= Mean displacement magnitude of three-dimensional Brownian motion
{title2=$\mathbb E|X_t|=4\sqrt{Dt/\pi}$}
For isotropic <Brownian motion> started at zero with independent coordinate variances $2Dt$, the <Gaussian distribution> has density $(4\pi Dt)^{-3/2}e^{-|x|^2/(4Dt)}$. Integrating the radius with spherical volume element $4\pi r^2dr$ gives $\mathbb E|X_t|=4\sqrt{Dt/\pi}$. The <root mean square> displacement is instead $\sqrt{6Dt}$, and the mean vector is zero. None of these is the microscopic path length.
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