Gauge fixing and transformation

Electric and magnetic fields are related to potentials this way

$$vec{B}=vec{nabla}timesvec{A} tag{1}$$ $$vec{E}=-vec{nabla}phi-frac{partial vec{A}}{partial t}. tag{2}$$

A gauge transformation for those potentials is

$$phi´=phi-frac{partial f(vec{x},t)}{partial t} tag{3}$$ $$vec{A}´=vec{A}+vec{nabla}f(vec{x},t), tag{4}$$

which leaves the fields $(1)$ and $(2)$ invariant.

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Suppose that we have the potentials $(phi,vec{A})$, and we perform a gauge transformation $(phi,vec{A})rightarrow(phi´,vec{A}´)$ so that the new potentials satisfy

$$vec{nabla}cdotvec{A}´=g(vec{x},t).tag{5}$$

Is this posible? That is, is there any $f(vec{x},t)$ connecting $(phi,vec{A})$ and $(phi´,vec{A}´)$ through $(3)$ and $(4)$ that makes $vec{A}´$ satisfy $(5)$?.

Taking the divergence of $(4)$ gives us

$$g(vec{x},t)=vec{nabla}cdotvec{A}+nabla ^2f(vec{x},t),$$ which is Poisson's equation for $f(vec{x},t)$, and the solution (appart from solutions to the homogeneous equation, i.e. Laplace's equation) is

$$f(vec{x},t)=-int d^3vec{x}'frac{g(vec{x}')-vec{nabla}'cdotvec{A}}{4pi|vec{x}-vec{x}'|},tag{6}$$

where $vec{nabla}'$ indicates derivatives with respect to $x', y', z'$.

So our new vector potential $vec{A}´$ will satisfy $(5)$ and will describe the same fields that our old potential did.

Edit:

Note that $phi$ must also be transformed via $(3)$ with the same $f$ in order to keep the fields invariant.