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A Emission of light from a surface

  1. Nov 21, 2018 #1
    I came across someone in the lighting industry who insists that because of Gauss's divergence theorem and Maxwell's Laws that when light is emitted from a surface that it is only emitted orthogonal to the surface. I have tried to point out numerous real world examples that contradict the premise, but this person keeps going back to the equation and insists they are right.

    Does anyone have a good mathematical argument I can present? The concern I have is this person presents themselves as an expert and there are people who will be awed by the mention of Gauss and Maxwell and won't follow the argument to logical (and impossible) conclusions.
     
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  3. Nov 21, 2018 #2

    ZapperZ

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    But what exactly does it mean to say that it is ".... only emitted orthogonal to the surface.. "? Does it mean that the wavevector is normal to the surface? Or does he mean that the Poynting vector is normal to the surface? While those two are often in the same direction, this is not true for all cases, especially when the medium of propagation is involved.

    Zz.
     
  4. Nov 21, 2018 #3
    The Poynting vector is normal to the surface. They equate emission from the surface of an LED chip to emission from a semiconductor laser. It is all related to the intensity distribution of light from LEDs vs other light sources.

    They have dismissed obvious counter examples and continue to focus on their mathematical derivations.
     
  5. Nov 21, 2018 #4

    Nugatory

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    Mentors' note: An unhelpful and off-topic argument has been removed from this thread. All members are reminded that we're hear to advance people's understanding of physics.
     
  6. Nov 21, 2018 #5
    What I am hoping for is a good argument as to why Gauss's divergence theorem combined with Maxwell's laws don't apply to generation/emission of photons. There seems to be a leap of faith between electric field lines and the path a photon would take.
     
    Last edited: Nov 21, 2018
  7. Nov 21, 2018 #6

    tech99

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    If light is reflected from a surface it can radiate in any direction. So if the electrons of the surface are excited somehow with some particular phase relation, the surface will presumably radiate a beam in any direction. And if they are excited with random phase, as perhaps by heating, then presumably the light will go in all directions.
     
  8. Nov 21, 2018 #7
    What frustrates me the most is that the following this person's argument to its logical conclusion means that a sphere heated to incandescence would appear as a point source. The sun should similarly be a point source.

    That is why I am looking for the flaw in the logic of combining Gauss and Maxwell.
     
  9. Nov 21, 2018 #8

    phinds

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    Seems to me that that's a very good qualitative argument right there. Have you asked him what he thinks of that?
     
  10. Nov 22, 2018 #9
    They ignore that and argue that I don't understand Gauss and Maxwell.

    I am struggling to see how they could be combined to explain emission of photons from a surface. Gauss describes the direction of electric field lines. Maxwell links electrical and magnetic fields and does describe photons.

    I am guessing that someone made a huge leap when they combined Gauss and Maxwell and derived a result that requires photons to be emitted in the direction of the electric field. If there is nothing else I'm missing then I will continue on.
     
  11. Nov 22, 2018 #10

    phinds

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    Well you could always quote Feynman. "If a theory doesn't fit the experiments, it's wrong", and his theory doesn't fit experiments.
     
  12. Nov 22, 2018 #11

    Dale

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    They are making a mistake in their derivation. Gauss divergence theorem and Maxwell’s law do not imply that all emission is normal to the surface.

    If they were correct then you could not get a penumbra from a light source with a smooth surface.
     
  13. Nov 22, 2018 #12

    ZapperZ

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    While I'm waiting for the turkey to get to room temperature before I stick it into the oven (don't worry, I've washed my hands and there's no risk of salmonella contamination), I'll address a few of the things that I'm puzzled here.

    1. Gauss's Divergence Theorem, as stated in the first post, is a mathematical operation. It gives us the ability to take the volume integral of a divergence, and relate it to an area integral over a closed surface. When the divergence theorem is applied to the differential form of Gauss's Law, we get the integral form of Gauss's law. So

    Divergence theorem ≠ Gauss's Law

    But in your posts, it appears that you have used the two terms to mean the same thing.

    2. Remember I asked earlier on what exactly is going out orthogonally from the surface. You clarified that it is the Poynting vector. If this is true, and if they are also insisting that the E-field is orthogonal to the surface, then these two are contradicting each other. Since the Poynting vector is E x H, it is impossible for both the Poynting vector and E to be normal to the surface.

    The problem here is that we are seeing all of this only through your lens. All we have is your interpretation of it, based on your understanding. This can easily be inaccurate, and in fact, unfair to these folks. So unless you have references to their actual derivation, I do not think it is worthwhile to tackle this any further, because we can all easily be in a tizzy over a misunderstanding.

    Zz.
     
  14. Nov 27, 2018 #13
    I am really trying to make this as simple as possible. Is there a way to combine Gauss and Maxwell to come up with photons only being emitted in the direction perpendicular to the surface? This is equivalent to saying that light emission from a flat surface is virtually the same as light emission from a laser (well collimated).

    It is disturbing that this is being taught to college level students.

    https://www.crcpress.com/Understanding-LED-Illumination/Khan/p/book/9781466507722
     
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