Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook