# d'Alembertian and wave equation.

by yungman
Tags: dalembertian, equation, wave
 P: 3,830 I am studying Coulomb and Lorentz gauge. Lorentz gauge help produce wave equation: $$\nabla^2 V-\mu_0\epsilon_0\frac{\partial^2V}{\partial t^2}=-\frac{\rho}{\epsilon_0},\;and\;\nabla^2 \vec A-\mu_0\epsilon_0\frac{\partial^2\vec A}{\partial t^2}=-\mu_0\vec J$$ Where the 4 dimensional d'Alembertian operator: $$\square^2=\nabla^2-\mu_0\epsilon_0\frac{\partial^2}{\partial t^2}$$ $$\Rightarrow\;\square^2V=-\frac{\rho}{\epsilon_0},\; and\;\square^2\vec A=-\mu_0\vec J$$ So the wave equations are really 4 dimensional d'Alembertian equations?
 Sci Advisor Thanks P: 2,104 Your equations hold for Lorenz (NOT Lorentz!) gauge but not for Coulomb gauge. Otherwise it's indeed the d'Alembert operator. Note further that $1/(\epsilon_0 \mu_0)=c^2$ is the speed of light squared which is (contrary to the conversion factors $\epsilon_0$ and $\mu_0$) a fundamental constant of nature.
P: 3,830
 Quote by vanhees71 Your equations hold for Lorenz (NOT Lorentz!) gauge but not for Coulomb gauge. Otherwise it's indeed the d'Alembert operator. Note further that $1/(\epsilon_0 \mu_0)=c^2$ is the speed of light squared which is (contrary to the conversion factors $\epsilon_0$ and $\mu_0$) a fundamental constant of nature.
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}##

C. Spirit