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If I understand correctly polarisation of electromagnetic wave, two opposite phase wave of same polarisation can cancel each over, so the question is : As white light have all polarisation, all the wave are in equally direction around 360°, and as there is all phase, it should cancel all each over and should be black ?

If it's because all electromagnetic wave are not exactly well oppositely synchronized and they "stay alive", is it possible to estimate the "loose" of the synchronized one (there must be some) or is it negligible ?

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    $\begingroup$ Note that phase and polarisation are not the same thing. Polarisation is the direction of the $E$ field and can be pictured as the plane in which the oscillation is occuring. Phase, on the other hand, is the "step" by which the peaks are shifted. $\endgroup$
    – Robbie
    Dec 5 '21 at 8:55
  • $\begingroup$ Of course, what I mean (I adjusted the question) : 1 - there is all polarisation 2 - there is all phase --> it should cancel to a non negligible amount...? $\endgroup$
    – Fabien
    Dec 5 '21 at 9:04
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    $\begingroup$ White light cannot contain "all phases", otherwise it would cancel and be black! More to the point, a light source which attempted to generate all phases, would simply generate nothing. $\endgroup$
    – Steve
    Dec 5 '21 at 9:36
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this is an animation of polarized light:

electromag

Electromagnetic waves can be imagined as a self-propagating transverse oscillating wave of electric and magnetic fields. This 3D animation shows a plane linearly polarized wave propagating from left to right. The electric and magnetic fields in such a wave are in-phase with each other, reaching minima and maxima together

In order to have a situation where two such waves would cancel and produce your "black":

  1. the waves should have the exact frequency and be at the exact opposite direction. This experiment can only be done with monochromatic sources, as are lasers,it has been done and can be viewed here. . The energy of the wave goes back to the source of the wave.

  2. White light depends on our perception of color and is composed by a large number of frequencies , see color perception here.

  3. To get a situation as in the experiment linked in 1) two white beams against each other are needed. The fact that white light is composed by innumerable frequencies means that the probability of overlap of two same frequency and polarization parts of the white beams is very small, so blackness (total extinction of white light) cannot result.

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  • $\begingroup$ Ok so the answer is : it's negligible ? $\endgroup$
    – Fabien
    Dec 5 '21 at 10:03
  • $\begingroup$ @Fab within experimental errors not measurable $\endgroup$
    – anna v
    Dec 5 '21 at 10:08
  • $\begingroup$ Based on this, would it be correct to say that white unpolarized light (e.g. sunlight), seen as an electromagnetic field resulting from summing up all the constituent components (with the resulting (summed) $\vec{E}$ and $\vec{B}$ "attached" at each point), is in general a smooth (wavelike) function with, say, $\vec{E}$ arbitrarily changing orientation and magnitude across space? And if so, are (the resultant) $\vec{E}$ and $\vec{B}$ in this picture at arbitrary angles, even though all constituent components (different frequencies, etc.) have them at the right angle? Or is this picture wrong? $\endgroup$ Dec 5 '21 at 15:51
  • $\begingroup$ @FilipMilovanović no, if one wants to model the white light with the myriads of components one maybe should use statistical methods, not addition of fields in the way you envisage. The E and B fields of the various components consisting the white light are incoherent and cannot be added. $\endgroup$
    – anna v
    Dec 5 '21 at 18:43
  • $\begingroup$ Oh, I see, thank you. The need for statistical methods suggests that maybe quantum mechanics comes into play, and these components cannot be "assembled" in a classical way? $\endgroup$ Dec 5 '21 at 19:03
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First, white light consists of a mixture of EM waves of many different frequencies as well as different phases (and, in the unpolarized case, polarizations, although white light can also be polarized). You can't have a superposition of waves of different frequencies be out of phase everywhere - if they're out of phase somewhere, then they'll be in phase somewhere else because they'll have gone through a different number of cycles.

Second, even if you only considered components of a single frequency (i.e. color), white light is an incoherent mixture of random phases and polarizations, not ones that are precisely uniformly distributed around the circle (either in real space for polarizations or the circle of phases). And $N$ randomly chosen (i.e. independent and uniformly distributed) phases don't perfectly cancel out when added; instead, they only partially cancel out and the magnitude of the sum is $\sqrt{N}$ times the magnitude of each term by the central limit theorem. It's actually very difficult to arrange it so that the phases precisely cancel each other out, as that requires strong correlation between the constituent phases.

So the overall amplitude still grows as you add in more contributions - just more slowly than if they were added coherently, as with a laser. This is also why having more people sing in unison (but with randomly offset phases) makes the total sound louder rather than quieter.

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  • $\begingroup$ So we can say that cancelation condition are so rare that they are negligible ? $\endgroup$
    – Fabien
    Dec 5 '21 at 18:20
  • $\begingroup$ @Fab I'd say that the probability of perfect cancellation is negligible. You (almost) always get partial cancellation. $\endgroup$
    – tparker
    Dec 5 '21 at 19:49
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I think it's an interesting question.

One perspective on this problem is that the waves we're talking about represent global oscillations of the EM fields and energy transfer, which solve the wave equation with given boundary conditions.

Waves that are mathematically complementary, which would cancel out, represent a zero solution to the wave equation. Such a situation would arise only when the boundary conditions are such that there is no energy propagation.

If there were any waves cancelling each other out, well, then they wouldn't contribute to the general solution. In a way, other waves would have to "take their place".

To put it slightly differently, if the light source is not producing any light, then where can the energy be going?

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  • $\begingroup$ The energy of canceled waves is still fully there, under an unknown form, we cannot detect them but they are there. We can prove this with two wave crossing their path at an angle of let say one degree in a perfect opposite phase and frequency : at the crossing path point you can't detect anything and you think there is no energy there but a little further the two wave fully reappear, so we can say they "was" still there at the crossing path point. $\endgroup$
    – Fabien
    Dec 5 '21 at 18:35
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The simplest answer is that white light has nothing to do with polarity or phase. It has everything to do with frequency and the way our eyes perceive a mixture of photons with different frequencies. In other words you can have white light thats polarized or not polarized and you can have white light thats in or out of phase.

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    $\begingroup$ It’s true that you can have white light that polarized or unpolarized, but you can’t have the different frequency components stay in phase over a period of time. $\endgroup$
    – tparker
    Dec 5 '21 at 21:47
  • $\begingroup$ @tparker no they cannot stay in phase but it has nothing to do with creating white light. $\endgroup$ Dec 6 '21 at 0:00
  • $\begingroup$ Oh, you’re saying that the different components can be in phase at one particular instant? Sure, but I don’t think that’s what most people mean when they say “in phase”. $\endgroup$
    – tparker
    Dec 6 '21 at 0:23
  • $\begingroup$ @tparker your thinking of amplitude modulation or beats but again white light has nothing to do with phase. It’s a a mixture of photons with different frequencies. $\endgroup$ Dec 6 '21 at 0:35
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    $\begingroup$ I agree with you that the definition of white light has nothing to do with phase. I'm just pointing out that since white light is a mixture of different frequencies, it isn't possible for those its components to remain in phase over time. So your statement that "you can have white light that's in phase" (over time) is incorrect, as a consequence of the fact that white light mixes different frequencies. $\endgroup$
    – tparker
    Dec 6 '21 at 5:03

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