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Normal modes of electromagnetic field

  1. Oct 15, 2014 #1
    Hey guys,

    I'm trying to understand the properties of normal modes of the electromagnetic field inside an arbitrary cavity, but I'm having some trouble.

    By definition, for a normal mode we have [itex]\mathbf{E}(\mathbf{x},t) = \mathbf{E}_0 (\mathbf{x}) e^{i \omega_1 t}[/itex] and [itex]\mathbf{B}(\mathbf{x},t) = \mathbf{B}_0 (\mathbf{x}) e^{i \omega_2t}[/itex]. It's easy to show that [itex]\omega_1 = \omega_2[/itex]. Substituting these two equations into the wave equation gives the standard equation for the normal modes of a system, namely
    [tex]\nabla^2 \mathbf{E}_0 + \frac{\omega^2}{c^2}\mathbf{E}_0 = 0[/tex]
    and
    [tex]\nabla^2 \mathbf{B}_0 + \frac{\omega^2}{c^2}\mathbf{B}_0 = 0.[/tex]

    For any allowed value of [itex]\omega[/itex], there will be three linearly independent solutions to each of these equations -- unless we have degeneracy, of course. The conditions
    [tex]\nabla \cdot \mathbf{E}_0 = 0[/tex]
    and
    [tex]\nabla \cdot \mathbf{B}_0 = 0[/tex]
    reduce this number to two.

    In the case of a rectangular cavity, it's easy to show that
    1) the electric and magnetic fields are perpendicular; and
    2) the two linearly independent solutions mentioned above correspond to the two possible polarization states.

    My intuition tells me that these properties extend to the case of an arbitrarily shaped cavity, but I can't prove that they do. In particular, I'm not sure how polarization is defined if our electromagnetic field solution doesn't have a well-defined direction of propagation.

    Any ideas?
     
  2. jcsd
  3. Oct 20, 2014 #2
    Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Oct 21, 2014 #3

    clem

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    I think the E and B fields will be orthogonal only if the cavity shape allows separation of coordinates in an orthogonal coordinate system.
    Two independent solutions for E always corresponds to two different 'polarization states', but the polarization may not be as simple as plane or elliptical.
     
  5. Oct 21, 2014 #4

    Andy Resnick

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    2016 Award

    Waveguide modes (TE, TM, and hybrid modes) can violate your result. Also, when you derived the wave equations, you made some assumptions (for example, linear constitutive relations between E and D, B and H; homogeneous materials, dielectric materials...) that can be violated. To be sure, Ampere's and Faraday's law are always valid, but if you don't have plane wave solutions then E and B are not orthogonal to each other, and neither are orthogonal to the propagation direction.
     
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