzip die Energie eines erzeugten Energiequants nicht größer sein als die eines erzeugenden Lichtquants; es muß also die Bezeichnung gelten: $\frac{R}{N} \beta \nu_2 \leqq \frac{R}{N} \beta \nu_1$ Gas constantAvogadro's numberSpectral constant BFrequency of the emitted lightFrequency of the absorbed lightExplain this equation oder $\nu_2 \leqq \nu_1 .$ Frequency of the emitted lightFrequency of the absorbed lightExplain this equation Dies ist die bekannte Stokessche Regel.

Text on page 144 of the German source, § 7, paragraph 1

In modern notation

h\,\nu_2 \le h\,\nu_1

Premise: The photoluminescent substance is not to be regarded as a permanent source of energy; only then does the energy principle bound the produced quantum.
Premise: Each exciting quantum acts on its own, at least when the exciting quanta are sparse enough; the absorption may also produce quanta of other frequencies and energy of another kind, such as heat.
Conditional: The paper names when deviations are conceivable: when so many quanta are being transformed at once that one produced quantum can take its energy from several, and when the exciting or produced light is not of the kind Wien's law describes, for instance light from a body so hot that Wien's law fails at the wavelength in question.

Decoder

$\nu_1$: The frequency of the exciting light, absorbed by the photoluminescent substance.
$\nu_2$: The frequency of the light the substance produces.
$\frac{R}{N}\beta\nu$: The energy of one quantum of light of frequency $\nu$ , taken from § 6 for both the exciting and the produced light.

Try it

LQ-07: Choose the exciting frequency: can the frequency of the emitted light exceed it while each quantum acts on its own? Choose the preset “Stokes's Rule (§7 as printed)” there.

LQ-07: Let two quanta act together: which of the paper's two conceivable deviations is this? Choose the preset “Deviation Case 1 (k = 2 multi-quantum)” there.

Where this stops: 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. Read the argument

Argument: A consequence of the quantum picture and the energy principle
Model: Holds when each exciting quantum acts on its own and the substance supplies no energy
Evidence: Accounts for a rule already known from experiment
History: The rule was known before 1905, and the paper calls it known; its derivation from energy quanta is introduced here

Pages 145, 146

The energy of the electrons that light sets free

If a light quantum gives its whole energy to one electron and the electron must do the work $P$ to leave the body, the fastest electrons leave with the kinetic energy $\frac{R}{N}\beta\nu - P$ , so the potential $\Pi$ that just stops them should be a straight line in the frequency with a slope that does not depend on the substance; for ultraviolet light at the edge of the solar spectrum the paper estimates about 4.3 volts, of the order of Lenard's results.

