# D'Alembertian and wave equation.

1. Aug 6, 2013

### yungman

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?

Last edited: Aug 6, 2013
2. Aug 6, 2013

### 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.

3. Aug 6, 2013

### yungman

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 $\Rightarrow\;\nabla\cdot\vec A=\mu_0\epsilon_0\frac{\partial V}{\partial t}$

4. Aug 6, 2013

### WannabeNewton

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