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

E = -grad Phi - &A/&t

  1. Jan 10, 2004 #1
    I would like your opinion regarding an explanation I gave elsewhere. I hold that the explanation below is straight forward. However it appears as if some were confused by it.

    In a certain frame of referance, for a particular electromagnetic field, the relation [tex]\partial A/\partial t} = 0[/tex] holds true. Such a condition will hold in the case of a time independant magnetic field. The equation

    [tex]E = - \nabla \Phi - \displaystyle{\frac{\partial A}{\partial t}}[/tex]

    in this example and in this frame reduces to

    [tex]E = - \nabla \Phi[/tex]

    Does anyone think that this is relativistically incorrect?

    I know this seems like a dumb question but some people claim that this is relativistically incorrect. Such a claim is obviously wrong. However I can't understand why they're having such a difficult time understanding this. Is it what I explained above confusing?

    The 4-potential, [tex]A^{\alpha}[/tex], is defined in terms of the Coulomb potential, [tex]\Phi[/tex], and the magnetic vector potential, A as

    [tex]A^{\alpha} = (\Phi/c, A) = (\Phi/c, A_x, A_y, A_z)[/tex]

    The Faraday tensor, [tex]F^{\alpha \beta}[/tex], is defined as

    [tex]F^{\alpha \beta} = \partial^{\alpha} A^{\beta} - \partial^{\beta} A^{\alpha} [/tex]

    [See "Classical Electrodynamics - 2nd Ed.," J. D. Jackson, page 551, Eq. (11.136). I'm using different units]

    The [tex]F^{0k}[/tex] components of this relationship for k = 1,2,3 are, respectively

    [tex]\displaystyle{\frac{E_{x}}{c}} = \partial^{0} A^{1} - \partial^{1} A^{0} = - \displaystyle{\frac{1}{c}} \displaystyle{\frac{\partial A_{x}}{\partial t}} - \displaystyle{\frac{1}{c}} \displaystyle{\frac{\partial \Phi}{\partial x}}[/tex]

    [tex]\displaystyle{\frac{E_{y}}{c}} = \partial^{0} A^{2} - \partial^{2} A^{0} = - \displaystyle{\frac{1}{c}} \displaystyle{\frac{\partial A_{y}}{\partial t}} - \displaystyle{\frac{1}{c}} \displaystyle{\frac{\partial \Phi}{\partial y}}[/tex]

    [tex]\displaystyle{\frac{E_{z}}{c}} = \partial^{0} A^{3} - \partial^{3} A^{0} = - \displaystyle{\frac{1}{c}} \displaystyle{\frac{\partial A_{z}}{\partial t}} - \displaystyle{\frac{1}{c}} \displaystyle{\frac{\partial \Phi}{\partial z}}[/tex]

    These can be expressed as the single equation

    [tex]E = -\nabla \Phi - \displaystyle{\frac{\partial A}{\partial t}}[/tex]

    This equation and the equation B = curl A are equation (11.134) in Jackson on page 551. In fact Jackson uses these two equations to define [tex]F^{\alpha \beta} = \partial^{\alpha} A^{\beta} - \partial^{\beta} A^{\alpha} [/tex]

    In the example stated above [tex]\partial A/\partial t = 0[/tex] so that

    [tex]E = -\nabla \Phi [/tex]

    Does anyone find that confusing?
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted