Can anyone help me prove that if $\vec{B}$ is a differentiable vector field with $\nabla \cdot \vec{B}=0$, then there exists an $\vec{A}$ such that $\vec{B} = \nabla \wedge \vec{A}$?
Thank you in advance!
Can anyone help me prove that if $\vec{B}$ is a differentiable vector field with $\nabla \cdot \vec{B}=0$, then there exists an $\vec{A}$ such that $\vec{B} = \nabla \wedge \vec{A}$?
Thank you in advance!
A high-level view is that you are asking whether there are any closed $2$-forms that are not exact. (See the Wikipedia article on closed and exact differential forms.) Since $\mathbb{R}^3$ is contractible, all its cohomology vanishes, so all closed forms are exact.