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Fourier components of vector potential

  1. Jun 18, 2011 #1
    Hi all,

    I was reading the http://en.wikipedia.org/wiki/Quanti...d#Electromagnetic_field_and_vector_potential" and I am a little bit confused.

    In this equation defining the vector potential

    [itex]\mathbf{A}(\mathbf{r}, t) = \sum_\mathbf{k}\sum_{\mu=-1,1} \left( \mathbf{e}^{(\mu)}(\mathbf{k}) a^{(\mu)}_\mathbf{k}(t) \, e^{i\mathbf{k}\cdot\mathbf{r}} + \bar{\mathbf{e}}^{(\mu)}(\mathbf{k}) \bar{a}^{(\mu)}_\mathbf{k}(t) \, e^{-i\mathbf{k}\cdot\mathbf{r}} \right)[/itex]

    are the [itex]a^{(\mu)}_\mathbf{k}(t)[/itex] and [itex]\bar{a}^{(\mu)}_\mathbf{k}(t)[/itex] complex or real?

    On my first reading I assumed complex as the text says the bar indicates complex conjugates but the vector potential is later defined as a linear addition with complex [itex]\mathbf{e}^{(1)}(\mathbf{k})[/itex] and [itex]\mathbf{e}^{(-1)}(\mathbf{k})[/itex] basis vectors so why would these vectors have complex multipliers? They are also not in bold font which would indicate scalers.

    Is the bar in this case simply a label for the scaler?

    Please help!
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Jun 18, 2011 #2

    atyy

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    Fourier coefficients of a function are generally complex, because they are the Fourier transform of the function.
     
  4. Jun 18, 2011 #3
    Thanks for the reply, I think the fourier coefficients will be complex regardless as the [itex]\mathbf{e}^{(1)}(\mathbf{k})[/itex] and [itex]\mathbf{e}^{(-1)}(\mathbf{k})[/itex] vectors are complex therefore [itex]\mathbf{e}^{(1)}(\mathbf{k}) a^{(1)}_\mathbf{k}(t)[/itex] should be complex regardless of whether [itex]a^{(1)}_\mathbf{k}(t)[/itex] is.. I think.
     
    Last edited: Jun 18, 2011
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