brownian motion

The equation explained

f(x,t+τ)=f(x,t)+τ∂f∂t.f(x, t + \tau) = f(x, t) + \tau \frac{\partial f}{\partial t}.
In wordsThe number of particles per unit volume one short interval after a given time is its value at that time plus the interval times the rate at which it is changing.OverviewBecause the interval is very small, the density a moment later is the density now plus the rate of change times the length of the moment.Full explanationThe first of the two expansions. The paper justifies it in one clause, that τ\tau is very small, and writes that we can therefore set this. It is the first-order Taylor expansion in the time, and what it drops is of the order of τ2\tau^2. The point of making it is that the left side of the balance then contains the time derivative that the diffusion equation needs, multiplied by the small interval. The interval cannot be shrunk to zero to make the step exact, because the derivation also needs τ\tau to be long enough that successive shifts are independent. That tension is in the paper's own description of τ\tau: very small against observable intervals, yet not too small.Every stepf(x,t+τ)=f(x,t)+τ∂f∂t+τ22∂2f∂t2…f(x, t + \tau) = f(x, t) + \tau \frac{\partial f}{\partial t} + \frac{\tau^2}{2} \frac{\partial^2 f}{\partial t^2} \ldotsThe Taylor expansion of the density in the time, about the earlier time.f(x,t+τ)=f(x,t)+τ∂f∂t.f(x, t + \tau) = f(x, t) + \tau \frac{\partial f}{\partial t} .Keep the first order and drop the rest, because τ\tau is very small: the printed line. What was dropped is smaller than what was kept by a further factor of τ\tau.Also

Read it in the paper