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Electromagnetic plane waves in vacuo

  1. Apr 15, 2009 #1
    Consider a plane wave f = cos(kz - wt). Applying Maxwell's equation (divE=0 in vacuo) gives
    kcos(kz - wt) = 0 which means that k = 0. This surely doesn't make sense?
  2. jcsd
  3. Apr 15, 2009 #2
    Well, mathematically,

    [tex] \bigtriangledown \cdot E_0 \cos(kz - wt) = E_0 \sin(kz - wt) = 0 [/tex]

    [tex] \Longrightarrow k = 0[/tex] or [tex]sin(kz - wt) = 0[/tex] [tex]\forall{ z, t }[/tex]

    EDIT: But yes, addressed below is the conceptual issue here...
    Last edited: Apr 15, 2009
  4. Apr 15, 2009 #3


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    Well the reason you get that is that EM waves are transverse waves, and you have oversimplified the situation. The more general result is:
    Consider a plane wave
    [tex]\vec{E} = \vec{E_0}cos(\vec{k}\cdot\vec{r} - \omega t})[/tex]
    [tex]\nabla\cdot\vec{E} = \vec{k}\cdot\vec{E} = 0[/tex]

    Which shows that the wave vector is perpendicular to the field.
  5. Apr 15, 2009 #4
    I think I see my mistake. I thought all i was doing was choosing my coordinates such that my z axis coincided with the wave vector k. The way i did it missed out the fact that k and E point in different directions.
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