Annus MirabilisEinstein in 1905
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brownian motion

The equation explained

1τ∫−∞+∞Δ22φ(Δ)dΔ=D\frac{1}{\tau} \int_{-\infty}^{+\infty} \frac{\Delta^2}{2} \varphi(\Delta) d \Delta = D
In wordsThe mean of half the squared displacement under the frequency law, divided by the short interval, is written as the diffusion coefficient.OverviewThe definition that turns the step law into a single number: half the mean square step per unit time. The paper sets this, and the next line shows the number is the diffusion coefficient.Full explanationThe paper introduces this with the words and by setting, which marks it as a definition rather than a derivation. The quantity on the left is a property of the stepping alone: half the mean square displacement in one interval, divided by the length of that interval. For it to name a constant of the substance, the mean square displacement in one interval must be proportional to the length of the interval, so that the quotient does not depend on how τ\tau was chosen. The paper does not say this and does not argue it; the proportionality is what the finished equation then asserts, since its solution has a mean square that grows in proportion to the time. Calling the number DD is at this point only a name, and the identification with the diffusion coefficient of section 3 is made on the next line by the form of the equation it produces.Every step∫−∞+∞Δ22φ(Δ)dΔ\int_{-\infty}^{+\infty} \frac{\Delta^2}{2} \varphi(\Delta) d \DeltaThe integral standing in the third term of the balance. It is the second moment of the step law, halved: half the mean square of the displacement made in one interval.1τ∫−∞+∞Δ22φ(Δ)dΔ=D\frac{1}{\tau} \int_{-\infty}^{+\infty} \frac{\Delta^2}{2} \varphi(\Delta) d \Delta = DDivide by the length of the interval and give the quotient a name: the printed line. Written the other way round it says that the mean square step is 2Dτ2 D \tau, which is the same relation between mean square and time that the result at the end of the section carries.Also

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