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Time averaged poynting vector

  1. Apr 26, 2010 #1
    Hello.

    I am reading in my book about the Poynting vector for monochromatic plane waves. It includes a cosine term: cos^2 (kz - omega t + phi). My book states that the time average of this term is 1/2. Can anyone explain this? I don't understand how they work that out.

    Thank you
     
  2. jcsd
  3. Apr 26, 2010 #2

    tiny-tim

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    Hello ronaldoshaky! :smile:

    (have an omega: ω and a phi: φ and try using the X2 tag just above the Reply box :wink:)

    Use one of the standard trigonometric identities …

    cos2x = 1/2 (1 + cos2x) :wink:
     
  4. Apr 27, 2010 #3
    Hi tiny-tim.

    Does finding the time average have something to do with integrating the cos^2 term?

    Thanks again
     
  5. Apr 27, 2010 #4

    gabbagabbahey

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    The time average of any periodic function, [itex]f(t)[/itex], with period [itex]T[/itex] is given by

    [tex]\langle f\rangle_t=\frac{\int_{t_0}^{t_0+T}f(t)dt}{\int_{t_0}^{t_0+T}dt}=\frac{1}{T}\int_{t_0}^{t_0+T}f(t)dt[/tex]

    Apply that to your [itex]\cos^2[/itex] term
     
  6. Apr 27, 2010 #5

    tiny-tim

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    Hi ronaldoshaky! :smile:
    "integrating" is a very technical word to use …

    can't you tell the average of cosx (or of cos2x = (1 + cos2x)/2) just by looking at the graph?! :smile:
     
  7. Apr 27, 2010 #6
    Thanks to all who replied. I will do both the graph and the integration. This has helped me a lot!
     
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