Why is $\operatorname{Div}\big(\operatorname{Curl} F\big) = 0$? Intuition?

Question

Why is $\operatorname{Div}\big(\operatorname{Curl} F\big) = 0$? Is there an intuitive explanation to what this means as well as an algebraic proof?

Also I understand that $\operatorname{Curl} \operatorname{Grad}F =0$, I interpet this as the gradient not having any rotation since its the direction of steepest ascent and if this formed some sort of loop this would not make sense? Is this somewhat correct? Any explanations would be appreciated.

Answer 1

Here's the algebraic proof: not sure if your question is looking for it.

It follows immediately from the definition of curl and div.

In cartesian coordinates for example the curl operator acting on a scalar function is

$curl F = (\partial /\partial y - \partial /\partial z), (\partial /\partial z - \partial /\partial x), (\partial /\partial x - \partial /\partial y) F $

The div operator acting on a vector function G is

$ div G = (\partial /\partial x , \partial /\partial y, \partial / \partial z).G$

So that div.curl acting on a scalar function H is

$div.curl H = (\partial^2 /\partial x.\partial y - \partial^2 / \partial x. \partial z + \partial^2 /\partial y.\partial z - \partial^2 /\partial y.\partial x + \partial ^2 /\partial z . \partial x - \partial^2 /\partial z.\partial y) H $

All you need to know then if that the partial derivative $\partial^2 /\partial x.\partial y $ is equal to $\partial^2 /\partial y.\partial x $ and you can then see that all the terms cancel out leaving zero.

Answer 2

Algebraically, both of these are simply based on the equality of mixed partials: $$ \begin{align} \nabla\cdot\nabla\times u &=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\cdot\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\times\left(u_1,u_2,u_3\right)\\ &=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\cdot\left(\frac{\partial u_3}{\partial y}-\frac{\partial u_2}{\partial z},\frac{\partial u_1}{\partial z}-\frac{\partial u_3}{\partial x},\frac{\partial u_2}{\partial x}-\frac{\partial u_1}{\partial y}\right)\\ &=\color{#0000FF}{\frac{\partial^2u_3}{\partial x\partial y}}-\color{#00A000}{\frac{\partial^2u_2}{\partial x\partial z}}+\color{#C00000}{\frac{\partial^2u_1}{\partial y\partial z}}-\color{#0000FF}{\frac{\partial^2u_3}{\partial y\partial x}}+\color{#00A000}{\frac{\partial^2u_2}{\partial z\partial x}}-\color{#C00000}{\frac{\partial^2u_1}{\partial z\partial y}}\\[9pt] &=0\tag{1}\\[20pt] \nabla\times\nabla u &=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\times\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)u\\ &=\left(\frac{\partial^2}{\partial y\partial z}-\frac{\partial^2}{\partial z\partial y},\frac{\partial^2}{\partial z\partial x}-\frac{\partial^2}{\partial x\partial z},\frac{\partial^2}{\partial x\partial y}-\frac{\partial^2}{\partial y\partial x}\right)u\\[6pt] &=(0,0,0)\,u\tag{2}\\[20pt] \end{align} $$ Intuitively(?), $(1)$ says that a flow induced perpendicular to the rotation of a smooth vector field has no source, and $(2)$ says that the flow induced by the gradient of a smooth scalar field has no rotation.