How do you prove that a vector field $\mathbf E$ that is irrotational ($\nabla \times \mathbf E =\mathbf0 $) may be written as $-\nabla \phi$ for a scalar field $\phi$? I have been trying to use the identity about the expansion of $\nabla(\mathbf {A\cdot B})$ but can't thing of a suitable second vector field. I'm sure I remember it being something like $\phi \mathbf A$ for some scalar field $\phi$ and fixed vector $\mathbf A$. I have also tried having the second vector as $\nabla \mathbf A$ but that doesn't seem to work either. Maybe it must be some function of $\mathbf E$? I can't find this proof anywhere!
EDIT: everything here is in simply connected space
Best Answer
Let's $\mathbf E=(E_x,E_y,E_z)$. For every closed curve $l$ by http://en.wikipedia.org/wiki/Stokes%27_theorem we have $$ \int_l E_xdx+E_ydy+E_zdz=\iint_S\nabla \times \mathbf E\cdot d\mathbf S=0, $$ where $S$ is any surface bounded by $l$. Therefore $E_xdx+E_ydy+E_zdz$ is complete differential for some function $\phi$. So $$ \mathbf E =-\nabla (-\phi). $$