Coulomb gauge

  • #1

Homework Statement



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:


Homework 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?

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.
 

Answers and Replies

  • #2
185
4
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

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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