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Thermal Emission and Scattering (Mie or Rayleigh)

  1. Jul 27, 2015 #1

    K41

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    From a book intro I've been reading, Scattering, Absorption and Emission of Light by Small Particles by Mishchenko, it states that when incident light hits a particle, that particle may absorb light, scatter light or emit the light.

    Excerpt:
    Can anyone explain to me, on a physical level, the difference between emission (referred in the book as thermal emission if the particle is above absolute zero) and scattering (whether its described by Mie or Rayleigh, I am looking for physical intuition rather than mathematical theory).

    I'm also confused by the scattering process in the sense that many authors describe what is happening by (as above):
    extinction = absorption_s + scattering.

    Is this "absorption_s" term related to the absorption you get during emission like in spontaneous and stimulated emission or is it describing something else? There was one source that said it referred energy being absorbed by the particle but lost during collisions, but this didn't explain how the energy was absorbed.

    Any help much appreciated
     
  2. jcsd
  3. Jul 31, 2015 #2
    No one answered you. Each particle has a number of atoms and a number of electrons. When the wave passes, the electrons are caused to move, initially by the electric field. They take energy from the incoming wave, which is weakened. As they accelerate, they re-radiate some of the energy. If they are constrained to move only in a certain direction, at an angle to the incoming field, the new radiation will have a different polarization. If the electrons collide with atoms when they move, their energy is donated as heat. If, before the wave arrives, the electrons are already moving around due to the temperature, they will radiate EM waves due to their accelerations. This is sometimes observed as thermal noise. I am sure this is very superficial but maybe someone can improve my description.
     
  4. Aug 1, 2015 #3

    K41

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    I've found out, in layman's terns, the difference between emission and scattering is dependent on the duration that the electron is in the "excited state". If the electron instantaneously falls back to a lower state, then that is scattering. Emission would then be described as any prolonged period of time to the lower state. When we refer to these time periods, I believe it takes 1x10^-15s for an electron to absorb the light. Scattering would then be defined as going to a lower state in that same period. Other forms of emission, including luminescence, will take orders of magnitude longer.

    The above mechanisms were separated by duration as defined in a paper by some Russian scientist, can't remember the name.

    For people with further interest, the mechanisms I believe are further separated by the fact that there also exists "vibrational states" amongst the traditional taught views of electric states (e.g. ground state with two electrons etc). More information can be obtained from textbooks in Spectroscopy field.

    For my needs, the above simplification is enough, but I hope people with further interest/ requirements, will have a good place to start now.
     
  5. Aug 2, 2015 #4
    djpailo,

    I think you are a victim of a widespread confusion between continuum and line radiation. Continuum or black-body radiation as it is often called is emitted by solids, whereas line radiation is emitted by vapours and gases. If the small particle is a solid then it will either absorb the photon and emit black-body radiation based on its higher temperature, or the photon will be reflected (which could be considered as scattering.) If the small particle is a molecule then it will either be absorbed and the molecule will become excited, or if the molecule is already excited it will emit an additional photon(stimulated emission) and return to the ground state, or the photon will bounce off the molecule becoming polarised and be scattered, or it will just pass straight through the molecule.

    If the molecule becomes excited then eventually it will emit a photon by spontaneous emission. But it is more likely that it will collide with another unexcited molecule and transfer the energy to it in the form of kinetic energy. All gas molecules buzz about which takes energy. It is collisions with a thermometer which warms the glass and mercury in it causing it to rise. A molecule can also become excited if it collides with a high speed molecule.

    It seems to me that Mishchenko is confusing the behaviour photons with dust and with gas.
     
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