How do I imagine why divergence of curl and curl of gradient is $0$?

Question

I tried watching several videos on YouTube, but I failed to gain intuition. I tried to solve it by myself by imagining water flow but I was unsuccessful and got stuck. How do I imagine why divergence of curl and curl of gradient is $0$?

Answer 1

very roughly speaking, the gradient of a function gives you the direction in which the function increases most. The curl of a vector field describes how much the vector field "winds" around itself or whether the flow of it forms closed loops. now if curl(grad(f)) would not be zero the gradient of f would infinitesimally form closed loops but then f(x)>f(x) because f increases along the flow of the gradient, which can not be.

divergence is telling you how much a vector field flows through the walls of an infinitesimal box around the point you are looking, but the curl has no net flow, because everything that flows out also flows in, because the loops are closed

Answer 2

After applying the Fourier transform curl becomes proportional to the vector product with $\vec k$, div to the scalar product with $\vec k$, and gradient to the product by $\vec k$.

Thus for example $div \,(curl \,\vec A)$ is proportional to $\vec k \cdot (\vec k \times \tilde A)=0$, where $\tilde A$ is the Fourier transform of $A$.

Answer 3

Imagining water flows or looking at Maxwells equations is definitely a good idea. A video for visualizing I can strongly recommend is this one: https://www.youtube.com/watch?v=rB83DpBJQsE&ab_channel=3Blue1Brown

Lets say $\vec{F}$ is a vector field. First start by actually understanding what div($\vec{F}$) really means. It represents the sources and sinks of the flow. In terms of electrodynamics these are the electrons and protons as they create the field. The curl on the other hand is a measure how much a field (flow) rotates around a certain point (for visuals watch the video). A rotation does not have a starting or end point or phrased differently it does not have a source or a sink. That is why the divergence of curl of $\vec{F}$ must be zero.

The gradient of a scalar field points into the direction of the strongest change of the field. So it is perpendicular to isosurfaces of the scalar field and that already requires that the curl of the gradient field is zero. A good example to visualize is a temperature distribution.

Answer 4

Preliminary Geometric Observations

The conceptually simplest answer I can offer is using the integral theorems (Stokes and divergence (which is really a special case of a more generalized Stokes theorem)); but first we need some simple geometric preliminaries.

• Consider the 2-dimensional setting, where we have a disk (also called a 2-dimensional ball) $B_r=\{(x,y)\in\Bbb{R}^2:\,\, x^2+y^2\leq r^2\}$. This is the closed disk of radius $r$ centered at the origin. Now, if I ask you "what is the boundary" of this surface, then I think you'd immediately tell me that the boundary of this disk is simply the circle $S_r=\{(x,y)\in\Bbb{R}^2:\,\, x^2+y^2=r^2\}$. Now, suppose I ask you what is the boundary of the circle? Well the circle itself has no boundary, because it is a closed loop (try to draw some pictures to convince yourself). Contrast this with the case of a line segment: if you draw a straight line segment, you would obviously say that the endpoints of the line segment are the boundary of the line. But for the circle, there are NO endpoints and there is no beginning! That is why there is no boundary to the circle.

• Similarly, we can discuss the same situation one-dimension higher. Suppose now I consider a 3-dimensional solid ball $B_r = \{(x,y,z)\in\Bbb{R}^3:\,\, x^2+y^2+z^2\leq r^2\}$ (think of a full-solid watermelon if you wish). What is the boundary of this? Well of course we'd expect that the boundary of this is simply the sphere of radius $r$: $S_r=\{(x,y,z)\in\Bbb{R}^3\,: x^2+y^2+z^2=r^2\}$. What is the boundary of the sphere? Nothing! (again, to convince yourself of this, imagine a rectangular piece of paper: here the boundary of the piece of paper is its 4 edges... but a sphere has no such edges hence it has no boundary).

• In fact the same thing is true in any dimension, in general: whenever we have a region $M$ (eg a solid ball or a solid ellipsoid, or a solid donut (torus) etc), we can ask what is its boundary. This is usually denoted $\partial M$. Then we can ask the same question again: what is the boundary of $\partial M$. Well it turns out that $\partial(\partial M)=\emptyset$ i.e it is the empty set, which means "the boundary of a boundary is non-existent". I hope the two examples I gave you above give you sufficient reason to think this is "believable"; and since you asked for intuition I'll stop here.

Integral Calculus Theorems

So far, we've just been talking about simple geometric observations of shapes you can easily imagine (while reading, please draw figures to follow along). Now, let us relate the geometry above to calculus. Specifically, we need the theorems of integral calculus which tell us how derivatives and integrals are related.

Your Actual Question.

If you don't already know the Stokes theorem and Divergence theorem, I guess this answer won't make much sense, but anyway here is a very quick 4-minute video which describes the "big idea" behind these theorems. Assuming you've understood that, we can now answer your questions.

First, I'll show the divergence of a curl is always $0$. So, let $\mathbf{F}$ be a given vector field. We wish to show that $\nabla \cdot (\nabla \times \mathbf{F})=0$. One way to show a function is zero is to show that its integral over EVERY possible region is zero. So, let $M$ be a solid 3-dimensional region (think of a solid ball as above). Then \begin{align} \int_{M}\nabla \cdot (\nabla \times \mathbf{F}) \, dV &= \int_{\partial M}(\nabla \times \mathbf{F})\cdot d\mathbf{a} \tag{divergence theorem}\\ &= \int_{\partial (\partial M)} \mathbf{F}\cdot d\mathbf{l}\tag{Stokes theorem}\\ &= \int_{\emptyset} \mathbf{F}\cdot d\mathbf{l}\\ &=0 \end{align} where I've invoked the above discussion that the boundary of a boundary is always empty. If we're integrating over an empty set then the answer is $0$. Since this is true for every region $M$, it follows that $\nabla\cdot (\nabla \times \mathbf{F})=0$.

The proof that curl of gradient is $0$ is identical: start with an arbitrary function $f$, and take any surface $S$. Then,

\begin{align} \int_{S}\nabla \times (\nabla f) \cdot d\mathbf{a}&= \int_{\partial S}(\nabla f)\cdot d\mathbf{l}\tag{Stokes theorem}\\ &= f(\text{endpoint})-f(\text{startpoint})\tag{FTC}\\ &= 0. \end{align} Here $\partial S$ is a closed loop which is why the starting and ending points of the loop are the same hence the result is $0$. Since this is true for all surfaces $S$, it follows that $\nabla \times (\nabla f)=0$.

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Of course, the proofs I offered above are not the quickest or most direct (there are some steps above which need to be filled in to give a complete proof). The quickest proof is to just use the definition of divergence, curl and gradient, plug everything in and check that terms miraculously cancel out to give you $0$ (essentially it's because for sufficiently nicely behaved functions, the order of partial derivatives does not matter; this is called Schwarz's theorem in multivariable calculus).

However I hope this explanation allows you to think about these facts geometrically. Essentially there's only two ideas here. The first is a simple geometric fact

that the boundary of a boundary is always empty (which is fairly easy to imagine with some simple examples).

while the second is a very fundamental idea in calculus:

Stokes and Divergence theorem, which I hope conceptually make sense: "adding up the derivatives, i.e small changes, inside a region equals the total change on the boundary".