## Abstract

Using numerical solutions of Maxwell’s equations in conjunction with the Lorentz law of force, we compute the electromagnetic force distribution in and around a dielectric micro-sphere trapped by a focused laser beam. Dependence of the optical trap’s stiffness on the polarization state of the incident beam is analyzed for particles suspended in air or immersed in water, under conditions similar to those realized in practical optical tweezers. A comparison of the simulation results with available experimental data reveals the merit of one physical model relative to two competing models; the three models arise from different interpretations of the same physical picture.

© 2006 Optical Society of America

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### Equations (19)

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(A1)
$${\mathbf{F}}_{1}\left(r\right)=-(\nabla \xb7\mathbf{P})\mathbf{E}+\left(\frac{\partial \mathbf{P}}{\partial t}\right)\times \mathbf{B}$$
(A2)
$${\mathbf{F}}_{2}\left(r\right)=(\mathbf{P}\xb7\nabla )\mathbf{E}+\left(\frac{\partial \mathbf{P}}{\partial t}\right)\times \mathbf{B}$$
(A3)
$${\mathbf{F}}_{2}\left(r\right)=\frac{1}{4}{\epsilon}_{0}\left(\epsilon -1\right)\nabla \left({\mid {E}_{x}\mid}^{2}+{\mid {E}_{y}\mid}^{2}+{\mid {E}_{z}\mid}^{2}\right).$$
(A4)
$${F}_{1x}^{\mathit{surface}}(x=0)=-\frac{1}{2}{P}_{x}(0)\left[{E}_{0}+{E}_{x}\left(0\right)\right]=-\left(\frac{{D}_{0}}{2{\epsilon}_{0}}\right)\left[1+\frac{1}{\epsilon \left(0\right)}\right]{P}_{x}\left(0\right);\phantom{\rule{1em}{0ex}}x=0$$
(A5)
$${F}_{1x}^{\mathit{bulk}}\left(x\right)=-\frac{{E}_{x}\left(x\right)d{P}_{x}\left(x\right)}{\mathit{dx}}=\left(\frac{1}{2{\epsilon}_{0}}\right)d\frac{{\left[{D}_{0}-{P}_{x}\left(x\right)\right]}^{2}}{\mathit{dx}};\phantom{\rule{.9em}{0ex}}\phantom{\rule{.9em}{0ex}}\phantom{\rule{.9em}{0ex}}\phantom{\rule{.9em}{0ex}}0<x<L$$
(A6)
$${F}_{1x}^{\mathit{surface}}\left(x=L\right)=\left(\frac{{D}_{0}}{2{\epsilon}_{0}}\right)\left[1+\frac{1}{\epsilon \left(L\right)}\right]{P}_{x}\left(L\right);\phantom{\rule{0.9em}{0ex}}\phantom{\rule{0.9em}{0ex}}\phantom{\rule{0.9em}{0ex}}x=L$$
(A7)
$${F}_{2x}^{\mathit{surface}}\left(x=0\right)=\frac{1}{2}{P}_{x}(0)\left[{E}_{x}\left(0\right)-{E}_{0}\right]=-\left(\frac{{D}_{0}}{2{\epsilon}_{0}}\right)\left[1-\frac{1}{\epsilon \left(0\right)}\right]{P}_{x}\left(0\right);\phantom{\rule{1em}{0ex}}x=0$$
(A8)
$${F}_{2x}^{surface}(x)={P}_{x}(x)d{E}_{x}(x)/dx=-(1/2{\epsilon}_{0})d{P}_{x}^{2}(x)/dx;\phantom{\rule{0.9em}{0ex}}\phantom{\rule{0.9em}{0ex}}\phantom{\rule{0.9em}{0ex}}\phantom{\rule{0.9em}{0ex}}0<x<L$$
(A9)
$${F}_{2x}^{\mathit{surface}}\left(x=L\right)=\left(\frac{{D}_{0}}{2{\epsilon}_{0}}\right)\left[1-\frac{1}{\epsilon \left(L\right)}\right]{P}_{x}\left(L\right);\phantom{\rule{0.9em}{0ex}}\phantom{\rule{0.9em}{0ex}}\phantom{\rule{0.9em}{0ex}}\phantom{\rule{0.9em}{0ex}}x=L$$
