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Showing the components of a plane EM wave are perpendicular

  1. May 23, 2008 #1
    1. The problem statement, all variables and given/known data
    I've derived the EM wave equations from Maxwell's equations, and I now need to show that the E and B components are both perpendicular to each other and to the direction of propagation.

    The text book I've been using attempts to show why this is, but it isn't particularly clear and seems to assume that propagation and one of the components are perpendicular.


    2. Relevant equations
    [tex]\nabla^2E = \mu_0 \epsilon_0 \ddot{E}[/tex]
    [tex]\nabla^2B = \mu_0 \epsilon_0 \ddot{B}[/tex]
     
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  3. May 23, 2008 #2

    Physics Monkey

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    You might start by solving the wave equations for E and B i.e. writing down the most general form for E and B consistent with a wave of definite frequency and wavelength. Try plugging those guesses into the original Maxwell equations and see what you find.
     
  4. May 23, 2008 #3
    That's kind of what I've tried, but I end up assuming that the E component and direction of propagation are perpendicular.

    [tex] \vec{E} = (E_0, 0, 0) \sin(\omega(t + \frac{z}{v}))[/tex]

    [tex] \vec{\nabla} \times \vec{E} = -\dot{\vec{B}} = (0, E_0, 0) \frac{\omega}{v}\cos(\omega(t + \frac{z}{v}))[/tex]

    This shows that E and B are perpendicular, but in doing so I've assumed that E and the direction of the propagation are perpendicular.

    I thought about using the Poynting vector to show that the direction of propagation is perpendicular to E and B, but I wasn't sure as to whether this proved it or not.
     
  5. May 23, 2008 #4

    nicksauce

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    You can use
    [tex] \nabla \cdot \vec{E} = 0 [/tex] (in free space w/ no source) to show that the electric field and the wave vector are perpendicular.
     
  6. May 23, 2008 #5
    Ah, I've got it now (I ended up using the integral form of Gauss's law, but it boils down to the same thing).

    Thanks for your help.
     
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