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?
Best Answer
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.