§2 · what spectral constants determine · Light quanta: from entropy to an energy scale

§2 · what spectral constants determine

A spectral fit is not yet a free-light hypothesis

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

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

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

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

\frac{A}{B}=\frac{8\pi R}{Nc^3},\qquad N=\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, $R$ equals $N_A k_B$ . Inserting that identity and the modern spectral constants recovers the defined $N_A$ ; 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.

Return to matching the two spectral coefficients. (included in this file)

Return to the hydrogen atom's 1.62 × 10⁻²⁴ gram and the number N. (included in this file)

Letters as printed: the paper writes Planck's two constants as α and β, and this explanation names them A and B. Here β divides a frequency by a temperature inside an exponential, so it is a constant of the radiation law, Wien's constant; it is not the relativity paper's β. L, written c below, is the speed of light.

Planck's formula, which fitted all the observations then available, gives the spectral density as $\rho_\nu=\frac{A\nu^3}{\exp(B\nu/T)-1}$ (p. 136).
Name the ratio in the exponent $y=B\nu/T$ . It has no units: B is in seconds times kelvin, ν in inverse seconds, T in kelvin.
For small y, $\exp(y)=1+y+y^2/2+\ldots$ , so $\exp(y)-1=y(1+y/2+\ldots)$ , which approaches y as y becomes small. Small y means low frequency or high temperature, that is, large T/ν: long wavelengths and dense radiation.
Replace $\exp(B\nu/T)-1$ by $B\nu/T$ and divide: $A\nu^3$ divided by $B\nu/T$ is $\frac{A}{B}\nu^2T$ .

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

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

This has the same form as the classical allocation of §1, $\frac{R}{N}\frac{8\pi\nu^2}{L^3}T$ : both are a constant times $\nu^2T$ . For the two to agree at every low frequency and every temperature, the constants in front must be equal.
Set them equal, then multiply both sides by N and by B/A to leave N alone.

\frac{A}{B}=\frac{8\pi R}{Nc^3},\qquad N=\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.

Worked with the paper's inputs. The page prints β as $4{,}866\cdot10^{-11}$ and α as $6{,}10\cdot10^{-56}$ . The provenance receipt records the exponent of α as a misprint, and this edition computes with $A=6.10\times10^{-57}$ .
R and L are not printed as numbers in §2. With the standard values of 1905, $R=8.31\times10^{7}$ erg per kelvin per gram-molecule and $L=3\times10^{10}$ cm per second, supplied as inputs: B/A is $7.98\times10^{45}$ , $8\pi R/L^3$ is $7.74\times10^{-23}$ , and their product is $N=6.17\times10^{23}$ .
The page prints $N=6{,}17\cdot10^{23}$ , and adds that one atom of hydrogen weighs 1/N gram. Check it: $1/(6.17\times10^{23})=1.62\times10^{-24}$ gram, the number the page gives.
The misprinted exponent cannot be the one used: the same arithmetic with $6.10\times10^{-56}$ gives $6.17\times10^{22}$ , ten times smaller than the number printed.
The paper notes that this is the value Planck found, and that it agrees satisfactorily with values found by other routes. Its point is narrower than it may look: Planck's determination of N depends only to a degree on his theory of radiation, because only the long-wavelength form of the spectrum enters.
The conclusion the paper draws from §§1 and 2: its theoretical foundations serve better the greater the energy density and the wavelength of the radiation, and fail completely for short wavelengths and small densities. §4 works in exactly that region.
Keep the evidence straight. In today's SI, R is defined as $N_Ak_B$ , so putting modern constants into this formula returns the defined $N_A$ and tests nothing about molecules.
Keep the two claims apart. Planck's resonators belong to the matter that emits and absorbs; the energy elements the paper goes on to propose belong to the radiation itself. Getting from one to the other takes the argument of §§3 to 6.

Return to matching the two spectral coefficients. (included in this file)

Avogadro's number from the spectrum

Modern teaching equation, not the equation as printed.

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

N A equals B over A times eight pi R over the speed of light cubed.

Solving the coefficient match for N gives Avogadro's number from the constants of the radiation law. The paper's value, from Planck's printed constants, is 6.17 times ten to the 23 per gram-molecule.

Two coefficients that must agree

Modern teaching equation, not the equation as printed.

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

A over B equals eight pi R over N A times the speed of light cubed.

The low-frequency limit of Planck's law and the classical allocation of energy both grow as the frequency squared times the temperature, so their coefficients can be set equal.

Planck's law at low frequency

Modern teaching equation, not the equation as printed.

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

A times the frequency cubed over the exponential of B frequency over temperature minus one is approximately A over B times the frequency squared times the temperature.

Where B times the frequency is small compared with the temperature, the exponential is close to one plus its argument, and Planck's law becomes proportional to the temperature.

Assumptions

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.

Limits

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.

Foundations included in this file

Fractions and ratios

What does dividing one quantity by another tell you?

The fraction 6/2 is 3, because 2 fits into 6 three times; it also says that 6 is three times 2. Every fraction a/b works the same way. The bottom number can never be zero, since no count of zeros adds up to 6.

When the two quantities have different units, their ratio is a new quantity with a unit of its own: 12 micrometres travelled in 4 seconds is 3 micrometres per second. When they have the same unit, the units cancel and the ratio is a plain number: 2 particles out of 8 is 0.25 of the group.

Worked example: Fourteen metres in ten seconds

A steady walk covers 14 metres in 10 seconds.
Divide the distance by the time: 14 divided by 10 is 1.4, and the unit divides with it, so the rate is 1.4 metres per second.
After 60 seconds the same walk has covered 84 metres, and 84 divided by 60 is 1.4 again. The walk is steady, so the ratio does not depend on how long it is watched.
Metres and seconds are different units, so neither cancels and what comes out is a new quantity that keeps a unit of its own.
The drawing below is the other case: 2 particles out of 8 has the same unit above and below, they cancel, and 0.25 is left as a plain number.

