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Clarification regarding physical fields from Fourier amp's

  1. Aug 22, 2015 #1
    My professor in Classical Electrodynamics is great and all, but sometimes he has trouble understanding what it is that I don't understand. So here I am.

    Let's say we have the some sort of (monochromatic) radiating system generating a electric field with Fourier amplitude Eω(x) and want to retrieve the physical electric field from this. One then takes the inverse Fourier transform which in this case should look something like E(t,x)=Re{Eω(x)e-iωt'} where t'=t-kr/ω is the retarded time (this is at least how my professor does it, and I think this makes sense).

    But lets say Eω(x) =(constant) * (e^(ikr)/r)e1, then he will say that

    E(t,x)=Re{Eω(x)e-iωt'}=Re{Eω(x)ei(kr-ωt)}=(constant) * cos(kr-ωt).

    Here is where I get confused, what happens to the e^(ikr) in Eω(x)? If this in included, my brain tells me that E(t,x)=(constant) * cos(2kr-ωt), but this is never the case.

    I am probably misunderstanding something and any kind of clarification would be highly appreciated. I am including a picture of an example he gave to further explain my question. In the attached picture, I think the answer should have sin(2kr-ωt) instead of sin(kr-ωt).

    http://i.imgur.com/RJuHwMK.png
     
  2. jcsd
  3. Aug 22, 2015 #2

    blue_leaf77

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    You can't make such assumption since the complex exponential has been included in the retarded time. If you do this, the E field expression won't satisfy the Helmholtz equation.
     
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