Annus MirabilisEinstein in 1905
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The equation explained

∫0∞(∂φ∂ϱ−λ)δϱdν=0,\int_0^\infty \left( \frac{\partial \varphi}{\partial \varrho} - \lambda \right) \delta \varrho d \nu = 0,
In wordsThe sum over every frequency of the rate at which the entropy density grows with the energy density, less a number the same at every frequency, times any small change of the density, is zero.OverviewThe two conditions combine with a number λ, the same at every frequency, into one condition that holds for every change of the density. It follows that in black radiation the rate ∂φ/∂ρ does not depend on the frequency.Full explanationThe entropy condition must hold for every change δϱ\delta \varrho that keeps the energy fixed. The paper expresses this with a number λ\lambda that does not depend on ν\nu: with the energy condition multiplied by λ\lambda and subtracted, the integral of the bracket times δϱ\delta \varrho vanishes for every choice of δϱ\delta \varrho as a function of ν\nu, whether or not it keeps the energy fixed. An integral of the bracket times an arbitrary δϱ\delta \varrho can vanish for every δϱ\delta \varrho only if the bracket is zero at every frequency. So in black radiation ∂φ/∂ϱ\partial \varphi / \partial \varrho equals λ\lambda everywhere: as the paper concludes, it is independent of ν\nu. λ\lambda is what is now called a Lagrange multiplier.Every step∫0∞∂φ∂ϱδϱdν=0\int_0^\infty \frac{\partial \varphi}{\partial \varrho} \delta \varrho d \nu = 0The entropy condition written out: to first order the entropy changes by ∂φ∂ϱδϱ\frac{\partial \varphi}{\partial \varrho} \delta \varrho at each frequency, and this must add up to zero whenever the energy is unchanged.λ∫0∞δϱdν=0\lambda \int_0^\infty \delta \varrho d \nu = 0The energy condition, multiplied by a number λ\lambda that is the same at every frequency. It is still zero.∫0∞(∂φ∂ϱ−λ)δϱdν=0\int_0^\infty \left( \frac{\partial \varphi}{\partial \varrho} - \lambda \right) \delta \varrho d \nu = 0Subtract the second from the first. With λ\lambda suitably chosen this holds for every choice of δϱ\delta \varrho as a function of ν\nu, not only those that keep the energy fixed: the printed line.∂φ∂ϱ=λ\frac{\partial \varphi}{\partial \varrho} = \lambdaIf the integral vanishes for every δϱ\delta \varrho, the bracket must vanish at every frequency; a δϱ\delta \varrho concentrated where it did not would give an integral that is not zero. So in black radiation ∂φ/∂ϱ\partial \varphi / \partial \varrho has the same value at every frequency.Also

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