We know the beam of light oscillates in electric field and magnetic field, both perpendicular to both the wave of propagation and each other. What does, however, a non-polarised beam of light look like?

Is there always only one electric wave in a specific point of propagation? Is every consecutive wave positioned randomly? (With its magnetic counterpart perpendicular to itself). Or are there more waves overlapping, travelling in one wave of propagation? Are these overlaping waves always of the same wave-lenght?

Is this picture accurate? With single electric wave (and its magnetic counterpart), oscillating randomly so that every consecutive wave is positioned differently (but still perpendicular to the wave of propagation)?

enter image description here

Or is it rather this one, with more waves overlapping and travelling together?

enter image description here

  • $\begingroup$ projecting a lot of the instantaneous polarizations onto a single plane (i.e., your first figure) is not the same as having several (balanced) polarizations at one location (i.e., incident beam in your second figure). In your first figure, the field is just rotating and you are accumulating the results on the plane shown at right. In your second figure, the incident wave contains "all" polarizations at all points. $\endgroup$ Jan 1, 2015 at 15:39

2 Answers 2


At any particular point in space, there is only one value for the electric field. Of course, if multiple electromagnetic fields are overlapping at that point, then their electric field components are added together to yield the total electric field (this is because electromagnetic fields combine with linearity).

When multiple electromagnetic waves overlap in space, it is most useful to think about each wave individually. For a perfectly non-polarized wave that is monochromatic, it must be made up of an even distribution of differently polarized waves, each randomly out of phase (if they were all in phase, they would cancel one another out). So yes, the net electric field at any given point would appear to rapidly change both in magnitude and direction (always changing continuously, of course, since it is the sum of several continuously changing values). But this is generally not a useful view, since it is easier to handle the effects of each polarized wave separately.

There's really no difference between your two diagrams; the second diagram is simply looking at the polarized electromagnetic wave components that make up your non-polarized wave, while the first diagram is looking at the sum of these, which is exactly the total non-polarized wave. But the second diagram is more useful.

  • $\begingroup$ That's not true, what you say. Non-polarized light has no polarization, even at a certain point in the space and at a certain time. I mean, you can measure the projection of the polarization on some direction and if at the certain point and time, all the photons pass through the polarizer, then at the given point and time the beam is polarized. I would suggest you to reconsider your answer. $\endgroup$
    – Sofia
    Dec 19, 2014 at 23:00
  • $\begingroup$ By 'any particular point in space', do you mean point in space or point on the propagation line? $\endgroup$
    – Roll
    Dec 21, 2014 at 12:13
  • $\begingroup$ @Sofia: I agree that polarization is not well defined in non-polarized light, but you can look at how the electric field behaves instead. I'm not sure what your criticism is; I think it certainly is reasonable to think about any EM wave by decomposing it into polarized components, no? $\endgroup$ Dec 21, 2014 at 18:11
  • $\begingroup$ @Roll: I didn't say explicitly but I am considering infinite plane waves, which are the simplest EM waves, so the waves pass through all points in space. $\endgroup$ Dec 21, 2014 at 18:15
  • $\begingroup$ @HarryLevine: you say "you can look at how the electric field behaves instead." How can I look at the electric field? I have to measure it? How? Can you indicate a procedure? Shall I let an electron move and look in which direction it moves? It won't help, because the electric field at a given point in the space varies in time. The e.m. beam is travelling, until the electron moves, another part of the beam comes to the region of the electron. $\endgroup$
    – Sofia
    Dec 21, 2014 at 21:11

I see your point, interesting question.

As for your 1st question, the upper diagram would be more accurate, but... check that you can have one or many electric field, it doesn't matter. When you have many electric fields, they add up like vectors. The upper diagram is just showing the "resulting vectors", but that light can be made of many vectors or not.

In fact, actually, you can always split an electric field in as many components as you want. We usually split it into two perpendicualr components (x,y; left and right, or any other).

As for the 2nd picture... it is different. It is showing that, if you throw electric fields in any direction, then only vertical component would pass.

However, that would not be unpolarized light because unpolarized light does not have 3 components symmetrically distributed (otherwise the net field would be 0), it's jsut a representation that, if you throw several fields, only the verical component will pass. Check that the "vertical wave" is not the only one which passes. A half of the 45 degree wave would also pass.

On the other hand, if the amplitudes varied in the second diagram, it could also be unpolarized light, because you can always split your electric field into 3 components, or as many as you want. So acombination of three random electric fields would also give a net random field.


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