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D'Alembertian and wave equation. 
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#1
Aug613, 12:51 AM

P: 3,904

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? 


#2
Aug613, 02:37 AM

Sci Advisor
Thanks
P: 2,546

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.



#3
Aug613, 03:03 AM

P: 3,904

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}## 


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