(A10)
$${F}_{2}^{\mathit{surface}}\left(r,\theta ={\theta}_{0},1\right)=\pm \left(\frac{1}{2{\epsilon}_{0}}\right)\left(1-\frac{1}{{\epsilon}^{2}}\right){\left(\frac{{D}_{0}{r}_{0}}{r}\right)}^{2}\hat{\theta};\phantom{\rule{2em}{0ex}}\theta ={\theta}_{0},{\theta}_{1}$$
(A11)
$${F}_{2}^{\mathit{surface}}\left(r,\theta ={\theta}_{0},1\right)=\pm \left(\frac{1}{2{\epsilon}_{0}}\right){\left(1-\frac{1}{\epsilon}\right)}^{2}{\left(\frac{{D}_{0}{r}_{0}}{r}\right)}^{2}\hat{\theta};\phantom{\rule{2em}{0ex}}\theta ={\theta}_{0},{\theta}_{1}$$
(A12)
$${F}_{2x}^{\mathit{bulk}}(r,\theta )=-\left(\frac{1}{{\epsilon}_{0}\epsilon}\right)\left(1-\frac{1}{\epsilon}\right)\left(\frac{{D}_{0}^{2}{r}_{0}^{2}}{{r}^{3}}\right)\hat{\mathit{r}};\phantom{\rule{0.9em}{0ex}}\phantom{\rule{0.9em}{0ex}}\phantom{\rule{0.9em}{0ex}}\phantom{\rule{0.9em}{0ex}}{\theta}_{0}<\theta <{\theta}_{1}$$
(A13)
$${\mathbf{F}}^{\mathit{total}}=-\left(\frac{1}{{\epsilon}_{0}}\right)\left(1-\frac{1}{{\epsilon}^{2}}\right){\left(\frac{{D}_{0}{r}_{0}}{r}\right)}^{2}\mathrm{sin}\left[\frac{\left({\theta}_{1}-{\theta}_{0}\right)}{2}\right]\hat{\mathbf{r}}.$$
(A14)
$$\mathbf{\Delta F}\left(\mathbf{r}\right)={\mathbf{F}}_{2}\left(\mathbf{r}\right)-{\mathbf{F}}_{1}\left(\mathbf{r}\right)=\left(\mathbf{P}\bullet \nabla \right)\mathbf{E}+\left(\nabla \bullet \mathbf{P}\right)\mathbf{E}=\partial \frac{\left({P}_{x}\mathbf{E}\right)}{\partial x}+\partial \frac{\left({P}_{y}\mathbf{E}\right)}{\partial y}+\partial \frac{\left({P}_{z}\mathbf{E}\right)}{\partial z}$$
(A15)
$$\Delta {\mathbf{T}}^{\mathit{total}}=\iiint \mathbf{r}\times \mathbf{\Delta F}\left(r\right)\mathit{dxdydz}$$
(A15)
$$\phantom{\rule{2.5em}{0ex}}=\iiint \mathbf{r}\times [\partial \left({P}_{x}\mathbf{E}\right)+\partial x+\frac{\partial \left({P}_{y}\mathbf{E}\right)}{\partial y}+\frac{\partial \left({P}_{z}\mathbf{E}\right)}{\partial z}]\mathit{dxdydz}\text{}$$
(A15)
$$\phantom{\rule{2.5em}{0ex}}=\iiint \{[y\partial \frac{\left({P}_{y}{E}_{z}\right)}{\partial y}-z\partial \frac{\left({P}_{z}{E}_{y}\right)}{\partial z}]\hat{\mathbf{x}}+[z\partial \frac{\left({P}_{z}{E}_{x}\right)}{\partial z}-x\partial \frac{\left({P}_{x}{E}_{z}\right)}{\partial x}]\hat{\mathbf{y}}+[x\partial \frac{\left({P}_{x}{E}_{y}\right)}{\partial x}-y\partial \frac{\left({P}_{y}{E}_{y}\right)}{\partial z}]\hat{\mathbf{z}}\mathit{dxdydz}$$
(A15)
$$\phantom{\rule{2.5em}{0ex}}=-\iiint [\left({P}_{y}{E}_{z}-{P}_{z}{E}_{z}\right)\hat{\mathbf{x}}+\left({P}_{z}{E}_{x}-{P}_{x}{E}_{z}\right)\hat{\mathbf{y}}+\left({P}_{x}{E}_{y}-{P}_{y}{E}_{x}\right)\hat{\mathbf{z}}]\mathit{dxdydz}$$
(A15)
$$\phantom{\rule{2.5em}{0ex}}=\iiint (\mathbf{E}\times P)\mathit{dxdydz}$$