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Transverse and longitudinal electric Green function

  1. Jan 7, 2015 #1
    Hello everyone:

    This is what I read in a paper, the spontaneous emission rate written by Fermi's Golden Rule is just related to the local transverse electric field.

    Dose anyone can explain to me what's the meaning of "transverse mode" here? Why the emission is not related to longitudinal electric? Actually, to the best of my knowledge, in most cases people will only use the transverse Green function to analysis the system, such as in the calculation of local density of state.

    However, if I want to calculate the energy flowing, such as poynting vector, shall I use the transverse Green function or the entire Green function?

    Thanks very much
     
  2. jcsd
  3. Jan 8, 2015 #2

    vanhees71

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    Where did you get that from? I'd need more context to answer this question.

    If you deal with the emission of photons into free space, there are of course only transverse modes involved. That changes within a medium. Then you need a "medium modified" photon propagator. If you are interested in the latter, have a look at many-body QFT books like

    J. Kapusta, C. Gale, Finite-Temperature Field Theory, Cambridge University Press
    (use the 2nd edition, because there was a lot of important progress in this field since the 1st edition appeared (hard thermal loops) although the basic facts about your issue are already also in the 1st edition).
     
  4. Jan 8, 2015 #3

    naima

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    I only know that the longitudinal part of an electric field is its "electrostatic" part. it does not propagate at a c speed but is instantaneous.
    The tranverse part is the part that we use in Maxwell equations You vave ##E_{Tr} = E - E_{//}##
     
  5. Jan 8, 2015 #4

    I read this statement in this paper http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.68.3698.

    I just start at this area and have no enough basic knowledge. Could you please explain to me what's the meaning of the "transverse mode" here? For a dipole, does the transverse mode here meaning the mode with electric field parallel to the dipole moment? Does this transverse mode have the same meaning to the one we used in the dipole field? The longitudinal component in dipole field is the Er in spherical coordinate. It seems different, at least when r||u, here u is the dipole moment.

    Thanks for your help
     
  6. Jan 8, 2015 #5
    When the dipole emit EM field, it contains both the transverse and longitudinal component. It is true that at the far-field region, there is only transverse left and we can safely say that only the transverse component transfer energy.

    But how about at the near-field region? Actually, in the near field region the longitudinal component contains more energy than the transverse component. Dose that mean in the near-field region, the longitudinal component can deliver energy?

    To the best of my knowledge, the energy transfer can only be done in the "transverse mode" because the poynting vector is the cross product result of the electric field. So, it seems that if there is energy absorption in the near field region, we can always treat them as "transverse mode" according to the energy flow direction. It's a little bit confusing. Do you have any further comments?

    Thanks for your help
     
  7. Jan 9, 2015 #6

    DrDu

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    It would be much easier to help you if you could provide the reference. How can we know to what kind of transverse mode this paper is referring to?

    Edit: Sorry, I just saw that you provided the reference already in post #4.
     
    Last edited: Jan 9, 2015
  8. Jan 9, 2015 #7

    naima

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    I think that you will find what you want in this text
    skip to "contribution of the longitudinal electric field to the total energy"
     
  9. Jan 9, 2015 #8

    DrDu

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    In isotropic media, as also stated in the paper you cited, the longitudinal and transversal part of the field decouple.
    As far as the Poynting vector is concerned, you can easily convince yourself that the longitudinal field does not lead to a flow of energy as E_long is parallel to k while B is transversal (i.e. perpendicular to k). Hence the contribution of E_long to the Pointing vector S, which is proportional to E_long x B is also perpendicular to k and thus the divergence of S, which gives the change of energy density at that point, depends only on the transversal field.
     
  10. Jan 12, 2015 #9
    Thanks DrDu! This is what I'm thinking about. This consideration is safe and correct in the far field region. However, I still want to confirm that this is also true in the near field region. If we consider the absorption, or dissipation, of electromagnetic energy in the near field, I think we can still define the Poynting vector. The energy conservation will lead the same conclusion as you made in the above. Is this correct?
     
  11. Jan 12, 2015 #10
    Thanks for your book!
     
  12. Jan 13, 2015 #11

    DrDu

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    Maybe I should have made it explicit that I considered the Fourier components of the electric and magnetic fields ##E(k,\omega)## and ##B(k,\omega)##. Hence there is no splitting of the field into near field and far field involved.
    The pointing vector has two parts, a longitudinal one whose divergence yields the change of energy density with time and a transversal one which may be important for the calculation of angular momentum. The longitudinal electric field (from div B=0 it is clear that the magnetic field never has a longitudinal component) contributes only to the transversal part of S. Hence it plays no role in the calculation of the energy emitted by an atom or molecule.
     
  13. Jan 13, 2015 #12

    naima

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    You can read in the book
    We have shown that the ##H_I^l## (contribution of the longitudinal photons to the interaction hamiltonian) can be ignored in calculating the amplitudes between two physical states.page 431
     
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