Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]


    [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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook