This is closely related to this recent question

The vector Helmholtz equation is

\begin{align} (\nabla^2 + k^2)\boldsymbol{u} = 0 \end{align}

The scalar Helmholtz equation is

\begin{align} (\nabla^2 + k^2)u = 0 \end{align}

In the linked question and reference therein (Reitz, Milford "Foundations of Electromagnetic Theory") it is claimed that if $u$ satisfies the scalar Helmholtz equation then

\begin{align} \boldsymbol{u}_1 =& \boldsymbol{r}\times \nabla u\\ \boldsymbol{u}_2 =& \nabla \times\left(\boldsymbol{r} \times \nabla u\right) \end{align}

both satisfy the vector Helmholtz equation. I have confirmed this as follows (though I would appreciate a check for correctness)

\begin{align} \left[\left(\nabla^2 + k^2\right)\boldsymbol{u}_1\right]_i =& (\partial_j\partial_j+k^2) \epsilon_{ikl} r_k\partial_lu\\ =&\epsilon_{ijk}r_k\partial_l(\partial_j\partial_j + k^2)u\\ =& 0 \end{align}

and pretty similarly for $\boldsymbol{u}_2$

\begin{align} \left[(\nabla^2 + k^2)\boldsymbol{u}_2\right]_i =& (\partial_j\partial_j + k^2)\epsilon_{ikl}\partial_k \epsilon_{lmn}r_m\partial_nu\\ =& \epsilon_{ikl}\epsilon_{lmn}\partial_kr_m\partial_n(\partial_j\partial_j+k^2)u\\ =& 0 \end{align}

In index notation the proof of the claim seems pretty obvious. In fact, it looks like pretty much any vector quantity derived from $u$ would do the trick.

For example $\boldsymbol{u}_3 = \nabla u$.

\begin{align} \left[(\nabla^2 + k^2)\boldsymbol{u}_3\right]_i =& (\partial_j\partial_j + k^2)\partial_i u\\ =& \partial_i (\partial_j\partial_j + k^2) u\\ =& 0 \end{align}

It seems $\boldsymbol{u}_4 = \boldsymbol{r}u$ would also work.

Here then are my questions:

1) Is it correct that if $u$ satisfies the scalar Helmholtz equation that $\nabla u$ and $\boldsymbol{r}u$ satisfy the vector Helmholtz equation?

2) Are my proofs that $\boldsymbol{u}_1, \boldsymbol{u}_2$ and $\boldsymbol{u}_3$ satisfy the vector Helmholtz equation valid?

3) If the answer to question 1) is yes then why are $\boldsymbol{u}_1$ and $\boldsymbol{u}_2$ natural choices to extend complete families of solutions of the scalar Helmholtz equation to complete families of solutions of the vector Helmholtz equation? Or would other choices be natural and $\boldsymbol{u}_1$ and $\boldsymbol{u}_2$ just happen to be the choices that Reitz and Milford use?

  • $\begingroup$ How is this not a "check my work" question? $\endgroup$
    – Semoi
    Mar 19, 2020 at 22:02
  • 1
    $\begingroup$ I've added a third questions which I would say is the underlying question that motivated me to ask the other two. Basically I think the answer to question 1) is yes but I'm surprised about it. I think the answer to question 2) is yes but I'm not sure (I guess this is explicitly a check my work question) because it seemed too easy. question 3) is sort of an open question. $\endgroup$
    – Jagerber48
    Mar 19, 2020 at 22:12

1 Answer 1


Good question. The answer for question (1) is no. Because $$\nabla\cdot\mathbf{u_{3,4}}\neq 0.$$ They do not satisfy the boundary condition of the electromagnetic wave equation, i.e., $\nabla\cdot\mathbf{E}\neq 0$. That is the reason we have to keep the relation between the scalar solution and the vector solution such that, $\mathbf{u}=\nabla\times(\mathbf{c}u)$, such that $\nabla\cdot(\nabla \times (\mathbf{c}u))=0$.

  • $\begingroup$ To clarify, the answer to (1) is actually yes. OP is correct that any such vector quantity will work, because the operator in the Helmholtz equation just acts on each vector component separately. However, not all such vectors are valid physical solutions of Maxwell's equations. Am I interpreting your answer correctly? $\endgroup$
    – kaylimekay
    Dec 31, 2020 at 4:59
  • 1
    $\begingroup$ Yes, you interpreted it correctly. Thank you. I need to clarify why I answered no; the question is related to the reference, "Foundations of Electromagnetic Theory" by Reitz, so I believe the underlying answer is also related to EM waves. If that is the case, although they satisfy the vector Helmholtz equation, they do not satisfy the physical solutions of EM waves, as you pointed out. $\endgroup$ Jan 2, 2021 at 0:30

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