Vector calculus identities involving curl and gradient

Question

having a bit of trouble wrapping my head around why the following result is true. I cant come up with a proof and it wasn't shown to me in class and I can't find it in any books. Any clarification on it would be nice. Thank you!

Let u = $\nabla \phi$. Then, $\nabla \times (\phi \nabla \phi) = 0$

Answer 1

$\nabla \phi=(\phi_x,\phi_y,\phi_z)$

so $$\nabla \times (\phi\nabla \phi)=\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ \phi \phi_x&\phi \phi_y&\phi \phi_z\\ \end{vmatrix}$$

Take for example the coefficient of $\mathbf{k}$, it is $$\frac{\partial}{\partial x}\phi \phi_y-\frac{\partial}{\partial y}\phi \phi_x$$

Now using the product rule

$$\phi_x \phi_y +\phi\phi_{yx}-\phi_y \phi_x -\phi\phi_{xy}=0$$ since $\phi_{xy}=\phi_{yx}$.

Answer 2

Hint

$$\nabla\times (\nabla \phi) = 0$$

And if $\phi$ is a scalar field, $\nabla\times\phi = 0$.

The curl has meaning only when applied to a vector field.

Answer 3

I think you just have to grind this one out. I'll use $f$ instead of $\phi$.

$f\nabla f = \langle ff_x,ff_y,ff_z \rangle$. Now compute the curl of this. For instance the $x$ coordinate is $$\frac{\partial}{\partial y}(ff_z)-\frac{\partial}{\partial z}(ffy) = (ff_{zy}+f_yf_z) - (ff_{yz}+f_zf_y) = 0$$ (assuming that $f$ is $C^2$ so that mixed partials are equal.) The other two components will also be $0$.

Answer 4

$$\phi\nabla\phi = \frac 12\nabla(\phi^2)$$

Then just use the fact that the curl of the gradient is identically zero.