As printed

Die einfachste Vorstellung ist die, daß ein Lichtquant seine ganze Energie an ein einziges Elektron abgibt; wir wollen annehmen, daß dies vorkomme. Es soll jedoch nicht ausgeschlossen sein, daß Elektronen die Energie von Lichtquanten nur teilweise aufnehmen. Ein im Innern des Körpers mit kinetischer Energie versehenes Elektron wird, wenn es die Oberfläche erreicht hat, einen Teil seiner kinetischen Energie eingebüßt haben. Außerdem wird anzunehmen sein, daß jedes Elektron beim Verlassen des Körpers eine (für den Körper charakteristische) Arbeit $P$ zu leisten hat, wenn es den Körper verläßt. Mit der größten Normalgeschwindigkeit werden die unmittelbar an der Oberfläche normal zu dieser erregten Elektronen den Körper verlassen. Die kinetische Energie solcher Elektronen ist $\frac{R}{N} \beta \nu - P .$ Gas constantAvogadro's numberSpectral constant BFrequency of the lightEscape workExplain this equation
Ist der Körper zum positiven Potential $\Pi$ geladen und von Leitern vom Potential Null umgeben und ist $\Pi$ eben imstande, einen Elektrizitätsverlust des Körpers zu verhindern, so muß sein: $\Pi \varepsilon = \frac{R}{N} \beta \nu - P ,$ Stopping potentialCharge of one electronGas constantAvogadro's numberSpectral constant BFrequency of the lightEscape workExplain this equation wobei $\varepsilon$ die elektrische Masse des Elektrons bedeutet, oder $\Pi E = R \beta \nu - P' ,$ Stopping potentialGram-equivalent chargeGas constantSpectral constant BFrequency of the lightWork function per gram-equivalentExplain this equation wobei $E$ die Ladung eines Grammäquivalentes eines einwertigen Ions und $P'$ das Potential dieser Menge negativer Elektrizität in bezug auf den Körper bedeutet.¹⁾
Um zunächst zu sehen, ob die abgeleitete Beziehung der Größenordnung nach mit der Erfahrung übereinstimmt, setzen wir $P' = 0$ , $\nu = 1{,}03 \cdot 10^{15}$ (entsprechend der Grenze des Sonnenspektrums nach dem Ultraviolett hin) und $\beta = 4{,}866 \cdot 10^{-11}$ . Wir erhalten $\Pi \cdot 10^7 = 4{,}3$ Volt, welches Resultat der Größenordnung nach mit den Resultaten von Hrn. Lenard übereinstimmt.²⁾
Ist die abgeleitete Formel richtig, so muß $\Pi$ , als Funktion der Frequenz des erregenden Lichtes in kartesischen Koordinaten dargestellt, eine Gerade sein, deren Neigung von der Natur der untersuchten Substanz unabhängig ist.

Text on pages 145–146 of the German source, § 8, paragraph 2 · Text on page 146 of the German source, § 8, paragraph 3 · Text on page 146 of the German source, § 8, paragraph 5 · Text on page 146 of the German source, § 8, paragraph 6

In modern notation

K_{\mathrm{max}} = h\,\nu - \Phi

e\,V_{\mathrm{s}} = K_{\mathrm{max}} = h\,\nu - \Phi

Premise: That a light quantum gives its whole energy to a single electron is an assumption: the paper calls it the simplest picture, assumes that it occurs, and does not exclude that electrons take up only part of a quantum's energy.
A comparison, not a premise: The 4.3 volts is compared with Lenard's results only in order of magnitude, with the work $P'$ set to zero. It checks the size of the prediction, not the relation.
Conditional: The straight line, with a slope independent of the substance, is what must hold if the derived formula is right; the paper reports no such measurement.

Decoder

$P$: The work, characteristic of the body, that each electron must do to leave it.
$\Pi$: The positive potential to which the body is charged when it just prevents a loss of electricity: in later language, the stopping potential.
$\varepsilon$: The charge of the electron, which the paper calls its electric mass.
$E$ and $P'$: The charge of one gram-equivalent of a monovalent ion, and the potential of that amount of negative electricity with respect to the body: the same relation written per gram-equivalent.
4,3 Volt: The paper's order-of-magnitude check, with $P' = 0$ , $\nu = 1{,}03 \cdot 10^{15}$ (the edge of the solar spectrum toward the ultraviolet) and $\beta = 4{,}866 \cdot 10^{-11}$ .

Try it

LQ-08: Repeat the paper's order-of-magnitude check: what stopping potential does light at the ultraviolet edge of the solar spectrum give? Choose the preset “Einstein 1905 §8 check (Lenard spark order-of-magnitude)” there.

LQ-08: Raise the intensity at a fixed frequency: what changes, the number of electrons or the energy of each? Choose the preset “The intensity probe (rate vs energy)” there.

Where this stops: 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. Read the argument

Where this stops: 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. Read the argument

Common wrong turns

Einstein's photoelectric law was an established experimental fact in 1905.

