Vector calculus identities and maxwell equations

[Reloaded47]

so we have the identity
[itex]\nabla\times\nabla\phi = 0[/itex]

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

But we also have that
[itex]\textbf{E} = -\nabla\phi[/itex]


So the problem I'm having is this
[itex]-\textbf{E} = \nabla\phi[/itex]

which i substitute into the identity
[itex]\nabla\times -\textbf{E} = - ( \nabla\times\textbf{E} ) = 0[/itex]

But this should be
[itex]\nabla\times - \textbf{E} = \frac{d\textbf{B}}{dt}[/itex]
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

[itex]\textbf{E} = - \nabla\phi - \frac{d\textbf{A}}{dt}[/itex]

where

[itex]\textbf{B} = \nabla\times\textbf{A}[/itex]

 

[DiracRules]

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

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

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

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.