Circular Reflection

Question

How efficiently can light be turned into angular momentum?

Light can be contained for long paths via total internal reflection with some steady decline in intensity due to absorptivity of the medium. This is usually compensated for in telecommunications with a coupled amplifier and the effects of the light on the fiber itself are negligible.

If we reflect the light in a tight circle (via wound fiber or plexiglass cylinder), in the style of an optical resonator ring - we can concentrate many reflections over a small area. The force generated for light reflecting perfectly off a surface is $F={2*I}/c$, where c is the speed of light and I is the light intensity in $W/m^2$. Obviously this does not hold for shallow angle reflections that could get the cylinder turning.

How do we model the torque imparted on a reflective cylinder in this fashion?

Arrows not scaled.

The sum of red arrows cancels out and blue arrows produce a counter-clockwise torque. How do we estimate the fraction of force produced by one light 'circuit' of the circular cross section, if any?

Answer 1

Nice idea! But unfortunately, there is no angular momentum transferred to the resonator, unless there is loss (i.e. absorption of light). In that case, the best you could do is impart $I/c$ momentum to the outer edge of the ring, giving an angular momentum which you would calculate according to the moment of inertia of the structure.

One way to think of this is in terms of momentum balance. For a single reflection (any reflection), you can break up the components of force acting on the mirror, which recoils from interaction with: (a) the normal component of incident light, (b) in-plane component of incident light, (c) normal component of reflected light, and (d) in-plane component of reflected light.

For a normal incidence reflection, and no loss, (a)=(c), and the recoils add, giving you $F=2I/c$. For non-normal incidence (b),(d)$\ne$0. But they subtract since the in-plane recoils are in opposite directions for incident and reflected beams. So the net push is $(1-R) (I/c)\sin\theta$, where $R$ is the power reflection coefficient and $\theta$ is the angle of incidence. Thus in the lossless case, $R=1$, and the in-plane force is zero. If there is absorption of light at the point of reflection, (b)>(d) since $R<1$, so there is a net in-plane push.

Basically, in your diagram you just forgot the force arrows for a recoil from reflected light at each point of interaction, which ends up cancelling the blue arrows in the lossless case.

Answer 2

Once the light is coupled into your cylindrical resonator (probably by evanescent field coupling), I don't think the circulation of light imparts any torque to your resonator.

But since you are interested in angular momentum imparted by light, you probably want to search "orbital angular momentum of light". By transmitting a Gaussian beam through a phase plate, you can generate spiral pattern. The actual angular momentum comes from the intensity pattern, not the polarization.

But in some systems, it turns out you can impart torque using polarization. Look at this recent paper that came out in PRL (Phys. Rev. Lett. 121, 033603). A nano-dumbbell made from silica is rotated to ~GHz frequency (faster than any gas turbine made in history) with a circularly polarized light. I think what is important in this case is the anisotropic polarizability of the nano-dumbbell (that is, I'm not sure if you can impart torque to a perfect nanosphere with a circularly polarized light).