# Why can we always find $\vec A$ such that it satisfies Coulomb (or Lorenz) gauge and Maxwell's equations?

I have a short question about the Coulomb potential.

Let $$\vec{E}$$ and $$\vec{B}$$ be the electric field and magnetic field respectively. The electric field and magnetic fields are described by the scalar potential $$V$$ and vector potential $$\vec{A}$$ respectively. As we know, there are a lot of scalar potentials and vector potentials describing the same electric field and magnetic field. My question is, can we $$\textbf{always}$$ find, in this case, the vector potential $$\vec{A}$$ that describes the given field $$\vec{B}$$ and also satisfy $$\nabla \cdot \vec{A} = 0~?$$

And is this also the case for Lorenz gauge, $$\nabla \cdot \vec{A} = -\mu_{0}\varepsilon_{0}\frac{\partial V}{\partial t}~?$$

• That's not a question about the Coulomb potential. More about the Coulomb gauge. Feb 21 at 18:05

The short answer is yes. The general proof of this does not make any assumptions about the fields and thus is valid in general. In particular, if $$\vec{A}_1$$ is a vector potential for $$\vec{B}$$, i.e. a potential such that $$\nabla \times \vec{A}_1 = \vec{B}$$, then $$\vec{A}_2 = \vec{A}_1 + \nabla f$$ describes the same magnetic field $$\vec{B}$$ since $$\nabla \times \nabla f = \vec{0}$$ for any function $$f$$. However,$$\nabla \cdot \vec{A}_2 = \nabla\cdot \vec{A}_1 + \nabla^2 f,$$ and thus by appropriately choosing the function $$f$$ we can set the divergence of $$\vec{A}_2$$ to anything we may want!
• this argument does not prove that the boundary and continuity conditions imposed on the $E,B,H,D$ field intensities can be satisfied in an arbitrary gauge. Feb 21 at 12:46