= Passive microrheology
{title2=$\langle\Delta r^2(t)\rangle=k_BTJ(t)/(\pi a)$}
For a dilute spherical no-slip probe of radius $a$ in an isotropic incompressible linear medium at equilibrium, the bead's displacement per unit step force is $J_b(t)=J(t)/(6\pi a)$, where $J$ is the bulk shear <creep compliance>. The <fluctuation-dissipation theorem> gives one-coordinate displacement variance $2k_BTJ_b(t)$. Adding the three equal Cartesian contributions proves the displayed relation. In a <Newtonian fluid>, $J=t/\eta$, so it becomes $\langle\Delta r^2\rangle=6Dt$ with the <Stokes–Einstein relation> $D=k_BT/(6\pi\eta a)$.
The particle must sample a continuum response on its scale and remain in thermal equilibrium. A two-dimensional image records only two of the three displacement components; isotropy then gives $\langle\Delta r^2\rangle_{3d}=(3/2)\langle\Delta r^2\rangle_{2d}$. Localization noise, confinement and unresolved fast relaxation can otherwise be mistaken for elastic response.
Back to article page
