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What do we mean when we say an electromagnetic radiation/photons can be polarized in a physical sense? If it mean orienting the light along certain axes or 'poles', do these poles have any real world meaning? For example, its rather easy in a physical sense for me to understand what is meant by polarization of a rubber thread (i tried to understand it form the wiki article).

My aim from this is more to understand what is meant when we say the magnetic fields and electric fields are perpendicular to each other. Now if we look at the figure below that I took from wikipedia about electromagnetic waves, the description reads as: A linearly polarized sinusoidal electromagnetic wave, propagating in the direction z through a homogeneous, isotropic, dissipationless medium, such as vacuum. The electric field (blue arrows) oscillates in the x-direction, and the orthogonal magnetic field (red arrows) oscillates in phase with the electric field, but in the y-direction description

Clearly, this figure represents polarized light. But then what is unpolarized light? Does it mean that we rotate this structure along the z-axis by 360 degrees and we'll have red and blue overlapping cylindrical objects representing electric field and magnetic fields that go outwards in the entire xy plane from the line x=0, y=0, varying only in amplitude along the z-axes?

This question seems to be a bit related to another question that is here (Electromagnetic fields vs electromagnetic radiation)

Another very related questions is for this linearly polarized radiation, if I place a negative electronic charge in its path at some place later on the z-axes line, then will the charge react by moving along the x-axes, rather than the z-axes, because the polarized radiation has the electric field along the x-axes ?

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An electromagnetic wave in vacuum can have two polarizations. In your figure it is polarized along the $x$ axis, since the electric field is directed along this axis, while the magnetic field is along the $y$ axis. Another option for the wave propagating in the same $z$ direction is for the electric field to be along the $y$ axis and for the magnetic field along the $x$ axis. Any combination of two polarized waves will be also a polarized wave - in fact, one often chooses to work with specific combinations called circularly polarized waves rather than with $x$ and $y$ polarizations.

When speaking about unpolarized light, what is meant is that this light does not have a well-defined polarization, that is it is a mixture of waves running in different directions and randomly polarized. This is the light that comes from conventional life sources, such as light bulbs, as opposed to quantum light sources, such as lasers and masers. Light emitting diodes that we encounter in our daily life are, in principle, quantum light sources, but they are usually of too low quality for their light to have a well defined polarization.

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  • $\begingroup$ thank you. it does clarify my question a bit. can you also please make a comment, if you would like to, about my other question. i can't intuitively understand what will happen to a charge in its path - will it move in the transverse direction? $\endgroup$ – Abhishek S Khetan Mar 11 at 9:13
  • $\begingroup$ Indeed the charge would move along the electric field, since it is this field that exerts the force. Since this field is oscillating, the charge will oscillate as well. In fact, such motion of charges accounts for the attenuation of the electromagnetic field when it propagates via a medium, i.e. for the electric and magnetic permittivities $\epsilon$ and $\mu$. $\endgroup$ – Vadim Mar 11 at 9:17
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    $\begingroup$ cheers! this is so counter intuitive and i never read this in any textbook or online article. i actually came to the polarization question while i was trying to understand permittivity. $\endgroup$ – Abhishek S Khetan Mar 11 at 10:42
  • $\begingroup$ Also, consider that due to Lorenz force, the charge will drift in z direction. This due to the electric field the charge will gain velocity and the term $v \, \wedge B$ term will be not vanishing. $\endgroup$ – Grego_gc Mar 11 at 13:23
  • $\begingroup$ Indeed. In addition, if we are talking about permittivities, the electrons are not free. The magnetic effects are however quite small in most cases - one is frequently justified taking $\mu = 1$. $\endgroup$ – Vadim Mar 11 at 13:32

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