Argument: A consequence of the quantum picture, with a stated assumption
Model: For electrons that take a quantum's whole energy and leave from the surface
Evidence: A prediction, with an order-of-magnitude comparison
History: Introduced in this paper

Pages 147, 148

Ionisation of gases by ultraviolet light

If each absorbed light quantum ionises one gas molecule, the work of ionisation per gram-equivalent can be no greater than $R\beta\nu$ , which for Lenard's longest effective wavelength in air is $6{,}4 \cdot 10^{12}$ erg, nearly the upper limit found from Stark's ionisation voltage; and the number $j$ of gram-molecules that the absorbed light $L$ ionises must be $L/(R\beta\nu)$ .

As printed

Wir werden anzunehmen haben, daß bei der Ionisierung eines Gases durch ultraviolettes Licht je ein absorbiertes Lichtenergiequant zur Ionisierung je eines Gasmoleküles verwendet wird. Hieraus folgt zunächst, daß die Ionisierungsarbeit (d. h. die zur Ionisierung theoretisch nötige Arbeit) eines Moleküles nicht größer sein kann als die Energie eines absorbierten wirksamen Lichtenergiequantes. Bezeichnet man mit $J$ die (theoretische) Ionisierungsarbeit pro Grammäquivalent, so muß also sein: $R \beta \nu \geqq J .$ Gas constantSpectral constant BFrequency of the lightIonization work per gram-equivalentExplain this equation Nach Messungen Lenards ist aber die größte wirksame Wellenlänge für Luft ca. $1{,}9 \cdot 10^{-5}$ cm, also $R \beta \nu = 6{,}4 \cdot 10^{12} \text{ Erg} \geqq J .$ Gas constantSpectral constant BFrequency of the lightIonization work per gram-equivalentExplain this equation
Nach J. Stark¹⁾ ist die kleinste gemessene Ionisierungsspannung (an Platinanoden) für Luft ca. 10 Volt.²⁾ Es ergibt sich also für $J$ die obere Grenze $9{,}6 \cdot 10^{12}$ , welche nahezu gleich der eben gefundenen ist. Es ergibt sich noch eine andere Konsequenz, deren Prüfung durch das Experiment mir von großer Wichtigkeit zu sein scheint. Wenn jedes absorbierte Lichtenergiequant ein Molekül ionisiert, so muß zwischen der absorbierten Lichtmenge $L$ und der Anzahl $j$ der durch dieselbe ionisierten Grammoleküle die Beziehung bestehen: $j = \frac{L}{R \beta \nu} .$ Ionized gram-moleculesLight energy absorbedGas constantSpectral constant BFrequency of the lightExplain this equation Diese Beziehung muß, wenn unsere Auffassung der Wirklichkeit entspricht, für jedes Gas gelten, welches (bei der betreffenden Frequenz) keine merkliche nicht von Ionisation begleitete Absorption aufweist.

Text on pages 147–148 of the German source, § 9, paragraph 1 · Text on page 148 of the German source, § 9, paragraph 3

In modern notation

h\,\nu \ge I

n_{\mathrm{ions}} \le \frac{L}{h\,\nu}

Premise: That each absorbed quantum ionises one molecule is an assumption the paper states at the start of the section.
A comparison, not a premise: The two figures come from different measurements, Lenard's longest effective wavelength and Stark's smallest ionisation voltage; the paper finds $6{,}4 \cdot 10^{12}$ and $9{,}6 \cdot 10^{12}$ nearly equal.
Conditional: The relation $j = L/(R\beta\nu)$ must hold, if the paper's view corresponds to reality, for any gas that shows no appreciable absorption without ionisation at that frequency; the paper calls its test very important and reports none.

Decoder

$J$: The theoretical work of ionisation per gram-equivalent.
$R\beta\nu$: The energy of a gram-equivalent's worth of quanta: $N$ quanta of $R\beta\nu/N$ each.
$j$: The number of gram-molecules ionised by the absorbed light.
$L$: The amount of light absorbed. In § 9, L is this energy, not the speed of light it names in §§ 1 and 2.

Try it

LQ-09: Is one quantum of the ultraviolet light enough to pay for one ionisation? Choose the preset “Golden Standard (12 eV UV, 10 eV Ionization)” there.

