How to show this vector cross product/gradient result

Question

One of my books has that if $$\bar A= \phi \nabla \psi$$ then $$\nabla \times \bar A = \nabla \phi \times \nabla \psi$$

But I don't see why it is true. What is the proof of this?

Thanks

Answer 1

First of all $(\nabla \times \vec{v})_i = \epsilon_{ijk}\partial_j v_k$ with Einstein summation. Then for any scalar $\phi$ $$ (\nabla \times \phi\vec{v})_i = \epsilon_{ijk}\partial_j (\phi v_k)= \epsilon_{ijk}(\partial_j \phi) v_k + \phi\epsilon_{ijk} \partial_j v_k= (\nabla\phi \times \vec{v})_i + \phi (\nabla\times \vec{v})_i $$ the first one is because $(\vec{A}\times \vec{B})_i = \epsilon_{ijk}A_j B_k$. So in general $$ \nabla \times(\phi \vec{v}) = \nabla \phi \times \vec{v} + \phi \nabla\times \vec{v} $$ using the fact that any gradient has zero curl: $\nabla \times \nabla \psi=0$ you get what you want with $\vec{v}=\nabla \psi$.