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Scalar and vector potential

  1. Sep 1, 2010 #1

    [tex]\vec{E}=-\frac{\partial\vec{A}}{\partial t}-grad\varphi[/tex]

    If I define

    [tex]\varphi=\widetilde{\varphi}-\frac{\partial f}{\partial t}[/tex]




    I will get


    [tex]\vec{E}=-\frac{\partial\vec{A}}{\partial t}-grad\varphi=-\frac{\partial\widetilde{\vec{A}}}{\partial t}-grad\widetilde{\varphi}[/tex]

    But if I say

    [tex]\varphi=\widetilde{\varphi}+\frac{\partial f}{\partial t}[/tex]


    I wouldnt get that result. How I know how to take minus sign in this relations!
  2. jcsd
  3. Sep 2, 2010 #2


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    The transformation for A is clear, because you want B to stay the same and you simply exploit the fact that grad(rot F) = 0 for any vector field F.
    So then you can simply define [itex]\tilde\phi = \phi + \delta[/itex], write out
    [tex]-\frac{\partial \tilde{\vec A}}{\partial t} - \nabla \tilde\phi[/tex]
    and see what [itex]\delta[/itex] has to be to cancel the extra term from the [tex]\tilde A[/tex]-derivative so you get
    [tex]-\frac{\partial \vec A}{\partial t} - \nabla \phi[/tex]
  4. Sep 2, 2010 #3
    Another way to see it is with four vectors:

    Define the (contravariant) four potential putting together the scalar and vector potential:


    and define the (covariant) four derivative putting together the time derivative and the gradient:


    Then the gauge transformation is simply

    [tex]A^\mu\longmapsto A^\mu+\partial^\mu f[/tex]

    and the fact that you have to raise the index of [tex]\partial_\mu[/tex] yields the minus sign.
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