Why is shear missing in Helmholtz theorem?

Let $$F_i$$ be a well behaved vector field in $$\mathbb{R}^3$$ which rapidly vanishes at infinity. (I am using here the index notation.) By the Helmholtz theorem, knowledge of divergence $$\partial_i F_i$$, and the curl $$\partial_iF_j - \partial_j F_i$$ of $$F_i$$ provides enough information to reconstruct the original vector field $$F_i$$.

I am wondering how is it that one can reconstruct $$F_i$$ only from divergence and curl, while the complete Jacobian $$J_{ij} = \partial_j F_i$$ corresponds additional information. Divergence is the trace of $$J_{ij}$$, curl is skew-symmetric part, while the symmetric part $$\partial_j F_i + \partial_i F_j$$, sometimes known as shear, does not participate in the Helmholtz theorem. I would like to intuitively understand this behavior.

In the case of a scalar function $$f$$, one needs to know all of its derivatives $$\partial_i f$$ in order to reconstruct $$f$$ via line integral. However, in the case of Helmholtz theorem, we are reconstructing $$F_i$$ as a volume integral. Does this make any difference?

• Actually, in the case of a scalar function, you only need the derivative in one direction, along with boundary conditions.
– user856
Dec 10, 2019 at 15:02
• @Rahul Can you please provide an example for your claim? Dec 11, 2019 at 13:28
• If $f:\mathbb R^3\to\mathbb R$ is a well-behaved scalar field which rapidly vanishes at infinity, then $f(x,y,z)=\int_{-\infty}^z\partial_3f(x,y,\zeta)\,\mathrm d\zeta$.
– user856
Dec 11, 2019 at 15:14
• If you expand your comment into an answer, I will accept it and award you the bounty. Dec 16, 2019 at 15:18
• No thanks, from my perspective @timur has given a much more interesting answer.
– user856
Dec 16, 2019 at 16:07

• First, if $$\nabla\times F=0$$, then $$F$$ is a conservative field, i.e., there is a scalar field $$\phi$$ such that $$F=\nabla\phi$$. So roughly speaking, a curl-free field has 1 dof per point.
• Now if $$\nabla\cdot F=0$$, then there is a vector potential $$A$$ such that $$F=\nabla\times A$$. So it appears that a divergence-free vector field has 3 dof per point. However, we should remember that the curl of a gradient vanishes, meaning that $$A$$ is determined only up to a gradient. More specifically, if $$\psi$$ is any scalar field, then $$\nabla\times(A+\nabla\psi)=\nabla\times A$$. This means that a divergence-free vector field really has 2 dof per point.
In other words, the curl of $$F$$ "cannot see" the gradient in $$F$$, and so specifying the curl of $$F$$ means specifying only 2 "real" degrees of freedom. Then the remaining 1 degree of freedom is fixed when we give the divergence.