## Proof that a Cross Product of Gradients is Sinusoidal

Theorem
If
$f(x,y,z), g(x,y,z)$
are differentiable scalar fields then
$\mathbf{\nabla} f \times \mathbf{\nabla} f$
is sinusoidal so that
$\nabla \cdot (\mathbf{\nabla} f \times \mathbf{\nabla} f)=0$
.
Proof
\begin{aligned} \nabla \cdot ( \mathbf{\nabla} f \times \mathbf{\nabla} f) &= \frac{\partial}{\partial x}(\frac{\partial f}{\partial y}\frac{\partial g}{\partial z} - \frac{\partial f}{\partial z}\frac{\partial g}{\partial y}) \mathbf{i} + \frac{\partial}{\partial y}(\frac{\partial f}{\partial z}\frac{\partial g}{\partial x} - \frac{\partial f}{\partial x}\frac{\partial g}{\partial z}) \mathbf{j} + \frac{\partial}{\partial z}((\frac{\partial f}{\partial x}\frac{\partial g}{\partial y} - \frac{\partial f}{\partial y}\frac{\partial g}{\partial x})) \mathbf{k} \\ &= \frac{\partial^2 f}{\partial x \partial y} \frac{\partial g}{\partial z} +\frac{\partial f}{\partial y} \frac{\partial^2 g}{\partial x \partial z} - \frac{\partial^2 f}{\partial x \partial z} \frac{\partial g}{\partial y} - \frac{\partial^2 g}{\partial x \partial y} \frac{\partial f}{\partial z} \\ & + \frac{\partial^2 f}{\partial y \partial z} \frac{\partial g}{\partial x} +\frac{\partial f}{\partial z} \frac{\partial^2 g}{\partial y \partial x} - \frac{\partial^2 f}{\partial y \partial x} \frac{\partial g}{\partial z} -\frac{\partial f}{\partial x} \frac{\partial^2 g}{\partial y \partial z} \\ & + \frac{\partial^2 f}{\partial z \partial x} \frac{\partial g}{\partial x} +\frac{\partial f}{\partial x} \frac{\partial^2 g}{\partial z \partial y} - \frac{\partial^2 f}{\partial z \partial y} \frac{\partial g}{\partial x} -\frac{\partial f}{\partial y} \frac{\partial^2 g}{\partial z \partial x} =0 \end{aligned}