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Vector calculus identities and maxwell equations

  1. Oct 8, 2011 #1
    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 im getting confused

    i think there is a good chance ive done somehing silly, i just need someone to point it out
     
  2. jcsd
  3. Oct 8, 2011 #2
    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]
     
  4. Oct 8, 2011 #3
    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.
     
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