Notation question for Maxwell's Equations.

[neutrino33]

I know I may be in the wrong place, but I think I'll get a quicker and better response here. My question is:

How do
$$\nabla \cdot \textbf{B} = 0$$
and
$$\nabla \times \textbf{E} + \frac{\partial \textbf{B}}{\partial t} = 0$$
derive from
$$\Box^2 A^\mu -\partial^\mu(\partial_\nu A^\nu) = j^\mu$$?

 

[ParticleGrl]

The equations you are thinking of are implicit in the fact that you are using a 4-potential.

Because

$$B = \nabla x A$$

then

$$\nabla \cdot B = \nabla \cdot \nabla x A$$

Because the divergence of a curl is 0 (or there is no boundary of a boundary, or the exterior derivative is nilpotent, or whatever your favorite math expression is)

Using

$$E = -\nabla A^0 -\frac{\partial}{\partial t} A$$

You can take the curl to get the standard

$$\nabla x E = -\frac{\partial}{\partial t} B$$

 

[sougata_chat]

del.B=0,is nonexistence of magnetic monopole, and the second equation is ampere's law where the time varying magnetic field depends on circulation of electric field.This all equation can be found by using wave equation.

 

[neutrino33]

The equations you are thinking of are implicit in the fact that you are using a 4-potential.

Because

$$B = \nabla x A$$

then

$$\nabla \cdot B = \nabla \cdot \nabla x A$$

Because the divergence of a curl is 0 (or there is no boundary of a boundary, or the exterior derivative is nilpotent, or whatever your favorite math expression is)

Using

$$E = -\nabla A^0 -\frac{\partial}{\partial t} A$$

You can take the curl to get the standard

$$\nabla x E = -\frac{\partial}{\partial t} B$$

I buy that. I don't see, however, what happens to the $$j^\mu$$.

 

[dextercioby]

It would be useful to know that the LaTex code for vector/cross product would be: $\times$ [ itex] \times [/itex].

 

[neutrino33]

This is what I'm working on. It is from Halzen Martin.