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Is there something as instanteneous intensity of a wave?

  1. Mar 13, 2015 #1
    At the moment I'm revising some interference and diffraction basics, and there is something that bothers me slightly and I can't quite figure it out.

    The intensity of a wave over some area ##dA## is in general is ##I=\frac{1}{dA} \frac{dE}{dt}##. Clearly for an electromagnetic wave falling on a surface, the part ##\frac{dE}{dt}## is not constant and depends on time. So intensity should be a function of time.

    In every text I encounter they seem to DEFINE the intensity as being the average over 1 period of ##c|A(t)|^{2}## where ##A## is the deviation of the wave at the area of interest. Are they simply using more practical definitions, and technically I'm correct above in a general sense OR am I missing something?
     
  2. jcsd
  3. Mar 14, 2015 #2

    andrewkirk

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    The instantaneous rate of energy transport is going to be a function of the instantaneous electric and magnetic fields.

    Here's a link with more:
    http://hyperphysics.phy-astr.gsu.edu/hbase/waves/emwv.html

    It gives the energy transport vector as

    [itex]\vec{S}=\frac{1}{\mu_0}\vec{E}\times\vec{B}[/itex] where [itex]\vec{E},\ \vec{B}[/itex] are the electric and magnetic field vectors. Note that the direction of energy transport, being a cross product, is perpendicular to both, as one would expect.

    By the way, that's the formula at a point. To get the energy transfer rate over an area, you'd need to integrate that formula over the area.
     
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