Why is the divergence of curl expected to be zero?

Question

I was reading the proof of Helmholtz decomposition theorem where I found the relation between the rotational and the irrotational fields are not symmetric. And by that I mean if the divergence of the gradient field is supposed to be zero except where the particle exist aka:

$$\nabla\cdot\nabla(1/r)=\delta^3(\bf r)$$

Then why is the divergence of the curl always zero?

$$\nabla\cdot\nabla\times \bf A = 0$$

and not: $$\nabla\cdot\nabla\times \bf A =\delta^3(\bf c)$$

where $\bf c $ is a vector with speed of light magnitude.

Edit: This question is about asymmetry of electromagnetic potential fields and I have no idea why it was migrated to math forum. Anyways, if mathematicians can see this question my question is:

Suppose there is a vector field $ F=\nabla(1/r)+\nabla \times \bf A $ made out of a scalar potential $1/r$ and a vector potential $\bf A$ where these relations hold: $$\nabla\cdot\nabla(1/r)=\delta^3(\bf r)$$ and: $$\nabla\cdot\nabla\times \bf A =\delta^3(\bf c)$$

So both potential fields have critical points, considering $\bf F$ should have been sufficiently smooth, can we still apply Helmholtz decomposition theorem?

Answer 1

That the divergence of a curl is zero, and that the curl of a gradient is zero are exact mathematical identities, which can be easily proven by writing these operations explicitly in terms of components and derivatives.

On the other hand, a Laplacian (divergence of gradient) of a function is not necessarily zero. Equating it to a charge or another source is a matter of specific physical interpretation, which goes beyond pure math.

Answer 2

This is a good question and the answer lies in the misuse of the notation $\nabla$, and it is also why I like to write $\operatorname{\mathbf{grad}}, \operatorname{\mathbf{curl}}, \operatorname{div}$ instead of $\nabla, \nabla \times, \nabla \cdot$, resp. That both $\operatorname{div} \operatorname{\mathbf{curl}} \mathbf{v}=0$ and $\operatorname{\mathbf{curl}}\operatorname{\mathbf{grad}} f=0$ hold for arbitrary $f, \mathbf{v}$ scalar or vector fields, resp., is nothing but the equality of mixed partial derivatives $\frac{\partial^2}{\partial x \partial y}=\frac{\partial^2}{\partial y \partial x} $as it is easily seen by writing them out in components. But this means neither $\nabla \cdot \nabla \equiv 0$ nor $\nabla \times \nabla \equiv \bf{0}$ for such repeated operator is not defined; $\nabla$ is not a vector and it is neither perpendicular to nor parallel with itself in any sense.

There is a formulation of vector analysis using (exterior) differential forms where a differentiation operator is introduced that does act as you would like $\nabla$ behave. It can replace $\operatorname{\mathbf{grad}}, \operatorname{\bf{curl}}, \operatorname{div}$ with a single operator and can also repeatedly act on its argument with the result that you would expect. Differential forms are the natural generalization of vector analysis (including the vector product) for higher than 3 dimensional space, and there are attempts to replace and teach all of vector analysis in that way including the 3D case. You can find much written on this subject here in phys.stack.

Answer 3

From the comments you've written, I think that what you're asking is this:

Math tells us that the divergence of a curl is always zero. But suppose we go beyond pure math. What happens when the divergence of the curl is allowed to be nonzero?

And the answer is that it's just not possible. The divergence of the curl is zero, always, everywhere, under all circumstances, in theory and in practice, in the real world and in imaginary worlds.

Asking what happens when the divergence of the curl is nonzero is like asking what happens when two people are both taller than each other. It just can't happen.

Answer 4

There are actually cases when $\frac{\partial^2}{\partial x \partial y} \ne\frac{\partial^2}{\partial y \partial x}$ for functions with asymmetric Hessian matrices. Such functions have $\nabla\cdot\nabla\times {\vec A} \ne 0$. Such functions are sometimes mentioned because of an unmotivated assumption that h > 0 in the definition of the derivative

$$ k = \frac{f(x+h) - f(x)}{h}. $$

h can also be < 0.