2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ ... and rotation has no source/drain. For an algebraic explanation, write everything with $\nabla$ and trust (like they do in physics) that the usual rules for vectors persist. $\endgroup$ – Hagen von Eitzen Apr 26 '14 at 11:50
  • $\begingroup$ Imagine a tiny ball $B$ centered at $x$. By the divergence theorem, $\int_B \text{div} \, \,\text{curl} \,F \,dV = \int_{\partial B} \text{curl} \,F \cdot dA$. But $\partial B$ is a closed surface, so by Stokes' theorem $\int_{\partial B} \text{curl} \, F \cdot dA = 0$. $\endgroup$ – littleO Apr 26 '14 at 13:52
4
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

Here's an informal way of interpreting it. Taking the curl of a function is a cross product operation: curl. Now let's try to take the divergence of this curl, however, recall that cross products produce orthogonal vectors. Since the divergence is a dot product operation divergence, and the dot product of orthogonal vectors is 0, the divergence of a curl is 0.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.