$∇ \cdot( ∇\times u)=0$ What is the concept of this.

Question

Why is it $=0$ . I understand the equation. But I don’t understand the concept of it. Is it because the divergence of it or the curl?

Answer 1

It is a general result that can be proven easily.

If $F(x,y,z)=F_{1}(x,y,z)\hat{i}+F_{2}(x,y,z)\hat{j}+F_{3}(x,y,z)\hat{k}$ is a smooth vector field then $\text{div}(\text{curl}(F))=0$ . That is $\nabla\cdot\bigg(\nabla\times F\bigg)=0$ .

The proof is very easy just by computation.

$$\nabla\times F= \bigg(\frac{\partial{F_{3}}}{\partial y}- \frac{\partial{F_{2}}}{\partial z}\bigg)\hat{i}-\bigg(\frac{\partial{F_{3}}}{{\partial x}}- \frac{\partial{F_{1}}}{\partial z}\bigg)\hat{j}+\bigg(\frac{\partial{F_{2}}}{\partial x}- \frac{\partial{F_{1}}}{\partial y}\bigg)\hat{k}$$ .

Now $$\nabla\cdot (\nabla\times F) = \frac{\partial^{2} F_{3}}{\partial y \partial x} - \frac{\partial^{2} F_{2}}{\partial z \partial x} - \frac{\partial^{2} F_{3}}{\partial y \partial x}+\frac{\partial^{2} F_{1}}{\partial z \partial y} + \frac{\partial^{2} F_{2}}{\partial z \partial x} - \frac{\partial^{2} F_{1}}{\partial y \partial z} = 0 $$

Answer 2

If you have three vectors $a$, $b$, $c$ then $a\cdot b\times c$ is equal to the volume of the shape formed by the three vectors, so if $a$ and $b$ say are parallel then $a\cdot b\times c=0$. Like multiplication, differentiation is commutative, so $\nabla\cdot\nabla\times u$ is identically zero. $$ \begin{align} \nabla\cdot\nabla\times u &=D_1(D_3 u_2-D_2 u_3)+D_2(D_3 u_1-D_1 u_3)+D_3(D_1 u_2-D_2 u_1)\\ &=(D_2 D_3-D_3 D_2)u_1+....\\ &=0+0+0 \end{align} $$

There isn't any "obvious" geometric interpretation of this since $\nabla\cdot$ and $\nabla\times$ are not actually vectors but operators.

But the divergence measures how much of the vector field is going in and out of a volume (Gauss's divergence theorem), whereas the curl measures how much the vector field is rotating on a surface (Stoke's theorem), so you can view the divergence as perpendicular to a surface (in and out), and the curl as parallel to the surface (rotating around the surface). I think that is the intuitive picture most people have.