Can the electric field vector of an EM wave oscillate in the propagation direction?
In text books the polarisation is always orthogonal to the propagation direction.
I'm wondering specifically because of this paper:
https://doi.org/10.1039/D2CP01009G
On page 17 of the PDF-reader the authors present the concept of synthetic chiral light.
In Fig. 10 c) they show the sum of a linearly polarised EM-field with an elliptically polarised EM field. The propagation direction apparently lies INSIDE the polarisation plane of the ellipticaly polarised field.
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4$\begingroup$ Plane wave solutions to the homogenous wave equation require E and B are orthogonal to the wave vector k, particularly to satisfy the Faraday maxwell equation $\endgroup$– jensen paullCommented Feb 8 at 15:29
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2$\begingroup$ @jensenpaull That's true in vacuum, but things can be trickier in materials. $\endgroup$– Buzz ♦Commented Feb 8 at 15:52
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$\begingroup$ I'm looking at this paper: doi.org/10.1039/D2CP01009G On page 17 of the PDF reader the authors present the concept of synthetic chiral light. In Fig. 10 c) they show the sum of a linearly polarised EM-field with an elliptically polarised EM field. The propagation direction of both apparently lie INSIDE the polarisation plane of the the latter field. $\endgroup$– JorgeCommented Feb 8 at 16:15
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1$\begingroup$ Does this answer your question? Why no longitudinal electromagnetic waves? $\endgroup$– John RennieCommented Feb 8 at 16:45
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$\begingroup$ Thanks for all the comments. The questions you link sound related to my question, but I can't make the bridge to the field mentioned on page 17 of the PDF-reader, Fig. 10c). That's an elliptically polarised field propagating INSIDE the polarisation plane.. :/ $\endgroup$– JorgeCommented Feb 10 at 10:35
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1 Answer
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In dielectric, homogeneous, isotropic, linear, charge-free, and non-dispersive media, the electric and magnetic fields are always perpendicular to the direction of wave propagation, so no polarisation can be parallel to the direction of propagation.
$$\hat{s}\times\vec{E}=v\vec{B} , \hat{s}\times\vec{B}=\vec{E}/v, \vec{S}=\vec{E}\times\vec{H}$$
These equations can be derived from maxwell's equations.
If you go outside this type of media, things could change, for example in anisotropic media, the energy and direction of wave propagation is different.
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3$\begingroup$ This is incorrect as stated. The TE and TM fields of homogeneous waveguides completely filled with lossless dielectric have $H_z$ and $E_z$ longitudinal components, resp., and these non-zero components are parallel with the axis along which the wave propagates. These are actually dispersive waves despite the medium being ideal completely non-dispersive. The reason for dispersion is the restriction of the wave by the conductive (metallic) enclosure (boundary condition). $\endgroup$ Commented Feb 8 at 17:25
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$\begingroup$ Yes one should include some comments on waveguides when answering this kind of question. One might add that when you have two plane waves propagating in directions ${\bf k}_1$ and ${\bf k}_2$, there is some sort of propagation in the direction $({\bf k}_1 + {\bf k}_2)/2$, and this direction is not perpendicular to the fields. The waveguide case can be viewed as an example of this. $\endgroup$ Commented Feb 10 at 14:14