Gauging fixing in the classical $U(1)$ gauge theory
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.
1. 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?
2. Furthermore, we can also shift to $$(\phi, \vec A) \mapsto (\phi + \phi_0, \vec A + \vec A')$$ such that the followings are satisfied: $$
3. 298
\vec \nabla \phi_0 -\frac{\partial}{\partial t} \vec A'=0 $$ $$ \vec \nabla \times \vec A'=0 $$ 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?
4. In a classical differential equation, we have $$
5. d * F = J $$ $$ dF=0 $$ 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: $$
6. d * d (A-A')=0 $$ 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? 418