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Dielectric constants in different directions - does this make sense?

  1. Apr 12, 2012 #1
    Dielectric constants in different directions -- does this make sense?

    So I'm trying to analyze a scenario in which I have a plane wave incident (from, say, some medium with permittivity and permeability [itex]ε_1[/itex] and [itex]μ_1[/itex]), on a plane of a dielectric material that has an anisotropic permittivity: in directions parallel to the plane (of the interface, not the plane of incidence), it is [itex]ε_{para}[/itex] and in the direction perpendicular to the plane, it's [itex]ε_{perp}[/itex].

    So I break the problem down into two parts as it is usually done: When the E component of the wave is perpendicular to the plane of incidence (POI), and when it is parallel. Anything else is just a linear combination of these.

    The boundary conditions at the interface are:

    [itex]\hat{n}\cdot(\vec{D_I} + \vec{D_R} - \vec{D_T}) = 0[/itex] (1)
    [itex]\hat{n}\cdot(\vec{B_I} + \vec{B_R} - \vec{B_T}) = 0[/itex] (2)
    [itex]\hat{n}\times(\vec{E_I} + \vec{E_R} - \vec{E_T}) = 0[/itex] (3)
    [itex]\hat{n}\times(\vec{H_I} + \vec{H_R} - \vec{H_T}) = 0[/itex] (4)

    Where [itex]B = \mu H[/itex] and [itex]D = εE[/itex] (they are vectors) and [itex]\hat{n}[/itex] is a unit vector normal to the interface. The angle of incidence is θ and the angle of transmission is [itex]\phi[/itex]

    B is related to E through [tex]\vec{B} = \frac{\hat{k} \times \vec E}{v}[/tex]

    Where [itex]\hat{k}[/itex] is a unit vector pointing in the direction of the wave's propagation ([itex]\vec k /k[/itex] if you wish, where k is the wave number). v is the velocity of the wave, equal to [tex]v = \frac{1}{\sqrt{εμ}}[/tex]

    So this is all very fine and dandy for the normal scenario. But now I'm running into some problems and I would love to know if I'm on the right track.

    For the E parallel to the POI, in the boundary conditions, (1) seems to be pretty simple: We just use [itex]ε_{perp}[/itex] because that's the direction of the electric field:
    [itex]ε_1 sin(\theta)(E_I + E_R) - ε_{perp}sin(\phi)E_T = 0[/itex]

    (2) tells us nothing because B has no components perpendicular to the interface.

    (3) is just the tangential E equation: [itex]cos(\theta)(E_I - E_R) - cos(\phi)E_T = 0[/itex]

    (4) is the tangential H equation: [itex]\frac{\sqrt{\mu_1\epsilon_1}}{\mu_1}(E_I+E_R) - \frac{\sqrt{\mu_2\epsilon_{para}}}{\mu_2}E_T = 0[/itex]

    So, in Jackson, for the normal simple case (not my anisotropic mess), he points out that of the 4 boundary conditions, two are actually equivalent to each other if you use Snell's law.

    But here, Snell's law doesn't make immediate sense: [itex]n = \sqrt{\frac{\epsilon\mu}{\epsilon_0\mu_0}}[/itex], but in the anisotropic medium, ε depends on the direction, so it seems like n does too.

    So how can I proceed from here? I know from Jackson which two of the boundary condition equations are supposed to be equivalent ((1) and (4)), so I could just used one of them and the other non equivalent one ((3)). But I was trying to prove their equivalence, and besides, I still don't have the angle of transmission.

    Anyone have any ideas??

    Thanks!
     
  2. jcsd
  3. Apr 12, 2012 #2

    DrDu

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    Re: Dielectric constants in different directions -- does this make sense?

    Yes, in an anisotropic medium n is a tensor. Your problem is that of light propagation in an uniaxial crystal. Jackson is not a good starting point. Landau Lifshetz "electrodynamics of continua" discusses this kind of problem extensively.
    Just some comments: In optics, usually all magnetic properties can either be neglected or encoded in the k dependence of the dielectric constant. Hence H=B, what simplified the boundary conditions somewhat.
    In the crystal, you have two rays: The ordinary, in which the polarization is in the plane of the crystal but perpendicular to the plane of incidence. For this ray, epsilon is independent of the angle of incidence and you can proceed as usual.
    The extraordinary ray which is polarized in the plane of incidence. As epsilon is a tensor, not both E and D can be perpendicular to k. Try to calculate first which one of the two is transversal.
     
  4. Apr 12, 2012 #3
    Re: Dielectric constants in different directions -- does this make sense?

    Hey thanks for the response!

    Yeah, Jackson doesn't seem to have anything on it. I'll try searching online.

    Question though, I think I get what you're saying about the ordinary and extraordinary rays (though I hadn't heard those name before!). I did out the extraordinary one above, right?

    I see that in the ordinary the electric field only 'sees' the parallel epsilon so it is independent of incident angle (what you said). But the B field in the boundary conditions is proportional to E by a factor containing epsilon, right?

    E will always be perpendicular to k, right? But D is scaled by epsilon (now a tensor), so it will be deformed...is that correct?

    Thank you!
     
  5. Apr 12, 2012 #4

    DrDu

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  6. Apr 14, 2012 #5
    Re: Dielectric constants in different directions -- does this make sense?

    Oh man thanks so much! Just what I needed.
     
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