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Transform Maxwell Equations into k-space

  1. Jul 19, 2012 #1
    Dear fellow physicists,

    looking at the derivation for the maxwell equations into k-space, I've stumbled upon something that seems not so logical to me. It is concerning the two parts where they transform [itex]\nabla \times E [/itex] and [itex]\nabla \bullet E [/itex] on page 27 (on the sheets 14).
    http://www.scribd.com/doc/15466480/EMFT-Exercises#page=27 [Broken]

    After integrating by parts they just state that the first term goes to zero. But for negative infinity it appears more to me that this term is actually blowing up. Is there something that I am missing and someone could me to?

    Thank you very much in advance,
    spookyfw
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jul 20, 2012 #2
    The term is zero because they have assumed that E(t,x) -> 0 as x -> +/- infinity.
     
  4. Jul 20, 2012 #3

    vanhees71

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    I don't find the exercise in this manuscript you link, but it's pretty simple. Just write all the fields in terms of their Fourier transform. Using the HEP-communities convention this reads, e.g., for the electric field
    [tex]\vec{E}(t,\vec{x})=\int_{\mathbb{R}^4} \frac{\mathrm{d}^4 k}{(2 \pi)^4} \hat{\vec{E}}(k) \exp(-\mathrm{i} k \cdot x).[/tex]
    I use the west-coast convention, i.e.,
    [tex]k_{\mu} x^{\mu}=:k \cdot x=(k^0 x^0)-(\vec{k} \cdot \vec{x}).[/tex]
    Now consider Gauss's Law as an example. In time-position representation it reads
    [tex]\vec{\nabla} \cdot \vec{E}=\rho[/tex]
    (Heaviside-Lorentz units). Applying the differential operator on the Fourier transform gives
    [tex]\int_{\mathbb{R}^4} \frac{\mathrm{d}^4 k}{(2 \pi)^4} \mathrm{i} \vec{k} \cdot \hat{\vec{E}}(k) \exp(-\mathrm{i} k \cdot x)=\rho(t,\vec{x})=\int_{\mathbb{R}^4} \frac{\mathrm{d}^4 k}{(2 \pi)^4} \hat{\rho}(k) \exp(-\mathrm{i} k \cdot x).[/tex]
    Since the Fourier transformation is invertible this means that
    [tex]\mathrm{i} \vec{k} \cdot \hat{\vec{E}}(k)=\hat{\rho}(k).[/tex]
    In the same way you get all the other Maxwell equations in wave-number representation.
     
  5. Jul 20, 2012 #4
    ah. that makes perfect sense. Thank you very much :).
     
  6. Jul 20, 2012 #5
    It can also be very helpful to keep in mind Euler's identities which make it possible to visualize the waveform in exponential form:

    [itex]e^{jx} = cos \ x + j \ sin \ x[/itex]
    [itex]e^{-jx} = cos \ x - j \ sin \ x[/itex]

    [itex]cos \ x = \frac{e^{jx} + e^{-jx}}{2}[/itex]
    [itex]sin \ x = \frac{e^{jx} - e^{-jx}}{2j}[/itex]
     
    Last edited: Jul 20, 2012
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