# Showing a vector identity

Problem: If $\phi$ and $\psi$ are smooth scalar fields, show that \begin{align*} \nabla \times (\phi \nabla \psi) = -\nabla \times (\psi \nabla \phi ) = \nabla \phi \times \nabla \psi .\end{align*}

Attempt at solution: First, I would use the vector identity \begin{align*} \nabla \times (\phi \mathbf{F}) = (\nabla \phi ) \times \mathbf{F} + \phi (\nabla \times \mathbf{F}). \end{align*} In our case $\nabla \psi = \mathbf{F}$. Hence we have that \begin{align*} \nabla \times (\phi \nabla \psi) = (\nabla \phi) \times (\nabla \psi) + \phi(\nabla \times \nabla \psi). \end{align*} Then I'm not really sure what to do next. Can I conclude on the RHS that $\phi(\nabla \times \nabla \psi) = 0$, because it is a curl grad? If that's the case, I still don't know how to arrive at the other expressions.

• Things to always remember: (1) The curl of the gradient of a scalar field is always zero [$\nabla\times\nabla\phi=0$]. (2) The divergence of the curl of a vector field is always zero [$\nabla\cdot(\nabla\times\mathbf{F})$]. – Arturo don Juan Mar 18 '15 at 18:52
• Solving vector identities is (almost always) pretty straightforward with index notation. I'd recommend taking a look at how I worked this problem and see if you can apply the same methods for your vector calculus identity. – user137731 Mar 18 '15 at 18:55

Yes that's zero, since it's a curl of grad. For the other expansion, try to expand $0=\nabla\times\nabla(\phi\psi)$.

• Expanding your expression gives me: $\nabla \times (\phi \nabla \psi + \psi \nabla \phi)$. What now? Can I just distribute the cross product? – Kamil Mar 18 '15 at 20:41
• @Kamil Yes. Curls distribute over addition. – user137731 Mar 18 '15 at 21:14
• OK, I worked it out and got : $(\nabla \phi \times \nabla \psi) + (\nabla \psi + \nabla \phi)$. Then I don't know how to proceed. – Kamil Mar 18 '15 at 21:47

Just to show you the ease of index notation, lookie here:

$$[\nabla \times (\phi\nabla\psi)]_p = \epsilon_{pqr}\partial_q(\phi\partial_r\psi) = \epsilon_{pqr}(\partial_q\phi)\partial_r\psi + \epsilon_{pqr}\phi\partial_q\partial_r\psi \\ = [(\nabla \phi)\times(\nabla \psi)]_p + [\phi(\nabla\times(\nabla \psi))]_p$$

Then showing that $\phi(\nabla\times(\nabla \psi))=0$ is easy too. I don't want to carry around the $\phi$, though, so convince yourself that if $\nabla\times(\nabla \psi)=0$, then $\phi(\nabla\times(\nabla \psi))=0$. Now, for the proof:

$$\begin{array}\ [\nabla\times(\nabla \psi)]_p = \epsilon_{pqr}\partial_q\partial_r\psi \\ = \epsilon_{pqr}\partial_r\partial_q\psi & \text{Clairaut's theorem} \\ =\epsilon_{prq}\partial_q\partial_r\psi & \text{relabel dummy indices: q \leftrightarrow r} \\ = -\epsilon_{pqr}\partial_q\partial_r\psi & \epsilon_{pqr} = -\epsilon_{prq}\end{array}$$

And from here we can see that $[\nabla\times(\nabla \psi)]_p=0$ because it's equal to both $\epsilon_{pqr}\partial_q\partial_r\psi$ AND $-\epsilon_{pqr}\partial_q\partial_r\psi$ and the only scalar field identically equal to its negative is the $0$ field.

Therefore $\nabla \times (\phi\nabla\psi) = (\nabla \phi)\times(\nabla \psi)$.$\ \ \ \ \ \square$