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Polarization of plane wave

  1. Oct 24, 2013 #1
    Hello everybody :)
    I'm doing some electromagnetic exercises, but I got stuck in calculating the polarization of a plane wave.
    The complex field associated to the wave is the following
    [itex]\vec{E} = (\sqrt{2}\hat{x}+\hat{y}-\hat{z})e^{-2\pi 10^6(y+z)} = \hat{p}e^{-2\pi 10^6(y+z)}[/itex]
    It is easy to calculate that the direction of propagation is
    [itex]\hat{k}= \frac{1}{\sqrt{2}}(\hat{y}+\hat{z})[/itex]
    but now I want to calculate if the wave is polarized and how.

    If the propagation direction was along one of the axis was pretty easy, I just need to look at the components in front of the exponential. But now that I can't how can I do it?

    My idea was to make a change of the basis, from [itex]\{\hat{x},\hat{y},\hat{z}\}[/itex] to [itex]\{\hat{u},\hat{v},\hat{w}\}[/itex] where
    [itex]\hat{u} = \frac{1}{\sqrt{2}}(\hat{x}+\hat{z}) ,\quad \hat{v} = \frac{1}{\sqrt{2}}(\hat{x}+\hat{y}) ,\quad \hat{w} = \frac{1}{\sqrt{2}}(\hat{y}+\hat{z})[/itex]
    Then rewrite [itex]\hat{p}[/itex] in this base, which became (I don't consider the square root factor, it doesn't change the result since I'm just looking how the components are related to each other)
    [itex]\sqrt{2}\hat{x}+\hat{y}-\hat{z} = a\hat{u} + b\hat{v} + c\hat{w}= a (\hat{x}+\hat{z}) + b(\hat{x}+\hat{y}) + c(\hat{y}+\hat{z})[/itex]

    Which gives me values for all a,b and c. And that's strange (c should be zero since is a plane wave so there shouldn't be anything along this direction).

    So... What's my mistake? :)
     
  2. jcsd
  3. Oct 25, 2013 #2

    Claude Bile

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    Science Advisor

    The wave is polarized - I don't understand what you mean by "how" it is polarized: the polarization direction is quite clear from the vector that precedes the exponent.

    The polarization is indeed orthogonal to the propagation direction as required (easily checked by taking the dot-product). If you want to rotate p so that the propagation direction lies along a principal axis, you can always employ a coordinate rotation (which is more intuitive than employing a transformation to a non-orthogonal basis set IMO).

    Claude.
     
  4. Oct 25, 2013 #3
    [itex]\{\hat{u},\hat{v},\hat{w}\}[/itex] are not perpendicular to each other.
     
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