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Coulomb gauge

  1. Dec 15, 2013 #1
    1. The problem statement, all variables and given/known data

    We study the free electromagnetic field in a charge and current free cubic box with with edge length L and volume V. The vector potential in such a system is given via Fourier series:


    2. Relevant equations

    [itex]\vec{A}(\vec{r}, t) = \sum\limits_{k} \vec{A}_k(t) e^{i \vec{k} \vec{r}}[/itex]

    With: [itex]\vec{k} = 2 \pi \begin{pmatrix} \frac{n_x}{L} \\ \frac{n_y}{L} \\ \frac{n_z}{L} \end{pmatrix} [/itex]

    Question:

    Which 2 conditions must [itex]\vec{A}_k(t)[/itex] satisfy so that the Coulomb gauge applies?

    3. The attempt at a solution

    Coulomb gauge means: [itex]\nabla \cdot \vec{A}(\vec{r}, t) = 0[/itex]

    If I didn't miscalculate, [itex]\nabla \cdot \vec{A}(\vec{r}, t) = \sum\limits_{k} \vec{A}_k(t) \vec{k} e^{i \vec{k} \vec{r}}[/itex]

    That would mean that the sum of the Fourier coefficients [itex]\vec{A}_1(t) + \vec{A}_2(t) + \vec{A}_3(t) + ...[/itex] must be 0

    That would be 1 condition (if I did it correctly to begin with). But what is the second condition?

    Any help appriciated.
     
  2. jcsd
  3. Dec 18, 2013 #2
    I don't know about two conditions.

    but, transversality (div A=0) says not that the sum of
    all of the Fourier components is 0, but instead that for each k
    [tex] {\bf k} \cdot {\bf A}_{\bf k}(t) = 0.[/tex]

    just brainstorming:
    If your original A was real you also have a reality condition that A=A*,
    that'll give you another condition but has nothing to do with the Coulomb gauge.
     
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