LQ-09: Repeat the paper's check with Lenard's wavelength: how does the energy of the quanta compare with the work of ionisation? Choose the preset “Historical Lenard Check (190 nm UV, 6.459 eV)” there.

Where this stops: 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. Read the argument

Argument: A consequence of the quantum picture, with a stated assumption
Model: For a gas that absorbs light at the frequency only by being ionised
Evidence: A bound compared with two measurements, and a proposed test
History: Introduced in this paper

Introduction · a heuristic viewpoint

Keep what waves explain

Wave optics describes propagation and interference successfully. Einstein asks whether energy exchange with matter needs an additional, explicitly heuristic description.

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§1 · classical energy allocation

A finite window cannot cure an infinite total

At a positive temperature, classical allocation assigns more energy as the frequency cutoff rises. Its all-frequency total has no finite value; the plotting cutoff is not a physical explanation.

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Two assumptions under the equilibrium of section 1

The equilibrium of §1 rests on two assumptions its footnotes spell out: gas molecules and electrons have the same mean kinetic energy, as Drude assumed for metals, and the radiation is as disordered as possible, meaning its Fourier amplitudes and phases are statistically independent of one another.

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§2 · what spectral constants determine

A spectral fit is not yet a free-light hypothesis

Matching a spectral coefficient to classical thermal allocation constrains R/N. Quantized resonator energies and independently propagating light quanta are different claims.

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§3 · temperature and entropy

A spectrum can determine an entropy derivative

At fixed volume and frequency, the entropy increase per added energy is 1/T. Inverting an admitted spectrum therefore determines an entropy derivative, but not yet its integration constant.

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§4 · dilute radiation and the volume law

The constant cannot simply be dropped

Inverting Wien’s law and integrating gives the entropy density plus a frequency-dependent constant. Requiring zero entropy density at zero radiation fixes that constant; it does not cancel automatically.

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Compare two states, not a compression movie

For the same energy and narrow frequency band, Wien radiation gains entropy logarithmically with accessible volume. Both endpoint states must remain in the admitted regime.

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§5 · independent configurations

Which probability the entropy uses

Einstein uses the probability of a state in one definite sense: how likely it is to find the system in that state at a moment chosen at random. Read that way, Boltzmann's principle ties entropy to probability without first deciding by hypothesis which cases are equally likely.

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Independence supplies the exponent

For n independent uniformly distributed points, the chance that all lie in a fraction f of a volume is fⁿ. Perfectly locked positions instead give f; the exponent is a statement about independence.

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§6 · the heuristic correspondence

A coefficient suggests an energy element

In the Wien regime, the radiation entropy coefficient has the form associated with independent elements. Reading an energy scale from that match is a heuristic inference, not a proof of every property of light.

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§7 · fluorescence and its conditions

An energy budget has conditions

With one absorbed quantum, no extra energy, and a single emitted quantum, the outgoing energy cannot exceed the incoming energy. Extra thermal energy or multiple-quantum input changes those premises.

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§8 · photoelectric energy and counts

More electrons is not more energy per electron

In the one-quantum model, frequency sets the maximum electron energy after paying the escape cost. Increasing power at fixed frequency can increase the number emitted without increasing that maximum.

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Einstein's own limits on the photoelectric laws

Einstein qualifies his photoelectric laws the way he qualified Stokes's rule: they may fail when so many light quanta act at once that one electron gathers energy from several, or when the light is not like black radiation within the range of Wien's law.

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A stopping voltage is a magnitude with a sign convention

The ideal stopping-potential magnitude is Kmax divided by the positive elementary-charge magnitude. Partial transfer or internal losses produce an inequality rather than a larger endpoint energy.

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§9 · ionization bounds and closing scope

A threshold does not specify a yield

Under a one-quantum ionization assumption, a quantum must meet the required ionization energy. Absorbed energy bounds the count; equality requires that every absorbed quantum causes exactly one counted ionization.

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