I am aware of an intuitive explanation for curlgradF=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 divcurlF=0.

**Answer**

*Update:* I felt that in my original answer, the intuition got lost in all of the formalism, so I’ve rewritten it to make it clearer. The core idea is still much the same as Qiaochu’s answer on the linked MO question.

First off, I don’t think you’re going to get an explanation for ∇⋅(∇×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 ∇⋅(∇×F)=0 is quite easy to grasp. For a warm-up, let’s look at another explanation for why ∇×(∇f)=0. The curl of any vector field F describes how much F “spins around”, right? So take any closed loop in space; the net curl inside the loop is nothing but how much F “circulates” around the loop, i.e. ∮F⋅dx. But if F is a gradient of something, say f, then F⋅dx is actually how much f is going up or down as you walk along dx. 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 ∇⋅(∇×F). Intuitively, the divergence of a vector field G measures how much G is “spreading out” or “pulling in”. In other words, pick any region of space; what does the total divergence of G inside it tell you? It tells you exactly how much G is flowing out of the surface of the region, i.e. ∮G⋅dA. But if G is the curl of another vector field F, then ∫G⋅dA on a surface just measures how much 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 F to circulate around, and the circulation must be zero.

All I’ve shown here is that the integrals of ∇×(∇f) and ∇⋅(∇×F) are zero on any arbitrary region, but that should be enough to see that (at least for smooth f and F) their values must be zero everywhere.

**Attribution***Source : Link , Question Author : J. Foote , Answer Author : Community*