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EM fields in unpolarized light

  1. May 8, 2013 #1
    How is it that the electric and magnetic fields in unpolarized light are not cancelled out?

    The simplest example would be two photons of the same energy travelling coherently in time along exactly the same path in space, differing only by a rotation of 180 degrees around the axis of travel. Would the electric/magnetic field of one cancel the electric/magnetic field of the other?
     
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  3. May 8, 2013 #2

    cepheid

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    The thing about "unpolarized" light is that it is not a concept that makes sense for any particular EM wave. A given pure EM wave is always going to have some particular polarization, with the most general case being elliptical polarization. However, light from some source can be considered to be a superposition of many EM waves, each with its own polarization. So, obviously, in the instantaneous sense, the E-field from this source has some magnitude and is in some direction, but this direction and amplitude changes rapidly. So, when the light is considered to be unpolarized, I think this is meant in a time-averaged sense. Over the time interval of observation, the E-field does not consist of (or spend more time in) any one polarization mode (e.g. linear polarization in the +x direction or right-handed circular polarization) any more so than it does of the opposite mode. You still receive some total amount of power over that time interval, because this depends on the square of the magnitude of the E-field. But none of that power is in the form of net, uncancelled polarization in any particular direction.

    So, for light to be unpolarized means that the E-field does not *always* (or even preferentially) point in some particular direction or have some particular handedness, but rather consists of a mixture of all possible directions. It makes sense to me when I think of putting the light through a polarizing filter with some axis. If the light is entirely polarized along that axis, then all of the power will get through. If it is entirely polarized orthogonal to that axis, none will get through. If it is unpolarized (i.e. randomly polarized), then I would expect about half the power to get through, and the transmitted light would now be entirely linearly polarized in the direction of that axis. Edit: I checked to make sure I was right about that: http://en.wikipedia.org/wiki/Polarizer#Malus.27_law_and_other_properties
     
  4. May 9, 2013 #3

    sophiecentaur

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    This is all rather difficult. I think that considering "two photons" is a fundamentally flawed concept to use for 'explaining' anything about fields because photons are quantum objects and fields are not.
    When you are considering "unpolarised" EM waves, you are always dealing with multiple sources, radiating at (albeit slightly) different frequencies - otherwise, the sum would be a coherent, polarised wave front.

    When I have a problem like this to resolve, I always like to go down in frequency and consider RF and antennae. Imagine if you had ten (or a million) transmitters, radiating at the same nominal frequency but spread over a bandwidth of, say 100Hz and fed to ten (or a million) different dipoles, randomly oriented. The field at a distance from them would vary, more or less at the 'mean' frequency of the transmitters and, as it would be the resultant of all ten fields, it could be in any orientation at any given time. That, in my book, would be Unpolarised. You would actually pick up half the available power flowing past a receiving dipole (just like your polarising specs). If you used a very narrow band receiver and could only receive one of the transmissions, you would find that there was a preferred orientation for your receive antenna. That would constitute a Polarised wave.
     
  5. May 9, 2013 #4
    Okay - let's ignore the term unpolarized light - I'll concede that it's a little ambiguous.

    If I rephrase my two photon example with two EM beams of identical frequency travelling along an identical path in space (z), but differing in phase by pi, such that when the electric field of one is in the +x direction the other is in the -x direction and vice versa (and similarly for the magnetic field in the -y and y directions) - would the fields cancel each other out?
     
  6. May 9, 2013 #5

    Cthugha

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    Indeed, it does not make much sense on the level of pure states. There are few example of what is considered truly unpolarized light. A mixed state like thermal light can be unpolarized.


    In that case, they cancel out, but they necessarily do so everywhere. In other words there is nothing travelling anywhere and you do not have beams to start from. One can also show that it is impossible to create such a situation of complete cancellation, starting from an intial situation with two beams which travel different paths and are made to travel identical paths at some point in space using beam splitters or similar stuff. This gives rise to conservation of energy in destructive interference. Put differently: You cannot create destructive interference without creating constructive interference elsewhere.
     
  7. May 9, 2013 #6

    sophiecentaur

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    Two beams would indeed cancel out but (and a big but) these two 'beams' would have a finite width and would not overlap exactly. Their phase relationship could not be pi, everywhere. There could be cancellation in some places but there would be addition in other places. The total power would be the same - merely re-distributed.
    Even if you envisage doing it with a half silvered mirror, for instance, so that both beams would 'appear' to come from the same source and any phase-tilt due to the angle of the mirror were removed, there would be two possible directions for the power to emerge and one of them could be arranged to be zero - but all the power would emerge from the other direction.
    But I don't see what this has to do with the other issue of polarisation.
     
  8. May 9, 2013 #7
    What do you mean by finite width? Does all em radiation have the same amplitude?

    If the beams had the same frequency, how could their phase difference not be the same everywhere?
     
  9. May 9, 2013 #8

    sophiecentaur

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    You need to consider the geometry. For total cancellation, the two beams need to be arranged so that, over the whole of the beam, the amplitudes are identical (assumed, of course) and the phase is exactly opposite. How do you propose to achieve that? Try some sketches and the real problem may reveal itself because the path lengths will not be the same for both beams over their whole width. In fact, the only geometry that would solve the problem would be for both beams to originate in exactly the same place. That would involve them cancelling before they even started.

    You may need to start off this by believing that what you want is not achievable and it is this that has led you to a paradoxical conclusion. There is no paradox if the situation cannot arise.
     
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