Is there any physical interpretation for $\nabla\cdot(\nabla \times F)=0$? It is well known that the divergence of the curl is always 0. Mathematically I understand why this happens ($d^2=0$ where $d$ is the exterior derivative) but today I was wondering what is the physical meaning of this.
The divergence represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point (from Wikipedia) and the curl describes the infinitesimal rotation of a 3-dimensional vector field (also from Wikipedia).
Does this mean that the rotation of a vector field is always stable and doesn't go inwards or outwards? What is the physical meaning of the divergence of the curl equals 0? 
 A: I don't think there is any physical meaning. There is geometric meaning, however, which is almost as good, in this case.
Consider the integral theorems, namely the Kelvin-Stokes-theorem: $$ \int_\Sigma \nabla\times \mathbf{V}\cdot d\mathbf{A}=\int_{\partial\Sigma}\mathbf{V}\cdot d\mathbf{l}, $$ and the Gauss-theorem: $$ \int_U \nabla\cdot\mathbf{V}\ d^3x=\int_{\partial U}\mathbf{V}\cdot d\mathbf{A},$$ where $\Sigma$ is some smooth surface, $U$ is a 3-dimensional domain of integration, and $\partial$ is the boundary operator, eg. the one that maps a set to its boundary.
Now let $U$ be once again a volume, and let $\mathbf{V}$ be a completely arbitrary vector field, and apply these theorems twice in succession: $$ \int_{\partial\partial U}\mathbf{V}\cdot d\mathbf{l}=\int_{\partial U}\nabla\times\mathbf{V}\cdot d\mathbf{A}=\int_U\nabla\cdot(\nabla\times\mathbf{V})\ d^3x=0. $$ These integrals are all zero, because $\nabla\cdot(\nabla\times\mathbf{V})$ is zero for any $\mathbf{V}$, however, if you look at the first expression, $$ \int_{\partial\partial U}\mathbf{V}\cdot d\mathbf{l}=0 $$, this must be true for all vector fields $\mathbf{V}$. The only way it is true for all such vector fields is that if the domain of integration $\partial\partial U$ is a set of zero measure.
Therefore, the identity $\nabla\cdot(\nabla\times\mathbf{V})=0$ essentially says that for a three-dimensional volume $U$, the boundary of its boundary is always zero. If you think about it, this is pretty obvious, but this line of thought actually proves this.
This is even more elegant for differential forms, where you can repeat the same procedure by applying the generalized stokes theorem twice, and you will essentially get that $\mathrm{dd}=0$ corresponds to $\partial\partial=0$.
A: First I should say that it is a mathematical definition and does not carry any physical meaning necessarily. But as an example it relates the Maxwell equations:
$$\nabla\times E=-\frac{\partial B}{\partial t}$$
and
$$\nabla \cdot B=0$$
$$\Longrightarrow \nabla\cdot(\nabla\times E)=-\frac{\partial \nabla\cdot B}{\partial t}=0$$
and the continuity equation can be deduced from
$$\nabla\times H=J+\frac{\partial D}{\partial t}$$
and
$$\nabla \cdot D=\rho$$$$\Longrightarrow \nabla\cdot(\nabla\times H)=\nabla\cdot J+\frac{\partial \nabla\cdot D}{\partial t}=0$$
which is the continuity equation:
$$\nabla\cdot J=-\frac{\partial \rho}{\partial t}$$
If one day in future a magnetic monopole is discovered in reality, as some people are interested in it for some reasons, then $\nabla \cdot B$ is not zero anymore, but, mathematically $\nabla\cdot(\nabla\times E)$ has to be zero.
This means that a new entity has to be defined instead of $B$ in order to hold the $\nabla\cdot(\nabla\times E)=0$. This is how math can help physics.
A: It quite simply means that the radial direction is orthogonal to the tangential direction.
The universe in which we live behaves, as far as we can tell and measure, as a 3D volume (time aside for now:). A 3D volume is defined by 3 orthogonal directions. We can choose to look at a 2D cross-section of this volume, and we get a 2D plain, in which, every point is defined by 2 orthogonal direction. One can show these directions can be X\Y, but also R\Teata. R is the radial direction, and Theata is the tangential direction. 
Because these directions are orthogonal, BY DEFINITION, it's not hard to understand that if you measure the radial component of a purely tangential scale (the deliverance of the curl), of the other way around, you get nothing - because radialiry is not tangential, and tangentiality is not radial :) 
