I am not a physicist. I am buying some polariser for my camera. Circular polariser intrigues me. Basically you pass light through a linear polariser, then through a waveplate, you get circular polarisation.

Wikipedia says the following:

By adjusting the thickness of the wave plate one can control how much the horizontal component is delayed relative to vertical component before the light leaves the wave plate and they begin again to travel at the same speed.

Does that mean the electric field in an electromagnetic wave is a 2D vector field? I am a bit confused. I thought in 3D space, the electric field should be 3D vector field.

  • $\begingroup$ Photons are circularly polarized. So the fields of the photon rotate like a spiral rather than oscillate like a sine. To get the fields oscillating in a plane, you need an equal number of photons with the opposite spin. This would represent a linear polarization. $\endgroup$ – safesphere Aug 17 '17 at 1:24

Electromagnetic waves such as visible light are transverse waves. This means that the direction of oscillations of the fields is perpendicular to the direction of propagation of the wave. In general we can decompose the electrical field into two orthogonal components along the oscillation plane and the relative phase between these components gives rise to different polarizations. For example, a linearly polarized electromagnetic wave propagating along the $z$ direction has the following form

enter image description here

Note how the electric field (as well as the magnetic field) is fully contained to a plane.


Here is an annotated image from the Wikipedia article on circular polarisation.

enter image description here

All that I have done is added three spatial coordinate axes: x, y and z.

This is a very much simplified diagram of what is actually happening and shows "a wave" travelling in the z-direction with the spatial values of x and y constant.
The other parameter which is constant is the time.
This is a snapshot of the wave at one instant of time.

However it is not a snapshot which could ever be taken with a camera in that it shows how the electric field (vector) varies with z keeping x and y constant.

Since the wave is travelling in the z direction the electric field can only have components in the x and y direction but these components vary as the value of z changes.
So at one particular point in space (x,y,z) the electric field must be in the xy plane (orthogonal to the z direction) and the electric field can only vary with time in this plane.
That is the electric field can only have components in the x and y directions but not in the z direction.

I did say that the diagram is simplified because I could have drawn another parallel wave with a different value of x incident on the Polaroid as shown below

enter image description here.

but still the variation of the electric field with time can only be in the xy plane.

Instead of two just imagine more and more such waves until you have a continuum equivalent to a parallel beam of light travelling in the z-direction but with the electric field only varying in xy planes.

The addition of two simple harmonic motions at right angles to one another produces what is called a Lissajous figure.

Use this simulation to show the result of such an addition by making the x and y frequencies and amplitudes the same and having a zero phase difference to get a straight line (linear polarisation). Then change the phase difference to $90^\circ$ produced by the wave plate and obtain a circle (circular polarisation).


In an electromagnetic wave (light), the electric and magnetic fields are perpendicular to the direction of travel. If the wave is traveling in the $z$-direction, then $E_z = B_z = 0$. So, yes, the fields are two-dimensional.


protected by Qmechanic Aug 17 '17 at 9:03

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.