2
$\begingroup$

When are equivalent the Maxwell's harmonic equations: $$ \nabla\times\left(\nabla\times\mathbf{E}\right)=\mu\epsilon\omega^2\mathbf{E} $$ and the vector Helmholtz equations: $$ \nabla^2\mathbf{E}=\mu\epsilon\omega^2\mathbf{E} $$ and when aren't they?

I understand that $\nabla\times\left(\nabla\times\mathbf{u}\right)=\nabla\left(\nabla\cdot\mathbf{u}\right)-\nabla^2\mathbf{u}$ for any $\mathbf{u}$ smooth enough, and I know that $\nabla\cdot\mathbf{u}$ can be zero by Gauss's law or by taking the divergence of the first equation. So... are they only equivalent when there is no free charge density?

$\endgroup$
  • 4
    $\begingroup$ They are also equivalent when the free charge density is constant (its gradient is zero), but yes, you answered your own question. $\endgroup$ – Bob Knighton Jan 15 at 9:34
  • $\begingroup$ my question was more in the way of "are they only equivalent in that case? (constant free charge density)" $\endgroup$ – Manuel Pena Jan 15 at 9:39
  • 1
    $\begingroup$ The answer is still yes. $\nabla(\nabla\cdot E)$ vanishes if and only if the free charge density is constant. $\endgroup$ – Bob Knighton Jan 15 at 9:41

Your Answer

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

Browse other questions tagged or ask your own question.