Where this lesson stops

A ratio of two quantities in the same unit is a plain number. A ratio of different units, such as micrometres per second, is a new quantity and keeps its unit.

This lesson builds on

Adding and averaging

Return to the chapter

Powers of ten and physical units

How do you write a very small length without counting zeros?

The Brownian paper works its example, in §5, for particles 0.001 mm across. In metres that is 0.000001 m: easy to miscount by one zero, which is a factor of ten. Written as 10⁻⁶ m, the −6 says the decimal point has moved six places to the left, and that length has a name of its own, the micrometre (μm).

The unit goes through the arithmetic with the number. A micrometre is 10⁻⁶ m, so a square micrometre is 10⁻¹² m², not 10⁻⁶ m²: squaring a length squares its unit, power of ten included.

Worked example: Squaring six micrometres

1 micrometre (1 μm) is 10⁻⁶ metres, or 0.000001 m.
The paper's displacement after one minute, about 6 μm, is 6 × 10⁻⁶ m.
Squaring it squares the number and the unit: (6 × 10⁻⁶ m)² = 36 × 10⁻¹² m² = 3.6 × 10⁻¹¹ m².

Where this lesson stops

A measurement is a number and a unit together. Squaring it squares both: (6 × 10⁻⁶ m)² is 3.6 × 10⁻¹¹ m².

This lesson builds on

Fractions and ratios

Return to the chapter

Adding and averaging

What does an average keep, and what does it lose?

Four jars hold 3, 1, 1 and 3 marbles. Pour them together and there are 8. Share the 8 equally between the four jars and each holds 2. Two is the average, although no jar held 2 to begin with.

The average keeps the total and the count and nothing else, so 3, 1, 1, 3 and 2, 2, 2, 2 have the same average. Listing every value twice doubles both the total and the count, and the average stays at 2.

Worked example: Three mornings with the same average

One morning the four jars hold 3, 1, 1 and 3 marbles. Add them: 8. Divide by the four jars: an average of 2.
Another morning they hold 8, 0, 0 and 0. Add them: 8 again. Divide by four: 2 again, and now three jars are empty.
A third morning every jar holds 2, which is the average itself. Add them: 8. Divide by four: 2.
Write any of those three lists out twice over. The total doubles to 16 and the count doubles to 8 jars, and 16 divided by 8 is still 2.
Four descriptions and one average. It keeps the total and the count, and the differences between the mornings are what it leaves behind.

Where this lesson stops

An average is the total shared equally among the observations, and two very different lists can have the same one.

Return to the chapter

Orders of magnitude

How big, how fast and how energetic are the things in the 1905 papers, measured against each other?

Two numbers of the same order differ by less than a factor of ten. Numbers several orders apart belong to different worlds, and comparing them usually tells you which effect can be ignored.

Sizes. A water molecule is about 0.3 nanometres across and a Brownian grain about 1 micrometre: the grain is roughly 3,000 times larger. The light-quanta paper prints the mass of a hydrogen atom as 1/N gram, 1.62 × 10⁻²⁴ g.

Speeds against light. The ratio v/c is about 3.3 × 10⁻⁶ for a rifle bullet at 1,000 m/s, 8.3 × 10⁻⁷ for a jet at 250 m/s, and 9.9 × 10⁻⁵ for the Earth in its orbit at 29.8 km/s. The corrections of relativity go as the square of v/c, so for the Earth they are about 10⁻⁸. Only Kaufmann's fast electrons, a large fraction of the speed of light, reached speeds where they are not small.

Energies of light. One quantum hν of red light (650 nm) carries about 1.91 electron volts, green (530 nm) 2.34 eV, and ultraviolet (250 nm) 4.96 eV. The light paper's photoelectric estimate, about 4.3 volts, is of the same order.

Molecular kicks. An estimate of how often water molecules strike a grain 1 μm across gives about 1.6 × 10¹⁹ blows a second, and about 6 × 10¹⁹ for a grain of 1 μm radius. The assumptions are rough, so the honest statement is 10¹⁹ to 10²⁰ blows a second: far too many for any single blow to be seen.

Worked example: Two estimates, done with powers of ten

The Earth: v/c = 29,800 ÷ 299,792,458 ≈ 9.9 × 10⁻⁵. Squared, that is about 9.9 × 10⁻⁹, near 10⁻⁸.
Water's molecules: 1,000 kg per cubic metre divided by 0.018015 kg per gram-molecule, times 6.02 × 10²³, is 3.34 × 10²⁸ molecules per cubic metre.
Molecules near 600 m/s crossing a surface at a quarter of n times their mean speed give about 5 × 10³⁰ strikes per square metre each second.
A sphere 1 μm across has a surface of 3.1 × 10⁻¹² m², so it takes about 1.6 × 10¹⁹ strikes a second. The inputs are rough, so the result is an order of magnitude, 10¹⁹ to 10²⁰.

Where this lesson stops

This lesson places numbers on a scale of powers of ten. Converting the papers' printed units into SI units is a separate step, with its own conventions.

Return to the chapter

Static laboratory examples

lq-02

This laboratory is not included in this file. The chapter's authored foundation examples remain available above. Open lq-02 online

lq-03

This laboratory is not included in this file. The chapter's authored foundation examples remain available above. Open lq-03 online

Source references

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).

Light quanta: from entropy to an energy scale: §2 · what spectral constants determine