Gauge fixing in the classical $U(1)$ gauge theory

Question

My question concerns the gauge fixing in classical v.s. quantum $U(1)$ gauge theory. I will ask about the gauging fixing in quantum $U(1)$ gauge theory in a separated Phys-SE post.
For the classical $U(1)$ gauge theory,
we have the electric $vec E$ and magnetic $vec B$ written as the scalar $phi $ and vector $vec  A$ potentials:
$$
vec E = - vec nabla phi -frac{partial}{partial t} vec  A.
$$
$$
vec B = vec nabla  times vec  A.
$$
My understanding about the gauge fixing for this gauge theory is that we can chose for example, $(phi,  vec  A)$ to give a set of $vec E$ and $vec B$ fields.

But we can also shift to
$$(phi,  vec  A) mapsto (phi + C_0,  vec  A + vec  C)$$
where arbitrary choices of $(C_0, vec  C)$ still give the same solutions of  $vec E$ and $vec B$. So a certain but arbitrary choice of $(C_0, vec  C)$ can be regarded as a way of gauge fixing? correct?

Furthermore, we can also shift to
$$(phi,  vec  A) mapsto (phi + phi_0,  vec  A + vec  A')$$
such that the followings are satisfied:
$$
- vec nabla  phi_0 -frac{partial}{partial t} vec  A'=0
$$
$$
vec nabla  times vec   A'=0 
$$
Then we have a choice of $(phi,  vec  A) mapsto (phi + phi_0,  vec  A + vec  A')$ such that any choice is a way of gauge fixing? correct?

In a classical differential equation, we have
$$
* d * F = J
$$
$$
dF=0
$$
so $F=dA$ with $* d * dA = J$ is a solution. Say $F=dA'$ with $* d * dA' = J$ is also a solution. And the gauge fixing implies a different solutions of $F=dA$ and $F=dA'$ where $A$ and $A'$ are both valid solutions. What are the differential equation constraints then?
Is my understanding complete to include ONLY:
$$
* d * d (A-A')=0
$$
or do we need more constraints to do gauge fixing?

Am I correct to say that $A$ and $A'$ are in the same gauge profile thus should be regarded as the same gauge equivalent classes. Choose $A$ or choose $A'$ is simply a choice of gauge fixing?

Aron Beekman · Answer

A gauge transformation is any transformation on  $A = (phi, vec{A})$ that does not change any physical observable, c.q. $F$. So any transformation $A to A + d lambda$ with $lambda$ any scalar field.
A gauge fix is, within this gauge freedom, a particular choice for $A$. This choice can be either complete ($A$ completely fixed, no further gauge transformations allowed) or not (there is still some leftover freedom).

A gauge fix is usually applied to facilitate certain computations (you have to then make sure the end result is nonetheless gauge-invariant). Therefore, one is not actually interested in the exact particular form of $A$, but rather in certain gauge-fixing conditions that $A$ satisfies. Given any gauge potential $A$, one can apply a gauge transformation $A to A_g =  A + d lambda$ so that $A_g$ satisfies the gauge-fixing condition.
Some examples are

In the Lorenz gauge fix, the condition $d star A_g = (1/c^2)partial_t phi_g + nabla cdot vec{A}_g = 0$ is satisfied.
In the Coulomb gauge fix, the condition $nabla cdot vec{A}_g = 0$ is satisfied.

The Lorenz gauge fix is not complete, because instead of using $lambda$ to arrive at the gauge fix, we could have used any $lambda + psi$ as long as $psi$ satisfies $Box psi= 0$. In other words, instead of $A_g$ we can have $A_g + d psi$ and still satisfy the gauge-fixing condition. On the other hand, the Coulomb gauge fix is complete.
In your list, the transformation $C$ in item 1. is not entirely free to choose. Instead, it must be of the form $C = d lambda$ for any smooth scalar field $lambda$. Then, items 1. and 2. are actually the same. I would use a different wording than what you have written down: $C$ is a gauge transformation. A particular transformation $C$ to arrive at $A$ satisfying a gauge-fixing condition can be called a gauge-fix.
I don't understand your question item 3. If $F = d A$ and $F = d A'$ then $A$ and $A'$ differ by some gauge transformation. A gauge fix is some condition on $A$, but it cannot affect any physical observable, in particular not $F$. So $d (A - A') = 0$ is always satisfied. Moreover, physical observables like $F$ and $J$ can tell you nothing about the gauge fix. Instead, a gauge-fixing condition (which is a choice) can be for  instance $d star A = 0$ (Lorenz gauge).

Gauge fixing in the classical $U(1)$ gauge theory

For the classical $U(1)$ gauge theory,

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