# Coulomb gauge in special relativity (for QFT)

I don't totally understand the procedure of Coulomb Gauge that we do in special relativity.

Here is what I understood.

We have: $$F_{\mu \nu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$$

It is a quantity invariant under a gauge transformation:

$$A_\mu \rightarrow A_\mu + \partial_\mu \Lambda.$$

Now, we know that we can fix $A$ to respect the Coulomb gauge. It means that if we say $div(A)=0$ the equations of the fields will not change ($B=curl(A)$ will stay the same for example. Same for the electric field).

Thus we can impose in any choice of $A$ (as they can differ from $\partial_\mu \Lambda$) : $div(A)=0$ : the physics will not change by doing it.

First as $A_i \rightarrow A_i + \partial_i \Lambda$ we have the gauge that must satisfy : $\Delta \Lambda=0$

I use the maxwell equations:

$\partial^i F_{i0}=\partial^i(\partial_i A_0 - \partial_0 A_i)=0$

So : $\Delta A_0=0$.

At the stage, I see that If I use Coulomb gauge, I will have:

$$\Delta \Lambda=0$$ and $$\Delta A_0=0.$$

Now in my course it is written that by a nice choice of $\Lambda$ under Coulomb gauge it is possible to have

$$div(A)=0, A_0=0$$

I am not sure about how to obtain the second condition?

Is it something like:

$A_0'=A_0+\partial_0 \Lambda$.

I choose $\Lambda$ such as $\partial_0 \Lambda = - A_0'$.

Then if I work with $A'$ (and not $A$), I would have : $div(A')=0, A_0'=0$.

But if it is the right thing: how can we know that it is possible to have at the same time $\Delta \Lambda=0$ and $\partial_0 \Lambda=-A_0'$.

The result is a consequence of a theorem whose name I cannot remember at the moment. The statement of the theorem goes as follows.

If $\Delta f (x) = 0$ and $f(x)$ is a function that goes to a constant $f_0$ everywhere on the boundary $B$ of a volume $V$ then $f(x) = f_0$ everywhere inside $V$. (Note that this theorem only works in Euclidean spaces since it requires that $\Delta$ is a positive-definite operator. It can therefore not be applied to wave equations of the type $\Box \phi = - \partial_t^2 \phi + \Delta \phi=0$)

In gauge theories it is assumed that all the fields vanish at spatial infinity $x \to \infty$. Then we have the case that $\Delta A_0 = 0$ and $A_0$ goes to a constant $A_0 = 0$ everywhere on the boundary $B$ (is infinity) of space. Thus, according to the theorem, we must have $A_0 = 0$ everywhere.

• "The result is a consequence of a theorem whose name I cannot remember at the moment" - Earnshaw's Theorem? – Alfred Centauri Oct 4 '17 at 17:38

The question is somewhat unclear, but I will try to answer it based on my understanding. By Coulomb gauge, we mean that we set,

$$\nabla \cdot \vec A = 0.$$

What we mean by this, and the reason it is a valid gauge condition, is that given any field configuration $A^\mu$ which solves the equations of motion, we can always use a gauge transformation to force it to satisfy this constraint, and thus we take it a priori.

In this case, if I have $\partial_i A^i = f(x)$, then I can make the change, $A_\mu \to A'_\mu =A_\mu + \partial_\mu \phi$ where we choose $\phi$ such that,

$$\partial_i A'^i = \partial_i A^i + \partial_i \partial^i\phi = 0$$

or more clearly,

$$\partial_i\partial^i\phi = -f(x) \implies \nabla^2 \phi = -f(x).$$

The last equation is the Laplace equation which we know admits a solution, thus justifying our choice of gauge, since we can always find a gauge transformation for each solution to render it in a form which satisfies it.