= Optical trap relaxation time
{title2=$\tau=\zeta/k$}
For a bead in a stationary viscous liquid, <linear drag> often dominates <inertia> on the observed timescale. Neglecting both <inertia> and the random thermal <force>, the deterministic <force balance> gives
$$
\zeta\dot x=-kx,\qquad x(t)=x(0)e^{-t/\tau},\qquad \tau=\frac{\zeta}{k}.
$$
Here $\zeta>0$ is the drag coefficient and $\tau$ the <optical trap relaxation time>: the time to reduce the deterministic <displacement> by a factor $e$. The bead approaches <equilibrium> without oscillation in this overdamped model. The approximation requires times long compared with the inertial time $m/\zeta$ and, for separated timescales, $m/\zeta\ll\zeta/k$, where $m$ is the bead's <mass>. For a no-slip spherical bead of radius $a$ in an unbounded liquid of <dynamic viscosity> $\eta$, <Stokes drag law> gives $\zeta=6\pi\eta a$. Nearby walls change that drag coefficient.
For example, $a=0.5\,\mathrm{\mu m}$, $\eta=10^{-3}\,\mathrm{Pa\,s}$ and $k=0.01\,\mathrm{pN/nm}=10^{-5}\,\mathrm{N/m}$ give $\zeta\approx9.42\times10^{-9}\,\mathrm{N\,s/m}$ and $\tau\approx0.942\,\mathrm{ms}$. A <displacement> of $100\,\mathrm{nm}$ gives a <restoring force> of magnitude $1\,\mathrm{pN}$, and its deterministic <displacement> drops to $100/e\approx36.8\,\mathrm{nm}$ after one <optical trap relaxation time>. These are predictions for the stated model, not measured trajectories.
\Image[/optical-trap-restoring-force.png]
{title=Local harmonic optical trap: potential well, restoring force and deterministic overdamped relaxation}
The figure uses $q=x/\ell$, $\kappa=k/k_0$ and $s=t/(\zeta/k_0)$, with arbitrary positive reference length $\ell$ and stiffness $k_0$. Thus $U/(k_0\ell^2)=\kappa q^2/2$, $F/(k_0\ell)=-\kappa q$ and, from $q(0)=1$, $q(s)=e^{-\kappa s}$. At fixed drag, greater stiffness gives both a steeper <potential energy> well and faster relaxation. Thermal <Brownian motion> is omitted from these deterministic curves.
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