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Photon Energy question

  1. Nov 8, 2007 #1
    Where does the energy go when light undergoes destructive interference?
  2. jcsd
  3. Nov 8, 2007 #2


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    If it is a dark spot in diffraction, the energy goes to the bright spot.
    If it is the destructive interference in a non-reflective coating, the energy that didn't go into reflection goes into transmission.
  4. Nov 8, 2007 #3


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    First of all you don't need photons to understand interference. The prototypical example of interference is that of a standing wave. So this not only happens with light (as you might have known already) but also with a guitar string. Just to make sure you can imagine what I'm talking about, take a look at the animation in http://en.wikipedia.org/wiki/Standing_wave .

    For a guitar string that shows a standing wave, it is obvious how the energy is distributed: at the nodes there is always zero kinetic energy and and oscillating maximum tension (potential energy); at the antinodes there is an oscillating maximum kinetic energy and always zero tension. But as the string vibrates, energy moves from the nodes (potential) to the antinodes (kinetic) and back.

    Now what happens if we send two monochromatic polarized laser beams head-on to each other ? For a plane wave (one of the laser beams) the electric field E and the magnetic field B are related with the wave vector k by the formula [itex]B=k\times E[/itex], i.e. they are orthogonal and have a certain orientation.

    Now suppose that we look at a node of the electric field of the two colliding laser beams. This means [itex]E_1=-E_2[/itex], so the electric field annihilates at this point. But the wave vectors of the two plane waves are opposite as well, that is [itex]k_1=-k_2[/itex] (because they move in opposite directions), from which you immediately obtain [itex]B_1=B_2[/itex]. This means the magnetic field does not annihilate where the electric field does, and vice versa.

    More generally the electric field energy can be considered as some sort of kinetic energy, whereas the magnetic field energy is a kind of potential energy (don't confuse this with the potential energy of a charge in an field). The most obvious representation of this fact is the electromagnetic field Lagrangian density

    [tex]{\cal L} = E^2-B^2[/tex]

    which looks very similar to the Lagrangian of classical mechanics

    [tex]L = T-V[/tex]

    But in contrast to the guitar string, energy doesn't have to move spatially in order to be converted from its kinetic to potential form.

    So the short answer to your question is: Nothing happens to the energy in a standing electromagnetic wave - it just stays where it is, but is transformed from "kinetic" to "potential" and back again.
    Last edited: Nov 8, 2007
  5. Nov 8, 2007 #4


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    This really is a more general question on wave interference. A similar question was asked a while back and I had provided a short reply and a reference that I think adequately address this issue.


  6. Nov 8, 2007 #5

    Hans de Vries

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    Now let me for once be nasty :devil: and turn this into the following puzzle:

    Two parallel laser beams pointing in the same direction have exactly
    the same frequency but they are 180 degrees out of phase. Now let
    the beams overlap under an extremely small angle. There will be a
    zone where they completely overlap, say for 1 meter or so. In that
    region there will be no E nor B field, no Poynting vector, no nothing.
    It will be pitch dark....

    Somewhere behind this region the beams stop overlapping and they
    reappear again. The question is how were energy and momentum
    transferred through the dark zone?

    Regards, Hans
    Last edited: Nov 8, 2007
  7. Nov 8, 2007 #6


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    Hello Hans, no need to apologize, we all feel like doing something nasty from time to time. :smile:

    Extremely small angle is not the same as exactly zero angle. Thus there will be exact destructive interference only in the sagittal node plane between the lasers. Even the slightest distance away from this plane there will be a small but nonzero field strength.

    The Poynting vectors emitting from the laser sources will "move" towards the node plane but then they bend off until they are parallel. They remain parallel (Edit: I think this will actually be more smooth than my description indicates) along the "1 meter" you referred to and afterwards they will move away from the plane again. They will never cross the node plane and that's the reason why the Poynting vector can be exactly zero in the node plane at all.

    So if one follows the Poynting vectors, all energy coming from the left laser will remain on the left hand side of the node plane, whereas all energy from the right laser remains on the right hand side of the node plane.

    Is it this what you tried to indicate (presumably from a quantum mechanical perspective) ?

    PS: Just another addition - I don't even think that there will be exactly zero energy density in the node plane, because as I have explained in post #3, if the electrical fields of two plane waves annihilate at some point, the magnetic fields can't if the wave vectors of both waves are different.
    Last edited: Nov 8, 2007
  8. Nov 8, 2007 #7

    Hans de Vries

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    I found this very nice java applet:

    http://www.falstad.com/ripple/ex-2slit.html [Broken]

    With some effort you can make your own two laser beam
    by selecting the "2 plane Src, 1 Freq option" in the second
    menu and drawing some walls with your Mouse (3rd menu)
    You can drag the sourcesto any location.

    Regards, Hans

    PS. The 3D-View is very nice also.
    Last edited by a moderator: May 3, 2017
  9. Nov 8, 2007 #8

    Hans de Vries

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    This works nicely:

    1) select the "2 plane Src, 1 Freq" option in the second menu

    2) set the Phase Difference to maximal (bottom slide)

    3) Drag the end points of the plane sources to the top to
    form a "roof" so that each plane source has one point in
    the middle of the top line and the other point a bit lower
    on either the left or the right border.

    4) Then set the view to 3D

    Regards, Hans
    Last edited by a moderator: May 3, 2017
  10. Nov 8, 2007 #9

    Hans de Vries

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    Indeed, As the simulations show, this dark zone doesn't
    occur at all!

    I would say that the reason is that the beams need to
    have a width much larger as the wavelength, so there
    are always multiple bands of additive and subtractive
    interference in the direction of the beam. If one tries
    to reduce the width of the beams they'll fan out.

    Regards, Hans.
  11. Nov 9, 2007 #10


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    This is awesome ! Didn't know that this is possible with java, considering performance.

    You can even move the sources around with the mouse to see the effects of superluminal propagation... :wink:

    PS: the guy has done tons of other educational simulations at http://www.falstad.com/mathphysics.html . Strongly recommended !
    Last edited by a moderator: May 3, 2017
  12. Nov 9, 2007 #11


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    But also keep in mind that the program seems to simulate a real scalar wave, whereas the EM field behaves technically like a set of complex fields:


    [tex]u = F^\dagger F[/tex]

    [tex]\nabla \cdot F = 0[/tex]

    [tex]\frac{\partial F}{\partial t}=-i\nabla \times F[/tex]

    PS: I've just discovered that he also has a simulation TE and TM fields !
    Last edited: Nov 9, 2007
  13. Nov 9, 2007 #12
    For two EM waves which are interfering the total energy of the field remains constant. For photons - The pattern of photons is merely a probability distribution. It doesn't mean that two photons hit the same spot and destructively interfered with each other. Either a photon is detected at R or it is not. The chances are given by a probability distribution.

    Hope that helps.

  14. Nov 12, 2007 #13
    The energy of the photons it remains with the photons because the destructive interference does not mean an energy loss; maybe the photons are re-absorved later
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