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Can this in any way be considered to be a circularly polarized wave?

  1. Oct 27, 2012 #1
    Can this in any way be considered a circularly polarized wave?

    Can this wave in any way be considered circularly polarized?

    [itex]\overrightarrow {E}\left( z,t\right) =\left[\widehat {i}\cos \left( \omega t\right) +\widehat {j}\cos \left( wt-\dfrac {\pi } {2}\right)\right] E_{0}\sin \left( kz\right)[/itex]

    I say no, since there is no twisting along the direction of propagation. If you plug in t=0 for instance, the vectors will be pointing entirely in the x axis direction.

    My friend disagrees and mentions the phase shift in the y axis as evidence for circular polarization.
    Last edited: Oct 27, 2012
  2. jcsd
  3. Oct 27, 2012 #2
    Your friend is right - at any z location, field is circulating in direction with time. And that means a CP (circularly polarized) wave. You appear to be confusing CP with corkscrew motion that would apply to propagation down a twisted waveguide.
  4. Oct 27, 2012 #3
    Thanks for the quick reply!
  5. Oct 27, 2012 #4
    You're welcome - I just happened to be viewing at the time and felt the impulse to respond! :smile:
  6. Oct 27, 2012 #5
    I should add that your expression in #1 represents a standing wave CP (sum of two counter-propagating CP waves of opposite CP polarization) not propagating one. For a single propagating CP wave, the sin(kz) term is absent and the cos(ωt), cos(ωt-π/2) arguments are replaced with cos(ωt-kz-θ), cos(ωt-π/2-kz-θ) respectively, with θ determined by the given t=0 temporal phase conditions at z=0. That may represent a LCP wave (left circularly polarized) - counterclockwise rotation of field as viewed along propagation axis with wave receding. For a RCP wave, an extra phase angle of π is added to either one, but not both, of the above arguments for cos. Convention for handedness can vary, but I believe above is the more standard one.
    Last edited: Oct 27, 2012
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