# Decomposite a vector field into two parts

Let A be a region in $\mathbb R^3$, and suppose $\vec {\mathbf F}$ is a smooth vector field on A. I was asked to show that I can write $\vec {\mathbf F}=\vec {\mathbf F_1}+\vec {\mathbf F_2}$, s.t. $\operatorname{rot}(\vec {\mathbf F_1})=0, \operatorname{div}(\vec {\mathbf F_2})=0$. How can I show this? I think this has something to do with physics?

## 1 Answer

I think this is right--I hope it's right, but I'm digging into memory with no notes or books; anyway, here goes: Consider $\nabla \cdot F$; it is a scalar function on $A$. Consider the Poisson equation on $A$ $\nabla^2 \phi = \nabla \cdot F$. Find (or, better in the present context, assume the existence of) such a $\phi$. Then $\nabla \cdot (F - \nabla \phi) = 0$ on $A$. Invoke the classical result that a divergence-free field is a curl (a "rot" in the OP's terminology), thus there exists a vector field $C$ on $A$ such that $\nabla \times C = F - \nabla \phi$; then $F = \nabla \times C + \nabla \phi$, with $\nabla \cdot \nabla \times C = 0$ (since it's a curl--a "rot", if you will) and, of course, $\nabla \times \nabla \phi = 0$, since $\nabla \phi$ is a gradient. Whew! This stuff in fact has everything to do with physics; check out electromagntism and the equations of J. C. Maxwell; see Feynman's Lectures, vol. II.

This question and answer are deeply related to this one.

Best of success in these endeavors. Cheers.

• I think your answer is almost right, with a slight problem. I think a divergence-free field is locally a curl(because otherwise you have to assume A is simply connected, as we are actually saying that the kernel of the exterior differential operator from two-forms to three-forms must be in the image of the exterior differential operator from one-forms to two-forms.) But actually this changes nothing.
– lee
Jul 12, 2013 at 15:07
• Divergence free in $\mathbb{R}^3$ does not necessarily implies the underlying vector field is a curl of some vector potential $\mathbf{A}$. There exists non-trivial harmonic function $\phi$ in $\mathbb{R}^3\backslash\{0\}$ such that $\Delta \phi = 0$, which is $\nabla \cdot(\nabla \phi) = 0$. Otherwise +1. Jul 16, 2013 at 2:32