If we consider $\mathbf{F}$ as a vector field, then we say that $\mathrm{div}(\mathrm{curl}(\mathbf{F}))=0$. We can prove this in mathematics easily. But I' am not getting an intuitive explanation due to which it is zero.
Can someone explain intuitively why it is zero?
Thanks
On
Might not be the answer you are hoping for, but intuitively you can see that if you have a volume in 3-space, then you can take that volume's boundary, which will be a closed 2-surface. A closed 2-surface has an empty boundary, ergo there is no curve that would serve as its boundary.
Through the use of the usual integral theorems in 3-space, the identity $\mathrm{div}(\mathrm{curl}(\mathbf{X}))$ expresses this.
By Gauss' theorem, for any vector field $\mathbf{X}:\mathbb{R}^3\rightarrow\mathbb{R}^3$ $$ \int_V\mathrm{div}\mathbf{X}\ d^3x=\int_{\partial V}\langle \mathbf{X},d\mathbf{A}\rangle $$ and by Stokes' theorem $$ \int_{\Sigma}\langle\mathrm{curl}\mathbf{X},d\mathbf{A}\rangle=\int_{\partial\Sigma}\langle\mathbf{X},d\mathbf{s}\rangle, $$ where $V$ is a volume, $\Sigma$ is a surface, $\partial$ is boundary and $\langle,\rangle$ is the inner product.
Getting together these two, we apply Gauss' theorem to $\mathbf{Y}=\mathrm{curl}\mathbf{X}$, then $$ \int_V\mathrm{div}\ \mathrm{curl}\mathbf{X}\ d^3x=\int_{\partial V}\langle\mathrm{curl}\mathbf{X},d\mathbf{A}\rangle=\int_{\partial\partial V}\langle\mathbf{X},d\mathbf{s}\rangle=0.$$ Since this is zero for all vector fields $\mathbf{X}$, due to the identity $\mathrm{div}\ \mathrm{curl}\mathbf{X}=0$, this must mean that the domain of integration $\partial\partial V$ is always of measure zero, which means that the boundary of a boundary is empty.
On
Intuitively, the curl of ${\mathbf F}$ is a vector field which points perpendicularly to the plane of net rotation of ${\mathbf F}$ at each point.
Let's look at the analog in $\Bbb R^2$. If ${\mathbf F}=\langle P(x,y), Q(x,y) \rangle$, then, if we could define the curl, it would be curl${\mathbf F}=(Q_x-P_y) \mathbf{k}$. This is perpendicular to the vector field ${\mathbf F}$ everywhere and so the net component of the curl of ${\mathbf F}$ in the $xy-$plane is zero. Hence the divergence of curl${\mathbf F}$ naturally would be zero. No matter what region of $\Bbb R^2$ we look at, the net flow of curl${\mathbf F}$ out of that region is zero since curl${\mathbf F}$ only flows perpendicular to the plane. Of course, curl isn't formally defined in two dimensions. You can think of the curl of a two dimensional vector field simply being identically zero everywhere, since it would have to be perpendicular to the plane if it were embedded into a three dimensional space.
Now let's look at $\Bbb R^3$. The interpretation of curl still holds as it is the vector field that points perpendicularly to the plane of net rotation of $\mathbf F$ at each point. However, we can't look into a fourth dimension to make an analogous argument as we did for $\Bbb R^2$.
Think about space being discretized into small cubes. The net divergence for a small cube will be the sum of the orthogonal components on each face. The orthogonal component of curl on each face is given by the net rotation about the boundary of that face. However, adjacent faces always share an edge, so the net rotation piece for that edge cancels out when adding up the outflux for all 6 faces. Hence we get zero divergence for the cube.
Intuitively, you would have to have a vector field which changes direction of flow extremely fast in order to get its curl to have a non-zero divergence. It would have to be non-smooth so that you simply can't take the divergence of it as it wouldn't be differentiable -- the limiting argument I used above for little cubes wouldn't make sense if the limit doesn't exist, i.e. the vector field is not differentiable.
I know this is quite hand-wavey, but I hope it makes some sense.
Hint: Check the intuitive interpretation of curl and then that of divergence and combine them to obtain the interpretation of $\text{div}(\vec{\text{curl}} \vec{F})$