vector and curl identity

Question

This popped up in my notes and the author made no remarks about the properties used

$\bigtriangledown \times \left ( \vec{E}+\frac{\partial \vec{A}}{\partial t} \right )=\vec{0}$

Then,

$\vec{E}+\frac{\partial \vec{A}}{\partial t}=-\bigtriangledown V$

where V is a scalar function.

I am unable to understand how the second line of the argument follows from the first.

Any help is appreciated.

Thanks in advance

Answer 1

The first equation states that the vector field $$\vec{E} + \frac{\partial\vec{A}}{\partial t}$$ has no rotation ($\text{curl}$ of that thing is $\vec{0}$).

For vector-fields with $\text{curl}\; \vec{V} = \vec{0}$ have a scalar potential function $f$ for which $\vec{V} = \nabla f$, because for every gradient field $\text{curl}\; \nabla f = \vec{0}$. (https://en.wikipedia.org/wiki/Curl_(mathematics)#Identities, 3rd identiy). It follows that $\vec{E} + \frac{\partial\vec{A}}{\partial t}$ can be represented as $\nabla V$ or also as $-\nabla V'$ if you set $V' = -V$.

Answer 2

This is a consequence of Helmholtz' Theorem: Any sufficiently nice vector field can be decomposed like $$ F=\nabla\Phi+\nabla\times B $$ In your case, $F=E+\partial_t A$ and the rotation component is zero.