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Homework Help: Gauge Theory - Having Trouble with V' = 0

  1. Jan 28, 2010 #1
    1. The problem statement, all variables and given/known data
    Show that it is always possible to pick a gauge so V' = 0.

    2. Relevant equations
    We weren't given any, but I've been working with:
    (a)[tex]\vec{E} + \frac{\partial\vec{A}}{\partial{t}} = -\nabla{V}[/tex]
    (b)[tex]\vec{A'} = \vec{A} + \nabla{\psi}[/tex] (where [tex]\psi[/tex] is a scalar function)
    (c)[tex]V' = V - \frac{\partial{\psi}}{\partial{t}}[/tex]

    3. The attempt at a solution
    The way I've attempted it is not to choose the Lorentz or Coulomb gauge, but to make
    [tex]\nabla\cdot{\vec{A}} + \mu_{0}\epsilon_{0}\frac{\partial{V}}{\partial{t}} = x[/tex] (where [tex]x[/tex] is a scalar function)
    Furthermore, from Griffths solutions I found:
    [tex]\nabla\cdot{\vec{A'}} = - \mu_{0}\epsilon_{0}\frac{\partial{V'}}{\partial{t}}[/tex]
    In our case, the above equals 0 as V' = 0.

    I took the divergence of (b):
    [tex]\nabla\cdot\{vec{A'}} = \nabla\cdot\{\vec{A}} + \nabla^{2}{\psi}[/tex]
    [tex]0 = \nabla\cdot\{\vec{A}} + \nabla^{2}{\psi}[/tex]
    [tex]- \nabla\cdot\{\vec{A}} = \nabla^{2}{\psi}[/tex]
    [tex]- (x - \mu_{0}\epsilon_{0}\frac{\partial{V}}{\partial{t}}) = \nabla^{2}{\psi}[/tex]
    [tex]-x + \mu_{0}\epsilon_{0}\frac{\partial{V}}{\partial{t}} = \nabla^{2}{\psi}[/tex]
    [tex]\mu_{0}\epsilon_{0}\frac{\partial{V}}{\partial{t}} = \nabla^{2}{\psi} + x[/tex]
    [tex]\mu_{0}\epsilon_{0}\frac{\partial^{2}{\psi}}{\partial{t^{2}}} = \nabla^{2}{\psi} + x[/tex]
    ... which gives me an inequality (i.e. an extra x, and also an extra factor of mu_0 and epsilon_0).

    I've also been trying to use the solution to Griffths problem 10.7 as reference but I am really stuck on the steps to take.
  2. jcsd
  3. Jan 28, 2010 #2

    Ben Niehoff

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    Science Advisor
    Gold Member

    The easiest way to prove this would be to take your equations for a gauge transformation (b and c), set V' to zero, and show that given any A and V, there exists some psi which effects the desired gauge transformation.
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