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I understand that while using an optical stretcher experimental apparatus we attach a small dielectric ball to the specimen of study which can either be a DNA molecule, a Biological cell or any microscopic object that we are interested in. The way the concentrated Laser beam in the apparatus is refracted through the ball is responsible for a gradient force towards the center of the beam. How important is it for the ball to be dielectric? Now assuming we don't use any balls in the experiment, can a similar radial force be generated on Brownian particles suspended in the beam that are not necessarily spherical in shape?

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  • $\begingroup$ when you say "how important is it for the ball to be dielectric?" do you actually mean how important is it to be spherical? Being dieletric simply means it is a polarizable object (and hence can induce dipole moment). $\endgroup$ – wcc Sep 10 '18 at 14:03
  • $\begingroup$ Basically, there are 2 questions. Firstly, what is the purpose of the ball to be dielectric? or why can't a non-polarizable object work as well as a polarizable one? secondly, If the suspended particles are non-spherical and are not connected to a spherical reference ball is it fair to say that the force on them is random and unlike radial when they are connected to a spherical transparent ball? $\endgroup$ – bitbitbitter Sep 10 '18 at 14:11
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In general, there are two limits to consider. In both cases, the polarizability of the object (the bead) is important. As for the importance of the geometric shapes, I think it's more subtle.

Limit 1 is where the wavelength of the laser is much shorter than the size of the object. This is the case considered by @Persian_Gulf. A dielectric object has index of refraction $n = \sqrt{\varepsilon}$, where $\varepsilon$ is the dielectric constant of the object material. Due to the index of refraction, the incoming light undergoes refraction, and the momentum of the deflected light provides force. To see how the dielectric constant is related to atomic polarizability for macroscopic objects, consult the Wikipedia article on "Clausius-Mossotti relation". I think Griffith's EM book also mentions this topic as well.

Limit 2 is where the wavelength of the laser is much larger than the size of the object. In this limit, you can consider the object as an ideal dipole moment, that is induced by the external electric field (of the laser). Griffith's EM book has an example problem on the induced dipole moment of a polarizable sphere under external E-field, I think. The induced dipole moment $d = \alpha E$ ($\alpha$ is polarizability) interacts with the external electric field and feels dipole force. The dipole force is the gradient of the dipole potential $U_{dip} = -\frac{1}{2} \alpha E^2$.

So this answers the first part of your question. As for the geometric shape, I am less familiar but I can say the following. In Limit 1, it will definitely affect how the light is refracted. If the surface of the object has some aberration profile that is comparable to wavelength scale, it will strongly scatter the incoming light. In Limit 2, the incoming light will be less sensitive to any surface profile variation, because the light cannot "resolve" the features on the sub-wavelength scale object.

But anyhow, in literature there are abundant reports of trapping yeasts and bacteria with optical tweezers, so that should leave no doubt that you can trap non-spherical objects with light. The tricky thing is that those examples do not clearly fall under either of the two limits mentioned, and there is no simple theory, although there are computational tools to numerically calculate the forces. See this paper.

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  • $\begingroup$ Thank you! That's very informative, it makes a lot of sense! $\endgroup$ – bitbitbitter Sep 11 '18 at 11:35
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In optical tweezers changes in photon momentum during refractions exerts forces on particle. So a transparent medium is needed. enter image description here

Spherical shape matters, because the refraction pattern depends on the shape of the surface.

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  • $\begingroup$ Yes, you are right but you didn't answer my question. $\endgroup$ – bitbitbitter Sep 10 '18 at 14:02

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