What is the simplest way to understand $\nabla \cdot (\nabla \times \textbf{A} ) =0$?

Question

Of course one can prove it by brute force.

But how to Understand it, in the simplest way?

Another issue is $\nabla \times (\nabla f) =0 $.

Answer 1

It's just a way to remember/understand. Think that $\nabla $ is a vector. Then, by definition, $\nabla \times A$ is perpendicular to $\nabla $ and so the product scalar of $\nabla $ and $\nabla \times A$ is nul, therefore $$\nabla \cdot (\nabla \times A)=0$$

Answer 2

Here is an explanation based on Stokes' theorem, the Divergence theorem, and the fundamental theorem of line integrals. Recall that Stokes' theorem states that the line integral along a closed curve is equivalent to a certain surface integral of the surface enclosed:

$$\int_{\partial D} \mathbf{F}\cdot d\mathbf{r}=\int_{D} (\nabla\times \mathbf{F})\cdot d\mathbf{S}$$

But the fundamental theorem of line integrals states that $\int_a^b(\nabla f)\cdot d\mathbf{r}=f(b)-f(a)$. So if $\mathbf{F}=\nabla f$ for some function $f$, then

$$\int_{D} (\nabla\times (\nabla f))\cdot d\mathbf{S}=\int_{\partial D} (\nabla f)\cdot d\mathbf{r}=f(b)-f(a)=0$$ since the endpoints coincide (it's a closed loop). So the surface integral vanishes for any surface $D$; the only way this can make sense is if $\nabla\times \nabla f=0$.

Similarly, recall that the Divergence Theorem states that we may express the flux integral over the surface of a volume $V$ as

$$\int_{\partial V} \mathbf{F}\cdot d\mathbf{r}=\int_V (\nabla\cdot \mathbf{F})dV$$

Now, if $\mathbf{F}=\nabla\times \mathbf{A}$, then Stokes' theorem implies that we may write this as a line integral of $\mathbf{A}\cdot d\mathbf{r}$ along the boundary of the surface $\partial V$---but this surface has no boundary! So the integral vanishes immediately. (If that seems too quick, cut the surface integral in two and show that the two line integrals obtained from Stokes' theorem cancel.) Therefore the volume integral over $\nabla\cdot \mathbf{F}=\nabla\cdot(\nabla\times \mathbf{A})$ vanishes for any volume $V$, and so we must have $\nabla\cdot \mathbf{(\nabla\times \mathbf{A})}=0$.