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Orthogonally polarized beams do not interfere, do they beat?

by reasonableman
Tags: beams, beat, interfere, orthogonally, polarized
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reasonableman
#1
Mar14-12, 02:43 AM
P: 80
I'm trying to understand what happens with beating light when the componenets have differing polarisation.

I know that orthogonally polarised beams do not interfere spatially. I looked in Hecht and this arises in the maths due to the 'interference term' in which there is a [itex]E_{1}.E_{2}[/itex] term. Obviously if they are orthogonal the dot product is zero.

However when I look up the treatment for beating (which is before) the electric fields are just treated as scalars, so the polarisation is not considered. As the above, vector treatment, is more 'complete' I'm inclined to believe that one, however I'm not sure it's correct.

The reason I'm not sure is that in previous thinking about this problem I decided that two co-linear beams of slightly different frequency (as in the beating example) is equivalent to a single frequency beam of rotating polarisation, if you consider that then the changing polarisation will still produce intensity modulation. Hence I'm confused!

Any comments?
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chrisbaird
#2
Mar14-12, 12:52 PM
P: 617
A beating signal is just a special case of interference where two waves with the same polarization and close but different frequencies interfere. Orthogonally polarized waves do not interfere, therefore they can not produce a beat signal.

A linearly-polarized beating signal is entirely different from a circularly-polarized wave. The beat signal is just a traveling cosine wave contained in a lower-frequency cosine envelope. This envelope extends spatially along the direction of propagation. The polarization of such a signal (if it is the simplest case of a plane-wave, linear-polarization beat signal) is constant in time and space, whereas the amplitude is what's beating. In a circularly-polarized monochromatic plane wave, the amplitude is constant (no beating), and the polarization vector is rotating in space. The changing polarization of circularly-polarized light does not produce intensity modulation. The electric field vector changes direction only, not length - that is why it sweeps out a circle (a circle is defined as all points the same distance from a central point, but at different angles).
reasonableman
#3
Mar14-12, 01:53 PM
P: 80
Ok, I agree that orthogonally polarised waves do not interfere. However I believe in a certain case a beating system can be analogous to a single beam that is constantly changing polarisation. Specifically:

If you have two orthogonal linear polarised beams of the same magnitude but slightly different frequency the system - at a specific time - will appear identical to a elliptically polarised beam i.e. two beams of orthogonal polarisation with different phases. The phase difference between the two 'beams' will vary with time going from linearly polarised, to elliptically, to circular and back again.

However this just supports the point that orthogonal beams do not interfere, if the system can be considered as a single beam of varying polarisation it will not produce intensity modulation, it was a mistake on my part to think it would.


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