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given the gauge choice that div A = some value/function.

i am completely fine that in the context of electromagnetism that by setting the divergence of this to be anything it has no effect on the curl of A e.g the magnetic field. as letting A= A + grad(f) means curl of grad(f) = 0

however in doing so you are changing the electric field given by phi - d(a)/dt

so why can you make this transformation and not change anything?

wikipedia says that by doing a transformation like this you have to do the transformation phi=phi-df/dt to keep the E field the same

But... how does simply setting div a = a value encompass this second transformation

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Electric and magnetic fields are related to potentials this way

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

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.


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

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

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

$$g(\vec{x},t)=\vec{\nabla}\cdot\vec{A}+\nabla ^2f(\vec{x},t),$$ 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

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

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.

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