# Equivalence between Maxwell's equations and vector Helmholtz equations

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?

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