Considering the wave equation in spherical coordinates, if we know that $\psi(\vec{r})$ is a solution, then $\vec{r}\times \nabla \psi$ is also a solution. (The hint is to take the difference between $\psi(r,\theta,\phi)$ and $\psi(r,\theta ',\phi ')$)

If I interpreted it correctly, it says that if $\psi $ solves $$\nabla^2\psi - \frac{1}{c^2}\frac{\partial^2 \psi}{\partial t^2} =0$$

Then show that:

$$\nabla^2(\vec{r}\times \nabla\psi) - \frac{1}{c^2}\frac{\partial^2 (\vec{r}\times \nabla\psi)}{\partial t^2} =0$$

This question seems outright wrong. As the argument of the Laplacian is a vector. Or am I misinterpreting it?


1 Answer 1


Sure, you can have a vector Laplacian - it's just the Laplacian of each component (in Cartesian coordinates). Observe (with some notational sleuth, and using $\Delta$ in place of $\nabla^2$),

$$\frac{\partial}{\partial x_i} (\;\vec{r}\times\nabla\psi)=\vec{e}_i\times\nabla\psi+\vec{r}\times\left(\frac{\partial}{\partial x_i}\nabla\psi\right),$$ $$\implies\frac{\partial^2}{\partial x_i^2} (\;\vec{r}\times\nabla\psi)=2\,\left(\frac{\partial}{\partial x_i}\vec{e}_i\right)\times\nabla\psi+\vec{r}\times\left(\frac{\partial^2}{\partial x_i^2} \nabla\psi\right)$$ $$\implies \Delta(\;\vec{r}\times\nabla\psi)=2\nabla\times\nabla\psi+\vec{r}\times\nabla(\Delta\psi).$$ Note how we move around the partial derivatives in suggestive and loose but legal and meaningful ways. And $\nabla\times\nabla=\vec{0}$ (it kills any function - basic vector calculus identity), so that drops off.

Putting the vector function $\vec{r}\times\nabla\psi$ into the LHS of the differential equation gives $$\vec{r}\times\nabla(\Delta\psi)-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}(\;\vec{r}\times\nabla\psi)=\vec{r}\times\nabla\left(\Delta\psi-\frac{1}{c^2}\frac{\partial^2 \psi}{\partial t^2}\right)$$ $$=\vec{r}\times\nabla(0)=\vec{0}.$$ Thus the vector function does indeed satisfy the differential equation. This may not be the sort of derivation your text or homework desires - it might want you to exploit rules specific to polar coordinates, hence the hint, but I can't off the top of my head figure out a heading in that direction.


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