1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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