# What is an intuitive explanation for divcurlF=0\operatorname{div} \operatorname{curl} F = 0?

I am aware of an intuitive explanation for $\operatorname{curl} \operatorname{grad} F = 0$ (a block placed on a mountainous frictionless surface will slide to lower ground without spinning), and was wondering if there were a similar explanation for $\operatorname{div} \operatorname{curl} F = 0$.

First off, I don’t think you’re going to get an explanation for $\nabla\cdot (\nabla\times \mathbf F) = 0$ that’s on quite the same level as your sliding block example, because while the gravitational force is well-known to be the gradient of the negative potential, there aren’t any tangible vector fields that are the curl of something. The curl, like the cross product, is a little “unnatural” as it depends on the choice of handedness; classical physics, on the other hand, is independent of handedness, so you can never observe a curl directly, only hidden under an integral or cross product to make the handedness goes away.
Nevertheless, the most mathematically natural argument that $\nabla\cdot (\nabla\times \mathbf F) = 0$ is quite easy to grasp. For a warm-up, let’s look at another explanation for why $\nabla\times (\nabla f) = 0$. The curl of any vector field $\mathbf F$ describes how much $\mathbf F$ “spins around”, right? So take any closed loop in space; the net curl inside the loop is nothing but how much $\mathbf F$ “circulates” around the loop, i.e. $\oint \mathbf F \cdot d\mathbf x$. But if $\mathbf F$ is a gradient of something, say $f$, then $\mathbf F \cdot d\mathbf x$ is actually how much $f$ is going up or down as you walk along $d\mathbf x$. So if you walk along a closed loop and come back to where you started, the net change in $f$ has to be zero! Or as wzzx’s comment states: you can’t walk from home to school and back and have gone uphill both ways.
Now we can do the same thing for $\nabla\cdot(\nabla\times \mathbf F)$. Intuitively, the divergence of a vector field $\mathbf G$ measures how much $\mathbf G$ is “spreading out” or “pulling in”. In other words, pick any region of space; what does the total divergence of $\mathbf G$ inside it tell you? It tells you exactly how much $\mathbf G$ is flowing out of the surface of the region, i.e. $\oint \mathbf G \cdot d\mathbf A$. But if $\mathbf G$ is the curl of another vector field $\mathbf F$, then $\int \mathbf G \cdot d\mathbf A$ on a surface just measures how much $\mathbf F$ circulates around the boundary of that surface. What’s the boundary of a closed surface? Imagine taking a portion of the entire surface and growing it to cover the whole. Its boundary eventually gets smaller and smaller, and then disappears. So there is nothing for $\mathbf F$ to circulate around, and the circulation must be zero.
All I’ve shown here is that the integrals of $\nabla\times(\nabla f)$ and $\nabla\cdot(\nabla\times\mathbf F)$ are zero on any arbitrary region, but that should be enough to see that (at least for smooth $f$ and $\mathbf F$) their values must be zero everywhere.