Annus MirabilisEinstein in 1905
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The equation explained

dS=∂φ∂ϱdE.d S = \frac{\partial \varphi}{\partial \varrho} d E.
In wordsThe increase of the radiation's entropy equals the rate at which the entropy density grows with the energy density, times the increase of the radiation's energy.OverviewIn black radiation the rate ∂φ/∂ρ is the same at every frequency, so it comes out of the sum, and what is left is the increase of the radiation's energy.Full explanationIn black radiation ∂φ/∂ϱ\partial \varphi / \partial \varrho does not depend on ν\nu, the conclusion of the paragraph before, so it can be taken out of the integral. What remains, the increase of the density added up over all frequencies, is the increase dEd E of the radiation's energy in the unit volume. The entropy's increase is then that common rate times the energy's increase.Every stepdS=∫ν=0ν=∞∂φ∂ϱdϱdνd S = \int_{\nu = 0}^{\nu = \infty} \frac{\partial \varphi}{\partial \varrho} d \varrho d \nuThe preceding display.dS=∂φ∂ϱ∫ν=0ν=∞dϱdνd S = \frac{\partial \varphi}{\partial \varrho} \int_{\nu = 0}^{\nu = \infty} d \varrho d \nu∂φ/∂ϱ\partial \varphi / \partial \varrho has the same value at every frequency in black radiation, so it comes out of the integral.dE=∫ν=0ν=∞dϱdνd E = \int_{\nu = 0}^{\nu = \infty} d \varrho d \nuThe remaining integral is the increase of the energy in the unit volume, since that energy is ∫0∞ϱdν\int_0^\infty \varrho d \nu.dS=∂φ∂ϱdEd S = \frac{\partial \varphi}{\partial \varrho} d EPut the two together: the printed line.Also

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