Vector calculus identities and maxwell equations

[Reloaded47]

so we have the identity
$\nabla\times\nabla\phi = 0$

and from Maxwell's equations we have
$\nabla\times \textbf{E} = -\frac{d\textbf{B}}{dt}$

But we also have that
$\textbf{E} = -\nabla\phi$


So the problem I'm having is this
$-\textbf{E} = \nabla\phi$

which i substitute into the identity
$\nabla\times -\textbf{E} = - ( \nabla\times\textbf{E} ) = 0$

But this should be
$\nabla\times - \textbf{E} = \frac{d\textbf{B}}{dt}$
according to maxwell's equations, not zero
which is why I am getting confused

i think there is a good chance I've done somehing silly, i just need someone to point it out

 

[Reloaded47]

Never mind i figured it out

$\textbf{E} = - \nabla\phi - \frac{d\textbf{A}}{dt}$

where

$\textbf{B} = \nabla\times\textbf{A}$

 

[DiracRules]

Yours is a good question. The fact is that you define a potential $\phi$ if the field Eis conservative. But a field is conservative if its rotor is zero (irrotational).

This means that considering Maxwell's equation $\nabla \cross \vec{E}=-\frac{\partial \vec{B}}{\partial t}$, you can argue that IF $-\frac{\partial \vec{B}}{\partial t}=0$ then you can define $\phi\backepsilon' \vec{E}=-\nabla\phi$.

However, if you define $\vec{B}=\nabla\cross\vec{A}$, you could write $\nabla \cross \vec{E}=-\frac{\partial \vec{B}}{\partial t}=-\frac{\partial}{\partial t}\nabla\cross\vec{A}\Rightarrow \nabla\cross\{\vec{E}+\frac{\partial\vec{A}}{\partial t}\}=0$ so that the field $\vec{E}+\frac{\partial\vec{A}}{\partial t}$ is irrotational, then conservative, so that $\vec{E}+\frac{\partial\vec{A}}{\partial t}=-\nabla\phi\Rightarrow\vec{E}=-\nabla\phi-\frac{\partial\vec{A}}{\partial t}$

This is a more general definition of the potential of the electric field that fits with Maxwell's equations. However, there is a term missing in my last expression related to a gauge transformation, but since it is a term related to the frame of reference, setting proper condition it can be put to zero.