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?

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    $\begingroup$ 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
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    $\begingroup$ 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

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