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D'Alembertian and wave equation.

  1. Aug 6, 2013 #1
    I am studying Coulomb and Lorentz gauge. Lorentz gauge help produce wave equation:
    [tex]\nabla^2 V-\mu_0\epsilon_0\frac{\partial^2V}{\partial t^2}=-\frac{\rho}{\epsilon_0},\;and\;\nabla^2 \vec A-\mu_0\epsilon_0\frac{\partial^2\vec A}{\partial t^2}=-\mu_0\vec J[/tex]
    Where the 4 dimensional d'Alembertian operator:
    [tex]\square^2=\nabla^2-\mu_0\epsilon_0\frac{\partial^2}{\partial t^2}[/tex]
    [tex]\Rightarrow\;\square^2V=-\frac{\rho}{\epsilon_0},\; and\;\square^2\vec A=-\mu_0\vec J[/tex]

    So the wave equations are really 4 dimensional d'Alembertian equations?
    Last edited: Aug 6, 2013
  2. jcsd
  3. Aug 6, 2013 #2


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    Your equations hold for Lorenz (NOT Lorentz!) gauge but not for Coulomb gauge. Otherwise it's indeed the d'Alembert operator. Note further that [itex]1/(\epsilon_0 \mu_0)=c^2[/itex] is the speed of light squared which is (contrary to the conversion factors [itex]\epsilon_0[/itex] and [itex]\mu_0[/itex]) a fundamental constant of nature.
  4. Aug 6, 2013 #3
    Thanks for the reply. I am reading Griffiths p422. It specified Lorentz gauge( that's how Griffiths spell it) put the two in the same footing. Actually Griffiths said Coulomb gauge using ##\nabla\cdot\vec A=0## to simplify ##\nabla^2V=-\frac{\rho}{\epsilon_0}## but make it more complicate for the vector potential ##\vec A##. That's the reason EM use Lorentz Gauge. This is all in p421 to 422 of Griffiths.

    You cannot combine Coulomb and Lorentz Gauge together as

    Coulomb ##\Rightarrow\;\nabla\cdot\vec A=0##

    Lorentz ##\Rightarrow\;\nabla\cdot\vec A=\mu_0\epsilon_0\frac{\partial V}{\partial t}##
  5. Aug 6, 2013 #4


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    It's an extremely common mistake but it should be Lorenz not Lorentz. Yes even Griffiths made that mistake.
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