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Conditions on complex plane wave solutions to Maxwell's Equations

  1. Nov 22, 2008 #1
    1. The problem statement, all variables and given/known data

    What conditions need to be imposed on [tex]\vec{E}[/tex]0, [tex]\vec{B}[/tex]0, [tex]\vec{k}[/tex] and ω to ensure the following equations solve Maxwell's equations in a region with permittivity ε and permeability µ, where the charge density and the current density vanish:

    [tex]\vec{E}[/tex] = Re{ [tex]\vec{E}[/tex]0 exp[i([tex]\vec{k}[/tex]⋅[tex]\vec{x}[/tex] - ωt)] }

    [tex]\vec{B}[/tex] = Re{ [tex]\vec{B}[/tex]0 exp[i([tex]\vec{k}[/tex]⋅[tex]\vec{x}[/tex] - ωt)] }

    2. Relevant equations

    Maxwell's Equations and ω=kv

    3. The attempt at a solution

    I know from Gauss' Law for E & B the following is required:

    [tex]\vec{k}[/tex]⋅[tex]\vec{E}[/tex]0 = [tex]\vec{k}[/tex]⋅[tex]\vec{B}[/tex]0 = 0

    and from Faraday's Law:

    [tex]\vec{k}[/tex] x [tex]\vec{E}[/tex]0 = ω[tex]\vec{B}[/tex]0

    Now the Ampere-Maxwell law would suggest the following may be a requirement:

    [tex]\vec{k}[/tex] x [tex]\vec{B}[/tex]0 = -(1/c²)ω[tex]\vec{E}[/tex]0

    But is this final one really required, or does it in fact follow from the three previous requirements? I remember being told that the latter was correct, but I have no idea how to show this last requirement from the previous three.
  2. jcsd
  3. Nov 22, 2008 #2
    You can try choosing your axes cleverly. Let's say you set [tex]\vec{k} = k \hat{x}[/tex], without loss of generality. Now from Gauss' law, pick a choice of directions for either [tex]\vec{E_0}[/tex] or [tex]\vec{B_0}[/tex]. Can you derive the last requirement from the first three?
  4. Nov 22, 2008 #3
    Got it, thanks for your help.
  5. Nov 22, 2008 #4
    That was only one of the ways of doing it. The Ampere-Maxwell relation involves the vector product of k and B. Can you think of some way of evaluating that, based on what you know from the other equations?

    This way, your solution will not depend on your choice of axes, which is always nice.
  6. Nov 22, 2008 #5
    Does it involve using:

    k x B0 = |k||B|sinθ [tex]\hat{n}[/tex] , where n is perpendicular to k & B. But it should be in either the same direction or the negative direction of E0. I could do a similar thing for the previous cross product equation and relate the two?
  7. Nov 22, 2008 #6
    Yes and no. You want to relate the previous equation to this in some way. You actually know what B_0 is from the previous equation. Just try and use it in this equation :smile:
  8. Nov 22, 2008 #7
    I'm a little confused, from the previous equation [tex]\vec{B_0}[/tex] = 1/ω [tex]\vec{k}[/tex] x [tex]\vec{E_0}[/tex].

    But if I sub that in for [tex]\vec{B_0}[/tex] in the final equation I will get 0 on the LHS as [tex]\vec{k}[/tex] x [tex]\vec{k}[/tex] = 0
  9. Nov 22, 2008 #8
    Ah, but you don't have [tex]\vec{k}[/tex] x [tex]\vec{k}[/tex] on the RHS. You have [tex]\vec{k}[/tex] x ([tex]\vec{k}[/tex] x [tex]\vec{E_0}[/tex]). The two are very different. Look up vector triple products (I think it may also be called Lagrange's formula, not sure).
  10. Nov 22, 2008 #9
    I used A x (B x C) = B(A⋅C) - C(A⋅B) and got the result. Thanks for your help, would not have thought of the triple product.
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