In my recent exercise book I've derived the following equation that needs solving: $\nabla^2\vec{u} + k^2\vec{u} = 0.$ The deformation vectors points only in the $\hat{e}_r$ direction. I didn't want to write out the Laplace in spherical coordinates, so I tried using what I learned in my PDE course the previous semester. It turns out, the vector Helmholtz equation is quite different from scalar one we've studied.

Suppose I have basic knowledge in solving scalar Helmholtz in spherical (and other coordinate systems). Is there any analogy that translates over to the vector version? In other words, should I be able to solve vector Helmholtz if I can solve scalar versions?

  • $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Mar 13, 2020 at 5:00

1 Answer 1


Yes, indeed you can use your knowledge of the scalar Helmholtz equation. The difficulty with the vectorial Helmholtz equation is that the basis vectors $\mathbf{e}_i$ also vary from point to point in any other coordinate system other than the cartesian one, so when you act $\nabla^2$ on $\mathbf{u}$ the basis vectors also get differentiated. This forces you to calculate $\nabla^2 \mathbf{u}$ through the identity $$ \nabla^2 \mathbf{u} = \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{u}) - \boldsymbol{\nabla}\times (\boldsymbol{\nabla}\times \mathbf{u}) \tag{1} $$ which is really cumbersome to deal with by brute force. A smart way to avoid all the hassle is by using the ansatz $$ \mathbf{u} = \mathbf{r} \times (\boldsymbol{\nabla} \psi) \tag{2} $$ where $\psi$ satisfies the scalar Helmholtz equation $$ (\nabla^2 + k^2) \psi = 0. $$

To check that $(\nabla^2 + k^2) \mathbf{u} = 0$ yourself you have to plug the ansatz $(2)$ on $(1)$ and make use of many vector identities and the scalar Helmholtz equation. The calculation is quite involved, so I'll point you to check Reitz, Milford & Christy's Foundations of Electromagnetic Theory, there they do the full calculation. With ansatz $(2)$ proven, it's just a matter of plugging the relevant mode $\psi_{lm}$ in eq. $(2)$ that you get your solution $\mathbf{u}_{lm}$.

  • $\begingroup$ Yes I figured the non-constant basis vectors are the source of problems (as I've seen in the solutions where we just wrote out the operator in spherical). This is just what I needed, thank you very much! $\endgroup$
    – mb28025
    Mar 12, 2020 at 22:55

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