Coulomb Gauge misunderstanding

Question

If we have $\vec A(\vec r,t)$ and $\phi (\vec r,t)$ and we make the following gauge transformations:

$$\vec A(\vec r,t)'= \vec A(\vec r,t) + \nabla f(\vec r,t)$$

$$\phi(\vec r,t)'=\phi(\vec r,t) - \frac{\partial f(\vec r,t)}{\partial t}.$$

Coulomb Gauge: When we say this condition must be fulfilled: $\nabla \vec A(\vec r,t)=0$ does this also imply:

$$\nabla \vec A(\vec r,t)'=0$$

In our notes we wrote:

$$\nabla \vec A(\vec r,t)'=\nabla [\vec A(\vec r,t) + \nabla f(\vec r,t)]=0$$

$$\nabla \vec A(\vec r,t)'=\nabla\vec A(\vec r,t) + \Delta f(\vec r,t)=0$$

Assuming $\nabla \vec A(\vec r,t)=g(\vec r,t)$

$$\nabla \vec A(\vec r,t)'= g(\vec r,t)+ \Delta f(\vec r,t)=0 \rightarrow \Delta f(\vec r,t)=-g(\vec r,t) $$

But I also have seen this:

$$\nabla \vec A(\vec r,t)'=\nabla\vec A(\vec r,t) + \Delta f(\vec r,t)$$

Intital condition: $$\nabla\vec A(\vec r,t)=0$$

Then $$\Delta f(\vec r,t)=0$$

So I am really confused. When we consider the following condition $\nabla \vec A=0$ do we assume that the initial NOT TRANSFORMED vector potential must satisfy this condition or the transformed vector potential $\vec A'$ ?

EDIT:

Or we could also have:

$$\nabla \vec A(\vec r,t)'=h(\vec r,t)$$

And in the end you would get for $\Delta f(\vec r,t)$:

1. $\Delta f(\vec r,t)= h(\vec r,t)$ when you assume that $\nabla \vec A =0$
2. $\Delta f(\vec r,t)= h(\vec r,t) - g(\vec r,t)$ when you assume that $\nabla \vec A = g(\vec r,t)$.

As I mentioned earlier, I am thoroughly confused about what is going on?

Answer 1

First of all I suggest to correct your notation, you used the same symbol for the gradient and the divergence (omitting the dot). In the following I'll assume you are familiar with these concepts.

Gauge freedom is a property of electrodynamics equations that lets us impose additional conditions on the potentials (this is actually true for any generalised potential). The reason why this arises is that, at least in a classical formulation, physics is described by the fields i.e. $\boldsymbol{E}$ and $\boldsymbol{B}$. Any scalar and vector potential that give the correct field can be chosen, after all the definition of the potentials involves derivatives so we can already expect some degree of freedom in the choice. As you wrote, it can be proved that any gauge transformation $$\phi\longrightarrow\phi':=\phi-\partial_tf\\ \boldsymbol{A}\longrightarrow\boldsymbol{A}':=\boldsymbol{A}+\nabla f \tag{1}$$ leaves the fields unchanged. In general our potentials won't satisfy Coulomb gauge condition but we can find an appropriate gauge transformation i.e. and appropriate function $f$ such that the new potentials satisfy our condition. So, what we want to impose is that $\boldsymbol{A}'$ satisfies $$\nabla\cdot\boldsymbol{A}'=0\tag{2}$$ and using the definition of $\boldsymbol{A}'$ in $(1)$, w efind the function $f$ w looking for: $$\nabla\cdot\boldsymbol{A}+\Delta f=0\iff\Delta f=-\nabla\cdot\boldsymbol{A} \tag{3}$$ This is Poisson equation. If the divergence of $A$ goes to zero at infinity, $(3)$ admits solutions of the kind: $$f=\frac{1}{4\pi}\int_{\mathbb{R}^3}\frac{\nabla\cdot\boldsymbol{A}}{|r-r'|}d^3\boldsymbol{r}'\tag{4}$$ So, given a vector potential $\boldsymbol{A}$ and a scalar potential $\phi$, you compute $(4)$ and then take as your new potentials $\boldsymbol{A}'$ and $\phi$ as defined in $(1)$.

Answer 2

Since we are not doing spacetime translation I will write $\vec{A}(r,t),\vec{A}'(r,t),\vec{A}''(r,t) $ as $\vec{A},\vec{A}',\vec{A}''$

Now lets do Coulomb gauge on $\vec{A}\rightarrow\vec{A}'=\vec{A}+\nabla f$. I'll find a $f$ such that $$\nabla\vec{A}'=0\implies\nabla\vec{A}+\Delta f=0$$ It's existence and uniqueness can be argued. This is called a Coulomb gauge. Now we do another gauge transformation and we will make sure the new transformed vector potential satisfies $\nabla\vec{A}''=0$; that's the definition of Coulomb gauge.

If we do gauge transformation equation in Coulomb gauge and make sure we remain in this gauge we'll get $$\vec{A}'\rightarrow\vec{A}''=\vec{A}'+\nabla g$$ for Coulomb gauge we want $$\nabla\vec{A}''=0\implies\nabla\vec{A}'+\Delta g=0$$ but $\nabla\vec{A}'=0$ therefore $$\boxed{\Delta g=0}$$ Once we are in Coulomb gauge we can only do restricted gauge transformation, given by solution of boxed equation, if you want to preserve your Coulomb gauge.