If there exists a $2D$ vector field $u=u(x)=(u_1,u_2)$ such that $\nabla\cdot u=0$ is it equivalent to saying following?
$$\nabla\cdot u=?(\nabla\cdot u)^\bot=\nabla^\bot\cdot u^\bot=\nabla\times u^\bot=0$$
- It is very easy to show that each part is identically zero but I am unsure if just because they are each zero, whether that makes them equivalent statements.
$$\nabla\cdot u=(\partial_1,\partial_2)\cdot(u_1,u_2)=\partial_1u_1+\partial_2u_2\equiv0$$ $$\nabla^\bot\cdot u^\bot=(\partial_2,-\partial_1)\cdot(u_2,-u_1)=\partial_2u_2+\partial_1u_1=\nabla\cdot u$$ $$\nabla\times u^\bot=\begin{vmatrix}e_1 & e_2 & e_3 \\\partial_1 & \partial_2 & 0 \\u_2 & -u_1 & 0 \\ \end{vmatrix}=(-\partial_1u_1-\partial_2u_2)e_3=-(\nabla\cdot u)e_3$$
- Since $\nabla\times u^\bot=0$ that means there must exist a potential $\phi$ such that $u^\bot=\nabla \phi$. Is this equivalent to saying $u=\nabla^\bot \phi$?