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I need some intuition for various EM phenomena

  1. May 28, 2013 #1
    I am using a computational electromagnetics software that solves for the resultant electric and magnetic fields around materials represented as lattices of discrete dipoles. Some of the results are the extinction, absorption, and scattering efficiency factors of EM waves, represented as a graph across a sweep of different wavelengths.

    According to the documentation:

    Absorption efficiency factor = absorption cross section / (pi*(characteristic dimension)^2)
    Scattering efficiency factor = scattering cross section / (pi*(characteristic dimension)^2)
    Extinction efficiency factor = absorption efficiency factor + scattering efficiency factor

    The characteristic dimension represents the radius of a sphere with equal volume to the object being calculated.

    When I plot out some of these graphs from different geometry I input, I get distinct peaks at certain wavelengths. If one of these peaks happens in the absorption graph, what does this mean? Does it mean that a lot more light is absorbed at that wavelength than what would be expected at the geometry's size?

    I have a sense for what extinction means, however. It is just the amount of light that is not transmitted. Is this right?
     
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  3. May 29, 2013 #2

    Simon Bridge

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    Sort of - it means your object has resonant absorption at that wavelength. It absorbs a lot more compared with other wavelengths for the same size.

    Looking at the definition above, that appears to be correct. Properly, the extinction efficiency would be the proportion (per characteristic area) that is either scattered or absorbed.
     
  4. May 29, 2013 #3
    This is starting to make sense. I am running tests on an infinite periodic structure with incident monochromatic plane waves of different polarizations (same polarization per test run, different test runs) hitting it. So, extinction efficiency factor is the proportion of light that does not transmit through the material?

    This is how I am thinking about it. Imagine a 1 square meter wall of this periodic structure in a room, and this room has ambient monochromatic light that does a sweep across wavelengths from ultraviolet through visible to infrared. There would be certain wavelengths whereby I would not be able to see through the wall, or perhaps the wall would become translucent. Is this right?
     
  5. May 29, 2013 #4

    Simon Bridge

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    Only if that is the only alternative to scattering and absorption in your simulation.

    That is a correct statement for, pretty much, any material.
    Though "see through" may be a bit ambitious... does your model handle refraction?
     
  6. May 29, 2013 #5
    This simulation only deals with scattering and absorption. I should have specified what I meant regarding when a material becomes translucent or opaque. Let me try to reword it.

    If there is a peak in the absorption at a certain wavelength, then would it mean that a wall of this material would be its most opaque? And, if there is a peak in the scattering at a certain wavelength, then this material would be its most translucent (fuzzy received light, can perceive no detail through it)?

    I suppose, on a large scale, this simulation would definitely take into account for refraction. I suppose refraction would be an example of transmitted light, right?
     
  7. May 30, 2013 #6

    Simon Bridge

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    Absorbed light ain't getting through - this is correct - so the wall will just look "black" at that wavelength ... if you absorbed all the green, then the wall would look magenta.

    But scattering usually happens off the surface ... so it is like randomly reflected light.
    It is scattered light that gives us the diffuse color of a surface.

    Light can be scattered through a bunch of particles though - think: cloud.
    So your intuition is working in that regard.

    Refraction is the effect of the path of light being bent at a surface due to changes in the speed of light across the surface. Your simulation will take it into account if the speed of light (or the refractive index) is a parameter.

    Refraction is what makes you see look odd viwed through different shaped glass.

    It is starting to look like you are trying to use the software without a grounding in the background physics.
     
  8. May 30, 2013 #7

    Simon Bridge

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    Absorbed light ain't getting through - this is correct - so the wall will just look "black" at that wavelength ... if you absorbed all the green, then the wall would look magenta.

    But scattering usually happens off the surface ... so it is like randomly reflected light.
    It is scattered light that gives us the diffuse color of a surface.

    Light can be scattered through a bunch of particles though - think: cloud.
    So your intuition is working in that regard.

    Refraction is the effect of the path of light being bent at a surface due to changes in the speed of light across the surface. Your simulation will take it into account if the speed of light (or the refractive index) is a parameter.

    Refraction is what makes you see look odd viewed through different shaped glass.

    It is starting to look like you are trying to use the software without a grounding in the background physics.
     
  9. May 30, 2013 #8
    Though there is also phenomena such as subsurface scattering, right? I am imagining the incident light as a certain vector, hitting the material, resulting in many randomly oriented vectors. Some of them represent bouncing off in random directions (negative dot product), and others represent transmitted, albeit deflected, in random directions (positive dot product).

    I'd be worried if I didn't know this by now! ;)


    [EDIT]
    Upon further investigation, I understand your concern regarding my question about refraction. That was definitely not a question I thought through, since I clearly understand this topic. It's one of the defining properties of the materials I input into the program, gosh! I understand it both from a classical and quantum level. Sometimes things in my brain don't click immediately.
    [/EDIT]

    Do not be alarmed by the seemingly general nature of the questions. I do have the background. I am just finding alternate ways of explaining things, since I am writing a paper for those who do not have the same background. I know the math behind all of this, but I like to try to develop an intuition for it, a way to visualize it. I am completely aware of what refraction is. I am looking for ways to describe it without misleading the audience into thinking the physics is something it's not. This is more a language exercise for me.

    This discussion has helped me a lot, regardless. This paper will be much easier to write.
     
    Last edited: May 30, 2013
  10. May 30, 2013 #9

    Simon Bridge

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    Ah well ... context is everything.
    When getting a lay audience to appreciate some physical stuff there is a temptation to go too far with the intuitive visualizations - a lot of the confusions discussed in these forums come from something in a pop science show or misreported in a newspaper. You should stick as close to the actual physics as remotely possible and not be frightened to leave your audience bewildered if you have to - you have to figure how much it is up to them to try to understand and how much it is up to you to make it understandable to them. The balance depends on the audience.

    You know the audience - maybe they would benefit from a particle approach rather than a classical ray approach? ... see Richard Feynman's descriptions of reflection and refraction.
     
  11. May 31, 2013 #10
    I totally agree. Feynman was amazing at providing intuitive explanations while making it very clear to say what the physics concepts weren't. I want to be able to provide these explanations to other physics students without telling them the wrong thing. The math is definitely covered, but intuition is never touched in the documentation. It's taken me discussions such as this to begin to understand what is happening. I only understand this mathematically, which was a dangerous place for me to be in.
     
  12. May 31, 2013 #11

    Simon Bridge

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    If you google for "university of auckland feynman lectures" you'll find a set on QED all in one place, which show how to cover the kinds of things you want in the particle model. You don't have to go as far as he does to use the approach.

    It's tricky to understand the math until you've described it in words because words is what you are used to - math is a language, and, while you are learning it, you have to translate what it says into some other language you already know... people like Feynman see the diagrams he draws as the same thing as writing out an integral as algebraic symbols - it becomes as easy to write an integral as to pass the time of day - they have become fluent in math.

    But it makes it frustrating when talking to normal people who still view math as a kind of conjuring trick.

    One of the best ways to gain the ability to translate for others is to answer questions in PF ;)
     
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