Can a Divergent Free Vector Field be Expressed in a Certain Manner?

[member 428835]

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?

thanks for your time!

josh

 

[pasmith]

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?

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.