# Divergent free question

1. Nov 24, 2013

### joshmccraney

hey pf!

so if i have a vector field $\vec{V}$ and i know $\nabla \cdot \vec{V}=0$ would i be able to express $\vec{V}$ in the following manner: $\vec{V}= \nabla \times \vec{f}$ for some $\vec{f}$since we know this automatically satisfies the divergent free requirement?

if not, what must be assumed in order to claim that such an $\vec{f}$ exists?

Yes. You have some freedom in choosing $\vec f$ since $\nabla \times (\vec f + \nabla \phi) = \nabla \times \vec f = \vec V$ for any scalar field $\phi$.