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I Lorenz Gauge

  1. Apr 23, 2017 #1
    Can someone explain to me (or point me towards a source) how is the Lorenz Gauge derived? I am reading the Griffiths book and from what I understand we can do the transformation ##A' = A + \nabla \lambda## and at the same time ##V' = V - \frac{\partial \lambda}{\partial t}## and B and E remain unchanged. This is pretty obvious for the Coulomb gauge (in finding ##\lambda## to satisfy the condition). But for the Lorenz gauge if you try to find ##\lambda## from the first equation, it will be a function of V and when you plug it in ##V' = V - \frac{\partial \lambda}{\partial t}##, V' will be different from V so the V you used in ##\nabla A = -\mu_0 \epsilon_0 \frac{\partial V}{\partial t}## is not anymore the same? So how do you prove it?
     
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  3. Apr 23, 2017 #2

    Twigg

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    Suppose ##\vec{A}## and ##V## are the potentials in the Coulomb gauge, and that you have already solved them from the relevant form of Maxwell's equations. Now, you want to prove that you can choose a value of ##\lambda## in the gauge transformation formulas you gave such that the resulting ##\vec{A}'## and ##V'## satisfy the Lorenz gauge condition: ##div(\vec{A}') + \frac{1}{c^{2}} \frac{\partial V'}{\partial t} = 0##. If you substitute in the gauge transformation formulas and the Coulomb gauge condition, ##div(\vec{A}) = 0##, then you will find ##\Box \lambda = -\frac{1}{c^{2}} \frac{\partial V'}{\partial t}## if my algebra isn't messed up, where the ##\Box## is the wave equation operator.
     
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