Just a simple question, from the title of the thread, is a vector field = 0 if the divergence is 0 and the curl is 0? I had trouble finding an answer anywhere online, proof of why or why not would be helpful, thanks!
$$\nabla \cdot \vec{F} = 0$$ $$\nabla \times \vec{F} = 0$$ $$\vec{F} = ?$$
No. To get a feel for what's up, let $\phi$ be a non-constant harmonic function so $\nabla^2 \phi = 0$ but $\nabla \phi \ne 0$. Set $\mathbf F = \nabla \phi \ne 0$; then Then $\nabla \times \mathbf F = \nabla \times \nabla \phi = 0, \,$ since the curl of a gradient always vanishes. Also, $\nabla \cdot \mathbf F = \nabla \cdot \nabla \phi = \nabla^2 \phi = 0$. The divergence and curl of $\mathbf F$ both vanish, but not $\mathbf F$!
This line of reasoning can, like tape or film, be re-wound and run "backwards": if $\mathbf F \ne 0$ and $\nabla \times \mathbf F = 0$, then (locally at least) there is a function $\phi$ with $\mathbf F = \nabla \phi \ne 0$; if now we also have $\nabla \cdot \mathbf F = 0$, then $\nabla^2 \phi = \nabla \cdot \nabla \phi = \nabla \cdot \mathbf F = 0$, and $\phi$ is harmonic.
The classic examples of such a field may be found in the elementary theory of electromagnetism: in the absence of sources, that is, charges and currents, static (non -time varying) electric fields $\mathbf E$ and magnetic fields $\mathbf B$ have vanishing divergence and curl: $\nabla \times \mathbf B = \nabla \times \mathbf E = 0$, and $\nabla \cdot \mathbf B = \nabla \cdot \mathbf E = 0$; the electrostatic potential function $\phi$ such that $\mathbf E = -\nabla \phi$ with $\nabla^2 \phi = 0$ exists by virtue of these facts; a similar assertion holds for the $\mathbf B$ field.
Hope this helps. Cheerio,
and as always (as ol' James Clerk M. has taught us),
Fiat Lux!!!