Annus MirabilisEinstein in 1905
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light quanta

The equation explained

ϱν=RN8πν2L3T.\varrho_\nu = \frac{R}{N} \frac{8 \pi \nu^2}{L^3} T.
In wordsThe radiation's energy per unit volume and unit frequency equals the gas constant over the number of molecules in a gram-equivalent, times eight π times the frequency squared over the speed of light cubed, times the temperature.OverviewThe equilibrium condition fixes the radiation's density: it grows as the square of the frequency and in proportion to the temperature. Einstein says at once that this disagrees with experience.Full explanationSolve the chain of equalities for the density and ϱν\varrho_\nu grows with the square of the frequency and in proportion to the absolute temperature, with nothing to check the growth at high frequencies. Doubling the frequency multiplies the density by four; doubling the temperature doubles it. The paper says of this relation, found as the condition of dynamic equilibrium, that it 'not only lacks agreement with experience' but also means there can be no definite distribution of energy between ether and matter, which the next display shows.Every stepRNT=L38πν2ϱν\frac{R}{N} T = \frac{L^3}{8 \pi \nu^2} \varrho_\nuKeep the two ends of the chain of equalities: the gas's side and the radiation's side.8πν2L3⋅RNT=ϱν\frac{8 \pi \nu^2}{L^3} \cdot \frac{R}{N} T = \varrho_\nuMultiply both sides by 8πν2L3\frac{8 \pi \nu^2}{L^3}, which cancels the factor standing before ϱν\varrho_\nu.ϱν=RN8πν2L3T\varrho_\nu = \frac{R}{N} \frac{8 \pi \nu^2}{L^3} TTurn the equation round and order the factors as the paper does: the printed line.Historian's marginEinstein gives the relation no name. Rayleigh had given a density growing as the square of the frequency times the temperature for long waves in June 1900 (Philosophical Magazine (5) 49, 539-540), and Jeans corrected its coefficient in July 1905 (Philosophical Magazine (6) 10, 91-98), after this paper was received.Also

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