If I'm given a a vector field $\vec{E}=\vec{E}(\vec{p})$, and $\vec{E}$ is solenoidal ( I.E. $\vec{E}(\vec{p}) = 0\ \forall\ \vec{p}$), and I construct the vector potential $A = \vec{A}(\vec{p})$ such that $\vec{E} = \nabla \times \vec{A}$, does it nessecarily follow that the vector potential $\vec{A}$ is also solenoidal (I.E. $\nabla \cdot \vec{A}=0\ \forall\
\vec{p} $) ? If so, how can I prove this?
Is the Vector Potential of a Solenoidal Field also Solenoidal
multivariable-calculusVector Fields
Best Answer
No, it does not follow. For example, you can add any vector field of the sort $\vec F = (f(x),g(y),h(z))$ to your vector potential $\vec A$ and create all sorts of divergence. For a specific example, $\vec A = (x+z,x+y,y+z)$ is a vector potential for $\vec E = (1,1,1)$.