Three-dimensional optical trap stability ($F_i\simeq -k_i\delta r_i,k_i>0$)

At an equilibrium ${\mathbf{r}}_{\ast }$, the total force vanishes. A small transverse displacement must produce a transverse restoring force; a small axial displacement must also restore the bead. Downstream of a tight focus, the backward optical gradient force can oppose the forward optical scattering force. Equality at one point is insufficient: the total axial force must decrease as the bead moves downstream through that point. The equilibrium can therefore lie downstream of the intensity maximum.

For uncoupled overdamped coordinates, ${\zeta }_i\dot{\delta r_i}=-k_i\delta r_i$ with ${\zeta }_i>0$ decays only when each $k_i>0$. More generally, linearizing gives $\dot{\delta \mathbf{r}}=MJ\delta \mathbf{r}$, where $M$ is the hydrodynamic mobility matrix and $J_{ij}=\partial F_i/\partial r_j$ at equilibrium. All eigenvalues of $MJ$ must have negative real parts for linear asymptotic stability. For a conservative force with symmetric positive-definite mobility, a positive-definite Hessian matrix of the potential energy supplies this condition. A general optical force need not possess a global potential energy.

Increasing numerical aperture admits more strongly converging light rays and a steeper axial variation of the field. But a larger nominal numerical aperture alone does not establish trapping: illumination of the objective pupil, optical aberration, particle properties and the actual in-liquid focus also matter. A harmonic optical trap describes local confinement after this stability question has been answered; it does not predict capture from arbitrary positions or an infinite holding force.