Foundation · Explanatory preview

The Gaussian and its width

A Gaussian is a family of densities whose spread is set by its variance; being bell-shaped is not a proof of its origin.

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How do we normalize the curve and find its mean square?

p(x,t)=\frac{1}{\sqrt{4\pi Dt}}\exp\left(-\frac{x^2}{4Dt}\right)

For positive D and t, the density is an exponential bell with normalization one over the square root of four pi D t.

Here x is displacement, t is elapsed time, and D is a constant diffusivity. Put u = x divided by the square root of 4Dt. Then dx is the square root of 4Dt times du. The normalization reduces to the integral of exp(−u²), which is the square root of π.

One way to evaluate that integral is to square it: I² is the integral of exp(−u²−v²) over the plane. In polar coordinates, area is r dr dθ; angles cover 0 to 2π. The radial integral of r exp(−r²) is one half, so I² = π. Since the integral is positive, I = √π. This is a modern verification calculation, not an attributed reconstruction of Einstein’s reasoning.

For the second moment use integration by parts: choose u as the factor and integrate u exp(−u²), whose antiderivative is −exp(−u²)/2. The endpoint term vanishes at both infinities. What remains is one half the normalization integral.

One worked example

\frac{1}{\sqrt\pi}\int_{-\infty}^{\infty}u^2e^{-u^2}\,du=\frac12

The normalized integral of u squared times exp of minus u squared is one half.

Substituting back gives mean square 4Dt times one half, or 2Dt. About 68.27% of the probability lies within one RMS width of the centre, not all of it.

A stopping point: A normalized density and its moments describe an ensemble, not a required endpoint for one tracer.

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