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.
[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.