# How do gauge transformation imply gauge conditions?

In classical EM I understand the electric and magnetic fields are invariant under the potential transformations $\varphi\to\varphi - \partial_t\chi$ and $\mathbf{A}\to\mathbf{A} + \nabla\chi$.

From here people often say this gives us a freedom to do something like choose $\nabla\cdot\mathbf{A} = 0$. I don't understand how we go from the above transformations to specifying properties that $\varphi$ and $\mathbf{A}$ satisfy.

If $\boldsymbol A$ does not satisfy $\nabla\cdot \boldsymbol A=0$, then redefine $$\tilde{\boldsymbol A}\equiv\boldsymbol A+\nabla\chi$$ where $\chi$ is any solution of the PDE $$-\nabla^2\chi=\nabla\cdot \boldsymbol A$$
The vector $\tilde{\boldsymbol A}$ satisfies, by construction, $\nabla\cdot \tilde{\boldsymbol A}=0$.
• @0x90 the fact that $\boldsymbol A$ is a well-defined vector potential (with trivial topology) means that it has to be fast decaying at infinity. The boundary conditions are important for uniqueness, but here we only need some solution of the PDE. Any solution works. – AccidentalFourierTransform Mar 2 '17 at 15:48