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, $\nabla\cdot \tilde{\boldsymbol A}=0$.

  • $\begingroup$ How do we know there are solutions to that PDE? $\endgroup$ – Nate Stemen Mar 2 '17 at 15:02
  • $\begingroup$ @NateStemen see en.wikipedia.org/wiki/… and farside.ph.utexas.edu/teaching/em/lectures/node31.html (last equation). $\endgroup$ – AccidentalFourierTransform Mar 2 '17 at 15:03
  • $\begingroup$ @AccidentalFourierTransform, I think that your references discuss the uniqueness of the solution and not its existence. $\endgroup$ – 0x90 Mar 2 '17 at 15:45
  • $\begingroup$ @0x90 the first one discusses uniqueness. The second one gives the explicit solution (for a certain boundary conditions). $\endgroup$ – AccidentalFourierTransform Mar 2 '17 at 15:46
  • 2
    $\begingroup$ @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. $\endgroup$ – AccidentalFourierTransform Mar 2 '17 at 15:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.