When finding a potential vector for the $\vec{B}$ field I understand that we have certain freedom because if $\nabla \times \vec{A}=\vec{B}$ then $\vec{A'} = \vec{A} + \nabla \psi$ also satisfies $\nabla \times \vec{A'}=\vec{B}$

What I don't understand is why that gives us the freedom to choose $\nabla \cdot \vec{A}=0$, when you can only choose any scalar function $\psi$.

I thought that maybe it has something to do with the Helmholtz theorem but I got nowhere.

Thank you, in advance

  • $\begingroup$ Just a vocabulary remark, the correct term for "divergenless" is "solenoidal". Also, related: physics.stackexchange.com/questions/59315/… $\endgroup$ Aug 20, 2016 at 14:57
  • $\begingroup$ Thank you for the correction. Unfortunately I'm not familiar with the notation in that question since I'm just beginning the course on time-dependent electric and magnetic fields. $\endgroup$
    – gabyarg25
    Aug 20, 2016 at 15:08
  • $\begingroup$ "Divergence-free" is ok. I think divergenceless is ok, too, but it is a strange word. In decades as a physicist I've never heard "solenoidal" used in this context, although I'm sure it's correct. $\endgroup$
    – garyp
    Aug 20, 2016 at 15:23
  • $\begingroup$ I'm not a native English speaker so this happens to me often. Thank you both for your input. $\endgroup$
    – gabyarg25
    Aug 20, 2016 at 16:48

1 Answer 1


The proof is like this. Suppose you have some vector potential $\mathbf{A}$, not necessarily satisfying your gauge condition. Now choose some $\psi$ such that

$$\nabla^2 \psi = - \nabla \cdot \mathbf{A}$$

This is Poisson's equation for $\psi$, and it always has a solution (which is unique if you specify boundary conditions). Now, if we define

$$\mathbf{A}' = \mathbf{A} + \nabla\psi$$

Then it holds that

$$\nabla\cdot\mathbf{A}' = 0$$

i.e., we've found an equivalent vector potential that satisfies the gauge condition.


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