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This is a question I put to my supervisor during my PhD many years ago, and never really got a satisfactory answer to.

In an optical tweezers, assume that a beam of light is used to move a glass bead. My question is whether the outgoing light is red-shifted. If it is not I cannot reconcile how energy is conserved in the system, as work has been done to move the bead?

Answers appreciated as always...

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    $\begingroup$ +1 because now I wonder the same thing. I've seen optical tweezers in action and the outgoing light doesn't look red-shifted, but given the amount of energy lost it could be imperceptible. $\endgroup$
    – Jim
    Commented Apr 26, 2013 at 13:43
  • $\begingroup$ Thanks Jim - I found largely the same thing. I just haven't seen it formalized and would be keen to. Another group in my uni actually had an OT setup and I volunteered to hook my spectrometer up and measure it, but they didn't fancy the fuss! $\endgroup$
    – Dave Cunnah
    Commented Apr 26, 2013 at 13:45

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Sure - the relativistic doppler effect means that light which is scattered off a moving object can be redshifted or blueshifted. And there can be more redshifted photons than blueshifted photons, or vice-versa, depending on where the object is, and how it's moving, relative to the center of the trap.

But since the object is moving much much much less than the speed of light, the frequency shift is completely negligible.

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  • $\begingroup$ So you would say that the relativistic doppler effect is the mechanism that preserves energy in an optical trap? Very possibly correct, just want to make sure I understand what you are saying! $\endgroup$
    – Dave Cunnah
    Commented Apr 26, 2013 at 15:23
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In theory, yes, the light will be redshifted. In practice, it sounds like the glass bead is too large for any measurable red shift. This is actually used in Mossbauer spectroscopy. What happens is that if your $\gamma$-ray source is a free atom, the recoil of the atom will cause the resulting radiation to be red-shifted relative to the natural frequency of the transition. In order to use the $\gamma$-ray usefully, you have to embed the source atom inside a crystal. Then it's the entire crystal that is recoiling, and that reduces the resulting red shift to something you can't measure.

This is the exact same problem as a collision between two billiard balls in classical mechanics. The photon has energy $E_\mathrm{ph} = hc/\lambda$ and momentum $|\vec p_\mathrm{ph}| = h/\lambda$. The glass bead has a kinetic energy $K = \frac{1}{2}mv^2$, momentum $\vec{p} = m\vec{v}$, and possibly some internal energy if we allow it to change state. The glass bead must change its velocity in order to conserve momentum. That will probably involve a change of speed, in which case the photon must change wavelength in order to conserve energy. It's easy to think of this as a doppler shift, if you think of the photon as being temporarily absorbed and then re-emitted. The magnitude of this goes down as the mass of the glass bead increases, in a manner that should be easy to work out.

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