Given that $f(x, y, z$) and $g(x, y, z)$ are scalar functions, is it always true that $\nabla f \times \nabla g = 0$? (Assuming f and g have continuous partial derivatives)
I am working on a problem where I have to prove that $g\nabla f \cdot (\nabla \times g\nabla f)=0$
So far I have:
$g\nabla f \cdot (\nabla \times g\nabla f)= g\nabla f \cdot (g(\nabla \times \nabla f) + \nabla g \times \nabla f)$
I know $(\nabla \times \nabla f)=0$ since the curl of $\nabla f$ is zero.
So now I was wondering, can I say that $\nabla f \times \nabla g = 0$? because then that would finish my proof, as I would have:
$g\nabla f \cdot (\nabla \times g\nabla f)= g\nabla f \cdot (g(0) + 0) = g\nabla f \cdot 0 = 0$
Any help is greatly appreciated!
Take $f(x,y,z) = x$ and $g(x,y,z) = y$. Then $\nabla f(x,y,z) = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ and $\nabla g(x,y,z) = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$, so $\nabla f(x,y,z)\times\nabla g(x,y,z) = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$. In particular, $\nabla f\times\nabla g\ne 0$ in this case.
Nonetheless, you want to show that $$ g\nabla f\cdot (\nabla g\times\nabla f) = 0.$$ It's enough to show $\nabla f\cdot(\nabla g\times\nabla f) = 0$, or in other words, that $\nabla f$ and $\nabla g\times\nabla f$ are perpendicular. Is this always true? (Think about properties of cross products...)