# Coulomb gauge

## 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

$\vec{A}(\vec{r}, t) = \sum\limits_{k} \vec{A}_k(t) e^{i \vec{k} \vec{r}}$

With: $\vec{k} = 2 \pi \begin{pmatrix} \frac{n_x}{L} \\ \frac{n_y}{L} \\ \frac{n_z}{L} \end{pmatrix}$

Question:

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

## The Attempt at a Solution

Coulomb gauge means: $\nabla \cdot \vec{A}(\vec{r}, t) = 0$

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

That would mean that the sum of the Fourier coefficients $\vec{A}_1(t) + \vec{A}_2(t) + \vec{A}_3(t) + ...$ 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.

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

That would mean that the sum of the Fourier coefficients $\vec{A}_1(t) + \vec{A}_2(t) + \vec{A}_3(t) + ...$ 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
$${\bf k} \cdot {\bf A}_{\bf k}(t) = 0.$$

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.