D'Alembertian and wave equation.

  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. vanhees71

    vanhees71 4,321
    Science Advisor
    2014 Award

    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. 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. WannabeNewton

    WannabeNewton 5,859
    Science Advisor

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