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Complex representation of electric field

  1. Feb 23, 2012 #1
    Confused.. please help!

    Often when an electric field varies sinusoidally with time, it is represented as a complex number. Say,
    [itex]\vec{E}(t)=A\cos(t) \cdot \hat{k}[/itex]
    We know at any time, the magnitude of E is [itex]A\cos(t)[/itex].

    Alternatively the same vector E is understood to be the real part of the complex number
    [itex]\vec{E}(t)=A e^{i t}\cdot\hat{k}[/itex] (right?)

    But since we are now dealing with a complex vector, its magnitude is given by:
    [itex]||\vec{E}||=\sqrt{Ae^{it}\cdot Ae^{-it}}=A[/itex]

    What happened? How can we ensure the magnitude of our vector is preserved when we represent it in its complex form?
  2. jcsd
  3. Feb 23, 2012 #2


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    Of course you have to take the real part first and then square the electric field. Then you'll get the same magnitude as it must be. Just use

    [tex]\exp(-\mathrm{i} \omega t)=\cos(\omega t)-\mathrm{i} \sin(\omega t).[/tex]

    If [itex]\omega \in \mathbb{R}[/itex], which is the case for em. waves in free space, the real part simply [itex]\cos(\omega t)[/itex].
  4. Feb 23, 2012 #3
    I'm sorry vanhees, I'm not sure if I completely understand
    Even if we use [itex]e^{it}=\cos(t)+i\sin(t)[/itex]
    [itex]\sqrt{Ae^{it}\cdot Ae^{-it}}=\sqrt{A^2(\cos(t)+i\sin(t))(\cos(t)-i\sin(t))}=A\sqrt{\cos^2(t)+\sin^2(t)}=A


    Do you mean
  5. Feb 24, 2012 #4
    Can anyone help me out on this?
  6. Feb 24, 2012 #5


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    Again: If you want to calculate quantities, which do not depend linearly on the fields you have to take first the real part and then go on.

    Take your example (corrected for obvious errors concerning the argument of the exponential function) of a plane electromagnetic wave in free space,

    [tex]\vec{E}'(t,\vec{x})=\vec{E}_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x})[/tex]

    with [itex]\omega=c |\vec{k}|[/itex].

    Of course the physical electromagnetic field here is

    [tex]\vec{E}=\mathrm{Re} \vec{E}'=\mathrm{Re} \vec{E}_0 \cos (\omega t-\vec{k} \cdot \vec{x})+\mathrm{Im} \vec{E}_0 \sin(\omega t-\vec{k} \cdot \vec{x}).[/tex]

    Now you can use this expression to evaluate, e.g., the electric part of the energy density, i.e., [itex]\vec{E}^2/2[/itex], etc.
  7. Mar 3, 2012 #6
    Hi vanhees

    I keep reading your reply over and over again hoping to understand this problem better, but my uneducated mind is just not getting it.

    Is it true that
  8. Mar 5, 2012 #7
    Apteronotus, you are confusing two very different things: the magnitude of a complex number (which is the length of the number's vector in the complex plane), and the magnitude of a vector (which is the length of the actual vector in real space).

    The electric field is both a complex number and a vector, so you have to specify which magnitude you mean. The complex-number magnitude of the electric field is the peak component strength, whereas the vector-length magnitude of the electric field is the instantaneous total strength. If you take both the complex-number magnitude and the vector-length magnitude of the electric field, you end up with the peak total strength. For instance:

    If E = (Ex [itex]\hat{\textbf{x}}[/itex] + Ey [itex]\hat{\textbf{y}}[/itex])eikz-iωt


    |E|complex-number = Ex [itex]\hat{\textbf{x}}[/itex] + Ey [itex]\hat{\textbf{y}}[/itex]
    |E|vector-length = [itex]\sqrt{(E_x^2 + E_y^2)cos^2(kz-\omega t)}[/itex]
    |E|both = [itex]\sqrt{E_x^2 + E_y^2}[/itex]
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